Optimal. Leaf size=153 \[ \frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 b d} \]
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Rubi [A] time = 0.453871, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4092, 4082, 3998, 3770, 3831, 2659, 208} \[ \frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^3 d \sqrt{a-b} \sqrt{a+b}}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \tan (c+d x) \sec (c+d x)}{2 b d} \]
Antiderivative was successfully verified.
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Rule 4092
Rule 4082
Rule 3998
Rule 3770
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac{C \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{\int \frac{\sec (c+d x) \left (a C+b (2 A+C) \sec (c+d x)+2 (b B-a C) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b}\\ &=\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{\int \frac{\sec (c+d x) \left (a b C+\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 b^2}\\ &=\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \sec (c+d x) \tan (c+d x)}{2 b d}+\frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \int \sec (c+d x) \, dx}{2 b^3}-\frac{\left (a \left (A b^2-a (b B-a C)\right )\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^3}\\ &=\frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \sec (c+d x) \tan (c+d x)}{2 b d}-\frac{\left (a \left (A b^2-a (b B-a C)\right )\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^4}\\ &=\frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \sec (c+d x) \tan (c+d x)}{2 b d}-\frac{\left (2 a \left (A b^2-a (b B-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 )}{b^4 d}\\ &=\frac{\left (b^2 (2 A+C)-2 a (b B-a C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 b^3 d}-\frac{2 a \left (A b^2-a (b B-a C)\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^3 \sqrt{a+b} d}+\frac{(b B-a C) \tan (c+d x)}{b^2 d}+\frac{C \sec (c+d x) \tan (c+d x)}{2 b d}\\ \end{align*}
Mathematica [C] time = 2.5958, size = 472, normalized size = 3.08 \[ \frac{\cos (c+d x) (a \cos (c+d x)+b) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-2 \left (2 a^2 C-2 a b B+2 A b^2+b^2 C\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \left (2 a^2 C-2 a b B+2 A b^2+b^2 C\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{8 a (\sin (c)+i \cos (c)) \left (a (a C-b B)+A b^2\right ) \tan ^{-1}\left (\frac{(\sin (c)+i \cos (c)) \left (\tan \left (\frac{d x}{2}\right ) (a \cos (c)-b)+a \sin (c)\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}\right )}{\sqrt{a^2-b^2} \sqrt{(\cos (c)-i \sin (c))^2}}+\frac{4 b \sin \left (\frac{d x}{2}\right ) (b B-a C)}{\left (\cos \left (\frac{c}{2}\right )-\sin \left (\frac{c}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{4 b \sin \left (\frac{d x}{2}\right ) (b B-a C)}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{b^2 C}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 C}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{2 b^3 d (a+b \sec (c+d x)) (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.083, size = 499, normalized size = 3.3 \begin{align*} \text{result too large to display} \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 = 37.0882, size = 1465, normalized size = 9.58 \begin{align*} \text{result too large to display} \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{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{a + b \sec{\left (c + d x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.28547, size = 387, normalized size = 2.53 \begin{align*} \frac{\frac{{\left (2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{3}} - \frac{{\left (2 \, C a^{2} - 2 \, B a b + 2 \, A b^{2} + C b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{3}} - \frac{4 \,{\left (C a^{3} - B a^{2} b + A a b^{2}\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 )}}{\sqrt{-a^{2} + b^{2}} b^{3}} + \frac{2 \,{\left (2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{2} b^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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