Integrand size = 23, antiderivative size = 87 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\left (a^2 c-b^2 c-2 a b d\right ) x-\frac {\left (2 a b c+a^2 d-b^2 d\right ) \log (\cos (e+f x))}{f}+\frac {b (b c+a d) \tan (e+f x)}{f}+\frac {d (a+b \tan (e+f x))^2}{2 f} \]
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Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3609, 3606, 3556} \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=-\frac {\left (a^2 d+2 a b c-b^2 d\right ) \log (\cos (e+f x))}{f}+x \left (a^2 c-2 a b d-b^2 c\right )+\frac {b (a d+b c) \tan (e+f x)}{f}+\frac {d (a+b \tan (e+f x))^2}{2 f} \]
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Rule 3556
Rule 3606
Rule 3609
Rubi steps \begin{align*} \text {integral}& = \frac {d (a+b \tan (e+f x))^2}{2 f}+\int (a+b \tan (e+f x)) (a c-b d+(b c+a d) \tan (e+f x)) \, dx \\ & = \left (a^2 c-b^2 c-2 a b d\right ) x+\frac {b (b c+a d) \tan (e+f x)}{f}+\frac {d (a+b \tan (e+f x))^2}{2 f}+\left (2 a b c+a^2 d-b^2 d\right ) \int \tan (e+f x) \, dx \\ & = \left (a^2 c-b^2 c-2 a b d\right ) x-\frac {\left (2 a b c+a^2 d-b^2 d\right ) \log (\cos (e+f x))}{f}+\frac {b (b c+a d) \tan (e+f x)}{f}+\frac {d (a+b \tan (e+f x))^2}{2 f} \\ \end{align*}
Result contains complex when optimal does not.
Time = 0.48 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.10 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {(a+i b)^2 (-i c+d) \log (i-\tan (e+f x))+(a-i b)^2 (i c+d) \log (i+\tan (e+f x))+2 b (b c+2 a d) \tan (e+f x)+b^2 d \tan ^2(e+f x)}{2 f} \]
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Time = 0.19 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.03
method | result | size |
norman | \(\left (a^{2} c -2 a b d -b^{2} c \right ) x +\frac {b \left (2 a d +b c \right ) \tan \left (f x +e \right )}{f}+\frac {b^{2} d \left (\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {\left (a^{2} d +2 a b c -b^{2} d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f}\) | \(90\) |
derivativedivides | \(\frac {\frac {b^{2} d \left (\tan ^{2}\left (f x +e \right )\right )}{2}+2 a b d \tan \left (f x +e \right )+b^{2} c \tan \left (f x +e \right )+\frac {\left (a^{2} d +2 a b c -b^{2} d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a^{2} c -2 a b d -b^{2} c \right ) \arctan \left (\tan \left (f x +e \right )\right )}{f}\) | \(97\) |
default | \(\frac {\frac {b^{2} d \left (\tan ^{2}\left (f x +e \right )\right )}{2}+2 a b d \tan \left (f x +e \right )+b^{2} c \tan \left (f x +e \right )+\frac {\left (a^{2} d +2 a b c -b^{2} d \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}+\left (a^{2} c -2 a b d -b^{2} c \right ) \arctan \left (\tan \left (f x +e \right )\right )}{f}\) | \(97\) |
parts | \(a^{2} c x +\frac {\left (2 a b d +b^{2} c \right ) \left (\tan \left (f x +e \right )-\arctan \left (\tan \left (f x +e \right )\right )\right )}{f}+\frac {\left (a^{2} d +2 a b c \right ) \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f}+\frac {b^{2} d \left (\frac {\left (\tan ^{2}\left (f x +e \right )\right )}{2}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2}\right )}{f}\) | \(98\) |
parallelrisch | \(\frac {2 a^{2} c f x -4 a b d f x -2 b^{2} c f x +b^{2} d \left (\tan ^{2}\left (f x +e \right )\right )+\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a^{2} d +2 \ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a b c -\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) b^{2} d +4 a b d \tan \left (f x +e \right )+2 b^{2} c \tan \left (f x +e \right )}{2 f}\) | \(115\) |
risch | \(\frac {2 i a^{2} d e}{f}+i a^{2} d x -i b^{2} d x +a^{2} c x -2 a b d x -b^{2} c x -\frac {2 i b^{2} d e}{f}+2 i a b c x +\frac {4 i a b c e}{f}+\frac {2 i b \left (2 a d \,{\mathrm e}^{2 i \left (f x +e \right )}+b c \,{\mathrm e}^{2 i \left (f x +e \right )}-i b d \,{\mathrm e}^{2 i \left (f x +e \right )}+2 a d +b c \right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a^{2} d}{f}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a b c}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) b^{2} d}{f}\) | \(204\) |
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Time = 0.25 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {b^{2} d \tan \left (f x + e\right )^{2} - 2 \, {\left (2 \, a b d - {\left (a^{2} - b^{2}\right )} c\right )} f x - {\left (2 \, a b c + {\left (a^{2} - b^{2}\right )} d\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right ) + 2 \, {\left (b^{2} c + 2 \, a b d\right )} \tan \left (f x + e\right )}{2 \, f} \]
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Time = 0.11 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.64 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\begin {cases} a^{2} c x + \frac {a^{2} d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {a b c \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{f} - 2 a b d x + \frac {2 a b d \tan {\left (e + f x \right )}}{f} - b^{2} c x + \frac {b^{2} c \tan {\left (e + f x \right )}}{f} - \frac {b^{2} d \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {b^{2} d \tan ^{2}{\left (e + f x \right )}}{2 f} & \text {for}\: f \neq 0 \\x \left (a + b \tan {\left (e \right )}\right )^{2} \left (c + d \tan {\left (e \right )}\right ) & \text {otherwise} \end {cases} \]
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Time = 0.35 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.06 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {b^{2} d \tan \left (f x + e\right )^{2} - 2 \, {\left (2 \, a b d - {\left (a^{2} - b^{2}\right )} c\right )} {\left (f x + e\right )} + {\left (2 \, a b c + {\left (a^{2} - b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right ) + 2 \, {\left (b^{2} c + 2 \, a b d\right )} \tan \left (f x + e\right )}{2 \, f} \]
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Leaf count of result is larger than twice the leaf count of optimal. 811 vs. \(2 (85) = 170\).
Time = 0.69 (sec) , antiderivative size = 811, normalized size of antiderivative = 9.32 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\text {Too large to display} \]
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Time = 6.22 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.05 \[ \int (a+b \tan (e+f x))^2 (c+d \tan (e+f x)) \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (c\,b^2+2\,a\,d\,b\right )}{f}-x\,\left (-c\,a^2+2\,d\,a\,b+c\,b^2\right )+\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )\,\left (\frac {d\,a^2}{2}+c\,a\,b-\frac {d\,b^2}{2}\right )}{f}+\frac {b^2\,d\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,f} \]
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