Optimal. Leaf size=251 \[ -\frac {a \sqrt {\pi } \text {Erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{4 b^3 e}-\frac {a \sqrt {\pi } \text {Erfi}\left (1+\sinh ^{-1}(a+b x)\right )}{4 b^3 e}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (-3+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}}-\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 \sqrt [4]{e}}+\frac {a^2 \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^3 \sqrt [4]{e}}-\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 \sqrt [4]{e}}+\frac {a^2 \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^3 \sqrt [4]{e}}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (3+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}} \]
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Rubi [A]
time = 0.42, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps
used = 27, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {5873, 6873,
12, 6874, 5624, 2266, 2235, 5625} \begin {gather*} \frac {\sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{4 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{4 \sqrt [4]{e} b^3}-\frac {\sqrt {\pi } a \text {Erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{4 e b^3}-\frac {\sqrt {\pi } a \text {Erfi}\left (\sinh ^{-1}(a+b x)+1\right )}{4 e b^3}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-3\right )\right )}{16 e^{9/4} b^3}-\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)-1\right )\right )}{16 \sqrt [4]{e} b^3}-\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+1\right )\right )}{16 \sqrt [4]{e} b^3}+\frac {\sqrt {\pi } \text {Erfi}\left (\frac {1}{2} \left (2 \sinh ^{-1}(a+b x)+3\right )\right )}{16 e^{9/4} b^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2235
Rule 2266
Rule 5624
Rule 5625
Rule 5873
Rule 6873
Rule 6874
Rubi steps
\begin {align*} \int e^{\sinh ^{-1}(a+b x)^2} x^2 \, dx &=\frac {\text {Subst}\left (\int e^{x^2} \cosh (x) \left (-\frac {a}{b}+\frac {\sinh (x)}{b}\right )^2 \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {e^{x^2} \cosh (x) (a-\sinh (x))^2}{b^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int e^{x^2} \cosh (x) (a-\sinh (x))^2 \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int \left (a^2 e^{x^2} \cosh (x)-2 a e^{x^2} \cosh (x) \sinh (x)+e^{x^2} \cosh (x) \sinh ^2(x)\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int e^{x^2} \cosh (x) \sinh ^2(x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \text {Subst}\left (\int e^{x^2} \cosh (x) \sinh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \text {Subst}\left (\int e^{x^2} \cosh (x) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{8} e^{-3 x+x^2}-\frac {1}{8} e^{-x+x^2}-\frac {e^{x+x^2}}{8}+\frac {1}{8} e^{3 x+x^2}\right ) \, dx,x,\sinh ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \text {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^3}+\frac {a^2 \text {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^3}\\ &=\frac {\text {Subst}\left (\int e^{-3 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3}-\frac {\text {Subst}\left (\int e^{-x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3}-\frac {\text {Subst}\left (\int e^{x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3}+\frac {\text {Subst}\left (\int e^{3 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3}+\frac {a \text {Subst}\left (\int e^{-2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}-\frac {a \text {Subst}\left (\int e^{2 x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \text {Subst}\left (\int e^{-x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \text {Subst}\left (\int e^{x+x^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3}\\ &=\frac {\text {Subst}\left (\int e^{\frac {1}{4} (-3+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3 e^{9/4}}+\frac {\text {Subst}\left (\int e^{\frac {1}{4} (3+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3 e^{9/4}}+\frac {a \text {Subst}\left (\int e^{\frac {1}{4} (-2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3 e}-\frac {a \text {Subst}\left (\int e^{\frac {1}{4} (2+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3 e}-\frac {\text {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3 \sqrt [4]{e}}-\frac {\text {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{8 b^3 \sqrt [4]{e}}+\frac {a^2 \text {Subst}\left (\int e^{\frac {1}{4} (-1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3 \sqrt [4]{e}}+\frac {a^2 \text {Subst}\left (\int e^{\frac {1}{4} (1+2 x)^2} \, dx,x,\sinh ^{-1}(a+b x)\right )}{2 b^3 \sqrt [4]{e}}\\ &=-\frac {a \sqrt {\pi } \text {erfi}\left (1-\sinh ^{-1}(a+b x)\right )}{4 b^3 e}-\frac {a \sqrt {\pi } \text {erfi}\left (1+\sinh ^{-1}(a+b x)\right )}{4 b^3 e}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-3+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 \sqrt [4]{e}}+\frac {a^2 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^3 \sqrt [4]{e}}-\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 \sqrt [4]{e}}+\frac {a^2 \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (1+2 \sinh ^{-1}(a+b x)\right )\right )}{4 b^3 \sqrt [4]{e}}+\frac {\sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (3+2 \sinh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}}\\ \end {align*}
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Mathematica [A]
time = 0.12, size = 138, normalized size = 0.55 \begin {gather*} -\frac {\sqrt {\pi } \left (\left (-1+4 a^2\right ) e^2 \text {Erfi}\left (\frac {1}{2}-\sinh ^{-1}(a+b x)\right )+4 a e^{5/4} \text {Erfi}\left (1-\sinh ^{-1}(a+b x)\right )+\text {Erfi}\left (\frac {3}{2}-\sinh ^{-1}(a+b x)\right )+e^2 \text {Erfi}\left (\frac {1}{2}+\sinh ^{-1}(a+b x)\right )-4 a^2 e^2 \text {Erfi}\left (\frac {1}{2}+\sinh ^{-1}(a+b x)\right )+4 a e^{5/4} \text {Erfi}\left (1+\sinh ^{-1}(a+b x)\right )-\text {Erfi}\left (\frac {3}{2}+\sinh ^{-1}(a+b x)\right )\right )}{16 b^3 e^{9/4}} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{\arcsinh \left (b x +a \right )^{2}} x^{2}\, dx\]
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 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
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 \begin {gather*} \int x^{2} e^{\operatorname {asinh}^{2}{\left (a + b x \right )}}\, dx \end {gather*}
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 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,{\mathrm {e}}^{{\mathrm {asinh}\left (a+b\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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