Optimal. Leaf size=112 \[ \frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}+\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 b^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]
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Rubi [A] time = 0.20, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3109, 2565, 30, 2564, 2637, 2638, 3074, 206} \[ \frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}+\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a^2 b^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]
Antiderivative was successfully verified.
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Rule 30
Rule 206
Rule 2564
Rule 2565
Rule 2637
Rule 2638
Rule 3074
Rule 3109
Rubi steps
\begin {align*} \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx &=\frac {a \int \cos (x) \sin ^2(x) \, dx}{a^2+b^2}+\frac {b \int \cos ^2(x) \sin (x) \, dx}{a^2+b^2}-\frac {(a b) \int \frac {\cos (x) \sin (x)}{a \cos (x)+b \sin (x)} \, dx}{a^2+b^2}\\ &=-\frac {\left (a^2 b\right ) \int \sin (x) \, dx}{\left (a^2+b^2\right )^2}-\frac {\left (a b^2\right ) \int \cos (x) \, dx}{\left (a^2+b^2\right )^2}+\frac {\left (a^2 b^2\right ) \int \frac {1}{a \cos (x)+b \sin (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac {a \operatorname {Subst}\left (\int x^2 \, dx,x,\sin (x)\right )}{a^2+b^2}-\frac {b \operatorname {Subst}\left (\int x^2 \, dx,x,\cos (x)\right )}{a^2+b^2}\\ &=\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac {\left (a^2 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,b \cos (x)-a \sin (x)\right )}{\left (a^2+b^2\right )^2}\\ &=-\frac {a^2 b^2 \tanh ^{-1}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )}\\ \end {align*}
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Mathematica [A] time = 0.63, size = 115, normalized size = 1.03 \[ \frac {2 a^2 b^2 \tanh ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )-b}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {\left (3 b^3-9 a^2 b\right ) \cos (x)+b \left (a^2+b^2\right ) \cos (3 x)+2 a \sin (x) \left (\left (a^2+b^2\right ) \cos (2 x)-a^2+5 b^2\right )}{12 \left (a^2+b^2\right )^2} \]
Antiderivative was successfully verified.
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fricas [B] time = 0.77, size = 215, normalized size = 1.92 \[ \frac {3 \, \sqrt {a^{2} + b^{2}} a^{2} b^{2} \log \left (-\frac {2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \relax (x) - a \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) + {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + b^{2}}\right ) - 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \relax (x)^{3} + 6 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \relax (x) + 2 \, {\left (a^{5} - a^{3} b^{2} - 2 \, a b^{4} - {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{6 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 8.00, size = 192, normalized size = 1.71 \[ -\frac {a^{2} b^{2} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} - 4 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) - 2 \, a^{2} b + b^{3}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.09, size = 168, normalized size = 1.50 \[ \frac {8 a^{2} b^{2} \arctanh \left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (4 a^{4}+8 a^{2} b^{2}+4 b^{4}\right ) \sqrt {a^{2}+b^{2}}}+\frac {-2 a \,b^{2} \left (\tan ^{5}\left (\frac {x}{2}\right )\right )-2 b^{3} \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+2 \left (\frac {4}{3} a^{3}-\frac {2}{3} a \,b^{2}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+4 a^{2} b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 a \,b^{2} \tan \left (\frac {x}{2}\right )+\frac {4 a^{2} b}{3}-\frac {2 b^{3}}{3}}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [B] time = 0.46, size = 281, normalized size = 2.51 \[ -\frac {a^{2} b^{2} \log \left (\frac {b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (2 \, a^{2} b - b^{3} - \frac {3 \, a b^{2} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {6 \, a^{2} b \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - \frac {3 \, b^{3} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} - \frac {3 \, a b^{2} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {2 \, {\left (2 \, a^{3} - a b^{2}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}}\right )}} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 1.00, size = 277, normalized size = 2.47 \[ -\frac {\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (a\,b^2-2\,a^3\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,b\,\left (2\,a^2-b^2\right )}{3\,{\left (a^2+b^2\right )}^2}+\frac {2\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{a^4+2\,a^2\,b^2+b^4}-\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {2\,a^2\,b^2\,\mathrm {atanh}\left (\frac {2\,a^4\,b+2\,b^5+4\,a^2\,b^3-2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{2\,{\left (a^2+b^2\right )}^{5/2}}\right )}{{\left (a^2+b^2\right )}^{5/2}} \]
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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