Optimal. Leaf size=67 \[ -\frac{a e^{2 \sinh ^{-1}(a+b x)}}{4 b^2}-\frac{a \sinh ^{-1}(a+b x)}{2 b^2}+\frac{e^{-\sinh ^{-1}(a+b x)}}{4 b^2}+\frac{e^{3 \sinh ^{-1}(a+b x)}}{12 b^2} \]
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Rubi [A] time = 0.070877, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {5898, 2282, 12, 1628} \[ -\frac{a e^{2 \sinh ^{-1}(a+b x)}}{4 b^2}-\frac{a \sinh ^{-1}(a+b x)}{2 b^2}+\frac{e^{-\sinh ^{-1}(a+b x)}}{4 b^2}+\frac{e^{3 \sinh ^{-1}(a+b x)}}{12 b^2} \]
Antiderivative was successfully verified.
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Rule 5898
Rule 2282
Rule 12
Rule 1628
Rubi steps
\begin{align*} \int e^{\sinh ^{-1}(a+b x)} x \, dx &=\frac{\operatorname{Subst}\left (\int e^x \cosh (x) \left (-\frac{a}{b}+\frac{\sinh (x)}{b}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1-x^2\right ) \left (1+2 a x-x^2\right )}{4 b x^2} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (-1-x^2\right ) \left (1+2 a x-x^2\right )}{x^2} \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{4 b^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{x^2}-\frac{2 a}{x}-2 a x+x^2\right ) \, dx,x,e^{\sinh ^{-1}(a+b x)}\right )}{4 b^2}\\ &=\frac{e^{-\sinh ^{-1}(a+b x)}}{4 b^2}-\frac{a e^{2 \sinh ^{-1}(a+b x)}}{4 b^2}+\frac{e^{3 \sinh ^{-1}(a+b x)}}{12 b^2}-\frac{a \sinh ^{-1}(a+b x)}{2 b^2}\\ \end{align*}
Mathematica [A] time = 0.0987441, size = 73, normalized size = 1.09 \[ \frac{1}{6} \left (\frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (-a^2+a b x+2 b^2 x^2+2\right )}{b^2}-\frac{3 a \sinh ^{-1}(a+b x)}{b^2}+3 a x^2+2 b x^3\right ) \]
Warning: Unable to verify antiderivative.
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Maple [A] time = 0.003, size = 138, normalized size = 2.1 \begin{align*}{\frac{1}{3\,{b}^{2}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{ax}{2\,b}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{a}^{2}}{2\,{b}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{a}{2\,b}\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{b{x}^{3}}{3}}+{\frac{a{x}^{2}}{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 2.60614, size = 212, normalized size = 3.16 \begin{align*} \frac{2 \, b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (2 \, b^{2} x^{2} + a b x - a^{2} + 2\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{6 \, b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x \left (a + b x + \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}\right )\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A] time = 1.34604, size = 143, normalized size = 2.13 \begin{align*} \frac{1}{3} \, b x^{3} + \frac{1}{2} \, a x^{2} + \frac{1}{6} \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left ({\left (2 \, x + \frac{a}{b}\right )} x - \frac{a^{2} b - 2 \, b}{b^{3}}\right )} + \frac{a \log \left (-a b -{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}{\left | b \right |}\right )}{2 \, b{\left | b \right |}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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