Integrand size = 23, antiderivative size = 53 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {(a+b)^2 \sin (e+f x)}{f}-\frac {2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac {a^2 \sin ^5(e+f x)}{5 f} \]
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Time = 0.08 (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 = {4232, 200} \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^2 \sin ^5(e+f x)}{5 f}-\frac {2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac {(a+b)^2 \sin (e+f x)}{f} \]
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Rule 200
Rule 4232
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (a+b-a x^2\right )^2 \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (a^2 \left (1+\frac {b (2 a+b)}{a^2}\right )-2 a^2 \left (1+\frac {b}{a}\right ) x^2+a^2 x^4\right ) \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {(a+b)^2 \sin (e+f x)}{f}-\frac {2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac {a^2 \sin ^5(e+f x)}{5 f} \\ \end{align*}
Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.00 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {b^2 \cos (f x) \sin (e)}{f}+\frac {b^2 \cos (e) \sin (f x)}{f}+\frac {a^2 \sin (e+f x)}{f}+\frac {2 a b \sin (e+f x)}{f}-\frac {2 a^2 \sin ^3(e+f x)}{3 f}-\frac {2 a b \sin ^3(e+f x)}{3 f}+\frac {a^2 \sin ^5(e+f x)}{5 f} \]
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Time = 0.47 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26
method | result | size |
derivativedivides | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )}{5}+\frac {2 a b \left (\cos \left (f x +e \right )^{2}+2\right ) \sin \left (f x +e \right )}{3}+\sin \left (f x +e \right ) b^{2}}{f}\) | \(67\) |
default | \(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )}{5}+\frac {2 a b \left (\cos \left (f x +e \right )^{2}+2\right ) \sin \left (f x +e \right )}{3}+\sin \left (f x +e \right ) b^{2}}{f}\) | \(67\) |
parallelrisch | \(\frac {150 \sin \left (f x +e \right ) a^{2}+360 \sin \left (f x +e \right ) a b +240 \sin \left (f x +e \right ) b^{2}+3 \sin \left (5 f x +5 e \right ) a^{2}+25 \sin \left (3 f x +3 e \right ) a^{2}+40 \sin \left (3 f x +3 e \right ) a b}{240 f}\) | \(80\) |
risch | \(\frac {5 a^{2} \sin \left (f x +e \right )}{8 f}+\frac {3 \sin \left (f x +e \right ) a b}{2 f}+\frac {\sin \left (f x +e \right ) b^{2}}{f}+\frac {a^{2} \sin \left (5 f x +5 e \right )}{80 f}+\frac {5 a^{2} \sin \left (3 f x +3 e \right )}{48 f}+\frac {\sin \left (3 f x +3 e \right ) a b}{6 f}\) | \(92\) |
norman | \(\frac {-\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{15}}{f}+\frac {2 \left (5 a^{2}+2 a b -3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}-\frac {2 \left (5 a^{2}+2 a b -3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{13}}{3 f}-\frac {2 \left (43 a^{2}-50 a b -45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}+\frac {2 \left (43 a^{2}-50 a b -45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{15 f}+\frac {2 \left (109 a^{2}+10 a b +45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{15 f}-\frac {2 \left (109 a^{2}+10 a b +45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{15 f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}\) | \(271\) |
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Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (3 \, a^{2} \cos \left (f x + e\right )^{4} + 2 \, {\left (2 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} \sin \left (f x + e\right )}{15 \, f} \]
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Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\text {Timed out} \]
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Time = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, a^{2} \sin \left (f x + e\right )^{5} - 10 \, {\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (f x + e\right )}{15 \, f} \]
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Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.43 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, a^{2} \sin \left (f x + e\right )^{5} - 10 \, a^{2} \sin \left (f x + e\right )^{3} - 10 \, a b \sin \left (f x + e\right )^{3} + 15 \, a^{2} \sin \left (f x + e\right ) + 30 \, a b \sin \left (f x + e\right ) + 15 \, b^{2} \sin \left (f x + e\right )}{15 \, f} \]
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Time = 18.54 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\sin \left (e+f\,x\right )\,{\left (a+b\right )}^2+\frac {a^2\,{\sin \left (e+f\,x\right )}^5}{5}-\frac {2\,a\,{\sin \left (e+f\,x\right )}^3\,\left (a+b\right )}{3}}{f} \]
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