Optimal. Leaf size=131 \[ \text {Li}_2\left (\frac {e^{\cosh ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )+\text {Li}_2\left (\frac {e^{\cosh ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )-\frac {1}{2} \cosh ^{-1}(a+b x)^2 \]
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Rubi [A] time = 0.25, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, Rules used = {5866, 5800, 5562, 2190, 2279, 2391} \[ \text {PolyLog}\left (2,\frac {e^{\cosh ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )+\text {PolyLog}\left (2,\frac {e^{\cosh ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{a-\sqrt {a^2-1}}\right )+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{\sqrt {a^2-1}+a}\right )-\frac {1}{2} \cosh ^{-1}(a+b x)^2 \]
Antiderivative was successfully verified.
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Rule 2190
Rule 2279
Rule 2391
Rule 5562
Rule 5800
Rule 5866
Rubi steps
\begin {align*} \int \frac {\cosh ^{-1}(a+b x)}{x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\cosh ^{-1}(x)}{-\frac {a}{b}+\frac {x}{b}} \, dx,x,a+b x\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x \sinh (x)}{-\frac {a}{b}+\frac {\cosh (x)}{b}} \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{2} \cosh ^{-1}(a+b x)^2+\frac {\operatorname {Subst}\left (\int \frac {e^x x}{-\frac {a}{b}-\frac {\sqrt {-1+a^2}}{b}+\frac {e^x}{b}} \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}+\frac {\operatorname {Subst}\left (\int \frac {e^x x}{-\frac {a}{b}+\frac {\sqrt {-1+a^2}}{b}+\frac {e^x}{b}} \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}\\ &=-\frac {1}{2} \cosh ^{-1}(a+b x)^2+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )-\operatorname {Subst}\left (\int \log \left (1+\frac {e^x}{\left (-\frac {a}{b}-\frac {\sqrt {-1+a^2}}{b}\right ) b}\right ) \, dx,x,\cosh ^{-1}(a+b x)\right )-\operatorname {Subst}\left (\int \log \left (1+\frac {e^x}{\left (-\frac {a}{b}+\frac {\sqrt {-1+a^2}}{b}\right ) b}\right ) \, dx,x,\cosh ^{-1}(a+b x)\right )\\ &=-\frac {1}{2} \cosh ^{-1}(a+b x)^2+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )-\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{\left (-\frac {a}{b}-\frac {\sqrt {-1+a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{\cosh ^{-1}(a+b x)}\right )-\operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{\left (-\frac {a}{b}+\frac {\sqrt {-1+a^2}}{b}\right ) b}\right )}{x} \, dx,x,e^{\cosh ^{-1}(a+b x)}\right )\\ &=-\frac {1}{2} \cosh ^{-1}(a+b x)^2+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )+\cosh ^{-1}(a+b x) \log \left (1-\frac {e^{\cosh ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )+\text {Li}_2\left (\frac {e^{\cosh ^{-1}(a+b x)}}{a-\sqrt {-1+a^2}}\right )+\text {Li}_2\left (\frac {e^{\cosh ^{-1}(a+b x)}}{a+\sqrt {-1+a^2}}\right )\\ \end {align*}
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Mathematica [A] time = 0.01, size = 153, normalized size = 1.17 \[ \text {Li}_2\left (-\frac {e^{\cosh ^{-1}(a+b x)}}{\sqrt {a^2-1}-a}\right )+\text {Li}_2\left (\frac {e^{\cosh ^{-1}(a+b x)}}{a+\sqrt {a^2-1}}\right )+\cosh ^{-1}(a+b x) \log \left (\frac {e^{\cosh ^{-1}(a+b x)}}{b \left (-\frac {\sqrt {a^2-1}}{b}-\frac {a}{b}\right )}+1\right )+\cosh ^{-1}(a+b x) \log \left (\frac {e^{\cosh ^{-1}(a+b x)}}{b \left (\frac {\sqrt {a^2-1}}{b}-\frac {a}{b}\right )}+1\right )-\frac {1}{2} \cosh ^{-1}(a+b x)^2 \]
Antiderivative was successfully verified.
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fricas [F] time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {arcosh}\left (b x + a\right )}{x}, x\right ) \]
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 \[ \int \frac {\operatorname {arcosh}\left (b x + a\right )}{x}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.23, size = 436, normalized size = 3.33 \[ -\frac {\mathrm {arccosh}\left (b x +a \right )^{2}}{2}+\frac {a \,\mathrm {arccosh}\left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}-b x -\sqrt {b x +a -1}\, \sqrt {b x +a +1}}{a +\sqrt {a^{2}-1}}\right )}{\sqrt {a^{2}-1}}-\frac {a \,\mathrm {arccosh}\left (b x +a \right ) \ln \left (\frac {\sqrt {a^{2}-1}+b x +\sqrt {b x +a -1}\, \sqrt {b x +a +1}}{-a +\sqrt {a^{2}-1}}\right )}{\sqrt {a^{2}-1}}-\frac {\left (a^{2}-1+a \sqrt {a^{2}-1}\right ) \mathrm {arccosh}\left (b x +a \right ) \left (2 \ln \left (\frac {\sqrt {a^{2}-1}-b x -\sqrt {b x +a -1}\, \sqrt {b x +a +1}}{a +\sqrt {a^{2}-1}}\right ) a^{2}-\ln \left (\frac {\sqrt {a^{2}-1}-b x -\sqrt {b x +a -1}\, \sqrt {b x +a +1}}{a +\sqrt {a^{2}-1}}\right )-\ln \left (\frac {\sqrt {a^{2}-1}+b x +\sqrt {b x +a -1}\, \sqrt {b x +a +1}}{-a +\sqrt {a^{2}-1}}\right )-2 a \sqrt {a^{2}-1}\, \ln \left (\frac {\sqrt {a^{2}-1}-b x -\sqrt {b x +a -1}\, \sqrt {b x +a +1}}{a +\sqrt {a^{2}-1}}\right )\right )}{a^{2}-1}+\dilog \left (\frac {\sqrt {a^{2}-1}+b x +\sqrt {b x +a -1}\, \sqrt {b x +a +1}}{-a +\sqrt {a^{2}-1}}\right )+\dilog \left (\frac {\sqrt {a^{2}-1}-b x -\sqrt {b x +a -1}\, \sqrt {b x +a +1}}{a +\sqrt {a^{2}-1}}\right ) \]
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 \[ \int \frac {\operatorname {arcosh}\left (b x + a\right )}{x}\,{d x} \]
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 \[ \int \frac {\mathrm {acosh}\left (a+b\,x\right )}{x} \,d x \]
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 \[ \int \frac {\operatorname {acosh}{\left (a + b x \right )}}{x}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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