Optimal. Leaf size=122 \[ \frac {a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac {2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac {b c x}{2 d^2 \sqrt {1-c^2 x^2}}+\frac {i b \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2} \]
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Rubi [A] time = 0.17, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4705, 4679, 4419, 4183, 2279, 2391, 191} \[ \frac {i b \text {PolyLog}\left (2,-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {i b \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}+\frac {a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac {2 \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{d^2}-\frac {b c x}{2 d^2 \sqrt {1-c^2 x^2}} \]
Antiderivative was successfully verified.
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Rule 191
Rule 2279
Rule 2391
Rule 4183
Rule 4419
Rule 4679
Rule 4705
Rubi steps
\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )^2} \, dx &=\frac {a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac {(b c) \int \frac {1}{\left (1-c^2 x^2\right )^{3/2}} \, dx}{2 d^2}+\frac {\int \frac {a+b \sin ^{-1}(c x)}{x \left (d-c^2 d x^2\right )} \, dx}{d}\\ &=-\frac {b c x}{2 d^2 \sqrt {1-c^2 x^2}}+\frac {a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}+\frac {\operatorname {Subst}\left (\int (a+b x) \csc (x) \sec (x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c x}{2 d^2 \sqrt {1-c^2 x^2}}+\frac {a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}+\frac {2 \operatorname {Subst}\left (\int (a+b x) \csc (2 x) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c x}{2 d^2 \sqrt {1-c^2 x^2}}+\frac {a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac {2 \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 i \sin ^{-1}(c x)}\right )}{d^2}-\frac {b \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}+\frac {b \operatorname {Subst}\left (\int \log \left (1+e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{d^2}\\ &=-\frac {b c x}{2 d^2 \sqrt {1-c^2 x^2}}+\frac {a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac {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) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {(i b) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}\\ &=-\frac {b c x}{2 d^2 \sqrt {1-c^2 x^2}}+\frac {a+b \sin ^{-1}(c x)}{2 d^2 \left (1-c^2 x^2\right )}-\frac {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 \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}-\frac {i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )}{2 d^2}\\ \end {align*}
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Mathematica [A] time = 0.41, size = 153, normalized size = 1.25 \[ \frac {\frac {a}{1-c^2 x^2}-a \log \left (1-c^2 x^2\right )+2 a \log (x)+b \left (-\frac {c x}{\sqrt {1-c^2 x^2}}+\frac {\sin ^{-1}(c x)}{1-c^2 x^2}+i \text {Li}_2\left (-e^{2 i \sin ^{-1}(c x)}\right )-i \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+2 \sin ^{-1}(c x) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-2 \sin ^{-1}(c x) \log \left (1+e^{2 i \sin ^{-1}(c x)}\right )\right )}{2 d^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \arcsin \left (c x\right ) + a}{c^{4} d^{2} x^{5} - 2 \, c^{2} d^{2} x^{3} + d^{2} x}, x\right ) \]
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 \[ \int \frac {b \arcsin \left (c x\right ) + a}{{\left (c^{2} d x^{2} - d\right )}^{2} x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.29, size = 335, normalized size = 2.75 \[ \frac {a}{4 d^{2} \left (c x +1\right )}-\frac {a \ln \left (c x +1\right )}{2 d^{2}}+\frac {a \ln \left (c x \right )}{d^{2}}-\frac {a}{4 d^{2} \left (c x -1\right )}-\frac {a \ln \left (c x -1\right )}{2 d^{2}}-\frac {i b \,c^{2} x^{2}}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b c x \sqrt {-c^{2} x^{2}+1}}{2 d^{2} \left (c^{2} x^{2}-1\right )}-\frac {b \arcsin \left (c x \right )}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {i b}{2 d^{2} \left (c^{2} x^{2}-1\right )}+\frac {b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {i b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {b \arcsin \left (c x \right ) \ln \left (1+\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{d^{2}}+\frac {i b \polylog \left (2, -\left (i c x +\sqrt {-c^{2} x^{2}+1}\right )^{2}\right )}{2 d^{2}}+\frac {b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}}-\frac {i b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )}{d^{2}} \]
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 \[ -\frac {1}{2} \, a {\left (\frac {1}{c^{2} d^{2} x^{2} - d^{2}} + \frac {\log \left (c x + 1\right )}{d^{2}} + \frac {\log \left (c x - 1\right )}{d^{2}} - \frac {2 \, \log \relax (x)}{d^{2}}\right )} + b \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{c^{4} d^{2} x^{5} - 2 \, c^{2} d^{2} x^{3} + d^{2} x}\,{d x} \]
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 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x\,{\left (d-c^2\,d\,x^2\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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