Optimal. Leaf size=214 \[ -\frac {1}{4} i b \text {ArcSin}\left (c+d x^2\right )^2+\frac {1}{2} b \text {ArcSin}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{i c-\sqrt {1-c^2}}\right )+\frac {1}{2} b \text {ArcSin}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{i c+\sqrt {1-c^2}}\right )+a \log (x)-\frac {1}{2} i b \text {PolyLog}\left (2,\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{i c-\sqrt {1-c^2}}\right )-\frac {1}{2} i b \text {PolyLog}\left (2,\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{i c+\sqrt {1-c^2}}\right ) \]
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Rubi [A]
time = 0.27, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps
used = 12, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {6874, 4889,
4825, 4617, 2221, 2317, 2438} \begin {gather*} a \log (x)-\frac {1}{2} i b \text {Li}_2\left (\frac {e^{i \text {ArcSin}\left (d x^2+c\right )}}{i c-\sqrt {1-c^2}}\right )-\frac {1}{2} i b \text {Li}_2\left (\frac {e^{i \text {ArcSin}\left (d x^2+c\right )}}{i c+\sqrt {1-c^2}}\right )+\frac {1}{2} b \text {ArcSin}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{-\sqrt {1-c^2}+i c}\right )+\frac {1}{2} b \text {ArcSin}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{\sqrt {1-c^2}+i c}\right )-\frac {1}{4} i b \text {ArcSin}\left (c+d x^2\right )^2 \end {gather*}
Antiderivative was successfully verified.
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Rule 2221
Rule 2317
Rule 2438
Rule 4617
Rule 4825
Rule 4889
Rule 6874
Rubi steps
\begin {align*} \int \frac {a+b \sin ^{-1}\left (c+d x^2\right )}{x} \, dx &=\int \left (\frac {a}{x}+\frac {b \sin ^{-1}\left (c+d x^2\right )}{x}\right ) \, dx\\ &=a \log (x)+b \int \frac {\sin ^{-1}\left (c+d x^2\right )}{x} \, dx\\ &=a \log (x)+\frac {1}{2} b \text {Subst}\left (\int \frac {\sin ^{-1}(c+d x)}{x} \, dx,x,x^2\right )\\ &=a \log (x)+\frac {b \text {Subst}\left (\int \frac {\sin ^{-1}(x)}{-\frac {c}{d}+\frac {x}{d}} \, dx,x,c+d x^2\right )}{2 d}\\ &=a \log (x)+\frac {b \text {Subst}\left (\int \frac {x \cos (x)}{-\frac {c}{d}+\frac {\sin (x)}{d}} \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )}{2 d}\\ &=-\frac {1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2+a \log (x)+\frac {(i b) \text {Subst}\left (\int \frac {e^{i x} x}{-\frac {i c}{d}-\frac {\sqrt {1-c^2}}{d}+\frac {e^{i x}}{d}} \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )}{2 d}+\frac {(i b) \text {Subst}\left (\int \frac {e^{i x} x}{-\frac {i c}{d}+\frac {\sqrt {1-c^2}}{d}+\frac {e^{i x}}{d}} \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )}{2 d}\\ &=-\frac {1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2+\frac {1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c-\sqrt {1-c^2}}\right )+\frac {1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c+\sqrt {1-c^2}}\right )+a \log (x)-\frac {1}{2} b \text {Subst}\left (\int \log \left (1+\frac {e^{i x}}{\left (-\frac {i c}{d}-\frac {\sqrt {1-c^2}}{d}\right ) d}\right ) \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )-\frac {1}{2} b \text {Subst}\left (\int \log \left (1+\frac {e^{i x}}{\left (-\frac {i c}{d}+\frac {\sqrt {1-c^2}}{d}\right ) d}\right ) \, dx,x,\sin ^{-1}\left (c+d x^2\right )\right )\\ &=-\frac {1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2+\frac {1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c-\sqrt {1-c^2}}\right )+\frac {1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c+\sqrt {1-c^2}}\right )+a \log (x)+\frac {1}{2} (i b) \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{\left (-\frac {i c}{d}-\frac {\sqrt {1-c^2}}{d}\right ) d}\right )}{x} \, dx,x,e^{i \sin ^{-1}\left (c+d x^2\right )}\right )+\frac {1}{2} (i b) \text {Subst}\left (\int \frac {\log \left (1+\frac {x}{\left (-\frac {i c}{d}+\frac {\sqrt {1-c^2}}{d}\right ) d}\right )}{x} \, dx,x,e^{i \sin ^{-1}\left (c+d x^2\right )}\right )\\ &=-\frac {1}{4} i b \sin ^{-1}\left (c+d x^2\right )^2+\frac {1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c-\sqrt {1-c^2}}\right )+\frac {1}{2} b \sin ^{-1}\left (c+d x^2\right ) \log \left (1-\frac {e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c+\sqrt {1-c^2}}\right )+a \log (x)-\frac {1}{2} i b \text {Li}_2\left (\frac {e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c-\sqrt {1-c^2}}\right )-\frac {1}{2} i b \text {Li}_2\left (\frac {e^{i \sin ^{-1}\left (c+d x^2\right )}}{i c+\sqrt {1-c^2}}\right )\\ \end {align*}
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Mathematica [A]
time = 0.02, size = 230, normalized size = 1.07 \begin {gather*} -\frac {1}{4} i b \text {ArcSin}\left (c+d x^2\right )^2+\frac {1}{2} b \text {ArcSin}\left (c+d x^2\right ) \log \left (1+\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{\left (-\frac {i c}{d}-\frac {\sqrt {1-c^2}}{d}\right ) d}\right )+\frac {1}{2} b \text {ArcSin}\left (c+d x^2\right ) \log \left (1+\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{\left (-\frac {i c}{d}+\frac {\sqrt {1-c^2}}{d}\right ) d}\right )+a \log (x)-\frac {1}{2} i b \text {PolyLog}\left (2,-\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{-i c+\sqrt {1-c^2}}\right )-\frac {1}{2} i b \text {PolyLog}\left (2,\frac {e^{i \text {ArcSin}\left (c+d x^2\right )}}{i c+\sqrt {1-c^2}}\right ) \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {a +b \arcsin \left (d \,x^{2}+c \right )}{x}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
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 [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {asin}{\left (c + d x^{2} \right )}}{x}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\mathrm {asin}\left (d\,x^2+c\right )}{x} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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