Optimal. Leaf size=117 \[ \frac {\sqrt {\pi } \text {erfi}\left (1-\cosh ^{-1}(a+b x)\right )}{8 e b^2}+\frac {\sqrt {\pi } \text {erfi}\left (\cosh ^{-1}(a+b x)+1\right )}{8 e b^2}+\frac {\sqrt {\pi } a \text {erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^2}-\frac {\sqrt {\pi } a \text {erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^2} \]
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Rubi [A] time = 0.28, antiderivative size = 117, normalized size of antiderivative = 1.00, 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 = {5899, 6741, 12, 6742, 5512, 2234, 2204, 5514} \[ \frac {\sqrt {\pi } \text {Erfi}\left (1-\cosh ^{-1}(a+b x)\right )}{8 e b^2}+\frac {\sqrt {\pi } \text {Erfi}\left (\cosh ^{-1}(a+b x)+1\right )}{8 e b^2}+\frac {\sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^2}-\frac {\sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} \left (2 \cosh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2204
Rule 2234
Rule 5512
Rule 5514
Rule 5899
Rule 6741
Rule 6742
Rubi steps
\begin {align*} \int e^{\cosh ^{-1}(a+b x)^2} x \, dx &=\frac {\operatorname {Subst}\left (\int e^{x^2} \left (-\frac {a}{b}+\frac {\cosh (x)}{b}\right ) \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e^{x^2} (-a+\cosh (x)) \sinh (x)}{b} \, dx,x,\cosh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} (-a+\cosh (x)) \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a e^{x^2} \sinh (x)+e^{x^2} \cosh (x) \sinh (x)\right ) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cosh (x) \sinh (x) \, dx,x,\cosh ^{-1}(a+b x)\right )}{b^2}-\frac {a \operatorname {Subst}\left (\int e^{x^2} \sinh (x) \, dx,x,\cosh ^{-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,\cosh ^{-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,\cosh ^{-1}(a+b x)\right )}{b^2}\\ &=-\frac {\operatorname {Subst}\left (\int e^{-2 x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{4 b^2}+\frac {\operatorname {Subst}\left (\int e^{2 x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{4 b^2}+\frac {a \operatorname {Subst}\left (\int e^{-x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^2}-\frac {a \operatorname {Subst}\left (\int e^{x+x^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (-2+2 x)^2} \, dx,x,\cosh ^{-1}(a+b x)\right )}{4 b^2 e}+\frac {\operatorname {Subst}\left (\int e^{\frac {1}{4} (2+2 x)^2} \, dx,x,\cosh ^{-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,\cosh ^{-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,\cosh ^{-1}(a+b x)\right )}{2 b^2 \sqrt [4]{e}}\\ &=\frac {\sqrt {\pi } \text {erfi}\left (1-\cosh ^{-1}(a+b x)\right )}{8 b^2 e}+\frac {\sqrt {\pi } \text {erfi}\left (1+\cosh ^{-1}(a+b x)\right )}{8 b^2 e}+\frac {a \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \cosh ^{-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 \cosh ^{-1}(a+b x)\right )\right )}{4 b^2 \sqrt [4]{e}}\\ \end {align*}
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Mathematica [A] time = 0.14, size = 76, normalized size = 0.65 \[ \frac {\sqrt {\pi } \left (-2 e^{3/4} a \text {erfi}\left (\frac {1}{2}-\cosh ^{-1}(a+b x)\right )+\text {erfi}\left (1-\cosh ^{-1}(a+b x)\right )-2 e^{3/4} a \text {erfi}\left (\cosh ^{-1}(a+b x)+\frac {1}{2}\right )+\text {erfi}\left (\cosh ^{-1}(a+b x)+1\right )\right )}{8 e b^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x e^{\left (\operatorname {arcosh}\left (b x + a\right )^{2}\right )}, x\right ) \]
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 \[ \int x e^{\left (\operatorname {arcosh}\left (b x + a\right )^{2}\right )}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.01, size = 0, normalized size = 0.00 \[ \int {\mathrm e}^{\mathrm {arccosh}\left (b x +a \right )^{2}} x\, dx \]
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 \[ \int x e^{\left (\operatorname {arcosh}\left (b x + a\right )^{2}\right )}\,{d x} \]
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 \[ \int x\,{\mathrm {e}}^{{\mathrm {acosh}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int x e^{\operatorname {acosh}^{2}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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