Optimal. Leaf size=270 \[ \frac {\sqrt {2} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac {1}{2};\frac {1}{2},-n;m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt {1-\sin (e+f x)}}+\frac {\sqrt {2} B \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac {3}{2};\frac {1}{2},-n;m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt {1-\sin (e+f x)}} \]
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Rubi [A] time = 0.43, antiderivative size = 270, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2987, 2788, 140, 139, 138} \[ \frac {\sqrt {2} (A-B) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac {1}{2};\frac {1}{2},-n;m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{f (2 m+1) \sqrt {1-\sin (e+f x)}}+\frac {\sqrt {2} B \cos (e+f x) (a \sin (e+f x)+a)^{m+1} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n} F_1\left (m+\frac {3}{2};\frac {1}{2},-n;m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a f (2 m+3) \sqrt {1-\sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 138
Rule 139
Rule 140
Rule 2788
Rule 2987
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c+d \sin (e+f x))^n \, dx &=(A-B) \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \, dx+\frac {B \int (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \, dx}{a}\\ &=\frac {\left (a^2 (A-B) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m} (c+d x)^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {(a B \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m} (c+d x)^n}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (a^2 (A-B) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m} (c+d x)^n}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a B \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m} (c+d x)^n}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (a^2 (A-B) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} (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 {(a+a x)^{-\frac {1}{2}+m} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^n}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}+\frac {\left (a B \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} (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 {(a+a x)^{\frac {1}{2}+m} \left (\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}\right )^n}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\sqrt {2} (A-B) F_1\left (\frac {1}{2}+m;\frac {1}{2},-n;\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{f (1+2 m) \sqrt {1-\sin (e+f x)}}+\frac {\sqrt {2} B F_1\left (\frac {3}{2}+m;\frac {1}{2},-n;\frac {5}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) \sqrt {1-\sin (e+f x)} (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^n \left (\frac {c+d \sin (e+f x)}{c-d}\right )^{-n}}{f (3+2 m) (a-a \sin (e+f x))}\\ \end {align*}
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Mathematica [B] time = 6.08, size = 682, normalized size = 2.53 \[ \frac {6 (c+d) \cot \left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \sin ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right )^{\frac {1}{2}-m} \cos ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )^{m-\frac {1}{2}} (a (\sin (e+f x)+1))^m (c+d \sin (e+f x))^n \left (\frac {(B c-A d) F_1\left (\frac {1}{2};\frac {1}{2}-m,-n;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \left (-4 d n F_1\left (\frac {3}{2};\frac {1}{2}-m,1-n;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac {3}{2};\frac {3}{2}-m,-n;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,-n;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}-\frac {B (c+d \sin (e+f x)) F_1\left (\frac {1}{2};\frac {1}{2}-m,-n-1;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}{\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ) \left (-4 d (n+1) F_1\left (\frac {3}{2};\frac {1}{2}-m,-n;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )-(2 m-1) (c+d) F_1\left (\frac {3}{2};\frac {3}{2}-m,-n-1;\frac {5}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )\right )+3 (c+d) F_1\left (\frac {1}{2};\frac {1}{2}-m,-n-1;\frac {3}{2};\cos ^2\left (\frac {1}{4} (2 e+2 f x+\pi )\right ),\frac {2 d \sin ^2\left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{c+d}\right )}\right )}{d f} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.35, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c +d \sin \left (f x +e \right )\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sin \left (f x + e\right ) + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{n}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^n \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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