Optimal. Leaf size=178 \[ -\frac{\left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{3 b^3 d}-\frac{2 a^3 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (2 a^2+b^2\right ) (A b-a B)}{2 b^4}+\frac{(A b-a B) \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{B \sin (c+d x) \cos ^2(c+d x)}{3 b d} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.493183, antiderivative size = 178, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {2990, 3049, 3023, 2735, 2659, 205} \[ -\frac{\left (-3 a^2 B+3 a A b-2 b^2 B\right ) \sin (c+d x)}{3 b^3 d}-\frac{2 a^3 (A b-a B) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{b^4 d \sqrt{a-b} \sqrt{a+b}}+\frac{x \left (2 a^2+b^2\right ) (A b-a B)}{2 b^4}+\frac{(A b-a B) \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac{B \sin (c+d x) \cos ^2(c+d x)}{3 b d} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 2990
Rule 3049
Rule 3023
Rule 2735
Rule 2659
Rule 205
Rubi steps
\begin{align*} \int \frac{\cos ^3(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx &=\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{\cos (c+d x) \left (2 a B+2 b B \cos (c+d x)+3 (A b-a B) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{3 b}\\ &=\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{3 a (A b-a B)+b (3 A b+a B) \cos (c+d x)-2 \left (3 a A b-3 a^2 B-2 b^2 B\right ) \cos ^2(c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^2}\\ &=-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \sin (c+d x)}{3 b^3 d}+\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 b d}+\frac{\int \frac{3 a b (A b-a B)+3 \left (2 a^2+b^2\right ) (A b-a B) \cos (c+d x)}{a+b \cos (c+d x)} \, dx}{6 b^3}\\ &=\frac{\left (2 a^2+b^2\right ) (A b-a B) x}{2 b^4}-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \sin (c+d x)}{3 b^3 d}+\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac{\left (a^3 (A b-a B)\right ) \int \frac{1}{a+b \cos (c+d x)} \, dx}{b^4}\\ &=\frac{\left (2 a^2+b^2\right ) (A b-a B) x}{2 b^4}-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \sin (c+d x)}{3 b^3 d}+\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 b d}-\frac{\left (2 a^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 )}{b^4 d}\\ &=\frac{\left (2 a^2+b^2\right ) (A b-a B) x}{2 b^4}-\frac{2 a^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 )}{\sqrt{a-b} b^4 \sqrt{a+b} d}-\frac{\left (3 a A b-3 a^2 B-2 b^2 B\right ) \sin (c+d x)}{3 b^3 d}+\frac{(A b-a B) \cos (c+d x) \sin (c+d x)}{2 b^2 d}+\frac{B \cos ^2(c+d x) \sin (c+d x)}{3 b d}\\ \end{align*}
Mathematica [A] time = 0.434912, size = 152, normalized size = 0.85 \[ \frac{6 \left (2 a^2+b^2\right ) (c+d x) (A b-a B)+3 b \left (4 a^2 B-4 a A b+3 b^2 B\right ) \sin (c+d x)-\frac{24 a^3 (a B-A b) \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+3 b^2 (A b-a B) \sin (2 (c+d x))+b^3 B \sin (3 (c+d x))}{12 b^4 d} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.115, 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.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 1.82331, size = 1164, normalized size = 6.54 \begin{align*} \left [-\frac{3 \,{\left (2 \, B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} d x - 3 \,{\left (B a^{4} - A a^{3} b\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{2 \, a b \cos \left (d x + c\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (d x + c\right ) + b\right )} \sin \left (d x + c\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + a^{2}}\right ) -{\left (6 \, B a^{4} b - 6 \, A a^{3} b^{2} - 2 \, B a^{2} b^{3} + 6 \, A a b^{4} - 4 \, B b^{5} + 2 \,{\left (B a^{2} b^{3} - B b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )} d}, -\frac{3 \,{\left (2 \, B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2} + A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} d x - 6 \,{\left (B a^{4} - A a^{3} b\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (d x + c\right )}\right ) -{\left (6 \, B a^{4} b - 6 \, A a^{3} b^{2} - 2 \, B a^{2} b^{3} + 6 \, A a b^{4} - 4 \, B b^{5} + 2 \,{\left (B a^{2} b^{3} - B b^{5}\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (B a^{3} b^{2} - A a^{2} b^{3} - B a b^{4} + A b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\left (a^{2} b^{4} - b^{6}\right )} d}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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.
[In]
[Out]
________________________________________________________________________________________
Giac [B] time = 1.54545, size = 486, normalized size = 2.73 \begin{align*} -\frac{\frac{3 \,{\left (2 \, B a^{3} - 2 \, A a^{2} b + B a b^{2} - A b^{3}\right )}{\left (d x + c\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.
[In]
[Out]