Integral of $$$\left(c + d \sec{\left(x \right)}\right)^{2}$$$ with respect to $$$x$$$

Find $$$\int \left(c + d \sec{\left(x \right)}\right)^{2}\, dx$$$.

Expand the expression:

$${\color{red}{\int{\left(c + d \sec{\left(x \right)}\right)^{2} d x}}} = {\color{red}{\int{\left(c^{2} + 2 c d \sec{\left(x \right)} + d^{2} \sec^{2}{\left(x \right)}\right)d x}}}$$

Integrate term by term:

$${\color{red}{\int{\left(c^{2} + 2 c d \sec{\left(x \right)} + d^{2} \sec^{2}{\left(x \right)}\right)d x}}} = {\color{red}{\left(\int{c^{2} d x} + \int{d^{2} \sec^{2}{\left(x \right)} d x} + \int{2 c d \sec{\left(x \right)} d x}\right)}}$$

Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=c^{2}$$$:

$$\int{d^{2} \sec^{2}{\left(x \right)} d x} + \int{2 c d \sec{\left(x \right)} d x} + {\color{red}{\int{c^{2} d x}}} = \int{d^{2} \sec^{2}{\left(x \right)} d x} + \int{2 c d \sec{\left(x \right)} d x} + {\color{red}{c^{2} x}}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=d^{2}$$$ and $$$f{\left(x \right)} = \sec^{2}{\left(x \right)}$$$:

$$c^{2} x + \int{2 c d \sec{\left(x \right)} d x} + {\color{red}{\int{d^{2} \sec^{2}{\left(x \right)} d x}}} = c^{2} x + \int{2 c d \sec{\left(x \right)} d x} + {\color{red}{d^{2} \int{\sec^{2}{\left(x \right)} d x}}}$$

The integral of $$$\sec^{2}{\left(x \right)}$$$ is $$$\int{\sec^{2}{\left(x \right)} d x} = \tan{\left(x \right)}$$$:

$$c^{2} x + d^{2} {\color{red}{\int{\sec^{2}{\left(x \right)} d x}}} + \int{2 c d \sec{\left(x \right)} d x} = c^{2} x + d^{2} {\color{red}{\tan{\left(x \right)}}} + \int{2 c d \sec{\left(x \right)} d x}$$

Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=2 c d$$$ and $$$f{\left(x \right)} = \sec{\left(x \right)}$$$:

$$c^{2} x + d^{2} \tan{\left(x \right)} + {\color{red}{\int{2 c d \sec{\left(x \right)} d x}}} = c^{2} x + d^{2} \tan{\left(x \right)} + {\color{red}{\left(2 c d \int{\sec{\left(x \right)} d x}\right)}}$$

Rewrite the secant as $$$\sec\left(x\right)=\frac{1}{\cos\left(x\right)}$$$:

$$c^{2} x + 2 c d {\color{red}{\int{\sec{\left(x \right)} d x}}} + d^{2} \tan{\left(x \right)} = c^{2} x + 2 c d {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}} + d^{2} \tan{\left(x \right)}$$

Rewrite the cosine in terms of the sine using the formula $$$\cos\left(x\right)=\sin\left(x + \frac{\pi}{2}\right)$$$ and then rewrite the sine using the double angle formula $$$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)$$$:

$$c^{2} x + 2 c d {\color{red}{\int{\frac{1}{\cos{\left(x \right)}} d x}}} + d^{2} \tan{\left(x \right)} = c^{2} x + 2 c d {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} + d^{2} \tan{\left(x \right)}$$

Multiply the numerator and denominator by $$$\sec^2\left(\frac{x}{2} + \frac{\pi}{4} \right)$$$:

$$c^{2} x + 2 c d {\color{red}{\int{\frac{1}{2 \sin{\left(\frac{x}{2} + \frac{\pi}{4} \right)} \cos{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} + d^{2} \tan{\left(x \right)} = c^{2} x + 2 c d {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} + d^{2} \tan{\left(x \right)}$$

Let $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$.

Then $$$du=\left(\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}\right)^{\prime }dx = \frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2} dx$$$ (steps can be seen »), and we have that $$$\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)} dx = 2 du$$$.

The integral can be rewritten as

$$c^{2} x + 2 c d {\color{red}{\int{\frac{\sec^{2}{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}{2 \tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}} d x}}} + d^{2} \tan{\left(x \right)} = c^{2} x + 2 c d {\color{red}{\int{\frac{1}{u} d u}}} + d^{2} \tan{\left(x \right)}$$

The integral of $$$\frac{1}{u}$$$ is $$$\int{\frac{1}{u} d u} = \ln{\left(\left|{u}\right| \right)}$$$:

$$c^{2} x + 2 c d {\color{red}{\int{\frac{1}{u} d u}}} + d^{2} \tan{\left(x \right)} = c^{2} x + 2 c d {\color{red}{\ln{\left(\left|{u}\right| \right)}}} + d^{2} \tan{\left(x \right)}$$

Recall that $$$u=\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}$$$:

$$c^{2} x + 2 c d \ln{\left(\left|{{\color{red}{u}}}\right| \right)} + d^{2} \tan{\left(x \right)} = c^{2} x + 2 c d \ln{\left(\left|{{\color{red}{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}}}\right| \right)} + d^{2} \tan{\left(x \right)}$$

Therefore,

$$\int{\left(c + d \sec{\left(x \right)}\right)^{2} d x} = c^{2} x + 2 c d \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + d^{2} \tan{\left(x \right)}$$

Add the constant of integration:

$$\int{\left(c + d \sec{\left(x \right)}\right)^{2} d x} = c^{2} x + 2 c d \ln{\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right| \right)} + d^{2} \tan{\left(x \right)}+C$$

$$$\int \left(c + d \sec{\left(x \right)}\right)^{2}\, dx = \left(c^{2} x + 2 c d \ln\left(\left|{\tan{\left(\frac{x}{2} + \frac{\pi}{4} \right)}}\right|\right) + d^{2} \tan{\left(x \right)}\right) + C$$$A