Integrand size = 15, antiderivative size = 49 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {3 \sin (a+b x)}{2 b}+\frac {\sin (a+b x) \tan ^2(a+b x)}{2 b} \]
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Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {2672, 294, 327, 212} \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {3 \sin (a+b x)}{2 b}+\frac {\sin (a+b x) \tan ^2(a+b x)}{2 b} \]
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Rule 212
Rule 294
Rule 327
Rule 2672
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (a+b x)\right )}{b} \\ & = \frac {\sin (a+b x) \tan ^2(a+b x)}{2 b}-\frac {3 \text {Subst}\left (\int \frac {x^2}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b} \\ & = \frac {3 \sin (a+b x)}{2 b}+\frac {\sin (a+b x) \tan ^2(a+b x)}{2 b}-\frac {3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\sin (a+b x)\right )}{2 b} \\ & = -\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {3 \sin (a+b x)}{2 b}+\frac {\sin (a+b x) \tan ^2(a+b x)}{2 b} \\ \end{align*}
Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \text {arctanh}(\sin (a+b x))}{2 b}+\frac {3 \sec (a+b x) \tan (a+b x)}{2 b}-\frac {\sin (a+b x) \tan ^2(a+b x)}{b} \]
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Time = 0.15 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.18
method | result | size |
derivativedivides | \(\frac {\frac {\sin ^{5}\left (b x +a \right )}{2 \cos \left (b x +a \right )^{2}}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{2}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(58\) |
default | \(\frac {\frac {\sin ^{5}\left (b x +a \right )}{2 \cos \left (b x +a \right )^{2}}+\frac {\left (\sin ^{3}\left (b x +a \right )\right )}{2}+\frac {3 \sin \left (b x +a \right )}{2}-\frac {3 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(58\) |
parallelrisch | \(\frac {\left (3 \cos \left (2 b x +2 a \right )+3\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (-3 \cos \left (2 b x +2 a \right )-3\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+3 \sin \left (b x +a \right )+\sin \left (3 b x +3 a \right )}{2 b \left (1+\cos \left (2 b x +2 a \right )\right )}\) | \(89\) |
risch | \(-\frac {i {\mathrm e}^{i \left (b x +a \right )}}{2 b}+\frac {i {\mathrm e}^{-i \left (b x +a \right )}}{2 b}-\frac {i \left ({\mathrm e}^{3 i \left (b x +a \right )}-{\mathrm e}^{i \left (b x +a \right )}\right )}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}+\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{2 b}-\frac {3 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{2 b}\) | \(108\) |
norman | \(\frac {\frac {3 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )}{b}-\frac {2 \left (\tan ^{3}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}+\frac {3 \left (\tan ^{5}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{b}}{\left (1+\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right ) \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{2}}+\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{2 b}-\frac {3 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{2 b}\) | \(114\) |
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Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.51 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3 \, \cos \left (b x + a\right )^{2} \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \cos \left (b x + a\right )^{2} \log \left (-\sin \left (b x + a\right ) + 1\right ) - 2 \, {\left (2 \, \cos \left (b x + a\right )^{2} + 1\right )} \sin \left (b x + a\right )}{4 \, b \cos \left (b x + a\right )^{2}} \]
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\[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=\int \sin ^{4}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \]
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Time = 0.27 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.14 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {\frac {2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + 3 \, \log \left (\sin \left (b x + a\right ) + 1\right ) - 3 \, \log \left (\sin \left (b x + a\right ) - 1\right ) - 4 \, \sin \left (b x + a\right )}{4 \, b} \]
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Time = 0.34 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.18 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {\frac {2 \, \sin \left (b x + a\right )}{\sin \left (b x + a\right )^{2} - 1} + 3 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right ) - 3 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right ) - 4 \, \sin \left (b x + a\right )}{4 \, b} \]
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Time = 3.88 (sec) , antiderivative size = 98, normalized size of antiderivative = 2.00 \[ \int \sin (a+b x) \tan ^3(a+b x) \, dx=-\frac {3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )\right )}{b}-\frac {3\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^3+3\,\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}{b\,\left (-{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {a}{2}+\frac {b\,x}{2}\right )}^2-1\right )} \]
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