Optimal. Leaf size=265 \[ -\frac {a e \sqrt {\pi } \text {Erf}(1-i \text {ArcSin}(a+b x))}{4 b^3}-\frac {a e \sqrt {\pi } \text {Erf}(1+i \text {ArcSin}(a+b x))}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (-i+2 \text {ArcSin}(a+b x))\right )}{16 b^3}+\frac {a^2 \sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (-i+2 \text {ArcSin}(a+b x))\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (i+2 \text {ArcSin}(a+b x))\right )}{16 b^3}+\frac {a^2 \sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (i+2 \text {ArcSin}(a+b x))\right )}{4 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (-3 i+2 \text {ArcSin}(a+b x))\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (3 i+2 \text {ArcSin}(a+b x))\right )}{16 b^3} \]
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Rubi [A]
time = 0.36, antiderivative size = 265, 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 = {4920, 6873,
12, 6874, 4561, 2266, 2235, 4562} \begin {gather*} \frac {\sqrt [4]{e} \sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)-i)\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } a^2 \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)+i)\right )}{4 b^3}-\frac {e \sqrt {\pi } a \text {Erf}(1-i \text {ArcSin}(a+b x))}{4 b^3}-\frac {e \sqrt {\pi } a \text {Erf}(1+i \text {ArcSin}(a+b x))}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)-i)\right )}{16 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)+i)\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)-3 i)\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)+3 i)\right )}{16 b^3} \end {gather*}
Antiderivative was successfully verified.
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Rule 12
Rule 2235
Rule 2266
Rule 4561
Rule 4562
Rule 4920
Rule 6873
Rule 6874
Rubi steps
\begin {align*} \int e^{\sin ^{-1}(a+b x)^2} x^2 \, dx &=\frac {\text {Subst}\left (\int e^{x^2} \cos (x) \left (-\frac {a}{b}+\frac {\sin (x)}{b}\right )^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {e^{x^2} \cos (x) (a-\sin (x))^2}{b^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int e^{x^2} \cos (x) (a-\sin (x))^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int \left (a^2 e^{x^2} \cos (x)-2 a e^{x^2} \cos (x) \sin (x)+e^{x^2} \cos (x) \sin ^2(x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int e^{x^2} \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \text {Subst}\left (\int e^{x^2} \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \text {Subst}\left (\int e^{x^2} \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{8} e^{-i x+x^2}+\frac {1}{8} e^{i x+x^2}-\frac {1}{8} e^{-3 i x+x^2}-\frac {1}{8} e^{3 i x+x^2}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac {(2 a) \text {Subst}\left (\int \left (\frac {1}{4} i e^{-2 i x+x^2}-\frac {1}{4} i e^{2 i x+x^2}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}+\frac {a^2 \text {Subst}\left (\int \left (\frac {1}{2} e^{-i x+x^2}+\frac {1}{2} e^{i x+x^2}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac {\text {Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {\text {Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {\text {Subst}\left (\int e^{-3 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {\text {Subst}\left (\int e^{3 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {(i a) \text {Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {(i a) \text {Subst}\left (\int e^{2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \text {Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {a^2 \text {Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}\\ &=\frac {\sqrt [4]{e} \text {Subst}\left (\int e^{\frac {1}{4} (-i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {\sqrt [4]{e} \text {Subst}\left (\int e^{\frac {1}{4} (i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac {\left (a^2 \sqrt [4]{e}\right ) \text {Subst}\left (\int e^{\frac {1}{4} (-i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {\left (a^2 \sqrt [4]{e}\right ) \text {Subst}\left (\int e^{\frac {1}{4} (i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}-\frac {(i a e) \text {Subst}\left (\int e^{\frac {1}{4} (-2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac {(i a e) \text {Subst}\left (\int e^{\frac {1}{4} (2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}-\frac {e^{9/4} \text {Subst}\left (\int e^{\frac {1}{4} (-3 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac {e^{9/4} \text {Subst}\left (\int e^{\frac {1}{4} (3 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}\\ &=-\frac {a e \sqrt {\pi } \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac {a e \sqrt {\pi } \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^3}+\frac {a^2 \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^3}+\frac {\sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^3}+\frac {a^2 \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-3 i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^3}-\frac {e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (3 i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.14, size = 161, normalized size = 0.61 \begin {gather*} -\frac {\sqrt {\pi } \left (4 a e \text {Erf}(1-i \text {ArcSin}(a+b x))+i \sqrt [4]{e} \left (-\left (\left (1+4 a^2\right ) \text {Erf}\left (\frac {1}{2}-i \text {ArcSin}(a+b x)\right )\right )+e^2 \text {Erf}\left (\frac {3}{2}-i \text {ArcSin}(a+b x)\right )+\text {Erf}\left (\frac {1}{2}+i \text {ArcSin}(a+b x)\right )+4 a^2 \text {Erf}\left (\frac {1}{2}+i \text {ArcSin}(a+b x)\right )-4 i a e^{3/4} \text {Erf}(1+i \text {ArcSin}(a+b x))-e^2 \text {Erf}\left (\frac {3}{2}+i \text {ArcSin}(a+b x)\right )\right )\right )}{16 b^3} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.00, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{\arcsin \left (b x +a \right )^{2}} x^{2}\, dx\]
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 {asin}^{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 {asin}\left (a+b\,x\right )}^2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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