Optimal. Leaf size=257 \[ -\frac{2 a^3 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};\frac{d (1-\sin (e+f x))}{c+d}\right )}{d^2 f (2 n+3) (2 n+5) \sqrt{a \sin (e+f x)+a}}+\frac{2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d^2 f (2 n+3) (2 n+5) \sqrt{a \sin (e+f x)+a}}-\frac{2 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)} \]
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Rubi [A] time = 0.481308, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {2763, 2981, 2776, 70, 69} \[ -\frac{2 a^3 \left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};\frac{d (1-\sin (e+f x))}{c+d}\right )}{d^2 f (2 n+3) (2 n+5) \sqrt{a \sin (e+f x)+a}}+\frac{2 a^3 (3 c-d (4 n+11)) \cos (e+f x) (c+d \sin (e+f x))^{n+1}}{d^2 f (2 n+3) (2 n+5) \sqrt{a \sin (e+f x)+a}}-\frac{2 a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{n+1}}{d f (2 n+5)} \]
Antiderivative was successfully verified.
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Rule 2763
Rule 2981
Rule 2776
Rule 70
Rule 69
Rubi steps
\begin{align*} \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))^n \, dx &=-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac{2 \int \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^n \left (\frac{1}{2} a^2 (c+d (7+4 n))-\frac{1}{2} a^2 (3 c-11 d-4 d n) \sin (e+f x)\right ) \, dx}{d (5+2 n)}\\ &=\frac{2 a^3 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac{\left (a^2 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right )\right ) \int \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^n \, dx}{d^2 (3+2 n) (5+2 n)}\\ &=\frac{2 a^3 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac{\left (a^4 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x)\right ) \operatorname{Subst}\left (\int \frac{(c+d x)^n}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d^2 f (3+2 n) (5+2 n) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{2 a^3 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}+\frac{\left (a^4 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x) (c+d \sin (e+f x))^n \left (-\frac{a (c+d \sin (e+f x))}{-a c-a d}\right )^{-n}\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{c}{c+d}+\frac{d x}{c+d}\right )^n}{\sqrt{a-a x}} \, dx,x,\sin (e+f x)\right )}{d^2 f (3+2 n) (5+2 n) \sqrt{a-a \sin (e+f x)} \sqrt{a+a \sin (e+f x)}}\\ &=\frac{2 a^3 (3 c-d (11+4 n)) \cos (e+f x) (c+d \sin (e+f x))^{1+n}}{d^2 f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}-\frac{2 a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{1+n}}{d f (5+2 n)}-\frac{2 a^3 \left (3 c^2-2 c d (7+4 n)+d^2 \left (43+56 n+16 n^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};\frac{d (1-\sin (e+f x))}{c+d}\right ) (c+d \sin (e+f x))^n \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n}}{d^2 f (3+2 n) (5+2 n) \sqrt{a+a \sin (e+f x)}}\\ \end{align*}
Mathematica [A] time = 32.5375, size = 190, normalized size = 0.74 \[ \frac{a^2 (\sin (e+f x)-1) \sec (e+f x) \sqrt{a (\sin (e+f x)+1)} (c+d \sin (e+f x))^n \left (\left (3 c^2-2 c d (4 n+7)+d^2 \left (16 n^2+56 n+43\right )\right ) \left (\frac{c+d \sin (e+f x)}{c+d}\right )^{-n} \, _2F_1\left (\frac{1}{2},-n;\frac{3}{2};-\frac{d (\sin (e+f x)-1)}{c+d}\right )-(3 c-d (4 n+11)) (c+d \sin (e+f x))+d (2 n+3) (\sin (e+f x)+1) (c+d \sin (e+f x))\right )}{d^2 f \left (n+\frac{5}{2}\right ) (2 n+3)} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.179, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{n}\, 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 (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{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 (a^{2} \cos \left (f x + e\right )^{2} - 2 \, a^{2} \sin \left (f x + e\right ) - 2 \, a^{2}\right )} \sqrt{a \sin \left (f x + e\right ) + a}{\left (d \sin \left (f x + e\right ) + c\right )}^{n}, 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*} \mathit{sage}_{2} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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