Optimal. Leaf size=117 \[ \frac{\sqrt{\pi } \text{Erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{8 e b^2}+\frac{\sqrt{\pi } \text{Erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{8 e b^2}-\frac{\sqrt{\pi } a \text{Erfi}\left (\frac{1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^2}-\frac{\sqrt{\pi } a \text{Erfi}\left (\frac{1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^2} \]
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Rubi [A] time = 0.276839, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5898, 6741, 12, 6742, 5513, 2234, 2204, 5514} \[ \frac{\sqrt{\pi } \text{Erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{8 e b^2}+\frac{\sqrt{\pi } \text{Erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{8 e b^2}-\frac{\sqrt{\pi } a \text{Erfi}\left (\frac{1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^2}-\frac{\sqrt{\pi } a \text{Erfi}\left (\frac{1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^2} \]
Antiderivative was successfully verified.
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Rule 5898
Rule 6741
Rule 12
Rule 6742
Rule 5513
Rule 2234
Rule 2204
Rule 5514
Rubi steps
\begin{align*} \int e^{\sinh ^{-1}(a+b x)^2} x \, dx &=\frac{\operatorname{Subst}\left (\int e^{x^2} \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{e^{x^2} \cosh (x) (-a+\sinh (x))}{b} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int e^{x^2} \cosh (x) (-a+\sinh (x)) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (-a e^{x^2} \cosh (x)+e^{x^2} \cosh (x) \sinh (x)\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int e^{x^2} \cosh (x) \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int e^{x^2} \cosh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac{\operatorname{Subst}\left (\int \left (-\frac{1}{4} e^{-2 x+x^2}+\frac{1}{4} e^{2 x+x^2}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}-\frac{a \operatorname{Subst}\left (\int \left (\frac{1}{2} e^{-x+x^2}+\frac{e^{x+x^2}}{2}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{\operatorname{Subst}\left (\int e^{-2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^2}+\frac{\operatorname{Subst}\left (\int e^{2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^2}-\frac{a \operatorname{Subst}\left (\int e^{-x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^2}-\frac{a \operatorname{Subst}\left (\int e^{x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac{\operatorname{Subst}\left (\int e^{\frac{1}{4} (-2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^2 e}+\frac{\operatorname{Subst}\left (\int e^{\frac{1}{4} (2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{4 b^2 e}-\frac{a \operatorname{Subst}\left (\int e^{\frac{1}{4} (-1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^2 \sqrt [4]{e}}-\frac{a \operatorname{Subst}\left (\int e^{\frac{1}{4} (1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^2 \sqrt [4]{e}}\\ &=\frac{\sqrt{\pi } \text{erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{8 b^2 e}+\frac{\sqrt{\pi } \text{erfi}\left (1+\sinh ^{-1}(a+b x)\right )}{8 b^2 e}-\frac{a \sqrt{\pi } \text{erfi}\left (\frac{1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^2 \sqrt [4]{e}}-\frac{a \sqrt{\pi } \text{erfi}\left (\frac{1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^2 \sqrt [4]{e}}\\ \end{align*}
Mathematica [A] time = 0.0932882, size = 76, normalized size = 0.65 \[ \frac{\sqrt{\pi } \left (2 e^{3/4} a \text{Erfi}\left (\frac{1}{2}-\sinh ^{-1}(a+b x)\right )+\text{Erfi}\left (1-\sinh ^{-1}(a+b x)\right )-2 e^{3/4} a \text{Erfi}\left (\sinh ^{-1}(a+b x)+\frac{1}{2}\right )+\text{Erfi}\left (\sinh ^{-1}(a+b x)+1\right )\right )}{8 e b^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.01, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{ \left ({\it Arcsinh} \left ( bx+a \right ) \right ) ^{2}}}x\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x e^{\left (\operatorname{arsinh}\left (b x + a\right )^{2}\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (x e^{\left (\operatorname{arsinh}\left (b x + a\right )^{2}\right )}, x\right ) \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 e^{\operatorname{asinh}^{2}{\left (a + b x \right )}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int x e^{\left (\operatorname{arsinh}\left (b x + a\right )^{2}\right )}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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