Optimal. Leaf size=265 \[ \frac{\sqrt [4]{e} \sqrt{\pi } a^2 \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{4 b^3}+\frac{\sqrt [4]{e} \sqrt{\pi } a^2 \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{4 b^3}-\frac{e \sqrt{\pi } a \text{Erf}\left (1-i \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{e \sqrt{\pi } a \text{Erf}\left (1+i \sin ^{-1}(a+b x)\right )}{4 b^3}+\frac{\sqrt [4]{e} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{16 b^3}+\frac{\sqrt [4]{e} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{16 b^3}-\frac{e^{9/4} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-3 i\right )\right )}{16 b^3}-\frac{e^{9/4} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+3 i\right )\right )}{16 b^3} \]
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Rubi [A] time = 0.474104, antiderivative size = 265, normalized size of antiderivative = 1., 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 = {4836, 6741, 12, 6742, 4473, 2234, 2204, 4474} \[ \frac{\sqrt [4]{e} \sqrt{\pi } a^2 \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{4 b^3}+\frac{\sqrt [4]{e} \sqrt{\pi } a^2 \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{4 b^3}-\frac{e \sqrt{\pi } a \text{Erf}\left (1-i \sin ^{-1}(a+b x)\right )}{4 b^3}-\frac{e \sqrt{\pi } a \text{Erf}\left (1+i \sin ^{-1}(a+b x)\right )}{4 b^3}+\frac{\sqrt [4]{e} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{16 b^3}+\frac{\sqrt [4]{e} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{16 b^3}-\frac{e^{9/4} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)-3 i\right )\right )}{16 b^3}-\frac{e^{9/4} \sqrt{\pi } \text{Erfi}\left (\frac{1}{2} \left (2 \sin ^{-1}(a+b x)+3 i\right )\right )}{16 b^3} \]
Antiderivative was successfully verified.
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Rule 4836
Rule 6741
Rule 12
Rule 6742
Rule 4473
Rule 2234
Rule 2204
Rule 4474
Rubi steps
\begin{align*} \int e^{\sin ^{-1}(a+b x)^2} x^2 \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{Subst}\left (\int e^{x^2} \cos (x) (a-\sin (x))^2 \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{\operatorname{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{\operatorname{Subst}\left (\int e^{x^2} \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}-\frac{(2 a) \operatorname{Subst}\left (\int e^{x^2} \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}+\frac{a^2 \operatorname{Subst}\left (\int e^{x^2} \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^3}\\ &=\frac{\operatorname{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) \operatorname{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 \operatorname{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{\operatorname{Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}+\frac{\operatorname{Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac{\operatorname{Subst}\left (\int e^{-3 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac{\operatorname{Subst}\left (\int e^{3 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^3}-\frac{(i a) \operatorname{Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac{(i a) \operatorname{Subst}\left (\int e^{2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac{a^2 \operatorname{Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}+\frac{a^2 \operatorname{Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^3}\\ &=\frac{\sqrt [4]{e} \operatorname{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} \operatorname{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 ) \operatorname{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 ) \operatorname{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) \operatorname{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) \operatorname{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} \operatorname{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} \operatorname{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*}
Mathematica [A] time = 0.214028, size = 161, normalized size = 0.61 \[ -\frac{\sqrt{\pi } \left (i \sqrt [4]{e} \left (4 a^2 \text{Erf}\left (\frac{1}{2}+i \sin ^{-1}(a+b x)\right )-\left (4 a^2+1\right ) \text{Erf}\left (\frac{1}{2}-i \sin ^{-1}(a+b x)\right )-4 i e^{3/4} a \text{Erf}\left (1+i \sin ^{-1}(a+b x)\right )+e^2 \text{Erf}\left (\frac{3}{2}-i \sin ^{-1}(a+b x)\right )+\text{Erf}\left (\frac{1}{2}+i \sin ^{-1}(a+b x)\right )-e^2 \text{Erf}\left (\frac{3}{2}+i \sin ^{-1}(a+b x)\right )\right )+4 e a \text{Erf}\left (1-i \sin ^{-1}(a+b x)\right )\right )}{16 b^3} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.007, size = 0, normalized size = 0. \begin{align*} \int{{\rm e}^{ \left ( \arcsin \left ( bx+a \right ) \right ) ^{2}}}{x}^{2}\, 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^{2} e^{\left (\arcsin \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^{2} e^{\left (\arcsin \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^{2} e^{\operatorname{asin}^{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^{2} e^{\left (\arcsin \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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