Optimal. Leaf size=145 \[ \frac {2^{\frac {1}{2}+m} a^3 c^2 (B (2-m)-A (3+m)) \cos ^5(e+f x) \, _2F_1\left (\frac {5}{2},\frac {1}{2}-m;\frac {7}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-3+m}}{5 f (3+m)}-\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)} \]
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Rubi [A]
time = 0.23, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3046, 2939,
2768, 72, 71} \begin {gather*} \frac {a^3 c^2 2^{m+\frac {1}{2}} (B (2-m)-A (m+3)) \cos ^5(e+f x) (\sin (e+f x)+1)^{\frac {1}{2}-m} (a \sin (e+f x)+a)^{m-3} \, _2F_1\left (\frac {5}{2},\frac {1}{2}-m;\frac {7}{2};\frac {1}{2} (1-\sin (e+f x))\right )}{5 f (m+3)}-\frac {a^2 B c^2 \cos ^5(e+f x) (a \sin (e+f x)+a)^{m-2}}{f (m+3)} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 2768
Rule 2939
Rule 3046
Rubi steps
\begin {align*} \int (a+a \sin (e+f x))^m (A+B \sin (e+f x)) (c-c \sin (e+f x))^2 \, dx &=\left (a^2 c^2\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^{-2+m} (A+B \sin (e+f x)) \, dx\\ &=-\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\left (a^2 c^2 \left (A-\frac {B (2-m)}{3+m}\right )\right ) \int \cos ^4(e+f x) (a+a \sin (e+f x))^{-2+m} \, dx\\ &=-\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\frac {\left (a^4 c^2 \left (A-\frac {B (2-m)}{3+m}\right ) \cos ^5(e+f x)\right ) \text {Subst}\left (\int (a-a x)^{3/2} (a+a x)^{-\frac {1}{2}+m} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2} (a+a \sin (e+f x))^{5/2}}\\ &=-\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}+\frac {\left (2^{-\frac {1}{2}+m} a^4 c^2 \left (A-\frac {B (2-m)}{3+m}\right ) \cos ^5(e+f x) (a+a \sin (e+f x))^{-3+m} \left (\frac {a+a \sin (e+f x)}{a}\right )^{\frac {1}{2}-m}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {x}{2}\right )^{-\frac {1}{2}+m} (a-a x)^{3/2} \, dx,x,\sin (e+f x)\right )}{f (a-a \sin (e+f x))^{5/2}}\\ &=-\frac {2^{\frac {1}{2}+m} a^3 c^2 \left (A-\frac {B (2-m)}{3+m}\right ) \cos ^5(e+f x) \, _2F_1\left (\frac {5}{2},\frac {1}{2}-m;\frac {7}{2};\frac {1}{2} (1-\sin (e+f x))\right ) (1+\sin (e+f x))^{\frac {1}{2}-m} (a+a \sin (e+f x))^{-3+m}}{5 f}-\frac {a^2 B c^2 \cos ^5(e+f x) (a+a \sin (e+f x))^{-2+m}}{f (3+m)}\\ \end {align*}
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Mathematica [F]
time = 180.01, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 1.24, size = 0, normalized size = 0.00 \[\int \left (a +a \sin \left (f x +e \right )\right )^{m} \left (A +B \sin \left (f x +e \right )\right ) \left (c -c \sin \left (f x +e \right )\right )^{2}\, dx\]
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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} c^{2} \left (\int A \left (a \sin {\left (e + f x \right )} + a\right )^{m}\, dx + \int \left (- 2 A \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\right )\, dx + \int A \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}\, dx + \int B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin {\left (e + f x \right )}\, dx + \int \left (- 2 B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{2}{\left (e + f x \right )}\right )\, dx + \int B \left (a \sin {\left (e + f x \right )} + a\right )^{m} \sin ^{3}{\left (e + f x \right )}\, dx\right ) \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 = -0.01 \begin {gather*} \int \left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (a+a\,\sin \left (e+f\,x\right )\right )}^m\,{\left (c-c\,\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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