Optimal. Leaf size=277 \[ \frac{a^2 (2 A (n+3)+B (2 n+5)) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{d^2 f (n+2) (n+3) \sqrt{\cos ^2(e+f x)}}+\frac{a^2 (A (2 n+3)+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{d f (n+1) (n+2) \sqrt{\cos ^2(e+f x)}}-\frac{a^2 (A (n+3)+B (n+4)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2) (n+3)}-\frac{B \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right ) (d \sin (e+f x))^{n+1}}{d f (n+3)} \]
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Rubi [A] time = 0.49217, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used = {2976, 2968, 3023, 2748, 2643} \[ \frac{a^2 (2 A (n+3)+B (2 n+5)) \cos (e+f x) (d \sin (e+f x))^{n+2} \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{d^2 f (n+2) (n+3) \sqrt{\cos ^2(e+f x)}}+\frac{a^2 (A (2 n+3)+2 B (n+1)) \cos (e+f x) (d \sin (e+f x))^{n+1} \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{d f (n+1) (n+2) \sqrt{\cos ^2(e+f x)}}-\frac{a^2 (A (n+3)+B (n+4)) \cos (e+f x) (d \sin (e+f x))^{n+1}}{d f (n+2) (n+3)}-\frac{B \cos (e+f x) \left (a^2 \sin (e+f x)+a^2\right ) (d \sin (e+f x))^{n+1}}{d f (n+3)} \]
Antiderivative was successfully verified.
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Rule 2976
Rule 2968
Rule 3023
Rule 2748
Rule 2643
Rubi steps
\begin{align*} \int (d \sin (e+f x))^n (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \, dx &=-\frac{B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}+\frac{\int (d \sin (e+f x))^n (a+a \sin (e+f x)) (a d (B (1+n)+A (3+n))+a d (A (3+n)+B (4+n)) \sin (e+f x)) \, dx}{d (3+n)}\\ &=-\frac{B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}+\frac{\int (d \sin (e+f x))^n \left (a^2 d (B (1+n)+A (3+n))+\left (a^2 d (B (1+n)+A (3+n))+a^2 d (A (3+n)+B (4+n))\right ) \sin (e+f x)+a^2 d (A (3+n)+B (4+n)) \sin ^2(e+f x)\right ) \, dx}{d (3+n)}\\ &=-\frac{a^2 (A (3+n)+B (4+n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n)}-\frac{B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}+\frac{\int (d \sin (e+f x))^n \left (a^2 d^2 (3+n) (2 B (1+n)+A (3+2 n))+a^2 d^2 (2+n) (2 A (3+n)+B (5+2 n)) \sin (e+f x)\right ) \, dx}{d^2 (2+n) (3+n)}\\ &=-\frac{a^2 (A (3+n)+B (4+n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n)}-\frac{B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}+\frac{\left (a^2 (2 B (1+n)+A (3+2 n))\right ) \int (d \sin (e+f x))^n \, dx}{2+n}+\frac{\left (a^2 (2 A (3+n)+B (5+2 n))\right ) \int (d \sin (e+f x))^{1+n} \, dx}{d (3+n)}\\ &=-\frac{a^2 (A (3+n)+B (4+n)) \cos (e+f x) (d \sin (e+f x))^{1+n}}{d f (2+n) (3+n)}+\frac{a^2 (2 B (1+n)+A (3+2 n)) \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}}{d f (1+n) (2+n) \sqrt{\cos ^2(e+f x)}}+\frac{a^2 (2 A (3+n)+B (5+2 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}}{d^2 f (2+n) (3+n) \sqrt{\cos ^2(e+f x)}}-\frac{B \cos (e+f x) (d \sin (e+f x))^{1+n} \left (a^2+a^2 \sin (e+f x)\right )}{d f (3+n)}\\ \end{align*}
Mathematica [A] time = 1.50093, size = 204, normalized size = 0.74 \[ \frac{a^2 \sin (e+f x) \cos (e+f x) (d \sin (e+f x))^n \left (\sin (e+f x) \left (\frac{(2 A+B) \, _2F_1\left (\frac{1}{2},\frac{n+2}{2};\frac{n+4}{2};\sin ^2(e+f x)\right )}{n+2}+\sin (e+f x) \left (\frac{(A+2 B) \, _2F_1\left (\frac{1}{2},\frac{n+3}{2};\frac{n+5}{2};\sin ^2(e+f x)\right )}{n+3}+\frac{B \sin (e+f x) \, _2F_1\left (\frac{1}{2},\frac{n+4}{2};\frac{n+6}{2};\sin ^2(e+f x)\right )}{n+4}\right )\right )+\frac{A \, _2F_1\left (\frac{1}{2},\frac{n+1}{2};\frac{n+3}{2};\sin ^2(e+f x)\right )}{n+1}\right )}{f \sqrt{\cos ^2(e+f x)}} \]
Antiderivative was successfully verified.
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Maple [F] time = 2.673, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sin \left ( fx+e \right ) \right ) ^{2} \left ( A+B\sin \left ( fx+e \right ) \right ) \, 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 \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\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 ({\left (A + 2 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + B\right )} a^{2} +{\left (B a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (A + B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \left (d \sin \left (f x + e\right )\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*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{2} \left (d \sin \left (f x + e\right )\right )^{n}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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