Integrand size = 13, antiderivative size = 120 \[ \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx=\frac {a \left (3 a^4+10 a^2 b^2+15 b^4\right ) x}{8 \left (a^2+b^2\right )^3}-\frac {b^5 \log (b \cos (x)+a \sin (x))}{\left (a^2+b^2\right )^3}-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )} \]
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Time = 0.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {3587, 755, 837, 815, 649, 209, 266} \[ \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx=-\frac {\sin ^4(x) (a \cot (x)+b)}{4 \left (a^2+b^2\right )}-\frac {b^5 \log (a \sin (x)+b \cos (x))}{\left (a^2+b^2\right )^3}-\frac {\sin ^2(x) \left (a \left (3 a^2+7 b^2\right ) \cot (x)+4 b^3\right )}{8 \left (a^2+b^2\right )^2}+\frac {a x \left (3 a^4+10 a^2 b^2+15 b^4\right )}{8 \left (a^2+b^2\right )^3} \]
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Rule 209
Rule 266
Rule 649
Rule 755
Rule 815
Rule 837
Rule 3587
Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {1}{(a+x) \left (1+\frac {x^2}{b^2}\right )^3} \, dx,x,b \cot (x)\right )}{b} \\ & = -\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}+\frac {b \text {Subst}\left (\int \frac {-4-\frac {3 a^2}{b^2}-\frac {3 a x}{b^2}}{(a+x) \left (1+\frac {x^2}{b^2}\right )^2} \, dx,x,b \cot (x)\right )}{4 \left (a^2+b^2\right )} \\ & = -\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {b^5 \text {Subst}\left (\int \frac {\frac {3 a^4+7 a^2 b^2+8 b^4}{b^6}+\frac {a \left (3 a^2+7 b^2\right ) x}{b^6}}{(a+x) \left (1+\frac {x^2}{b^2}\right )} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^2} \\ & = -\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {b^5 \text {Subst}\left (\int \left (\frac {8}{\left (a^2+b^2\right ) (a+x)}+\frac {3 a^5+10 a^3 b^2+15 a b^4-8 b^4 x}{b^4 \left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^2} \\ & = -\frac {b^5 \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}-\frac {b \text {Subst}\left (\int \frac {3 a^5+10 a^3 b^2+15 a b^4-8 b^4 x}{b^2+x^2} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^3} \\ & = -\frac {b^5 \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )}+\frac {b^5 \text {Subst}\left (\int \frac {x}{b^2+x^2} \, dx,x,b \cot (x)\right )}{\left (a^2+b^2\right )^3}-\frac {\left (a b \left (3 a^4+10 a^2 b^2+15 b^4\right )\right ) \text {Subst}\left (\int \frac {1}{b^2+x^2} \, dx,x,b \cot (x)\right )}{8 \left (a^2+b^2\right )^3} \\ & = \frac {a \left (3 a^4+10 a^2 b^2+15 b^4\right ) x}{8 \left (a^2+b^2\right )^3}-\frac {b^5 \log (a+b \cot (x))}{\left (a^2+b^2\right )^3}-\frac {b^5 \log (\sin (x))}{\left (a^2+b^2\right )^3}-\frac {\left (4 b^3+a \left (3 a^2+7 b^2\right ) \cot (x)\right ) \sin ^2(x)}{8 \left (a^2+b^2\right )^2}-\frac {(b+a \cot (x)) \sin ^4(x)}{4 \left (a^2+b^2\right )} \\ \end{align*}
Time = 0.42 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx=\frac {12 a^5 x+40 a^3 b^2 x+60 a b^4 x+4 b \left (a^4+4 a^2 b^2+3 b^4\right ) \cos (2 x)-b \left (a^2+b^2\right )^2 \cos (4 x)-32 b^5 \log (b \cos (x)+a \sin (x))-8 a^5 \sin (2 x)-24 a^3 b^2 \sin (2 x)-16 a b^4 \sin (2 x)+a^5 \sin (4 x)+2 a^3 b^2 \sin (4 x)+a b^4 \sin (4 x)}{32 \left (a^2+b^2\right )^3} \]
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Time = 4.31 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.42
method | result | size |
default | \(-\frac {b^{5} \ln \left (\tan \left (x \right ) a +b \right )}{\left (a^{2}+b^{2}\right )^{3}}+\frac {\frac {\left (-\frac {7}{4} a^{3} b^{2}-\frac {9}{8} a \,b^{4}-\frac {5}{8} a^{5}\right ) \tan \left (x \right )^{3}+\left (\frac {1}{2} a^{4} b +\frac {3}{2} a^{2} b^{3}+b^{5}\right ) \tan \left (x \right )^{2}+\left (-\frac {3}{8} a^{5}-\frac {5}{4} a^{3} b^{2}-\frac {7}{8} a \,b^{4}\right ) \tan \left (x \right )+\frac {a^{4} b}{4}+a^{2} b^{3}+\frac {3 b^{5}}{4}}{\left (\tan \left (x \right )^{2}+1\right )^{2}}+\frac {b^{5} \ln \left (\tan \left (x \right )^{2}+1\right )}{2}+\frac {\left (3 a^{5}+10 a^{3} b^{2}+15 a \,b^{4}\right ) \arctan \left (\tan \left (x \right )\right )}{8}}{\left (a^{2}+b^{2}\right )^{3}}\) | \(170\) |
risch | \(\frac {9 i x a b}{8 \left (3 i a^{2} b -i b^{3}+a^{3}-3 a \,b^{2}\right )}+\frac {3 x \,a^{2}}{8 \left (3 i a^{2} b -i b^{3}+a^{3}-3 a \,b^{2}\right )}-\frac {x \,b^{2}}{3 i a^{2} b -i b^{3}+a^{3}-3 a \,b^{2}}-\frac {3 \,{\mathrm e}^{2 i x} b}{16 \left (2 i a b +a^{2}-b^{2}\right )}+\frac {i a \,{\mathrm e}^{2 i x}}{16 i a b +8 a^{2}-8 b^{2}}-\frac {3 \,{\mathrm e}^{-2 i x} b}{16 \left (-i b +a \right )^{2}}-\frac {i {\mathrm e}^{-2 i x} a}{8 \left (-i b +a \right )^{2}}+\frac {2 i b^{5} x}{a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}}-\frac {b^{5} \ln \left ({\mathrm e}^{2 i x}+\frac {i b -a}{i b +a}\right )}{a^{6}+3 b^{2} a^{4}+3 b^{4} a^{2}+b^{6}}+\frac {b \cos \left (4 x \right )}{-32 a^{2}-32 b^{2}}-\frac {a \sin \left (4 x \right )}{32 \left (-a^{2}-b^{2}\right )}\) | \(303\) |
