Optimal. Leaf size=182 \[ -\frac{\left (15 a^2+10 a b (1-2 p)+b^2 \left (4 p^2-8 p+3\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac{b \sec ^2(e+f x)}{a}+1\right )^{-p} \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a}\right )}{15 a^2 f}+\frac{(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a f} \]
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Rubi [A] time = 0.192125, antiderivative size = 182, 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 = {4134, 462, 453, 365, 364} \[ -\frac{\left (15 a^2+10 a b (1-2 p)+b^2 \left (4 p^2-8 p+3\right )\right ) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (\frac{b \sec ^2(e+f x)}{a}+1\right )^{-p} \, _2F_1\left (-\frac{1}{2},-p;\frac{1}{2};-\frac{b \sec ^2(e+f x)}{a}\right )}{15 a^2 f}+\frac{(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{p+1}}{5 a f} \]
Antiderivative was successfully verified.
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Rule 4134
Rule 462
Rule 453
Rule 365
Rule 364
Rubi steps
\begin{align*} \int \left (a+b \sec ^2(e+f x)\right )^p \sin ^5(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^2 \left (a+b x^2\right )^p}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{\left (-10 a-b (3-2 p)+5 a x^2\right ) \left (a+b x^2\right )^p}{x^4} \, dx,x,\sec (e+f x)\right )}{5 a f}\\ &=\frac{(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac{\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 a^2 f}\\ &=\frac{(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}+\frac{\left (\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 p^2\right )\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac{b \sec ^2(e+f x)}{a}\right )^{-p}\right ) \operatorname{Subst}\left (\int \frac{\left (1+\frac{b x^2}{a}\right )^p}{x^2} \, dx,x,\sec (e+f x)\right )}{15 a^2 f}\\ &=\frac{(10 a+b (3-2 p)) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{15 a^2 f}-\frac{\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{1+p}}{5 a f}-\frac{\left (15 a^2+10 a b (1-2 p)+b^2 \left (3-8 p+4 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}\right ) \left (a+b \sec ^2(e+f x)\right )^p \left (1+\frac{b \sec ^2(e+f x)}{a}\right )^{-p}}{15 a^2 f}\\ \end{align*}
Mathematica [A] time = 7.7896, size = 253, normalized size = 1.39 \[ \frac{2 \sin ^4(e+f x) \cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^p \left (4 \left (15 a^2+10 a b (1-2 p)+b^2 \left (4 p^2-8 p+3\right )\right ) \text{Hypergeometric2F1}\left (-\frac{1}{2},-p,\frac{1}{2},-\frac{b \sec ^2(e+f x)}{a}\right )+(a \cos (2 (e+f x))+a+2 b) (3 a \cos (2 (e+f x))-17 a+4 b p-6 b) \left (\frac{a+b \tan ^2(e+f x)+b}{a}\right )^p\right )}{15 a^2 f \left (4 \cos (2 (e+f x)) \left (\frac{a+b \tan ^2(e+f x)+b}{a}\right )^p-2^{-p} \left (2^p \cos (4 (e+f x)) \left (\frac{a+b \tan ^2(e+f x)+b}{a}\right )^p+3 \left (\frac{\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)}{a}\right )^p\right )\right )} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 1.09, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p} \left ( \sin \left ( fx+e \right ) \right ) ^{5}\, 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 \sec \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 \sec \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 \sec \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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