Optimal. Leaf size=413 \[ \frac {2 \left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {2} \left (c (A+C) d-d^2 (A-C+4 A m)-2 c^2 (C+2 C m)\right ) F_1\left (\frac {1}{2}+m;\frac {1}{2},\frac {1}{2};\frac {3}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^m \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}{d \left (c^2-d^2\right ) f (1+2 m) \sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {2} \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right ) F_1\left (\frac {3}{2}+m;\frac {1}{2},\frac {1}{2};\frac {5}{2}+m;\frac {1}{2} (1+\sin (e+f x)),-\frac {d (1+\sin (e+f x))}{c-d}\right ) \cos (e+f x) (a+a \sin (e+f x))^{1+m} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}{a d \left (c^2-d^2\right ) f (3+2 m) \sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}} \]
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Rubi [A]
time = 0.65, antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 6, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3123, 3066,
2867, 145, 144, 143} \begin {gather*} \frac {\sqrt {2} \cos (e+f x) \left (c d (A+C)-d^2 (4 A m+A-C)-2 c^2 (2 C m+C)\right ) (a \sin (e+f x)+a)^m \sqrt {\frac {c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac {1}{2};\frac {1}{2},\frac {1}{2};m+\frac {3}{2};\frac {1}{2} (\sin (e+f x)+1),-\frac {d (\sin (e+f x)+1)}{c-d}\right )}{d f (2 m+1) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {2} \cos (e+f x) \left (d^2 (2 A m+A-C)+2 c^2 C (m+1)\right ) (a \sin (e+f x)+a)^{m+1} \sqrt {\frac {c+d \sin (e+f x)}{c-d}} F_1\left (m+\frac {3}{2};\frac {1}{2},\frac {1}{2};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) \left (c^2-d^2\right ) \sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}+\frac {2 \left (A d^2+c^2 C\right ) \cos (e+f x) (a \sin (e+f x)+a)^m}{d f \left (c^2-d^2\right ) \sqrt {c+d \sin (e+f x)}} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 145
Rule 2867
Rule 3066
Rule 3123
Rubi steps
\begin {align*} \int \frac {(a+a \sin (e+f x))^m \left (A+C \sin ^2(e+f x)\right )}{(c+d \sin (e+f x))^{3/2}} \, dx &=\frac {2 \left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}-\frac {2 \int \frac {(a+a \sin (e+f x))^m \left (-\frac {1}{2} a \left (2 c C \left (\frac {d}{2}-c m\right )+2 A d \left (\frac {c}{2}-d m\right )\right )-\frac {1}{2} a \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right ) \sin (e+f x)\right )}{\sqrt {c+d \sin (e+f x)}} \, dx}{a d \left (c^2-d^2\right )}\\ &=\frac {2 \left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {\left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right ) \int \frac {(a+a \sin (e+f x))^{1+m}}{\sqrt {c+d \sin (e+f x)}} \, dx}{a d \left (c^2-d^2\right )}-\frac {\left (2 \left (\frac {1}{2} a^2 \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right )-\frac {1}{2} a^2 \left (2 c C \left (\frac {d}{2}-c m\right )+2 A d \left (\frac {c}{2}-d m\right )\right )\right )\right ) \int \frac {(a+a \sin (e+f x))^m}{\sqrt {c+d \sin (e+f x)}} \, dx}{a^2 d \left (c^2-d^2\right )}\\ &=\frac {2 \left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {\left (a \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m}}{\sqrt {a-a x} \sqrt {c+d x}} \, dx,x,\sin (e+f x)\right )}{d \left (c^2-d^2\right ) f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}-\frac {\left (2 \left (\frac {1}{2} a^2 \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right )-\frac {1}{2} a^2 \left (2 c C \left (\frac {d}{2}-c m\right )+2 A d \left (\frac {c}{2}-d m\right )\right )\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{\sqrt {a-a x} \sqrt {c+d x}} \, dx,x,\sin (e+f x)\right )}{d \left (c^2-d^2\right ) f \sqrt {a-a \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\\ &=\frac {2 \left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {\left (a \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right ) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m}}{\sqrt {\frac {1}{2}-\frac {x}{2}} \sqrt {c+d x}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}-\frac {\left (\sqrt {2} \left (\frac {1}{2} a^2 \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right )-\frac {1}{2} a^2 \left (2 c C \left (\frac {d}{2}-c m\right )+2 A d \left (\frac {c}{2}-d m\right )\right )\right ) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{\sqrt {\frac {1}{2}-\frac {x}{2}} \sqrt {c+d x}} \, dx,x,\sin (e+f x)\right )}{d \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)}}\\ &=\frac {2 \left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {\left (a \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right ) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} \sqrt {\frac {a (c+d \sin (e+f x))}{a c-a d}}\right ) \text {Subst}\left (\int \frac {(a+a x)^{\frac {1}{2}+m}}{\sqrt {\frac {1}{2}-\frac {x}{2}} \sqrt {\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}}} \, dx,x,\sin (e+f x)\right )}{\sqrt {2} d \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}-\frac {\left (\sqrt {2} \left (\frac {1}{2} a^2 \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right )-\frac {1}{2} a^2 \left (2 c C \left (\frac {d}{2}-c m\right )+2 A d \left (\frac {c}{2}-d m\right )\right )\right ) \cos (e+f x) \sqrt {\frac {a-a \sin (e+f x)}{a}} \sqrt {\frac {a (c+d \sin (e+f x))}{a c-a d}}\right ) \text {Subst}\left (\int \frac {(a+a x)^{-\frac {1}{2}+m}}{\sqrt {\frac {1}{2}-\frac {x}{2}} \sqrt {\frac {a c}{a c-a d}+\frac {a d x}{a c-a d}}} \, dx,x,\sin (e+f x)\right )}{d \left (c^2-d^2\right ) f (a-a \sin (e+f x)) \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\\ &=\frac {2 \left (c^2 C+A d^2\right ) \cos (e+f x) (a+a \sin (e+f x))^m}{d \left (c^2-d^2\right ) f \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {2} \left (c (A+C) d-d^2 (A-C+4 A m)-2 c^2 (C+2 C m)\right ) F_1\left (\frac {1}{2}+m;\frac {1}{2},\frac {1}{2};\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 \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}{d \left (c^2-d^2\right ) f (1+2 m) \sqrt {1-\sin (e+f x)} \sqrt {c+d \sin (e+f x)}}+\frac {\sqrt {2} \left (2 c^2 C (1+m)+d^2 (A-C+2 A m)\right ) F_1\left (\frac {3}{2}+m;\frac {1}{2},\frac {1}{2};\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} \sqrt {\frac {c+d \sin (e+f x)}{c-d}}}{d \left (c^2-d^2\right ) f (3+2 m) (a-a \sin (e+f x)) \sqrt {c+d \sin (e+f x)}}\\ \end {align*}
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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(19634\) vs. \(2(413)=826\).
time = 57.91, size = 19634, normalized size = 47.54 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 0.54, size = 0, normalized size = 0.00 \[\int \frac {\left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +C \left (\sin ^{2}\left (f x +e \right )\right )\right )}{\left (c +d \sin \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{m} \left (A + C \sin ^{2}{\left (e + f x \right )}\right )}{\left (c + d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \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 )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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