Integrand size = 23, antiderivative size = 53 \[ \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {a^2 \tan (e+f x)}{f}+\frac {2 a (a+b) \tan ^3(e+f x)}{3 f}+\frac {(a+b)^2 \tan ^5(e+f x)}{5 f} \]
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Time = 0.06 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3270, 200} \[ \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {a^2 \tan (e+f x)}{f}+\frac {(a+b)^2 \tan ^5(e+f x)}{5 f}+\frac {2 a (a+b) \tan ^3(e+f x)}{3 f} \]
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Rule 200
Rule 3270
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (a+(a+b) x^2\right )^2 \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (a^2+2 a (a+b) x^2+(a+b)^2 x^4\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {a^2 \tan (e+f x)}{f}+\frac {2 a (a+b) \tan ^3(e+f x)}{3 f}+\frac {(a+b)^2 \tan ^5(e+f x)}{5 f} \\ \end{align*}
Time = 1.33 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26 \[ \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {\left (8 a^2-4 a b+3 b^2+\left (4 a^2-2 a b-6 b^2\right ) \sec ^2(e+f x)+3 (a+b)^2 \sec ^4(e+f x)\right ) \tan (e+f x)}{15 f} \]
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Leaf count of result is larger than twice the leaf count of optimal. \(100\) vs. \(2(49)=98\).
Time = 1.01 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.91
method | result | size |
derivativedivides | \(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin ^{3}\left (f x +e \right )}{5 \cos \left (f x +e \right )^{5}}+\frac {2 \left (\sin ^{3}\left (f x +e \right )\right )}{15 \cos \left (f x +e \right )^{3}}\right )+\frac {b^{2} \left (\sin ^{5}\left (f x +e \right )\right )}{5 \cos \left (f x +e \right )^{5}}}{f}\) | \(101\) |
default | \(\frac {-a^{2} \left (-\frac {8}{15}-\frac {\left (\sec ^{4}\left (f x +e \right )\right )}{5}-\frac {4 \left (\sec ^{2}\left (f x +e \right )\right )}{15}\right ) \tan \left (f x +e \right )+2 a b \left (\frac {\sin ^{3}\left (f x +e \right )}{5 \cos \left (f x +e \right )^{5}}+\frac {2 \left (\sin ^{3}\left (f x +e \right )\right )}{15 \cos \left (f x +e \right )^{3}}\right )+\frac {b^{2} \left (\sin ^{5}\left (f x +e \right )\right )}{5 \cos \left (f x +e \right )^{5}}}{f}\) | \(101\) |
parallelrisch | \(-\frac {2 \left (a^{2} \left (\tan ^{8}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {4 a \left (a -2 b \right ) \left (\tan ^{6}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+\left (\frac {58}{15} a^{2}+\frac {16}{15} a b +\frac {16}{5} b^{2}\right ) \left (\tan ^{4}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-\frac {4 a \left (a -2 b \right ) \left (\tan ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{3}+a^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\) | \(125\) |
risch | \(\frac {2 i \left (15 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-60 a b \,{\mathrm e}^{6 i \left (f x +e \right )}+80 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+20 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+30 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+40 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-20 a b \,{\mathrm e}^{2 i \left (f x +e \right )}+8 a^{2}-4 a b +3 b^{2}\right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{5}}\) | \(130\) |
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Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.57 \[ \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {{\left ({\left (8 \, a^{2} - 4 \, a b + 3 \, b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (2 \, a^{2} - a b - 3 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}\right )} \sin \left (f x + e\right )}{15 \, f \cos \left (f x + e\right )^{5}} \]
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Timed out. \[ \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\text {Timed out} \]
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Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 10 \, {\left (a^{2} + a b\right )} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \]
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Time = 0.36 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.51 \[ \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {3 \, a^{2} \tan \left (f x + e\right )^{5} + 6 \, a b \tan \left (f x + e\right )^{5} + 3 \, b^{2} \tan \left (f x + e\right )^{5} + 10 \, a^{2} \tan \left (f x + e\right )^{3} + 10 \, a b \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right )}{15 \, f} \]
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Time = 13.78 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \sec ^6(e+f x) \left (a+b \sin ^2(e+f x)\right )^2 \, dx=\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )+\frac {{\mathrm {tan}\left (e+f\,x\right )}^5\,{\left (a+b\right )}^2}{5}+\frac {2\,a\,{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (a+b\right )}{3}}{f} \]
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