Optimal. Leaf size=249 \[ \frac{3 a^2 b (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{a^3 \cos (c+d x) (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{3 a b^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{b^3 (e \sin (c+d x))^{m+1} \text{Hypergeometric2F1}\left (2,\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )}{d e (m+1)} \]
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Rubi [A] time = 0.386978, antiderivative size = 249, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3872, 2912, 2643, 2564, 364, 2577} \[ \frac{3 a^2 b (e \sin (c+d x))^{m+1} \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{a^3 \cos (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac{1}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1) \sqrt{\cos ^2(c+d x)}}+\frac{3 a b^2 \sqrt{\cos ^2(c+d x)} \sec (c+d x) (e \sin (c+d x))^{m+1} \, _2F_1\left (\frac{3}{2},\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)}+\frac{b^3 (e \sin (c+d x))^{m+1} \, _2F_1\left (2,\frac{m+1}{2};\frac{m+3}{2};\sin ^2(c+d x)\right )}{d e (m+1)} \]
Antiderivative was successfully verified.
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Rule 3872
Rule 2912
Rule 2643
Rule 2564
Rule 364
Rule 2577
Rubi steps
\begin{align*} \int (a+b \sec (c+d x))^3 (e \sin (c+d x))^m \, dx &=-\int (-b-a \cos (c+d x))^3 \sec ^3(c+d x) (e \sin (c+d x))^m \, dx\\ &=-\int \left (-a^3 (e \sin (c+d x))^m-3 a^2 b \sec (c+d x) (e \sin (c+d x))^m-3 a b^2 \sec ^2(c+d x) (e \sin (c+d x))^m-b^3 \sec ^3(c+d x) (e \sin (c+d x))^m\right ) \, dx\\ &=a^3 \int (e \sin (c+d x))^m \, dx+\left (3 a^2 b\right ) \int \sec (c+d x) (e \sin (c+d x))^m \, dx+\left (3 a b^2\right ) \int \sec ^2(c+d x) (e \sin (c+d x))^m \, dx+b^3 \int \sec ^3(c+d x) (e \sin (c+d x))^m \, dx\\ &=\frac{a^3 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{3 a b^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac{\left (3 a^2 b\right ) \operatorname{Subst}\left (\int \frac{x^m}{1-\frac{x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}+\frac{b^3 \operatorname{Subst}\left (\int \frac{x^m}{\left (1-\frac{x^2}{e^2}\right )^2} \, dx,x,e \sin (c+d x)\right )}{d e}\\ &=\frac{a^3 \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m) \sqrt{\cos ^2(c+d x)}}+\frac{3 a^2 b \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac{b^3 \, _2F_1\left (2,\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) (e \sin (c+d x))^{1+m}}{d e (1+m)}+\frac{3 a b^2 \sqrt{\cos ^2(c+d x)} \, _2F_1\left (\frac{3}{2},\frac{1+m}{2};\frac{3+m}{2};\sin ^2(c+d x)\right ) \sec (c+d x) (e \sin (c+d x))^{1+m}}{d e (1+m)}\\ \end{align*}
Mathematica [A] time = 0.315501, size = 182, normalized size = 0.73 \[ \frac{\tan (c+d x) (e \sin (c+d x))^m \left (b \left (3 a^2 \cos (c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )+b \left (3 a \sqrt{\cos ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{3}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )+b \cos (c+d x) \text{Hypergeometric2F1}\left (2,\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )\right )\right )+a^3 \sqrt{\cos ^2(c+d x)} \text{Hypergeometric2F1}\left (\frac{1}{2},\frac{m+1}{2},\frac{m+3}{2},\sin ^2(c+d x)\right )\right )}{d (m+1)} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.569, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( dx+c \right ) \right ) ^{3} \left ( e\sin \left ( dx+c \right ) \right ) ^{m}\, 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 (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{m}\,{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 (b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}\right )} \left (e \sin \left (d x + c\right )\right )^{m}, 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 (d x + c\right ) + a\right )}^{3} \left (e \sin \left (d x + c\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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