Optimal. Leaf size=60 \[ \frac {\text {Li}_2\left (e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}-\frac {\sinh ^{-1}(a+b x)^2}{2 d}+\frac {\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d} \]
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Rubi [A] time = 0.10, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5865, 12, 5659, 3716, 2190, 2279, 2391} \[ \frac {\text {PolyLog}\left (2,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}-\frac {\sinh ^{-1}(a+b x)^2}{2 d}+\frac {\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2190
Rule 2279
Rule 2391
Rule 3716
Rule 5659
Rule 5865
Rubi steps
\begin {align*} \int \frac {\sinh ^{-1}(a+b x)}{\frac {a d}{b}+d x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {b \sinh ^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sinh ^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac {\sinh ^{-1}(a+b x)^2}{2 d}-\frac {2 \operatorname {Subst}\left (\int \frac {e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac {\sinh ^{-1}(a+b x)^2}{2 d}+\frac {\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}-\frac {\operatorname {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{d}\\ &=-\frac {\sinh ^{-1}(a+b x)^2}{2 d}+\frac {\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}-\frac {\operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}\\ &=-\frac {\sinh ^{-1}(a+b x)^2}{2 d}+\frac {\sinh ^{-1}(a+b x) \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )}{d}+\frac {\text {Li}_2\left (e^{2 \sinh ^{-1}(a+b x)}\right )}{2 d}\\ \end {align*}
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Mathematica [A] time = 0.02, size = 52, normalized size = 0.87 \[ \frac {\text {Li}_2\left (e^{2 \sinh ^{-1}(a+b x)}\right )-\sinh ^{-1}(a+b x) \left (\sinh ^{-1}(a+b x)-2 \log \left (1-e^{2 \sinh ^{-1}(a+b x)}\right )\right )}{2 d} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b \operatorname {arsinh}\left (b x + a\right )}{b d x + a d}, x\right ) \]
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 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )}{d x + \frac {a d}{b}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.06, size = 125, normalized size = 2.08 \[ -\frac {\arcsinh \left (b x +a \right )^{2}}{2 d}+\frac {\arcsinh \left (b x +a \right ) \ln \left (1+b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )}{d}+\frac {\polylog \left (2, -b x -a -\sqrt {1+\left (b x +a \right )^{2}}\right )}{d}+\frac {\arcsinh \left (b x +a \right ) \ln \left (1-b x -a -\sqrt {1+\left (b x +a \right )^{2}}\right )}{d}+\frac {\polylog \left (2, b x +a +\sqrt {1+\left (b x +a \right )^{2}}\right )}{d} \]
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 \[ \int \frac {\operatorname {arsinh}\left (b x + a\right )}{d x + \frac {a d}{b}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {asinh}\left (a+b\,x\right )}{d\,x+\frac {a\,d}{b}} \,d x \]
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 \[ \frac {b \int \frac {\operatorname {asinh}{\left (a + b x \right )}}{a + b x}\, dx}{d} \]
Verification of antiderivative is not currently implemented for this CAS.
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