Optimal. Leaf size=392 \[ -\frac {a 2^{m+\frac {1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)-\left (c^2 (2 C m+C)\right )\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{d f (m+1) (c-d) (c+d)^2}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}+\frac {\sqrt {2} C \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m (c+d \sin (e+f x))^{-m} F_1\left (m+\frac {3}{2};\frac {1}{2},m+1;m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (c-d) \sqrt {1-\sin (e+f x)}} \]
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Rubi [A] time = 0.99, antiderivative size = 392, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3044, 2987, 2788, 132, 140, 139, 138} \[ -\frac {a 2^{m+\frac {1}{2}} \cos (e+f x) \left (c d (m+1) (A+C)+d^2 (-A m+C m+C)+c^2 (-(2 C m+C))\right ) (a \sin (e+f x)+a)^{m-1} \left (\frac {(c+d) (\sin (e+f x)+1)}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m} \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right )}{d f (m+1) (c-d) (c+d)^2}+\frac {\left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m (c+d \sin (e+f x))^{-m-1}}{d f (m+1) \left (c^2-d^2\right )}+\frac {\sqrt {2} C \cos (e+f x) (a \sin (e+f x)+a)^{m+1} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m (c+d \sin (e+f x))^{-m} F_1\left (m+\frac {3}{2};\frac {1}{2},m+1;m+\frac {5}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{a d f (2 m+3) (c-d) \sqrt {1-\sin (e+f x)}} \]
Antiderivative was successfully verified.
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Rule 132
Rule 138
Rule 139
Rule 140
Rule 2788
Rule 2987
Rule 3044
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-2-m} \left (A+C \sin ^2(e+f x)\right ) \, dx &=\frac {\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac {\int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \left (-a (A d (c+c m-d m)+c C (d-c m+d m))-a C \left (c^2-d^2\right ) (1+m) \sin (e+f x)\right ) \, dx}{a d \left (c^2-d^2\right ) (1+m)}\\ &=\frac {\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}+\frac {C \int (a+a \sin (e+f x))^{1+m} (c+d \sin (e+f x))^{-1-m} \, dx}{a d}+\frac {\left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \int (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m} \, dx}{d \left (c^2-d^2\right ) (1+m)}\\ &=\frac {\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}+\frac {(a C \cos (e+f x)) \operatorname {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m} (c+d x)^{-1-m}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}+\frac {\left (a^2 \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m} (c+d x)^{-1-m}}{\sqrt {a-a x}} \, dx,x,\sin (e+f x)\right )}{d \left (c^2-d^2\right ) f (1+m) \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac {2^{\frac {1}{2}+m} a \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac {(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c-d) d (c+d)^2 f (1+m)}+\frac {\left (a C \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)^{-1-m}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac {2^{\frac {1}{2}+m} a \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac {(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c-d) d (c+d)^2 f (1+m)}+\frac {\left (a^2 C \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} (c+d \sin (e+f x))^{-m} \left (\frac {a (c+d \sin (e+f x))}{a c-a d}\right )^m\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 )^{-1-m}}{\sqrt {\frac {1}{2}-\frac {x}{2}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d (a c-a d) f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {\left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m (c+d \sin (e+f x))^{-1-m}}{d \left (c^2-d^2\right ) f (1+m)}-\frac {2^{\frac {1}{2}+m} a \left (c (A+C) d (1+m)+d^2 (C-A m+C m)-c^2 (C+2 C m)\right ) \cos (e+f x) \, _2F_1\left (\frac {1}{2},\frac {1}{2}-m;\frac {3}{2};\frac {(c-d) (1-\sin (e+f x))}{2 (c+d \sin (e+f x))}\right ) (a+a \sin (e+f x))^{-1+m} \left (\frac {(c+d) (1+\sin (e+f x))}{c+d \sin (e+f x)}\right )^{\frac {1}{2}-m} (c+d \sin (e+f x))^{-m}}{(c-d) d (c+d)^2 f (1+m)}+\frac {\sqrt {2} C F_1\left (\frac {3}{2}+m;\frac {1}{2},1+m;\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))^{-m} \left (\frac {c+d \sin (e+f x)}{c-d}\right )^m}{(c-d) d f (3+2 m) (a-a \sin (e+f x))}\\ \end {align*}
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Mathematica [B] time = 59.05, size = 7530, normalized size = 19.21 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (C \cos \left (f x + e\right )^{2} - A - C\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [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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maple [F] time = 20.40, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{-2-m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )\, dx \]
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 \[ \int {\left (C \sin \left (f x + e\right )^{2} + A\right )} {\left (a \sin \left (f x + e\right ) + a\right )}^{m} {\left (d \sin \left (f x + e\right ) + c\right )}^{-m - 2}\,{d x} \]
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 \[ \int \frac {\left (C\,{\sin \left (e+f\,x\right )}^2+A\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{m+2}} \,d x \]
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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