Optimal. Leaf size=159 \[ \frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac {4 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}+\frac {i b c^2 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {i b c^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}} \]
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Rubi [A] time = 0.26, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {4701, 4705, 4679, 4419, 4183, 2279, 2391, 191, 271} \[ \frac {i b c^2 \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {i b c^2 \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac {4 c^2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}} \]
Antiderivative was successfully verified.
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Rule 191
Rule 271
Rule 2279
Rule 2391
Rule 4183
Rule 4419
Rule 4679
Rule 4701
Rule 4705
Rubi steps
\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{x^3 \left (d-c^2 d x^2\right )^2} \, dx &=-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\left (2 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^2} \, dx+\frac {(b c) \int \frac {1}{x^2 \left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {\left (2 c^2\right ) \int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx}{d}\\ &=-\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {\left (2 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \csc (x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}+\frac {\left (4 c^2\right ) \operatorname {Subst}\left (\int (a+b x) \csc (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac {4 c^2 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {\left (2 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac {\left (2 b c^2\right ) \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac {4 c^2 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\frac {\left (i b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {\left (i b c^2\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{d^2}\\ &=-\frac {b c}{2 d^2 x \sqrt {1-c^2 x^2}}+\frac {c^2 \left (a+b \sin ^{-1}(c x)\right )}{d^2 \left (1-c^2 x^2\right )}-\frac {a+b \sin ^{-1}(c x)}{2 d^2 x^2 \left (1-c^2 x^2\right )}-\frac {4 c^2 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}+\frac {i b c^2 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {i b c^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}\\ \end {align*}
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Mathematica [A] time = 0.80, size = 213, normalized size = 1.34 \[ -\frac {\frac {a c^2}{c^2 x^2-1}+2 a c^2 \log \left (1-c^2 x^2\right )-4 a c^2 \log (x)+\frac {a}{x^2}-2 i b c^2 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )+2 i b c^2 \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+\frac {b c \sqrt {1-c^2 x^2}}{x}+\frac {b c^2 \sin ^{-1}(c x)}{c^2 x^2-1}-4 b c^2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+4 b c^2 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )+\frac {b c^3 x}{\sqrt {1-c^2 x^2}}+\frac {b \sin ^{-1}(c x)}{x^2}}{2 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (c x\right ) + a}{c^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}, x\right ) \]
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 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{2} x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.27, size = 367, normalized size = 2.31 \[ \frac {c^{2} a}{4 d^{2} \left (c x +1\right )}-\frac {c^{2} a \ln \left (c x +1\right )}{d^{2}}-\frac {a}{2 d^{2} x^{2}}+\frac {2 c^{2} a \ln \left (c x \right )}{d^{2}}-\frac {c^{2} a}{4 d^{2} \left (c x -1\right )}-\frac {c^{2} a \ln \left (c x -1\right )}{d^{2}}-\frac {c^{2} b \arcsin \left (c x \right )}{d^{2} \left (c^{2} x^{2}-1\right )}+\frac {c b \sqrt {-c^{2} x^{2}+1}}{2 d^{2} x \left (c^{2} x^{2}-1\right )}+\frac {b \arcsin \left (c x \right )}{2 d^{2} x^{2} \left (c^{2} x^{2}-1\right )}-\frac {2 c^{2} b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}+\frac {2 c^{2} b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {2 c^{2} b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {2 i c^{2} b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {2 i c^{2} b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}+\frac {i b \,c^{2} \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}} \]
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 \[ -\frac {1}{2} \, a {\left (\frac {2 \, c^{2} \log \left (c x + 1\right )}{d^{2}} + \frac {2 \, c^{2} \log \left (c x - 1\right )}{d^{2}} - \frac {4 \, c^{2} \log \relax (x)}{d^{2}} + \frac {2 \, c^{2} x^{2} - 1}{c^{2} d^{2} x^{4} - d^{2} x^{2}}\right )} + b \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{4} d^{2} x^{7} - 2 \, c^{2} d^{2} x^{5} + d^{2} x^{3}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x^3\,{\left (d-c^2\,d\,x^2\right )}^2} \,d x \]
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 \[ \frac {\int \frac {a}{c^{4} x^{7} - 2 c^{2} x^{5} + x^{3}}\, dx + \int \frac {b \operatorname {asin}{\left (c x \right )}}{c^{4} x^{7} - 2 c^{2} x^{5} + x^{3}}\, dx}{d^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
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