Optimal. Leaf size=311 \[ -\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\sqrt {2} (a+b) d (a d-2 b c (2+m)) F_1\left (\frac {1}{2};\frac {1}{2},-1-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}}-\frac {\sqrt {2} \left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}} \]
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Rubi [A]
time = 0.30, antiderivative size = 311, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2870, 2835,
2744, 144, 143} \begin {gather*} -\frac {\sqrt {2} \cos (e+f x) \left (a d (a d-2 b c (m+2))+b^2 \left (c^2 (m+2)+d^2 (m+1)\right )\right ) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}+\frac {\sqrt {2} d (a+b) \cos (e+f x) (a d-2 b c (m+2)) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m} F_1\left (\frac {1}{2};\frac {1}{2},-m-1;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right )}{b^2 f (m+2) \sqrt {\sin (e+f x)+1}}-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{m+1}}{b f (m+2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 143
Rule 144
Rule 2744
Rule 2835
Rule 2870
Rubi steps
\begin {align*} \int (a+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx &=-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\int (a+b \sin (e+f x))^m \left (b \left (d^2 (1+m)+c^2 (2+m)\right )-d (a d-2 b c (2+m)) \sin (e+f x)\right ) \, dx}{b (2+m)}\\ &=-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}-\frac {(d (a d-2 b c (2+m))) \int (a+b \sin (e+f x))^{1+m} \, dx}{b^2 (2+m)}+\frac {\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \int (a+b \sin (e+f x))^m \, dx}{b^2 (2+m)}\\ &=-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}-\frac {(d (a d-2 b c (2+m)) \cos (e+f x)) \text {Subst}\left (\int \frac {(a+b x)^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}+\frac {\left (\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \cos (e+f x)\right ) \text {Subst}\left (\int \frac {(a+b x)^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\left ((-a-b) d (a d-2 b c (2+m)) \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac {a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^{1+m}}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}+\frac {\left (\left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (-\frac {a+b \sin (e+f x)}{-a-b}\right )^{-m}\right ) \text {Subst}\left (\int \frac {\left (-\frac {a}{-a-b}-\frac {b x}{-a-b}\right )^m}{\sqrt {1-x} \sqrt {1+x}} \, dx,x,\sin (e+f x)\right )}{b^2 f (2+m) \sqrt {1-\sin (e+f x)} \sqrt {1+\sin (e+f x)}}\\ &=-\frac {d^2 \cos (e+f x) (a+b \sin (e+f x))^{1+m}}{b f (2+m)}+\frac {\sqrt {2} (a+b) d (a d-2 b c (2+m)) F_1\left (\frac {1}{2};\frac {1}{2},-1-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}}-\frac {\sqrt {2} \left (a d (a d-2 b c (2+m))+b^2 \left (d^2 (1+m)+c^2 (2+m)\right )\right ) F_1\left (\frac {1}{2};\frac {1}{2},-m;\frac {3}{2};\frac {1}{2} (1-\sin (e+f x)),\frac {b (1-\sin (e+f x))}{a+b}\right ) \cos (e+f x) (a+b \sin (e+f x))^m \left (\frac {a+b \sin (e+f x)}{a+b}\right )^{-m}}{b^2 f (2+m) \sqrt {1+\sin (e+f x)}}\\ \end {align*}
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Mathematica [F]
time = 9.40, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b \sin (e+f x))^m (c+d \sin (e+f x))^2 \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.40, size = 0, normalized size = 0.00 \[\int \left (a +b \sin \left (f x +e \right )\right )^{m} \left (c +d \sin \left (f x +e \right )\right )^{2}\, dx\]
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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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 {\left (a+b\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^2 \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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