Integrand size = 15, antiderivative size = 38 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b} \]
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Time = 0.03 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2672, 308, 212} \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin ^3(a+b x)}{3 b}-\frac {\sin (a+b x)}{b} \]
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Rule 212
Rule 308
Rule 2672
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4}{1-x^2} \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (-1-x^2+\frac {1}{1-x^2}\right ) \, dx,x,\sin (a+b x)\right )}{b} \\ & = -\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b}+\frac {\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {\text {arctanh}(\sin (a+b x))}{b}-\frac {\sin (a+b x)}{b}-\frac {\sin ^3(a+b x)}{3 b} \]
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Time = 0.12 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.00
method | result | size |
derivativedivides | \(\frac {-\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{3}-\sin \left (b x +a \right )+\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}\) | \(38\) |
default | \(\frac {-\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{3}-\sin \left (b x +a \right )+\ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{b}\) | \(38\) |
parallelrisch | \(\frac {12 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )-12 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )-15 \sin \left (b x +a \right )+\sin \left (3 b x +3 a \right )}{12 b}\) | \(52\) |
risch | \(\frac {5 i {\mathrm e}^{i \left (b x +a \right )}}{8 b}-\frac {5 i {\mathrm e}^{-i \left (b x +a \right )}}{8 b}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{b}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{b}+\frac {\sin \left (3 b x +3 a \right )}{12 b}\) | \(81\) |
norman | \(\frac {-\frac {2 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}-\frac {20 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{3 b}-\frac {2 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )^{3}}+\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{b}-\frac {\ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{b}\) | \(98\) |
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Time = 0.30 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {2 \, {\left (\cos \left (b x + a\right )^{2} - 4\right )} \sin \left (b x + a\right ) + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (-\sin \left (b x + a\right ) + 1\right )}{6 \, b} \]
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Timed out. \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\text {Timed out} \]
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Time = 0.26 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.21 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=-\frac {2 \, \sin \left (b x + a\right )^{3} - 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 3 \, \log \left (\sin \left (b x + a\right ) - 1\right ) + 6 \, \sin \left (b x + a\right )}{6 \, b} \]
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Time = 0.31 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.26 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=-\frac {2 \, \sin \left (b x + a\right )^{3} - 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right ) + 6 \, \sin \left (b x + a\right )}{6 \, b} \]
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Time = 0.34 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.39 \[ \int \sin ^3(a+b x) \tan (a+b x) \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {a}{2}+\frac {b\,x}{2}\right )}{\cos \left (\frac {a}{2}+\frac {b\,x}{2}\right )}\right )}{b}-\frac {5\,\sin \left (a+b\,x\right )}{4\,b}+\frac {\sin \left (3\,a+3\,b\,x\right )}{12\,b} \]
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