Integrand size = 23, antiderivative size = 108 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{7/2} \sqrt {a+b} f}+\frac {\left (a^2-a b+b^2\right ) \sin (e+f x)}{a^3 f}-\frac {(2 a-b) \sin ^3(e+f x)}{3 a^2 f}+\frac {\sin ^5(e+f x)}{5 a f} \]
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Time = 0.12 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4232, 398, 214} \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{7/2} f \sqrt {a+b}}-\frac {(2 a-b) \sin ^3(e+f x)}{3 a^2 f}+\frac {\left (a^2-a b+b^2\right ) \sin (e+f x)}{a^3 f}+\frac {\sin ^5(e+f x)}{5 a f} \]
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Rule 214
Rule 398
Rule 4232
Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a^2-a b+b^2}{a^3}-\frac {(2 a-b) x^2}{a^2}+\frac {x^4}{a}-\frac {b^3}{a^3 \left (a+b-a x^2\right )}\right ) \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\left (a^2-a b+b^2\right ) \sin (e+f x)}{a^3 f}-\frac {(2 a-b) \sin ^3(e+f x)}{3 a^2 f}+\frac {\sin ^5(e+f x)}{5 a f}-\frac {b^3 \text {Subst}\left (\int \frac {1}{a+b-a x^2} \, dx,x,\sin (e+f x)\right )}{a^3 f} \\ & = -\frac {b^3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{a^{7/2} \sqrt {a+b} f}+\frac {\left (a^2-a b+b^2\right ) \sin (e+f x)}{a^3 f}-\frac {(2 a-b) \sin ^3(e+f x)}{3 a^2 f}+\frac {\sin ^5(e+f x)}{5 a f} \\ \end{align*}
Time = 0.58 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {120 b^3 \left (\log \left (\sqrt {a+b}-\sqrt {a} \sin (e+f x)\right )-\log \left (\sqrt {a+b}+\sqrt {a} \sin (e+f x)\right )\right )}{\sqrt {a+b}}+30 \sqrt {a} \left (5 a^2-6 a b+8 b^2\right ) \sin (e+f x)+5 a^{3/2} (5 a-4 b) \sin (3 (e+f x))+3 a^{5/2} \sin (5 (e+f x))}{240 a^{7/2} f} \]
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Time = 1.94 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.02
method | result | size |
derivativedivides | \(\frac {\frac {\frac {a^{2} \sin \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \sin \left (f x +e \right )^{3}}{3}+\frac {a \sin \left (f x +e \right )^{3} b}{3}+\sin \left (f x +e \right ) a^{2}-\sin \left (f x +e \right ) a b +\sin \left (f x +e \right ) b^{2}}{a^{3}}-\frac {b^{3} \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}}{f}\) | \(110\) |
default | \(\frac {\frac {\frac {a^{2} \sin \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \sin \left (f x +e \right )^{3}}{3}+\frac {a \sin \left (f x +e \right )^{3} b}{3}+\sin \left (f x +e \right ) a^{2}-\sin \left (f x +e \right ) a b +\sin \left (f x +e \right ) b^{2}}{a^{3}}-\frac {b^{3} \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{a^{3} \sqrt {a \left (a +b \right )}}}{f}\) | \(110\) |
risch | \(-\frac {5 i {\mathrm e}^{i \left (f x +e \right )}}{16 a f}+\frac {3 i {\mathrm e}^{i \left (f x +e \right )} b}{8 a^{2} f}-\frac {i {\mathrm e}^{i \left (f x +e \right )} b^{2}}{2 a^{3} f}+\frac {5 i {\mathrm e}^{-i \left (f x +e \right )}}{16 a f}-\frac {3 i {\mathrm e}^{-i \left (f x +e \right )} b}{8 a^{2} f}+\frac {i {\mathrm e}^{-i \left (f x +e \right )} b^{2}}{2 a^{3} f}+\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, f \,a^{3}}-\frac {b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{2 \sqrt {a^{2}+a b}\, f \,a^{3}}+\frac {\sin \left (5 f x +5 e \right )}{80 a f}+\frac {5 \sin \left (3 f x +3 e \right )}{48 a f}-\frac {\sin \left (3 f x +3 e \right ) b}{12 a^{2} f}\) | \(282\) |
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Time = 0.29 (sec) , antiderivative size = 305, normalized size of antiderivative = 2.82 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [\frac {15 \, \sqrt {a^{2} + a b} b^{3} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (3 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3} + {\left (4 \, a^{4} - a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{30 \, {\left (a^{5} + a^{4} b\right )} f}, \frac {15 \, \sqrt {-a^{2} - a b} b^{3} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) + {\left (3 \, {\left (a^{4} + a^{3} b\right )} \cos \left (f x + e\right )^{4} + 8 \, a^{4} - 2 \, a^{3} b + 5 \, a^{2} b^{2} + 15 \, a b^{3} + {\left (4 \, a^{4} - a^{3} b - 5 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{5} + a^{4} b\right )} f}\right ] \]
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Timed out. \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Timed out} \]
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Time = 0.27 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.08 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {15 \, b^{3} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{\sqrt {{\left (a + b\right )} a} a^{3}} + \frac {2 \, {\left (3 \, a^{2} \sin \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} - a b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a^{2} - a b + b^{2}\right )} \sin \left (f x + e\right )\right )}}{a^{3}}}{30 \, f} \]
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Time = 0.29 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.19 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\frac {15 \, b^{3} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{\sqrt {-a^{2} - a b} a^{3}} + \frac {3 \, a^{4} \sin \left (f x + e\right )^{5} - 10 \, a^{4} \sin \left (f x + e\right )^{3} + 5 \, a^{3} b \sin \left (f x + e\right )^{3} + 15 \, a^{4} \sin \left (f x + e\right ) - 15 \, a^{3} b \sin \left (f x + e\right ) + 15 \, a^{2} b^{2} \sin \left (f x + e\right )}{a^{5}}}{15 \, f} \]
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Time = 0.14 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.03 \[ \int \frac {\cos ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\sin \left (e+f\,x\right )\,\left (\frac {3}{a}+\frac {\left (a+b\right )\,\left (\frac {a+b}{a^2}-\frac {3}{a}\right )}{a}\right )}{f}+\frac {{\sin \left (e+f\,x\right )}^5}{5\,a\,f}+\frac {{\sin \left (e+f\,x\right )}^3\,\left (\frac {a+b}{3\,a^2}-\frac {1}{a}\right )}{f}-\frac {b^3\,\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )}{a^{7/2}\,f\,\sqrt {a+b}} \]
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