Optimal. Leaf size=208 \[ \frac {x^{1+m} (a+b \text {ArcSin}(c x))}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} (a+b \text {ArcSin}(c x))}{8 d^3 \left (1-c^2 x^2\right )}-\frac {b c (3-m) x^{2+m} \text {Hypergeometric2F1}\left (\frac {3}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{8 d^3 (2+m)}-\frac {b c x^{2+m} \text {Hypergeometric2F1}\left (\frac {5}{2},\frac {2+m}{2},\frac {4+m}{2},c^2 x^2\right )}{4 d^3 (2+m)}+\frac {(1-m) (3-m) \text {Int}\left (\frac {x^m (a+b \text {ArcSin}(c x))}{d-c^2 d x^2},x\right )}{8 d^2} \]
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Rubi [A]
time = 0.17, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {x^m (a+b \text {ArcSin}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {x^m \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^3} \, dx &=\frac {x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}-\frac {(b c) \int \frac {x^{1+m}}{\left (1-c^2 x^2\right )^{5/2}} \, dx}{4 d^3}+\frac {(3-m) \int \frac {x^m \left (a+b \sin ^{-1}(c x)\right )}{\left (d-c^2 d x^2\right )^2} \, dx}{4 d}\\ &=\frac {x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac {b c x^{2+m} \, _2F_1\left (\frac {5}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{4 d^3 (2+m)}-\frac {(b c (3-m)) \int \frac {x^{1+m}}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{8 d^3}+\frac {((1-m) (3-m)) \int \frac {x^m \left (a+b \sin ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx}{8 d^2}\\ &=\frac {x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{4 d^3 \left (1-c^2 x^2\right )^2}+\frac {(3-m) x^{1+m} \left (a+b \sin ^{-1}(c x)\right )}{8 d^3 \left (1-c^2 x^2\right )}-\frac {b c (3-m) x^{2+m} \, _2F_1\left (\frac {3}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{8 d^3 (2+m)}-\frac {b c x^{2+m} \, _2F_1\left (\frac {5}{2},\frac {2+m}{2};\frac {4+m}{2};c^2 x^2\right )}{4 d^3 (2+m)}+\frac {((1-m) (3-m)) \int \frac {x^m \left (a+b \sin ^{-1}(c x)\right )}{d-c^2 d x^2} \, dx}{8 d^2}\\ \end {align*}
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Mathematica [A]
time = 4.34, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^m (a+b \text {ArcSin}(c x))}{\left (d-c^2 d x^2\right )^3} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [A]
time = 0.44, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \left (a +b \arcsin \left (c x \right )\right )}{\left (-c^{2} d \,x^{2}+d \right )^{3}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
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 [A]
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 [A]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \frac {\int \frac {a x^{m}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx + \int \frac {b x^{m} \operatorname {asin}{\left (c x \right )}}{c^{6} x^{6} - 3 c^{4} x^{4} + 3 c^{2} x^{2} - 1}\, dx}{d^{3}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
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 [A]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^m\,\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )}{{\left (d-c^2\,d\,x^2\right )}^3} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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