Integrand size = 23, antiderivative size = 51 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=-\frac {1}{2} b (4 a+3 b) x+\frac {b^2 \cos (e+f x) \sin (e+f x)}{2 f}+\frac {(a+b)^2 \tan (e+f x)}{f} \]
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Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3270, 398, 393, 209} \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {(a+b)^2 \tan (e+f x)}{f}-\frac {1}{2} b x (4 a+3 b)+\frac {b^2 \sin (e+f x) \cos (e+f x)}{2 f} \]
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Rule 209
Rule 393
Rule 398
Rule 3270
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (a+(a+b) x^2\right )^2}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left ((a+b)^2-\frac {b (2 a+b)+2 b (a+b) x^2}{\left (1+x^2\right )^2}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {(a+b)^2 \tan (e+f x)}{f}-\frac {\text {Subst}\left (\int \frac {b (2 a+b)+2 b (a+b) x^2}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {b^2 \cos (e+f x) \sin (e+f x)}{2 f}+\frac {(a+b)^2 \tan (e+f x)}{f}-\frac {(b (4 a+3 b)) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{2 f} \\ & = -\frac {1}{2} b (4 a+3 b) x+\frac {b^2 \cos (e+f x) \sin (e+f x)}{2 f}+\frac {(a+b)^2 \tan (e+f x)}{f} \\ \end{align*}
Time = 0.91 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.94 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {-2 b (4 a+3 b) (e+f x)+b^2 \sin (2 (e+f x))+4 (a+b)^2 \tan (e+f x)}{4 f} \]
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Time = 1.33 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.27
method | result | size |
parallelrisch | \(\frac {b^{2} \sin \left (3 f x +3 e \right )-16 b f \left (a +\frac {3 b}{4}\right ) x \cos \left (f x +e \right )+8 \sin \left (f x +e \right ) \left (a^{2}+2 a b +\frac {9}{8} b^{2}\right )}{8 f \cos \left (f x +e \right )}\) | \(65\) |
derivativedivides | \(\frac {a^{2} \tan \left (f x +e \right )+2 a b \left (\tan \left (f x +e \right )-f x -e \right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )}{f}\) | \(87\) |
default | \(\frac {a^{2} \tan \left (f x +e \right )+2 a b \left (\tan \left (f x +e \right )-f x -e \right )+b^{2} \left (\frac {\sin ^{5}\left (f x +e \right )}{\cos \left (f x +e \right )}+\left (\sin ^{3}\left (f x +e \right )+\frac {3 \sin \left (f x +e \right )}{2}\right ) \cos \left (f x +e \right )-\frac {3 f x}{2}-\frac {3 e}{2}\right )}{f}\) | \(87\) |
risch | \(-2 a b x -\frac {3 b^{2} x}{2}-\frac {i b^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{8 f}+\frac {i b^{2} {\mathrm e}^{-2 i \left (f x +e \right )}}{8 f}+\frac {2 i a^{2}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {4 i a b}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {2 i b^{2}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(114\) |
norman | \(\frac {\left (2 a b +\frac {3}{2} b^{2}\right ) x +\left (-6 a b -\frac {9}{2} b^{2}\right ) x \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-2 a b -\frac {3}{2} b^{2}\right ) x \left (\tan ^{10}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (6 a b +\frac {9}{2} b^{2}\right ) x \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (-4 a b -3 b^{2}\right ) x \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+\left (4 a b +3 b^{2}\right ) x \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {8 \left (a^{2}+2 a b +b^{2}\right ) \left (\tan ^{3}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {8 \left (a^{2}+2 a b +b^{2}\right ) \left (\tan ^{7}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {\left (2 a^{2}+4 a b +3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}-\frac {\left (2 a^{2}+4 a b +3 b^{2}\right ) \left (\tan ^{9}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}-\frac {2 \left (6 a^{2}+12 a b +5 b^{2}\right ) \left (\tan ^{5}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{f}}{\left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \left (1+\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )^{4}}\) | \(305\) |
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Time = 0.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.33 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=-\frac {{\left (4 \, a b + 3 \, b^{2}\right )} f x \cos \left (f x + e\right ) - {\left (b^{2} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sin \left (f x + e\right )}{2 \, f \cos \left (f x + e\right )} \]
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\[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{2} \sec ^{2}{\left (e + f x \right )}\, dx \]
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Time = 0.33 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=-\frac {4 \, {\left (f x + e - \tan \left (f x + e\right )\right )} a b + {\left (3 \, f x + 3 \, e - \frac {\tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1} - 2 \, \tan \left (f x + e\right )\right )} b^{2} - 2 \, a^{2} \tan \left (f x + e\right )}{2 \, f} \]
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Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.51 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {2 \, a^{2} \tan \left (f x + e\right ) + 4 \, a b \tan \left (f x + e\right ) + 2 \, b^{2} \tan \left (f x + e\right ) - {\left (4 \, a b + 3 \, b^{2}\right )} {\left (f x + e\right )} + \frac {b^{2} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{2} + 1}}{2 \, f} \]
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Time = 13.84 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.45 \[ \int \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )\,{\left (a+b\right )}^2}{f}+\frac {b^2\,\sin \left (2\,e+2\,f\,x\right )}{4\,f}-\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (e+f\,x\right )\,\left (4\,a+3\,b\right )}{2\,\left (\frac {3\,b^2}{2}+2\,a\,b\right )}\right )\,\left (4\,a+3\,b\right )}{2\,f} \]
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