Optimal. Leaf size=164 \[ \frac {\left (2 a^2 A-3 a b B+A b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]
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Rubi [A] time = 0.19, antiderivative size = 164, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2754, 12, 2659, 205} \[ \frac {\left (2 a^2 A-3 a b B+A b^2\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d (a-b)^{5/2} (a+b)^{5/2}}-\frac {\left (a^2 (-B)+3 a A b-2 b^2 B\right ) \sin (c+d x)}{2 d \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac {(A b-a B) \sin (c+d x)}{2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 205
Rule 2659
Rule 2754
Rubi steps
\begin {align*} \int \frac {A+B \cos (c+d x)}{(a+b \cos (c+d x))^3} \, dx &=-\frac {(A b-a B) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\int \frac {-2 (a A-b B)+(A b-a B) \cos (c+d x)}{(a+b \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac {(A b-a B) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\int \frac {2 a^2 A+A b^2-3 a b B}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \int \frac {1}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac {(A b-a B) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}+\frac {\left (2 a^2 A+A b^2-3 a b 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 )}{\left (a^2-b^2\right )^2 d}\\ &=\frac {\left (2 a^2 A+A b^2-3 a b B\right ) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} (a+b)^{5/2} d}-\frac {(A b-a B) \sin (c+d x)}{2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^2}-\frac {\left (3 a A b-a^2 B-2 b^2 B\right ) \sin (c+d x)}{2 \left (a^2-b^2\right )^2 d (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [A] time = 0.67, size = 157, normalized size = 0.96 \[ \frac {\frac {\left (a^2 B-3 a A b+2 b^2 B\right ) \sin (c+d x)}{(a-b)^2 (a+b)^2 (a+b \cos (c+d x))}-\frac {2 \left (2 a^2 A-3 a b B+A b^2\right ) \tanh ^{-1}\left (\frac {(a-b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac {(a B-A b) \sin (c+d x)}{(a-b) (a+b) (a+b \cos (c+d x))^2}}{2 d} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.68, size = 742, normalized size = 4.52 \[ \left [-\frac {{\left (2 \, A a^{4} - 3 \, B a^{3} b + A a^{2} b^{2} + {\left (2 \, A a^{2} b^{2} - 3 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A a^{3} b - 3 \, B a^{2} b^{2} + A a b^{3}\right )} \cos \left (d x + c\right )\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 ) - 2 \, {\left (2 \, B a^{5} - 4 \, A a^{4} b - B a^{3} b^{2} + 5 \, A a^{2} b^{3} - B a b^{4} - A b^{5} + {\left (B a^{4} b - 3 \, A a^{3} b^{2} + B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}, \frac {{\left (2 \, A a^{4} - 3 \, B a^{3} b + A a^{2} b^{2} + {\left (2 \, A a^{2} b^{2} - 3 \, B a b^{3} + A b^{4}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (2 \, A a^{3} b - 3 \, B a^{2} b^{2} + A a b^{3}\right )} \cos \left (d x + c\right )\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 (2 \, B a^{5} - 4 \, A a^{4} b - B a^{3} b^{2} + 5 \, A a^{2} b^{3} - B a b^{4} - A b^{5} + {\left (B a^{4} b - 3 \, A a^{3} b^{2} + B a^{2} b^{3} + 3 \, A a b^{4} - 2 \, B b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{6} b^{2} - 3 \, a^{4} b^{4} + 3 \, a^{2} b^{6} - b^{8}\right )} d \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{7} b - 3 \, a^{5} b^{3} + 3 \, a^{3} b^{5} - a b^{7}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} - 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} - a^{2} b^{6}\right )} d\right )}}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.79, size = 390, normalized size = 2.38 \[ \frac {\frac {{\left (2 \, A a^{2} - 3 \, B a b + A b^{2}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}}} + \frac {2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 4 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 4 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b\right )}^{2}}}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.07, size = 886, normalized size = 5.40 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 3.54, size = 248, normalized size = 1.51 \[ \frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,B\,a^2-A\,b^2+2\,B\,b^2-4\,A\,a\,b+B\,a\,b\right )}{{\left (a+b\right )}^2\,\left (a-b\right )}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A\,b^2+2\,B\,a^2+2\,B\,b^2-4\,A\,a\,b-B\,a\,b\right )}{\left (a+b\right )\,\left (a^2-2\,a\,b+b^2\right )}}{d\,\left (2\,a\,b+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a^2-2\,b^2\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (a^2-2\,a\,b+b^2\right )+a^2+b^2\right )}+\frac {\mathrm {atan}\left (\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,a-2\,b\right )\,\left (a^2-2\,a\,b+b^2\right )}{2\,\sqrt {a+b}\,{\left (a-b\right )}^{5/2}}\right )\,\left (2\,A\,a^2-3\,B\,a\,b+A\,b^2\right )}{d\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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