3.25.17 \(\int \frac {1}{5} e^{x^2} (2+13 x+4 x^2+3 x^3+e^x (-2-2 x-4 x^2)) \, dx\)

Optimal. Leaf size=25 \[ e^{x^2} \left (1-\frac {1}{10} x \left (x-4 \left (1-e^x+x\right )\right )\right ) \]

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Rubi [B]  time = 0.23, antiderivative size = 53, normalized size of antiderivative = 2.12, number of steps used = 10, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 6742, 2204, 2209, 2212, 2288} \begin {gather*} \frac {3}{10} e^{x^2} x^2+\frac {2 e^{x^2} x}{5}+e^{x^2}-\frac {2 e^{x^2+x} \left (2 x^2+x\right )}{5 (2 x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^x^2*(2 + 13*x + 4*x^2 + 3*x^3 + E^x*(-2 - 2*x - 4*x^2)))/5,x]

[Out]

E^x^2 + (2*E^x^2*x)/5 + (3*E^x^2*x^2)/10 - (2*E^(x + x^2)*(x + 2*x^2))/(5*(1 + 2*x))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{5} \int e^{x^2} \left (2+13 x+4 x^2+3 x^3+e^x \left (-2-2 x-4 x^2\right )\right ) \, dx\\ &=\frac {1}{5} \int \left (2 e^{x^2}+13 e^{x^2} x+4 e^{x^2} x^2+3 e^{x^2} x^3-2 e^{x+x^2} \left (1+x+2 x^2\right )\right ) \, dx\\ &=\frac {2}{5} \int e^{x^2} \, dx-\frac {2}{5} \int e^{x+x^2} \left (1+x+2 x^2\right ) \, dx+\frac {3}{5} \int e^{x^2} x^3 \, dx+\frac {4}{5} \int e^{x^2} x^2 \, dx+\frac {13}{5} \int e^{x^2} x \, dx\\ &=\frac {13 e^{x^2}}{10}+\frac {2 e^{x^2} x}{5}+\frac {3}{10} e^{x^2} x^2-\frac {2 e^{x+x^2} \left (x+2 x^2\right )}{5 (1+2 x)}+\frac {1}{5} \sqrt {\pi } \text {erfi}(x)-\frac {2}{5} \int e^{x^2} \, dx-\frac {3}{5} \int e^{x^2} x \, dx\\ &=e^{x^2}+\frac {2 e^{x^2} x}{5}+\frac {3}{10} e^{x^2} x^2-\frac {2 e^{x+x^2} \left (x+2 x^2\right )}{5 (1+2 x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 24, normalized size = 0.96 \begin {gather*} \frac {1}{10} e^{x^2} \left (10-4 \left (-1+e^x\right ) x+3 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x^2*(2 + 13*x + 4*x^2 + 3*x^3 + E^x*(-2 - 2*x - 4*x^2)))/5,x]

[Out]

(E^x^2*(10 - 4*(-1 + E^x)*x + 3*x^2))/10

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fricas [A]  time = 0.65, size = 21, normalized size = 0.84 \begin {gather*} \frac {1}{10} \, {\left (3 \, x^{2} - 4 \, x e^{x} + 4 \, x + 10\right )} e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4*x^2-2*x-2)*exp(x)+3*x^3+4*x^2+13*x+2)*exp(x^2),x, algorithm="fricas")

[Out]

1/10*(3*x^2 - 4*x*e^x + 4*x + 10)*e^(x^2)

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giac [A]  time = 0.37, size = 26, normalized size = 1.04 \begin {gather*} -\frac {2}{5} \, x e^{\left (x^{2} + x\right )} + \frac {1}{10} \, {\left (3 \, x^{2} + 4 \, x + 10\right )} e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4*x^2-2*x-2)*exp(x)+3*x^3+4*x^2+13*x+2)*exp(x^2),x, algorithm="giac")

[Out]

-2/5*x*e^(x^2 + x) + 1/10*(3*x^2 + 4*x + 10)*e^(x^2)

