Optimal. Leaf size=187 \[ -\frac{\left (-3 a^2 B+3 a A b-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac{\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a^3 (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{(A b-a B) \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{B \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]
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Rubi [A] time = 0.675946, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {4033, 4092, 4082, 3998, 3770, 3831, 2659, 208} \[ -\frac{\left (-3 a^2 B+3 a A b-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac{\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a^3 (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{(A b-a B) \tan (c+d x) \sec (c+d x)}{2 b^2 d}+\frac{B \tan (c+d x) \sec ^2(c+d x)}{3 b d} \]
Antiderivative was successfully verified.
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Rule 4033
Rule 4092
Rule 4082
Rule 3998
Rule 3770
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\sec ^4(c+d x) (A+B \sec (c+d x))}{a+b \sec (c+d x)} \, dx &=\frac{B \sec ^2(c+d x) \tan (c+d x)}{3 b d}+\frac{\int \frac{\sec ^2(c+d x) \left (2 a B+2 b B \sec (c+d x)+3 (A b-a B) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 b}\\ &=\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac{B \sec ^2(c+d x) \tan (c+d x)}{3 b d}+\frac{\int \frac{\sec (c+d x) \left (3 a (A b-a B)+b (3 A b+a B) \sec (c+d x)-2 \left (3 a A b-3 a^2 B-2 b^2 B\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 b^2}\\ &=-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac{B \sec ^2(c+d x) \tan (c+d x)}{3 b d}+\frac{\int \frac{\sec (c+d x) \left (3 a b (A b-a B)+3 \left (2 a^2+b^2\right ) (A b-a B) \sec (c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 b^3}\\ &=-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac{B \sec ^2(c+d x) \tan (c+d x)}{3 b d}-\frac{\left (a^3 (A b-a B)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{b^4}+\frac{\left (\left (2 a^2+b^2\right ) (A b-a B)\right ) \int \sec (c+d x) \, dx}{2 b^4}\\ &=\frac{\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac{B \sec ^2(c+d x) \tan (c+d x)}{3 b d}-\frac{\left (a^3 (A b-a B)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{b^5}\\ &=\frac{\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac{B \sec ^2(c+d x) \tan (c+d x)}{3 b d}-\frac{\left (2 a^3 (A b-a B)\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^5 d}\\ &=\frac{\left (2 a^2+b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 b^4 d}-\frac{2 a^3 (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{\sqrt{a-b} b^4 \sqrt{a+b} d}-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \tan (c+d x)}{3 b^3 d}+\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 b^2 d}+\frac{B \sec ^2(c+d x) \tan (c+d x)}{3 b d}\\ \end{align*}
Mathematica [B] time = 2.36635, size = 422, normalized size = 2.26 \[ \frac{\frac{4 b \left (3 a^2 B-3 a A b+2 b^2 B\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b \left (3 a^2 B-3 a A b+2 b^2 B\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{24 a^3 (A b-a B) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+6 \left (2 a^2+b^2\right ) (a B-A b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-6 \left (2 a^2+b^2\right ) (a B-A b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{b^2 (B (b-3 a)+3 A b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^2 (B (b-3 a)+3 A b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 b^3 B \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 b^3 B \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}}{12 b^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.083, size = 688, normalized size = 3.7 \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 = 2.35656, size = 1650, normalized size = 8.82 \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 )}\right ) \sec ^{4}{\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 = 1.26332, size = 556, normalized size = 2.97 \begin{align*} -\frac{\frac{3 \,{\left (2 \, B a^{3} - 2 \, A a^{2} b + B a b^{2} - A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{b^{4}} - \frac{3 \,{\left (2 \, B a^{3} - 2 \, A a^{2} b + B a b^{2} - A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{b^{4}} - \frac{12 \,{\left (B a^{4} - A a^{3} b\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^{4}} + \frac{2 \,{\left (6 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B b^{2} \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 )}^{3} b^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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