Optimal. Leaf size=362 \[ \frac{(1-n) (n+1) (-4 A n+7 A+4 B n+3 B) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{15 a^3 d^2 f (n+2) \sqrt{\cos ^2(e+f x)}}-\frac{n \left (A \left (4 n^2-9 n+2\right )+B \left (-4 n^2-n+3\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{15 a^3 d f (n+1) \sqrt{\cos ^2(e+f x)}}+\frac{(1-n) (-4 A n+7 A+4 B n+3 B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{15 d f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{n+1}}{15 a d f (a \sin (e+f x)+a)^2}+\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{5 d f (a \sin (e+f x)+a)^3} \]
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Rubi [A] time = 0.846187, antiderivative size = 362, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {2978, 2748, 2643} \[ \frac{(1-n) (n+1) (-4 A n+7 A+4 B n+3 B) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{15 a^3 d^2 f (n+2) \sqrt{\cos ^2(e+f x)}}-\frac{n \left (A \left (4 n^2-9 n+2\right )+B \left (-4 n^2-n+3\right )\right ) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{15 a^3 d f (n+1) \sqrt{\cos ^2(e+f x)}}+\frac{(1-n) (-4 A n+7 A+4 B n+3 B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{15 d f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{n+1}}{15 a d f (a \sin (e+f x)+a)^2}+\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{n+1}}{5 d f (a \sin (e+f x)+a)^3} \]
Antiderivative was successfully verified.
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Rule 2978
Rule 2748
Rule 2643
Rubi steps
\begin{align*} \int \frac{(d \sin (e+f x))^n (A+B \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx &=\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac{\int \frac{(d \sin (e+f x))^n (a d (4 A+B-A n+B n)-a (A-B) d (1-n) \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx}{5 a^2 d}\\ &=\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac{(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 a d f (a+a \sin (e+f x))^2}+\frac{\int \frac{(d \sin (e+f x))^n \left (a^2 d^2 \left (B \left (3+n-2 n^2\right )+A \left (7-6 n+2 n^2\right )\right )+a^2 d^2 n (A (5-2 n)+2 B n) \sin (e+f x)\right )}{a+a \sin (e+f x)} \, dx}{15 a^4 d^2}\\ &=\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac{(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 a d f (a+a \sin (e+f x))^2}+\frac{(1-n) (7 A+3 B-4 A n+4 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 d f \left (a^3+a^3 \sin (e+f x)\right )}+\frac{\int (d \sin (e+f x))^n \left (-a^3 d^3 n \left (B \left (3-n-4 n^2\right )+A \left (2-9 n+4 n^2\right )\right )+a^3 d^3 (1-n) (1+n) (7 A+3 B-4 A n+4 B n) \sin (e+f x)\right ) \, dx}{15 a^6 d^3}\\ &=\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac{(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 a d f (a+a \sin (e+f x))^2}+\frac{(1-n) (7 A+3 B-4 A n+4 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 d f \left (a^3+a^3 \sin (e+f x)\right )}+\frac{((1-n) (1+n) (7 A+3 B-4 A n+4 B n)) \int (d \sin (e+f x))^{1+n} \, dx}{15 a^3 d}-\frac{\left (n \left (B \left (3-n-4 n^2\right )+A \left (2-9 n+4 n^2\right )\right )\right ) \int (d \sin (e+f x))^n \, dx}{15 a^3}\\ &=-\frac{n \left (B \left (3-n-4 n^2\right )+A \left (2-9 n+4 n^2\right )\right ) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{1+n}{2};\frac{3+n}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{1+n}}{15 a^3 d f (1+n) \sqrt{\cos ^2(e+f x)}}+\frac{(1-n) (1+n) (7 A+3 B-4 A n+4 B n) \cos (e+f x) \, _2F_1\left (\frac{1}{2},\frac{2+n}{2};\frac{4+n}{2};\sin ^2(e+f x)\right ) (d \sin (e+f x))^{2+n}}{15 a^3 d^2 f (2+n) \sqrt{\cos ^2(e+f x)}}+\frac{(A-B) \cos (e+f x) (d \sin (e+f x))^{1+n}}{5 d f (a+a \sin (e+f x))^3}+\frac{(A (5-2 n)+2 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 a d f (a+a \sin (e+f x))^2}+\frac{(1-n) (7 A+3 B-4 A n+4 B n) \cos (e+f x) (d \sin (e+f x))^{1+n}}{15 d f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 4.42434, size = 260, normalized size = 0.72 \[ \frac{(d \sin (e+f x))^n \left (\frac{2 \sin (e+f x) \cos (e+f x) \left (n \left (A \left (-4 n^2+9 n-2\right )+B \left (4 n^2+n-3\right )\right ) \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )+\frac{(n-1) (n+1)^2 (A (4 n-7)-B (4 n+3)) \sin (e+f x) \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{n+2}\right )}{(n+1) \sqrt{\cos ^2(e+f x)}}+\frac{(n-1) (A (4 n-7)-B (4 n+3)) \sin (2 (e+f x))}{\sin (e+f x)+1}+\frac{(A (5-2 n)+2 B n) \sin (2 (e+f x))}{(\sin (e+f x)+1)^2}+\frac{3 (A-B) \sin (2 (e+f x))}{(\sin (e+f x)+1)^3}\right )}{30 a^3 f} \]
Antiderivative was successfully verified.
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Maple [F] time = 1.724, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( A+B\sin \left ( fx+e \right ) \right ) }{ \left ( a+a\sin \left ( fx+e \right ) \right ) ^{3}}}\, 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 \frac{{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}\,{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 (-\frac{{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \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 \frac{{\left (B \sin \left (f x + e\right ) + A\right )} \left (d \sin \left (f x + e\right )\right )^{n}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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