Integrand size = 21, antiderivative size = 53 \[ \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {(a-b) \log (\cos (e+f x))}{f}+\frac {(a-b) \tan ^2(e+f x)}{2 f}+\frac {b \tan ^4(e+f x)}{4 f} \]
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Time = 0.05 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3712, 3554, 3556} \[ \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {(a-b) \tan ^2(e+f x)}{2 f}+\frac {(a-b) \log (\cos (e+f x))}{f}+\frac {b \tan ^4(e+f x)}{4 f} \]
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Rule 3554
Rule 3556
Rule 3712
Rubi steps \begin{align*} \text {integral}& = \frac {b \tan ^4(e+f x)}{4 f}+(a-b) \int \tan ^3(e+f x) \, dx \\ & = \frac {(a-b) \tan ^2(e+f x)}{2 f}+\frac {b \tan ^4(e+f x)}{4 f}+(-a+b) \int \tan (e+f x) \, dx \\ & = \frac {(a-b) \log (\cos (e+f x))}{f}+\frac {(a-b) \tan ^2(e+f x)}{2 f}+\frac {b \tan ^4(e+f x)}{4 f} \\ \end{align*}
Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.23 \[ \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {a \left (2 \log (\cos (e+f x))+\tan ^2(e+f x)\right )}{2 f}-\frac {b \left (4 \log (\cos (e+f x))+2 \tan ^2(e+f x)-\tan ^4(e+f x)\right )}{4 f} \]
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Time = 0.07 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\frac {b \tan \left (f x +e \right )^{4}}{4 f}+\frac {\left (a -b \right ) \tan \left (f x +e \right )^{2}}{2 f}-\frac {\left (a -b \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2 f}\) | \(55\) |
derivativedivides | \(\frac {\frac {b \tan \left (f x +e \right )^{4}}{4}+\frac {a \tan \left (f x +e \right )^{2}}{2}-\frac {b \tan \left (f x +e \right )^{2}}{2}+\frac {\left (-a +b \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}}{f}\) | \(57\) |
default | \(\frac {\frac {b \tan \left (f x +e \right )^{4}}{4}+\frac {a \tan \left (f x +e \right )^{2}}{2}-\frac {b \tan \left (f x +e \right )^{2}}{2}+\frac {\left (-a +b \right ) \ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}}{f}\) | \(57\) |
parallelrisch | \(-\frac {-b \tan \left (f x +e \right )^{4}-2 a \tan \left (f x +e \right )^{2}+2 b \tan \left (f x +e \right )^{2}+2 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) a -2 \ln \left (1+\tan \left (f x +e \right )^{2}\right ) b}{4 f}\) | \(68\) |
parts | \(\frac {a \left (\frac {\tan \left (f x +e \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}+\frac {b \left (\frac {\tan \left (f x +e \right )^{4}}{4}-\frac {\tan \left (f x +e \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (f x +e \right )^{2}\right )}{2}\right )}{f}\) | \(70\) |
risch | \(-i x a +i x b -\frac {2 i a e}{f}+\frac {2 i b e}{f}+\frac {2 a \,{\mathrm e}^{6 i \left (f x +e \right )}-4 b \,{\mathrm e}^{6 i \left (f x +e \right )}+4 a \,{\mathrm e}^{4 i \left (f x +e \right )}-4 b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-4 b \,{\mathrm e}^{2 i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) b}{f}\) | \(152\) |
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Time = 0.26 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.96 \[ \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {b \tan \left (f x + e\right )^{4} + 2 \, {\left (a - b\right )} \tan \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{4 \, f} \]
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Leaf count of result is larger than twice the leaf count of optimal. 88 vs. \(2 (42) = 84\).
Time = 0.12 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.66 \[ \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\begin {cases} - \frac {a \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a \tan ^{2}{\left (e + f x \right )}}{2 f} + \frac {b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {b \tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {b \tan ^{2}{\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \tan ^{2}{\left (e \right )}\right ) \tan ^{3}{\left (e \right )} & \text {otherwise} \end {cases} \]
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Time = 0.22 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.32 \[ \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {2 \, {\left (a - b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - \frac {2 \, {\left (a - 2 \, b\right )} \sin \left (f x + e\right )^{2} - 2 \, a + 3 \, b}{\sin \left (f x + e\right )^{4} - 2 \, \sin \left (f x + e\right )^{2} + 1}}{4 \, f} \]
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Leaf count of result is larger than twice the leaf count of optimal. 890 vs. \(2 (49) = 98\).
Time = 1.38 (sec) , antiderivative size = 890, normalized size of antiderivative = 16.79 \[ \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\text {Too large to display} \]
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Time = 11.75 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.08 \[ \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right ) \, dx=\frac {b\,{\mathrm {tan}\left (e+f\,x\right )}^4}{4\,f}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {a}{2}-\frac {b}{2}\right )}{f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (\frac {a}{2}-\frac {b}{2}\right )}{f} \]
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