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Time = 0.31 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.53 \[ \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx=-\frac {4 \, b^{5} \log \left (2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}\right ) + 2 \, {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{4} - 4 \, {\left (a^{4} b + 3 \, a^{2} b^{3} + 2 \, b^{5}\right )} \cos \left (x\right )^{2} - {\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x - {\left (2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{3} - {\left (5 \, a^{5} + 14 \, a^{3} b^{2} + 9 \, a b^{4}\right )} \cos \left (x\right )\right )} \sin \left (x\right )}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \]
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\[ \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx=\int \frac {\sin ^{4}{\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \]
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Leaf count of result is larger than twice the leaf count of optimal. 244 vs. \(2 (114) = 228\).
Time = 0.35 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.03 \[ \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx=-\frac {b^{5} \log \left (a \tan \left (x\right ) + b\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac {b^{5} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac {{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac {{\left (5 \, a^{3} + 9 \, a b^{2}\right )} \tan \left (x\right )^{3} - 2 \, a^{2} b - 6 \, b^{3} - 4 \, {\left (a^{2} b + 2 \, b^{3}\right )} \tan \left (x\right )^{2} + {\left (3 \, a^{3} + 7 \, a b^{2}\right )} \tan \left (x\right )}{8 \, {\left ({\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (x\right )^{4} + a^{4} + 2 \, a^{2} b^{2} + b^{4} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \tan \left (x\right )^{2}\right )}} \]
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Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (114) = 228\).
Time = 0.27 (sec) , antiderivative size = 273, normalized size of antiderivative = 2.28 \[ \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx=-\frac {a b^{5} \log \left ({\left | a \tan \left (x\right ) + b \right |}\right )}{a^{7} + 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} + a b^{6}} + \frac {b^{5} \log \left (\tan \left (x\right )^{2} + 1\right )}{2 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} + \frac {{\left (3 \, a^{5} + 10 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} - \frac {6 \, b^{5} \tan \left (x\right )^{4} + 5 \, a^{5} \tan \left (x\right )^{3} + 14 \, a^{3} b^{2} \tan \left (x\right )^{3} + 9 \, a b^{4} \tan \left (x\right )^{3} - 4 \, a^{4} b \tan \left (x\right )^{2} - 12 \, a^{2} b^{3} \tan \left (x\right )^{2} + 4 \, b^{5} \tan \left (x\right )^{2} + 3 \, a^{5} \tan \left (x\right ) + 10 \, a^{3} b^{2} \tan \left (x\right ) + 7 \, a b^{4} \tan \left (x\right ) - 2 \, a^{4} b - 8 \, a^{2} b^{3}}{8 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )} {\left (\tan \left (x\right )^{2} + 1\right )}^{2}} \]
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Time = 12.26 (sec) , antiderivative size = 263, normalized size of antiderivative = 2.19 \[ \int \frac {\sin ^4(x)}{a+b \cot (x)} \, dx=\frac {\frac {a^2\,b+3\,b^3}{4\,{\left (a^2+b^2\right )}^2}-\frac {{\mathrm {tan}\left (x\right )}^3\,\left (5\,a^3+9\,a\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (x\right )}^2\,\left (a^2\,b+2\,b^3\right )}{2\,{\left (a^2+b^2\right )}^2}-\frac {a\,\mathrm {tan}\left (x\right )\,\left (3\,a^2+7\,b^2\right )}{8\,\left (a^4+2\,a^2\,b^2+b^4\right )}}{{\mathrm {tan}\left (x\right )}^4+2\,{\mathrm {tan}\left (x\right )}^2+1}-\frac {b^5\,\ln \left (b+a\,\mathrm {tan}\left (x\right )\right )}{{\left (a^2+b^2\right )}^3}+\frac {\ln \left (\mathrm {tan}\left (x\right )-\mathrm {i}\right )\,\left (-3\,a^2+a\,b\,9{}\mathrm {i}+8\,b^2\right )}{16\,\left (-a^3\,1{}\mathrm {i}-3\,a^2\,b+a\,b^2\,3{}\mathrm {i}+b^3\right )}+\frac {\ln \left (\mathrm {tan}\left (x\right )+1{}\mathrm {i}\right )\,\left (-a^2\,3{}\mathrm {i}+9\,a\,b+b^2\,8{}\mathrm {i}\right )}{16\,\left (-a^3-a^2\,b\,3{}\mathrm {i}+3\,a\,b^2+b^3\,1{}\mathrm {i}\right )} \]
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