Optimal. Leaf size=208 \[ -\frac{\left (15 a^2-20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac{b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a-b}\right )}{15 f (a-b)^2}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{5 f (a-b)}+\frac{(10 a-2 b p-7 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{15 f (a-b)^2} \]
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Rubi [A] time = 0.222801, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3664, 462, 453, 365, 364} \[ -\frac{\left (15 a^2-20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac{b \sec ^2(e+f x)}{a-b}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sec ^2(e+f x)}{a-b}\right )}{15 f (a-b)^2}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{5 f (a-b)}+\frac{(10 a-2 b p-7 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{p+1}}{15 f (a-b)^2} \]
Antiderivative was successfully verified.
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Rule 3664
Rule 462
Rule 453
Rule 365
Rule 364
Rubi steps
\begin{align*} \int \sin ^5(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2 \left (a-b+b x^2\right )^p}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac{\operatorname{Subst}\left (\int \frac{\left (-10 a+b (7+2 p)+5 (a-b) x^2\right ) \left (a-b+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=\frac{(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac{\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \operatorname{Subst}\left (\int \frac{\left (a-b+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f}\\ &=\frac{(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}+\frac{\left (\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac{b \sec ^2(e+f x)}{a-b}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a-b}\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 (a-b)^2 f}\\ &=\frac{(10 a-7 b-2 b p) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{15 (a-b)^2 f}-\frac{\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{1+p}}{5 (a-b) f}-\frac{\left (15 a^2-20 a b (1+p)+4 b^2 \left (2+3 p+p^2\right )\right ) \cos (e+f x) \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sec ^2(e+f x)}{a-b}\right ) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac{b \sec ^2(e+f x)}{a-b}\right )^{-p}}{15 (a-b)^2 f}\\ \end{align*}
Mathematica [A] time = 7.42739, size = 283, normalized size = 1.36 \[ -\frac{2^{p+3} \sin ^4(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (\left (15 a^2-20 a b (p+1)+4 b^2 \left (p^2+3 p+2\right )\right ) \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a-b}\right )+\frac{1}{4} ((a-b) \cos (2 (e+f x))+a+b) (3 (a-b) \cos (2 (e+f x))-17 a+b (4 p+11)) \left (\frac{a+b \tan ^2(e+f x)}{a-b}\right )^p\right )}{15 f (a-b)^2 \left (-2^{p+2} \cos (2 (e+f x)) \left (\frac{a+b \tan ^2(e+f x)}{a-b}\right )^p+2^p \cos (4 (e+f x)) \left (\frac{a+b \tan ^2(e+f x)}{a-b}\right )^p+3 \left (\frac{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{a-b}\right )^p\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.885, size = 0, normalized size = 0. \begin{align*} \int \left ( \sin \left ( fx+e \right ) \right ) ^{5} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (\cos \left (f x + e\right )^{4} - 2 \, \cos \left (f x + e\right )^{2} + 1\right )}{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \sin \left (f x + e\right )^{5}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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