Optimal. Leaf size=125 \[ -\frac{2 b \left (2 a^2 A+a^2 C-A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{A x}{a^2} \]
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Rubi [A] time = 0.226777, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {4061, 3919, 3831, 2659, 208} \[ -\frac{2 b \left (2 a^2 A+a^2 C-A b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac{\left (a^2 C+A b^2\right ) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac{A x}{a^2} \]
Antiderivative was successfully verified.
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Rule 4061
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac{\int \frac{-A \left (a^2-b^2\right )+a b (A+C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac{A x}{a^2}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (b \left (A b^2-a^2 (2 A+C)\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac{A x}{a^2}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (A b^2-a^2 (2 A+C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac{A x}{a^2}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac{\left (2 \left (A b^2-a^2 (2 A+C)\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}+\left (1-\frac{a}{b}\right ) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^2 \left (a^2-b^2\right ) d}\\ &=\frac{A x}{a^2}-\frac{2 b \left (2 a^2 A-A b^2+a^2 C\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac{\left (A b^2+a^2 C\right ) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end{align*}
Mathematica [C] time = 1.97814, size = 270, normalized size = 2.16 \[ \frac{2 (a \cos (c+d x)+b) \left (A+C \sec ^2(c+d x)\right ) \left (\frac{\left (a^2 C+A b^2\right ) (a \sin (d x)-b \sin (c))}{d (a-b) (a+b) \left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right )}+\frac{2 b (\sin (c)+i \cos (c)) \left (a^2 (2 A+C)-A b^2\right ) (a \cos (c+d x)+b) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{d \left (a^2-b^2\right )^{3/2} \sqrt{(\cos (c)-i \sin (c))^2}}+A x (a \cos (c+d x)+b)\right )}{a^2 (a+b \sec (c+d x))^2 (A \cos (2 (c+d x))+A+2 C)} \]
Warning: Unable to verify antiderivative.
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Maple [B] time = 0.094, size = 328, normalized size = 2.6 \begin{align*} 2\,{\frac{A\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{2}}}-2\,{\frac{A\tan \left ( 1/2\,dx+c/2 \right ){b}^{2}}{ad \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}-2\,{\frac{a\tan \left ( 1/2\,dx+c/2 \right ) C}{d \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}-4\,{\frac{Ab}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{A{b}^{3}}{d{a}^{2} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-2\,{\frac{Cb}{d \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 0.599719, size = 1188, normalized size = 9.5 \begin{align*} \left [\frac{2 \,{\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \,{\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x -{\left ({\left (2 \, A + C\right )} a^{2} b^{2} - A b^{4} +{\left ({\left (2 \, A + C\right )} a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{a^{2} - b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) -{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt{a^{2} - b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + 2 \,{\left (C a^{5} +{\left (A - C\right )} a^{3} b^{2} - A a b^{4}\right )} \sin \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac{{\left (A a^{5} - 2 \, A a^{3} b^{2} + A a b^{4}\right )} d x \cos \left (d x + c\right ) +{\left (A a^{4} b - 2 \, A a^{2} b^{3} + A b^{5}\right )} d x -{\left ({\left (2 \, A + C\right )} a^{2} b^{2} - A b^{4} +{\left ({\left (2 \, A + C\right )} a^{3} b - A a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) +{\left (C a^{5} +{\left (A - C\right )} a^{3} b^{2} - A a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + C \sec ^{2}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.19606, size = 277, normalized size = 2.22 \begin{align*} -\frac{\frac{2 \,{\left (2 \, A a^{2} b + C a^{2} b - A b^{3}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{{\left (d x + c\right )} A}{a^{2}} + \frac{2 \,{\left (C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a - b\right )}}}{d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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