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maple [A]  time = 0.11, size = 22, normalized size = 0.88

method	result	size
risch	\(\frac {\left (5+2 x +\frac {3 x^{2}}{2}-2 \,{\mathrm e}^{x} x \right ) {\mathrm e}^{x^{2}}}{5}\)	\(22\)
default	\(\frac {2 \,{\mathrm e}^{x^{2}} x}{5}+\frac {3 x^{2} {\mathrm e}^{x^{2}}}{10}+{\mathrm e}^{x^{2}}-\frac {2 x \,{\mathrm e}^{x^{2}+x}}{5}\)	\(31\)
norman	\(\frac {3 x^{2} {\mathrm e}^{x^{2}}}{10}+\frac {2 \,{\mathrm e}^{x^{2}} x}{5}-\frac {2 x \,{\mathrm e}^{x} {\mathrm e}^{x^{2}}}{5}+{\mathrm e}^{x^{2}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/5*((-4*x^2-2*x-2)*exp(x)+3*x^3+4*x^2+13*x+2)*exp(x^2),x,method=_RETURNVERBOSE)

[Out]

1/5*(5+2*x+3/2*x^2-2*exp(x)*x)*exp(x^2)

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maxima [C]  time = 0.62, size = 176, normalized size = 7.04 \begin {gather*} \frac {1}{5} i \, \sqrt {\pi } \operatorname {erf}\left (i \, x + \frac {1}{2} i\right ) e^{\left (-\frac {1}{4}\right )} + \frac {1}{10} \, {\left (\frac {4 \, {\left (2 \, x + 1\right )}^{3} \Gamma \left (\frac {3}{2}, -\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}{\left (-{\left (2 \, x + 1\right )}^{2}\right )^{\frac {3}{2}}} - \frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} + 4 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} + \frac {1}{10} \, {\left (\frac {\sqrt {\pi } {\left (2 \, x + 1\right )} {\left (\operatorname {erf}\left (\frac {1}{2} \, \sqrt {-{\left (2 \, x + 1\right )}^{2}}\right ) - 1\right )}}{\sqrt {-{\left (2 \, x + 1\right )}^{2}}} - 2 \, e^{\left (\frac {1}{4} \, {\left (2 \, x + 1\right )}^{2}\right )}\right )} e^{\left (-\frac {1}{4}\right )} + \frac {3}{10} \, {\left (x^{2} - 1\right )} e^{\left (x^{2}\right )} + \frac {2}{5} \, x e^{\left (x^{2}\right )} + \frac {13}{10} \, e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4*x^2-2*x-2)*exp(x)+3*x^3+4*x^2+13*x+2)*exp(x^2),x, algorithm="maxima")

[Out]

1/5*I*sqrt(pi)*erf(I*x + 1/2*I)*e^(-1/4) + 1/10*(4*(2*x + 1)^3*gamma(3/2, -1/4*(2*x + 1)^2)/(-(2*x + 1)^2)^(3/
2) - sqrt(pi)*(2*x + 1)*(erf(1/2*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) + 4*e^(1/4*(2*x + 1)^2))*e^(-1/4)
+ 1/10*(sqrt(pi)*(2*x + 1)*(erf(1/2*sqrt(-(2*x + 1)^2)) - 1)/sqrt(-(2*x + 1)^2) - 2*e^(1/4*(2*x + 1)^2))*e^(-
1/4) + 3/10*(x^2 - 1)*e^(x^2) + 2/5*x*e^(x^2) + 13/10*e^(x^2)

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mupad [B]  time = 1.38, size = 21, normalized size = 0.84 \begin {gather*} \frac {{\mathrm {e}}^{x^2}\,\left (4\,x-4\,x\,{\mathrm {e}}^x+3\,x^2+10\right )}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^2)*(13*x - exp(x)*(2*x + 4*x^2 + 2) + 4*x^2 + 3*x^3 + 2))/5,x)

[Out]

(exp(x^2)*(4*x - 4*x*exp(x) + 3*x^2 + 10))/10

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sympy [A]  time = 0.21, size = 22, normalized size = 0.88 \begin {gather*} \frac {\left (3 x^{2} - 4 x e^{x} + 4 x + 10\right ) e^{x^{2}}}{10} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/5*((-4*x**2-2*x-2)*exp(x)+3*x**3+4*x**2+13*x+2)*exp(x**2),x)

[Out]

(3*x**2 - 4*x*exp(x) + 4*x + 10)*exp(x**2)/10

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