Optimal. Leaf size=121 \[ \frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{2} i b d \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{4} b d \sin ^{-1}(c x) \]
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Rubi [A] time = 0.12, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {4683, 4625, 3717, 2190, 2279, 2391, 195, 216} \[ -\frac {1}{2} i b d \text {PolyLog}\left (2,e^{2 i \sin ^{-1}(c x)}\right )+\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d \log \left (1-e^{2 i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{4} b d \sin ^{-1}(c x) \]
Antiderivative was successfully verified.
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Rule 195
Rule 216
Rule 2190
Rule 2279
Rule 2391
Rule 3717
Rule 4625
Rule 4683
Rubi steps
\begin {align*} \int \frac {\left (d-c^2 d x^2\right ) \left (a+b \sin ^{-1}(c x)\right )}{x} \, dx &=\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+d \int \frac {a+b \sin ^{-1}(c x)}{x} \, dx-\frac {1}{2} (b c d) \int \sqrt {1-c^2 x^2} \, dx\\ &=-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}+\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )+d \operatorname {Subst}\left (\int (a+b x) \cot (x) \, dx,x,\sin ^{-1}(c x)\right )-\frac {1}{4} (b c d) \int \frac {1}{\sqrt {1-c^2 x^2}} \, dx\\ &=-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{4} b d \sin ^{-1}(c x)+\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}-(2 i d) \operatorname {Subst}\left (\int \frac {e^{2 i x} (a+b x)}{1-e^{2 i x}} \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{4} b d \sin ^{-1}(c x)+\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-(b d) \operatorname {Subst}\left (\int \log \left (1-e^{2 i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )\\ &=-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{4} b d \sin ^{-1}(c x)+\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )+\frac {1}{2} (i b d) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 i \sin ^{-1}(c x)}\right )\\ &=-\frac {1}{4} b c d x \sqrt {1-c^2 x^2}-\frac {1}{4} b d \sin ^{-1}(c x)+\frac {1}{2} d \left (1-c^2 x^2\right ) \left (a+b \sin ^{-1}(c x)\right )-\frac {i d \left (a+b \sin ^{-1}(c x)\right )^2}{2 b}+d \left (a+b \sin ^{-1}(c x)\right ) \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-\frac {1}{2} i b d \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )\\ \end {align*}
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Mathematica [A] time = 0.13, size = 99, normalized size = 0.82 \[ -\frac {1}{4} d \left (2 a c^2 x^2-4 a \log (x)+b c x \sqrt {1-c^2 x^2}+b \sin ^{-1}(c x) \left (2 c^2 x^2-4 \log \left (1-e^{2 i \sin ^{-1}(c x)}\right )-1\right )+2 i b \text {Li}_2\left (e^{2 i \sin ^{-1}(c x)}\right )+2 i b \sin ^{-1}(c x)^2\right ) \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {a c^{2} d x^{2} - a d + {\left (b c^{2} d x^{2} - b d\right )} \arcsin \left (c x\right )}{x}, x\right ) \]
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 \[ \int -\frac {{\left (c^{2} d x^{2} - d\right )} {\left (b \arcsin \left (c x\right ) + a\right )}}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.24, size = 164, normalized size = 1.36 \[ -\frac {d a \,c^{2} x^{2}}{2}+d a \ln \left (c x \right )-\frac {i b d \arcsin \left (c x \right )^{2}}{2}+d b \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )+d b \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )-i d b \polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-i d b \polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )+\frac {d b \cos \left (2 \arcsin \left (c x \right )\right ) \arcsin \left (c x \right )}{4}-\frac {d b \sin \left (2 \arcsin \left (c x \right )\right )}{8} \]
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 \[ -\frac {1}{2} \, a c^{2} d x^{2} + a d \log \relax (x) - \int \frac {{\left (b c^{2} d x^{2} - b d\right )} \arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{x}\,{d x} \]
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 \[ \int \frac {\left (a+b\,\mathrm {asin}\left (c\,x\right )\right )\,\left (d-c^2\,d\,x^2\right )}{x} \,d x \]
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 \[ - d \left (\int \left (- \frac {a}{x}\right )\, dx + \int a c^{2} x\, dx + \int \left (- \frac {b \operatorname {asin}{\left (c x \right )}}{x}\right )\, dx + \int b c^{2} x \operatorname {asin}{\left (c x \right )}\, dx\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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