Integrand size = 20, antiderivative size = 112 \[ \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {a^2 b^2 \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )} \]
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Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3188, 2645, 30, 2644, 2717, 2718, 3153, 212} \[ \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {a^2 b^2 \text {arctanh}\left (\frac {b \cos (x)-a \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {a \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {b \cos ^3(x)}{3 \left (a^2+b^2\right )}+\frac {a^2 b \cos (x)}{\left (a^2+b^2\right )^2} \]
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Rule 30
Rule 212
Rule 2644
Rule 2645
Rule 2717
Rule 2718
Rule 3153
Rule 3188
Rubi steps \begin{align*} \text {integral}& = \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 \text {Subst}\left (\int x^2 \, dx,x,\sin (x)\right )}{a^2+b^2}-\frac {b \text {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 ) \text {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 \text {arctanh}\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*}
Time = 0.88 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {2 a^2 b^2 \text {arctanh}\left (\frac {-b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {\left (-9 a^2 b+3 b^3\right ) \cos (x)+b \left (a^2+b^2\right ) \cos (3 x)+2 a \left (-a^2+5 b^2+\left (a^2+b^2\right ) \cos (2 x)\right ) \sin (x)}{12 \left (a^2+b^2\right )^2} \]
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Time = 0.59 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.47
method | result | size |
default | \(-\frac {2 \left (a \,b^{2} \tan \left (\frac {x}{2}\right )^{5}+b^{3} \tan \left (\frac {x}{2}\right )^{4}+\left (-\frac {4}{3} a^{3}+\frac {2}{3} a \,b^{2}\right ) \tan \left (\frac {x}{2}\right )^{3}-2 \tan \left (\frac {x}{2}\right )^{2} a^{2} b +\tan \left (\frac {x}{2}\right ) a \,b^{2}-\frac {2 a^{2} b}{3}+\frac {b^{3}}{3}\right )}{\left (a^{4}+2 a^{2} b^{2}+b^{4}\right ) \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{3}}+\frac {8 a^{2} b^{2} \operatorname {arctanh}\left (\frac {2 a \tan \left (\frac {x}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (4 a^{4}+8 a^{2} b^{2}+4 b^{4}\right ) \sqrt {a^{2}+b^{2}}}\) | \(165\) |
risch | \(\frac {{\mathrm e}^{i x} b}{-16 i b a +8 a^{2}-8 b^{2}}-\frac {i {\mathrm e}^{i x} a}{8 \left (-2 i b a +a^{2}-b^{2}\right )}+\frac {{\mathrm e}^{-i x} b}{8 \left (i b +a \right )^{2}}+\frac {i {\mathrm e}^{-i x} a}{8 \left (i b +a \right )^{2}}-\frac {b^{2} a^{2} \ln \left ({\mathrm e}^{i x}-\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {b^{2} a^{2} \ln \left ({\mathrm e}^{i x}+\frac {i a^{5}+2 i a^{3} b^{2}+i a \,b^{4}-a^{4} b -2 a^{2} b^{3}-b^{5}}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {5}{2}}}+\frac {b \cos \left (3 x \right )}{-12 a^{2}-12 b^{2}}+\frac {a \sin \left (3 x \right )}{-12 a^{2}-12 b^{2}}\) | \(276\) |
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Leaf count of result is larger than twice the leaf count of optimal. 215 vs. \(2 (104) = 208\).
Time = 0.26 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.92 \[ \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\frac {3 \, \sqrt {a^{2} + b^{2}} a^{2} b^{2} \log \left (-\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a^{2} - b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cos \left (x\right ) - a \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) + {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + b^{2}}\right ) - 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{3} + 6 \, {\left (a^{4} b + a^{2} b^{3}\right )} \cos \left (x\right ) + 2 \, {\left (a^{5} - a^{3} b^{2} - 2 \, a b^{4} - {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \]
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Timed out. \[ \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx=\text {Timed out} \]
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Leaf count of result is larger than twice the leaf count of optimal. 281 vs. \(2 (104) = 208\).
Time = 0.30 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.51 \[ \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {a^{2} b^{2} \log \left (\frac {b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{b - \frac {a \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} + \frac {2 \, {\left (2 \, a^{2} b - b^{3} - \frac {3 \, a b^{2} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {6 \, a^{2} b \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} - \frac {3 \, b^{3} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} - \frac {3 \, a b^{2} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {2 \, {\left (2 \, a^{3} - a b^{2}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}}\right )}} \]
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Time = 0.32 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.71 \[ \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {a^{2} b^{2} \log \left (\frac {{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, a \tan \left (\frac {1}{2} \, x\right ) - 2 \, b + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{4} - 4 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 6 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right ) - 2 \, a^{2} b + b^{3}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \]
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Time = 23.26 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.47 \[ \int \frac {\cos ^2(x) \sin ^2(x)}{a \cos (x)+b \sin (x)} \, dx=-\frac {\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (a\,b^2-2\,a^3\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {2\,b\,\left (2\,a^2-b^2\right )}{3\,{\left (a^2+b^2\right )}^2}+\frac {2\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{a^4+2\,a^2\,b^2+b^4}-\frac {4\,a^2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {2\,a^2\,b^2\,\mathrm {atanh}\left (\frac {2\,a^4\,b+2\,b^5+4\,a^2\,b^3-2\,a\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{2\,{\left (a^2+b^2\right )}^{5/2}}\right )}{{\left (a^2+b^2\right )}^{5/2}} \]
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