Optimal. Leaf size=178 \[ \frac{\left (2 a^2 B-3 a b C+3 b^2 B\right ) \sin (c+d x)}{3 a^3 d}+\frac{2 b^3 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{x \left (a^2+2 b^2\right ) (b B-a C)}{2 a^4}-\frac{(b B-a C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{B \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]
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Rubi [A] time = 0.704363, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4034, 4104, 3919, 3831, 2659, 208} \[ \frac{\left (2 a^2 B-3 a b C+3 b^2 B\right ) \sin (c+d x)}{3 a^3 d}+\frac{2 b^3 (b B-a C) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^4 d \sqrt{a-b} \sqrt{a+b}}-\frac{x \left (a^2+2 b^2\right ) (b B-a C)}{2 a^4}-\frac{(b B-a C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}+\frac{B \sin (c+d x) \cos ^2(c+d x)}{3 a d} \]
Antiderivative was successfully verified.
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Rule 4072
Rule 4034
Rule 4104
Rule 3919
Rule 3831
Rule 2659
Rule 208
Rubi steps
\begin{align*} \int \frac{\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\int \frac{\cos ^3(c+d x) (B+C \sec (c+d x))}{a+b \sec (c+d x)} \, dx\\ &=\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac{\int \frac{\cos ^2(c+d x) \left (3 (b B-a C)-2 a B \sec (c+d x)-2 b B \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{3 a}\\ &=-\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\int \frac{\cos (c+d x) \left (2 \left (2 a^2 B+3 b^2 B-3 a b C\right )+a (b B+3 a C) \sec (c+d x)-3 b (b B-a C) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{6 a^2}\\ &=\frac{\left (2 a^2 B+3 b^2 B-3 a b C\right ) \sin (c+d x)}{3 a^3 d}-\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac{\int \frac{3 \left (a^2+2 b^2\right ) (b B-a C)+3 a b (b B-a C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{6 a^3}\\ &=-\frac{\left (a^2+2 b^2\right ) (b B-a C) x}{2 a^4}+\frac{\left (2 a^2 B+3 b^2 B-3 a b C\right ) \sin (c+d x)}{3 a^3 d}-\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (b^3 (b B-a C)\right ) \int \frac{\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4}\\ &=-\frac{\left (a^2+2 b^2\right ) (b B-a C) x}{2 a^4}+\frac{\left (2 a^2 B+3 b^2 B-3 a b C\right ) \sin (c+d x)}{3 a^3 d}-\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (b^2 (b B-a C)\right ) \int \frac{1}{1+\frac{a \cos (c+d x)}{b}} \, dx}{a^4}\\ &=-\frac{\left (a^2+2 b^2\right ) (b B-a C) x}{2 a^4}+\frac{\left (2 a^2 B+3 b^2 B-3 a b C\right ) \sin (c+d x)}{3 a^3 d}-\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 a d}+\frac{\left (2 b^2 (b B-a C)\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^4 d}\\ &=-\frac{\left (a^2+2 b^2\right ) (b B-a C) x}{2 a^4}+\frac{2 b^3 (b B-a C) \tanh ^{-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 (2 a^2 B+3 b^2 B-3 a b C\right ) \sin (c+d x)}{3 a^3 d}-\frac{(b B-a C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 a d}\\ \end{align*}
Mathematica [A] time = 0.490599, size = 152, normalized size = 0.85 \[ \frac{6 \left (a^2+2 b^2\right ) (c+d x) (a C-b B)+3 a \left (3 a^2 B-4 a b C+4 b^2 B\right ) \sin (c+d x)-\frac{24 b^3 (b B-a C) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+3 a^2 (a C-b B) \sin (2 (c+d x))+a^3 B \sin (3 (c+d x))}{12 a^4 d} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.12, size = 641, normalized size = 3.6 \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 = 0.636973, size = 1177, normalized size = 6.61 \begin{align*} \left [\frac{3 \,{\left (C a^{5} - B a^{4} b + C a^{3} b^{2} - B a^{2} b^{3} - 2 \, C a b^{4} + 2 \, B b^{5}\right )} d x - 3 \,{\left (C a b^{3} - B b^{4}\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 ) +{\left (4 \, B a^{5} - 6 \, C a^{4} b + 2 \, B a^{3} b^{2} + 6 \, C a^{2} b^{3} - 6 \, B a b^{4} + 2 \,{\left (B a^{5} - B a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (C a^{5} - B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - a^{4} b^{2}\right )} d}, \frac{3 \,{\left (C a^{5} - B a^{4} b + C a^{3} b^{2} - B a^{2} b^{3} - 2 \, C a b^{4} + 2 \, B b^{5}\right )} d x - 6 \,{\left (C a b^{3} - B b^{4}\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 (4 \, B a^{5} - 6 \, C a^{4} b + 2 \, B a^{3} b^{2} + 6 \, C a^{2} b^{3} - 6 \, B a b^{4} + 2 \,{\left (B a^{5} - B a^{3} b^{2}\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (C a^{5} - B a^{4} b - C a^{3} b^{2} + B a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{6} - a^{4} b^{2}\right )} d}\right ] \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 = 1.22358, size = 486, normalized size = 2.73 \begin{align*} \frac{\frac{3 \,{\left (C a^{3} - B a^{2} b + 2 \, C a b^{2} - 2 \, B b^{3}\right )}{\left (d x + c\right )}}{a^{4}} - \frac{12 \,{\left (C a b^{3} - B 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 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a^{2} \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} - 6 \, C a b \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} + 4 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, 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 ) + 3 \, C a^{2} \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 ) - 6 \, C a b \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} a^{3}}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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