Optimal. Leaf size=381 \[ \frac {e \sqrt {\pi } \text {Erf}(1-i \text {ArcSin}(a+b x))}{16 b^4}+\frac {3 a^2 e \sqrt {\pi } \text {Erf}(1-i \text {ArcSin}(a+b x))}{8 b^4}-\frac {e^4 \sqrt {\pi } \text {Erf}(2-i \text {ArcSin}(a+b x))}{32 b^4}+\frac {e \sqrt {\pi } \text {Erf}(1+i \text {ArcSin}(a+b x))}{16 b^4}+\frac {3 a^2 e \sqrt {\pi } \text {Erf}(1+i \text {ArcSin}(a+b x))}{8 b^4}-\frac {e^4 \sqrt {\pi } \text {Erf}(2+i \text {ArcSin}(a+b x))}{32 b^4}-\frac {3 a \sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (-i+2 \text {ArcSin}(a+b x))\right )}{16 b^4}-\frac {a^3 \sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (-i+2 \text {ArcSin}(a+b x))\right )}{4 b^4}-\frac {3 a \sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (i+2 \text {ArcSin}(a+b x))\right )}{16 b^4}-\frac {a^3 \sqrt [4]{e} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (i+2 \text {ArcSin}(a+b x))\right )}{4 b^4}+\frac {3 a e^{9/4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (-3 i+2 \text {ArcSin}(a+b x))\right )}{16 b^4}+\frac {3 a e^{9/4} \sqrt {\pi } \text {Erfi}\left (\frac {1}{2} (3 i+2 \text {ArcSin}(a+b x))\right )}{16 b^4} \]
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Rubi [A]
time = 0.49, antiderivative size = 381, normalized size of antiderivative = 1.00, number of steps
used = 37, 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^3 \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)-i)\right )}{4 b^4}-\frac {\sqrt [4]{e} \sqrt {\pi } a^3 \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)+i)\right )}{4 b^4}+\frac {3 e \sqrt {\pi } a^2 \text {Erf}(1-i \text {ArcSin}(a+b x))}{8 b^4}+\frac {3 e \sqrt {\pi } a^2 \text {Erf}(1+i \text {ArcSin}(a+b x))}{8 b^4}+\frac {e \sqrt {\pi } \text {Erf}(1-i \text {ArcSin}(a+b x))}{16 b^4}-\frac {e^4 \sqrt {\pi } \text {Erf}(2-i \text {ArcSin}(a+b x))}{32 b^4}+\frac {e \sqrt {\pi } \text {Erf}(1+i \text {ArcSin}(a+b x))}{16 b^4}-\frac {e^4 \sqrt {\pi } \text {Erf}(2+i \text {ArcSin}(a+b x))}{32 b^4}-\frac {3 \sqrt [4]{e} \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)-i)\right )}{16 b^4}-\frac {3 \sqrt [4]{e} \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)+i)\right )}{16 b^4}+\frac {3 e^{9/4} \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)-3 i)\right )}{16 b^4}+\frac {3 e^{9/4} \sqrt {\pi } a \text {Erfi}\left (\frac {1}{2} (2 \text {ArcSin}(a+b x)+3 i)\right )}{16 b^4} \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^3 \, dx &=\frac {\text {Subst}\left (\int e^{x^2} \cos (x) \left (-\frac {a}{b}+\frac {\sin (x)}{b}\right )^3 \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {e^{x^2} \cos (x) (-a+\sin (x))^3}{b^3} \, dx,x,\sin ^{-1}(a+b x)\right )}{b}\\ &=\frac {\text {Subst}\left (\int e^{x^2} \cos (x) (-a+\sin (x))^3 \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}\\ &=\frac {\text {Subst}\left (\int \left (-a^3 e^{x^2} \cos (x)+3 a^2 e^{x^2} \cos (x) \sin (x)-3 a e^{x^2} \cos (x) \sin ^2(x)+e^{x^2} \cos (x) \sin ^3(x)\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}\\ &=\frac {\text {Subst}\left (\int e^{x^2} \cos (x) \sin ^3(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}-\frac {(3 a) \text {Subst}\left (\int e^{x^2} \cos (x) \sin ^2(x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int e^{x^2} \cos (x) \sin (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}-\frac {a^3 \text {Subst}\left (\int e^{x^2} \cos (x) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{8} i e^{-2 i x+x^2}-\frac {1}{8} i e^{2 i x+x^2}-\frac {1}{16} i e^{-4 i x+x^2}+\frac {1}{16} i e^{4 i x+x^2}\right ) \, dx,x,\sin ^{-1}(a+b x)\right )}{b^4}-\frac {(3 a) \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^4}+\frac {\left (3 a^2\right ) \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^4}-\frac {a^3 \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^4}\\ &=-\frac {i \text {Subst}\left (\int e^{-4 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{16 b^4}+\frac {i \text {Subst}\left (\int e^{4 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{16 b^4}+\frac {i \text {Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {i \text {Subst}\left (\int e^{2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {(3 a) \text {Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {(3 a) \text {Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}+\frac {(3 a) \text {Subst}\left (\int e^{-3 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}+\frac {(3 a) \text {Subst}\left (\int e^{3 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}+\frac {\left (3 i a^2\right ) \text {Subst}\left (\int e^{-2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^4}-\frac {\left (3 i a^2\right ) \text {Subst}\left (\int e^{2 i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^4}-\frac {a^3 \text {Subst}\left (\int e^{-i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^4}-\frac {a^3 \text {Subst}\left (\int e^{i x+x^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{2 b^4}\\ &=-\frac {\left (3 a \sqrt [4]{e}\right ) \text {Subst}\left (\int e^{\frac {1}{4} (-i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {\left (3 a \sqrt [4]{e}\right ) \text {Subst}\left (\int e^{\frac {1}{4} (i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {\left (a^3 \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^4}-\frac {\left (a^3 \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^4}+\frac {(i e) \text {Subst}\left (\int e^{\frac {1}{4} (-2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {(i e) \text {Subst}\left (\int e^{\frac {1}{4} (2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}+\frac {\left (3 i a^2 e\right ) \text {Subst}\left (\int e^{\frac {1}{4} (-2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^4}-\frac {\left (3 i a^2 e\right ) \text {Subst}\left (\int e^{\frac {1}{4} (2 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{4 b^4}+\frac {\left (3 a e^{9/4}\right ) \text {Subst}\left (\int e^{\frac {1}{4} (-3 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}+\frac {\left (3 a e^{9/4}\right ) \text {Subst}\left (\int e^{\frac {1}{4} (3 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {\left (i e^4\right ) \text {Subst}\left (\int e^{\frac {1}{4} (-4 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{16 b^4}+\frac {\left (i e^4\right ) \text {Subst}\left (\int e^{\frac {1}{4} (4 i+2 x)^2} \, dx,x,\sin ^{-1}(a+b x)\right )}{16 b^4}\\ &=\frac {e \sqrt {\pi } \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )}{16 b^4}+\frac {3 a^2 e \sqrt {\pi } \text {erf}\left (1-i \sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {e^4 \sqrt {\pi } \text {erf}\left (2-i \sin ^{-1}(a+b x)\right )}{32 b^4}+\frac {e \sqrt {\pi } \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )}{16 b^4}+\frac {3 a^2 e \sqrt {\pi } \text {erf}\left (1+i \sin ^{-1}(a+b x)\right )}{8 b^4}-\frac {e^4 \sqrt {\pi } \text {erf}\left (2+i \sin ^{-1}(a+b x)\right )}{32 b^4}-\frac {3 a \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^4}-\frac {a^3 \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^4}-\frac {3 a \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^4}-\frac {a^3 \sqrt [4]{e} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (i+2 \sin ^{-1}(a+b x)\right )\right )}{4 b^4}+\frac {3 a e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (-3 i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^4}+\frac {3 a e^{9/4} \sqrt {\pi } \text {erfi}\left (\frac {1}{2} \left (3 i+2 \sin ^{-1}(a+b x)\right )\right )}{16 b^4}\\ \end {align*}
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Mathematica [A]
time = 0.22, size = 221, normalized size = 0.58 \begin {gather*} -\frac {\sqrt {\pi } \left (-2 \left (e+6 a^2 e\right ) \text {Erf}(1-i \text {ArcSin}(a+b x))+e^4 \text {Erf}(2-i \text {ArcSin}(a+b x))+\sqrt [4]{e} \left (-2 i a \left (3+4 a^2\right ) \text {Erf}\left (\frac {1}{2}+i \text {ArcSin}(a+b x)\right )-2 \left (1+6 a^2\right ) e^{3/4} \text {Erf}(1+i \text {ArcSin}(a+b x))+6 i a e^2 \text {Erf}\left (\frac {3}{2}+i \text {ArcSin}(a+b x)\right )+e^{15/4} \text {Erf}(2+i \text {ArcSin}(a+b x))+6 a \text {Erfi}\left (\frac {1}{2} (i+2 \text {ArcSin}(a+b x))\right )+8 a^3 \text {Erfi}\left (\frac {1}{2} (i+2 \text {ArcSin}(a+b x))\right )-6 a e^2 \text {Erfi}\left (\frac {1}{2} (3 i+2 \text {ArcSin}(a+b x))\right )\right )\right )}{32 b^4} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int {\mathrm e}^{\arcsin \left (b x +a \right )^{2}} x^{3}\, dx\]
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^{3} 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^3\,{\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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