Optimal. Leaf size=123 \[ \frac {e \sqrt {\pi } \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )}{8 b^2}+\frac {e \sqrt {\pi } \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )}{8 b^2}-\frac {\sqrt [4]{e} \sqrt {\pi } a \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{4 b^2}-\frac {\sqrt [4]{e} \sqrt {\pi } a \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{4 b^2} \]
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Rubi [A] time = 0.27, antiderivative size = 123, 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 = {4836, 6741, 12, 6742, 4473, 2234, 2204, 4474} \[ \frac {e \sqrt {\pi } \text {Erf}\left (1-i \sin ^{-1}(a+b x)\right )}{8 b^2}+\frac {e \sqrt {\pi } \text {Erf}\left (1+i \sin ^{-1}(a+b x)\right )}{8 b^2}-\frac {\sqrt [4]{e} \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )}{4 b^2}-\frac {\sqrt [4]{e} \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )}{4 b^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 2204
Rule 2234
Rule 4473
Rule 4474
Rule 4836
Rule 6741
Rule 6742
Rubi steps
\begin {align*} \int e^{\sin ^{-1}(a+b x)^2} x \, dx &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cos (x) \left (-\frac {a}{b}+\frac {\sin (x)}{b}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int \frac {e^{x^2} \cos (x) (-a+\sin (x))}{b} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cos (x) (-a+\sin (x)) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a e^{x^2} \cos (x)+e^{x^2} \cos (x) \sin (x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\operatorname {Subst}\left (\int e^{x^2} \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}-\frac {a \operatorname {Subst}\left (\int e^{x^2} \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^2}\\ &=\frac {\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^2}-\frac {a \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^2}\\ &=\frac {i \operatorname {Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}-\frac {i \operatorname {Subst}\left (\int e^{2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}-\frac {a \operatorname {Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}-\frac {a \operatorname {Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac {\left (a \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^2}-\frac {\left (a \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^2}+\frac {(i e) \operatorname {Subst}\left (\int e^{\frac {1}{4} (-2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}-\frac {(i e) \operatorname {Subst}\left (\int e^{\frac {1}{4} (2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^2}\\ &=\frac {e \sqrt {\pi } \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )}{8 b^2}+\frac {e \sqrt {\pi } \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )}{8 b^2}-\frac {a \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^2}-\frac {a \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^2}\\ \end {align*}
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Mathematica [A] time = 0.13, size = 93, normalized size = 0.76 \[ \frac {\sqrt {\pi } \left (e \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )+e \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )-2 \sqrt [4]{e} a \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)-i\right )\right )-2 \sqrt [4]{e} a \text {erfi}\left (\frac {1}{2} \left (2 \sin ^{-1}(a+b x)+i\right )\right )\right )}{8 b^2} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x e^{\left (\arcsin \left (b x + a\right )^{2}\right )}, x\right ) \]
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 \[ \int x e^{\left (\arcsin \left (b x + a\right )^{2}\right )}\,{d x} \]
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 \[ \int {\mathrm e}^{\arcsin \left (b x +a \right )^{2}} x\, dx \]
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 \[ \int x e^{\left (\arcsin \left (b x + a\right )^{2}\right )}\,{d x} \]
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 \[ \int x\,{\mathrm {e}}^{{\mathrm {asin}\left (a+b\,x\right )}^2} \,d x \]
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 \[ \int x e^{\operatorname {asin}^{2}{\left (a + b x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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