Optimal. Leaf size=187 \[ \frac{2 b^3 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{\left (2 a^2 A-3 a b B+3 A b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac{\left (a^2+2 b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{(A b-a B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x)}{3 a d} \]
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Rubi [A] time = 0.769869, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3000, 3055, 3001, 3770, 2659, 205} \[ \frac{2 b^3 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{\left (2 a^2 A-3 a b B+3 A b^2\right ) \tan (c+d x)}{3 a^3 d}-\frac{\left (a^2+2 b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}-\frac{(A b-a B) \tan (c+d x) \sec (c+d x)}{2 a^2 d}+\frac{A \tan (c+d x) \sec ^2(c+d x)}{3 a d} \]
Antiderivative was successfully verified.
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Rule 3000
Rule 3055
Rule 3001
Rule 3770
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{(A+B \cos (c+d x)) \sec ^4(c+d x)}{a+b \cos (c+d x)} \, dx &=\frac{A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (-3 (A b-a B)+2 a A \cos (c+d x)+2 A b \cos ^2(c+d x)\right ) \sec ^3(c+d x)}{a+b \cos (c+d x)} \, dx}{3 a}\\ &=-\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (2 \left (2 a^2 A+3 A b^2-3 a b B\right )+a (A b+3 a B) \cos (c+d x)-3 b (A b-a B) \cos ^2(c+d x)\right ) \sec ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^2}\\ &=\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \tan (c+d x)}{3 a^3 d}-\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\int \frac{\left (-3 \left (a^2+2 b^2\right ) (A b-a B)-3 a b (A b-a B) \cos (c+d x)\right ) \sec (c+d x)}{a+b \cos (c+d x)} \, dx}{6 a^3}\\ &=\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \tan (c+d x)}{3 a^3 d}-\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\left (b^3 (A b-a B)\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{a^4}-\frac{\left (\left (a^2+2 b^2\right ) (A b-a B)\right ) \int \sec (c+d x) \, dx}{2 a^4}\\ &=-\frac{\left (a^2+2 b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \tan (c+d x)}{3 a^3 d}-\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{A \sec ^2(c+d x) \tan (c+d x)}{3 a d}+\frac{\left (2 b^3 (A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{a+b+(a-b) x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{a^4 d}\\ &=\frac{2 b^3 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 \sqrt{a-b} \sqrt{a+b} d}-\frac{\left (a^2+2 b^2\right ) (A b-a B) \tanh ^{-1}(\sin (c+d x))}{2 a^4 d}+\frac{\left (2 a^2 A+3 A b^2-3 a b B\right ) \tan (c+d x)}{3 a^3 d}-\frac{(A b-a B) \sec (c+d x) \tan (c+d x)}{2 a^2 d}+\frac{A \sec ^2(c+d x) \tan (c+d x)}{3 a d}\\ \end{align*}
Mathematica [B] time = 2.07219, size = 422, normalized size = 2.26 \[ \frac{\frac{4 a \left (2 a^2 A-3 a b B+3 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 a \left (2 a^2 A-3 a b B+3 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}+\frac{24 b^3 (a B-A b) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}-6 \left (a^2+2 b^2\right ) (a B-A b) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+6 \left (a^2+2 b^2\right ) (a B-A b) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+\frac{a^2 (a (A+3 B)-3 A b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{a^2 (a (A+3 B)-3 A b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 a^3 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 a^3 A \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}}{12 a^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.175, 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 [A] time = 5.02031, size = 1634, normalized size = 8.74 \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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.73242, size = 556, normalized size = 2.97 \begin{align*} \frac{\frac{3 \,{\left (B a^{3} - A a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a^{4}} - \frac{3 \,{\left (B a^{3} - A a^{2} b + 2 \, B a b^{2} - 2 \, A b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{a^{4}} + \frac{12 \,{\left (B a b^{3} - A b^{4}\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}} a^{4}} - \frac{2 \,{\left (6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 4 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 6 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, A 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} a^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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