Optimal. Leaf size=98 \[ \frac {\sqrt {b} (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{7/2} f}-\frac {(a+b)^2 \cos (e+f x)}{a^3 f}+\frac {(2 a+b) \cos ^3(e+f x)}{3 a^2 f}-\frac {\cos ^5(e+f x)}{5 a f} \]
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Rubi [A] time = 0.11, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4133, 461, 205} \[ \frac {(2 a+b) \cos ^3(e+f x)}{3 a^2 f}-\frac {(a+b)^2 \cos (e+f x)}{a^3 f}+\frac {\sqrt {b} (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{7/2} f}-\frac {\cos ^5(e+f x)}{5 a f} \]
Antiderivative was successfully verified.
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Rule 205
Rule 461
Rule 4133
Rubi steps
\begin {align*} \int \frac {\sin ^5(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {x^2 \left (1-x^2\right )^2}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {\operatorname {Subst}\left (\int \left (\frac {(a+b)^2}{a^3}-\frac {(2 a+b) x^2}{a^2}+\frac {x^4}{a}+\frac {-a^2 b-2 a b^2-b^3}{a^3 \left (b+a x^2\right )}\right ) \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac {(a+b)^2 \cos (e+f x)}{a^3 f}+\frac {(2 a+b) \cos ^3(e+f x)}{3 a^2 f}-\frac {\cos ^5(e+f x)}{5 a f}+\frac {\left (b (a+b)^2\right ) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\cos (e+f x)\right )}{a^3 f}\\ &=\frac {\sqrt {b} (a+b)^2 \tan ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{7/2} f}-\frac {(a+b)^2 \cos (e+f x)}{a^3 f}+\frac {(2 a+b) \cos ^3(e+f x)}{3 a^2 f}-\frac {\cos ^5(e+f x)}{5 a f}\\ \end {align*}
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Mathematica [C] time = 3.26, size = 425, normalized size = 4.34 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-75 a^3 \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )-75 a^3 \tan ^{-1}\left (\frac {\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )+\sqrt {a}}{\sqrt {b}}\right )-8 \sqrt {a} \sqrt {b} \cos (e+f x) \left (3 a^2 \cos (4 (e+f x))+89 a^2-4 a (7 a+5 b) \cos (2 (e+f x))+220 a b+120 b^2\right )+15 \left (5 a^3+64 a^2 b+128 a b^2+64 b^3\right ) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+15 \left (5 a^3+64 a^2 b+128 a b^2+64 b^3\right ) \tan ^{-1}\left (\frac {\sin (e) \tan \left (\frac {f x}{2}\right ) \left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )\right )}{1920 a^{7/2} \sqrt {b} f \left (a+b \sec ^2(e+f x)\right )} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.72, size = 229, normalized size = 2.34 \[ \left [-\frac {6 \, a^{2} \cos \left (f x + e\right )^{5} - 10 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{30 \, a^{3} f}, -\frac {3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{15 \, a^{3} f}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [B] time = 0.27, size = 373, normalized size = 3.81 \[ -\frac {\frac {15 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (-\frac {a \cos \left (f x + e\right ) - b}{\sqrt {a b} \cos \left (f x + e\right ) + \sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {2 \, {\left (8 \, a^{2} + 25 \, a b + 15 \, b^{2} - \frac {40 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {110 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {60 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {80 \, a^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {160 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {90 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {90 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac {60 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, a b {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {15 \, b^{2} {\left (\cos \left (f x + e\right ) - 1\right )}^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{a^{3} {\left (\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 1\right )}^{5}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.90, size = 183, normalized size = 1.87 \[ -\frac {\cos ^{5}\left (f x +e \right )}{5 a f}+\frac {2 \left (\cos ^{3}\left (f x +e \right )\right )}{3 a f}+\frac {\left (\cos ^{3}\left (f x +e \right )\right ) b}{3 f \,a^{2}}-\frac {\cos \left (f x +e \right )}{a f}-\frac {2 b \cos \left (f x +e \right )}{f \,a^{2}}-\frac {\cos \left (f x +e \right ) b^{2}}{f \,a^{3}}+\frac {b \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{f a \sqrt {a b}}+\frac {2 b^{2} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{f \,a^{2} \sqrt {a b}}+\frac {b^{3} \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{f \,a^{3} \sqrt {a b}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.43, size = 102, normalized size = 1.04 \[ \frac {\frac {15 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} - \frac {3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )}{a^{3}}}{15 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [B] time = 4.31, size = 123, normalized size = 1.26 \[ \frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {b}{3\,a^2}+\frac {2}{3\,a}\right )}{f}-\frac {{\cos \left (e+f\,x\right )}^5}{5\,a\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {1}{a}+\frac {b\,\left (\frac {b}{a^2}+\frac {2}{a}\right )}{a}\right )}{f}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,{\left (a+b\right )}^2}{a^2\,b+2\,a\,b^2+b^3}\right )\,{\left (a+b\right )}^2}{a^{7/2}\,f} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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