Optimal. Leaf size=145 \[ -\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}+\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}} \]
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Rubi [A] time = 0.19, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {4713, 4709, 4183, 2279, 2391} \[ \frac {i b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \text {PolyLog}\left (2,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}-\frac {2 \sqrt {1-c^2 x^2} \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right ) \left (a+b \sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}} \]
Antiderivative was successfully verified.
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Rule 2279
Rule 2391
Rule 4183
Rule 4709
Rule 4713
Rubi steps
\begin {align*} \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {d-c^2 d x^2}} \, dx &=\frac {\sqrt {1-c^2 x^2} \int \frac {a+b \sin ^{-1}(c x)}{x \sqrt {1-c^2 x^2}} \, dx}{\sqrt {d-c^2 d x^2}}\\ &=\frac {\sqrt {1-c^2 x^2} \operatorname {Subst}\left (\int (a+b x) \csc (x) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}\\ &=-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1-e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \log \left (1+e^{i x}\right ) \, dx,x,\sin ^{-1}(c x)\right )}{\sqrt {d-c^2 d x^2}}\\ &=-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {\left (i b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}-\frac {\left (i b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}\\ &=-\frac {2 \sqrt {1-c^2 x^2} \left (a+b \sin ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}+\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}-\frac {i b \sqrt {1-c^2 x^2} \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )}{\sqrt {d-c^2 d x^2}}\\ \end {align*}
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Mathematica [A] time = 0.34, size = 146, normalized size = 1.01 \[ -\frac {a \log \left (\sqrt {d} \sqrt {-d \left (c^2 x^2-1\right )}+d\right )}{\sqrt {d}}+\frac {a \log (x)}{\sqrt {d}}+\frac {b \sqrt {1-c^2 x^2} \left (i \text {Li}_2\left (-e^{i \sin ^{-1}(c x)}\right )-i \text {Li}_2\left (e^{i \sin ^{-1}(c x)}\right )+\sin ^{-1}(c x) \left (\log \left (1-e^{i \sin ^{-1}(c x)}\right )-\log \left (1+e^{i \sin ^{-1}(c x)}\right )\right )\right )}{\sqrt {d \left (1-c^2 x^2\right )}} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-c^{2} d x^{2} + d} {\left (b \arcsin \left (c x\right ) + a\right )}}{c^{2} d x^{3} - d x}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.18, size = 180, normalized size = 1.24 \[ -\frac {a \ln \left (\frac {2 d +2 \sqrt {d}\, \sqrt {-c^{2} d \,x^{2}+d}}{x}\right )}{\sqrt {d}}-\frac {i b \sqrt {-c^{2} x^{2}+1}\, \sqrt {-d \left (c^{2} x^{2}-1\right )}\, \left (i \arcsin \left (c x \right ) \ln \left (1+i c x +\sqrt {-c^{2} x^{2}+1}\right )-i \arcsin \left (c x \right ) \ln \left (1-i c x -\sqrt {-c^{2} x^{2}+1}\right )+\polylog \left (2, -i c x -\sqrt {-c^{2} x^{2}+1}\right )-\polylog \left (2, i c x +\sqrt {-c^{2} x^{2}+1}\right )\right )}{d \left (c^{2} x^{2}-1\right )} \]
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 \[ \frac {b \int \frac {\arctan \left (c x, \sqrt {c x + 1} \sqrt {-c x + 1}\right )}{\sqrt {c x + 1} \sqrt {-c x + 1} x}\,{d x}}{\sqrt {d}} - \frac {a \log \left (\frac {2 \, \sqrt {-c^{2} d x^{2} + d} \sqrt {d}}{{\left | x \right |}} + \frac {2 \, d}{{\left | x \right |}}\right )}{\sqrt {d}} \]
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 \[ \int \frac {a+b\,\mathrm {asin}\left (c\,x\right )}{x\,\sqrt {d-c^2\,d\,x^2}} \,d x \]
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 \[ \int \frac {a + b \operatorname {asin}{\left (c x \right )}}{x \sqrt {- d \left (c x - 1\right ) \left (c x + 1\right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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