Optimal. Leaf size=30 \[ -\frac {a \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac {\text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2} \]
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Rubi [A]
time = 0.16, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps
used = 10, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {5859, 5830,
6873, 12, 6874, 3382, 5556, 3379} \begin {gather*} \frac {\text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac {a \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 3379
Rule 3382
Rule 5556
Rule 5830
Rule 5859
Rule 6873
Rule 6874
Rubi steps
\begin {align*} \int \frac {x}{\sinh ^{-1}(a+b x)} \, dx &=\frac {\text {Subst}\left (\int \frac {-\frac {a}{b}+\frac {x}{b}}{\sinh ^{-1}(x)} \, dx,x,a+b x\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\cosh (x) \left (-\frac {a}{b}+\frac {\sinh (x)}{b}\right )}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\cosh (x) (-a+\sinh (x))}{b x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\cosh (x) (-a+\sinh (x))}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\text {Subst}\left (\int \left (-\frac {a \cosh (x)}{x}+\frac {\cosh (x) \sinh (x)}{x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\text {Subst}\left (\int \frac {\cosh (x) \sinh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}-\frac {a \text {Subst}\left (\int \frac {\cosh (x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {a \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac {\text {Subst}\left (\int \frac {\sinh (2 x)}{2 x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {a \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac {\text {Subst}\left (\int \frac {\sinh (2 x)}{x} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {a \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac {\text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2}\\ \end {align*}
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Mathematica [A]
time = 0.04, size = 30, normalized size = 1.00 \begin {gather*} -\frac {a \text {Chi}\left (\sinh ^{-1}(a+b x)\right )}{b^2}+\frac {\text {Shi}\left (2 \sinh ^{-1}(a+b x)\right )}{2 b^2} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 2.69, size = 27, normalized size = 0.90
method | result | size |
derivativedivides | \(\frac {\frac {\hyperbolicSineIntegral \left (2 \arcsinh \left (b x +a \right )\right )}{2}-a \hyperbolicCosineIntegral \left (\arcsinh \left (b x +a \right )\right )}{b^{2}}\) | \(27\) |
default | \(\frac {\frac {\hyperbolicSineIntegral \left (2 \arcsinh \left (b x +a \right )\right )}{2}-a \hyperbolicCosineIntegral \left (\arcsinh \left (b x +a \right )\right )}{b^{2}}\) | \(27\) |
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 {x}{\operatorname {asinh}{\left (a + b x \right )}}\, 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 = -0.03 \begin {gather*} \int \frac {x}{\mathrm {asinh}\left (a+b\,x\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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