3.103.95 \(\int e^{x^2} (1+2 x+2 x^2+e^8 (2+4 x^2)) \, dx\)

Optimal. Leaf size=15 \[ e^{x^2} \left (1+x+2 e^8 x\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 20, normalized size of antiderivative = 1.33, number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2226, 2204, 2209, 2212} \begin {gather*} \left (1+2 e^8\right ) e^{x^2} x+e^{x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^x^2*(1 + 2*x + 2*x^2 + E^8*(2 + 4*x^2)),x]

[Out]

E^x^2 + E^x^2*(1 + 2*E^8)*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 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (e^{x^2} \left (1+2 e^8\right )+2 e^{x^2} x+2 e^{x^2} \left (1+2 e^8\right ) x^2\right ) \, dx\\ &=2 \int e^{x^2} x \, dx+\left (1+2 e^8\right ) \int e^{x^2} \, dx+\left (2 \left (1+2 e^8\right )\right ) \int e^{x^2} x^2 \, dx\\ &=e^{x^2}+e^{x^2} \left (1+2 e^8\right ) x+\frac {1}{2} \left (1+2 e^8\right ) \sqrt {\pi } \text {erfi}(x)+\left (-1-2 e^8\right ) \int e^{x^2} \, dx\\ &=e^{x^2}+e^{x^2} \left (1+2 e^8\right ) x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 1.00 \begin {gather*} e^{x^2} \left (1+x+2 e^8 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x^2*(1 + 2*x + 2*x^2 + E^8*(2 + 4*x^2)),x]

[Out]

E^x^2*(1 + x + 2*E^8*x)

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fricas [A]  time = 0.52, size = 14, normalized size = 0.93 \begin {gather*} e^{\left (x^{2} + \log \left (2 \, x e^{8} + x + 1\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+2)*exp(4)^2+2*x^2+2*x+1)*exp(log(2*x*exp(4)^2+x+1)+x^2)/(2*x*exp(4)^2+x+1),x, algorithm="fri
cas")

[Out]

e^(x^2 + log(2*x*e^8 + x + 1))

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giac [A]  time = 0.14, size = 20, normalized size = 1.33 \begin {gather*} 2 \, x e^{\left (x^{2} + 8\right )} + x e^{\left (x^{2}\right )} + e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+2)*exp(4)^2+2*x^2+2*x+1)*exp(log(2*x*exp(4)^2+x+1)+x^2)/(2*x*exp(4)^2+x+1),x, algorithm="gia
c")

[Out]

2*x*e^(x^2 + 8) + x*e^(x^2) + e^(x^2)

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maple [A]  time = 0.10, size = 14, normalized size = 0.93

method	result	size
risch	\(\left (2 x \,{\mathrm e}^{8}+x +1\right ) {\mathrm e}^{x^{2}}\)	\(14\)
gosper	\({\mathrm e}^{\ln \left (2 x \,{\mathrm e}^{8}+x +1\right )+x^{2}}\)	\(17\)
norman	\({\mathrm e}^{\ln \left (2 x \,{\mathrm e}^{8}+x +1\right )+x^{2}}\)	\(17\)
default	\({\mathrm e}^{x^{2}} x +{\mathrm e}^{x^{2}}+{\mathrm e}^{8} \sqrt {\pi }\, \erfi \relax (x )+4 \,{\mathrm e}^{8} \left (\frac {{\mathrm e}^{x^{2}} x}{2}-\frac {\sqrt {\pi }\, \erfi \relax (x )}{4}\right )\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^2+2)*exp(4)^2+2*x^2+2*x+1)*exp(ln(2*x*exp(4)^2+x+1)+x^2)/(2*x*exp(4)^2+x+1),x,method=_RETURNVERBOSE)

[Out]

(2*x*exp(8)+x+1)*exp(x^2)

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maxima [A]  time = 0.35, size = 20, normalized size = 1.33 \begin {gather*} 2 \, x e^{\left (x^{2} + 8\right )} + x e^{\left (x^{2}\right )} + e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^2+2)*exp(4)^2+2*x^2+2*x+1)*exp(log(2*x*exp(4)^2+x+1)+x^2)/(2*x*exp(4)^2+x+1),x, algorithm="max
ima")

[Out]

2*x*e^(x^2 + 8) + x*e^(x^2) + e^(x^2)

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mupad [B]  time = 0.54, size = 13, normalized size = 0.87 \begin {gather*} {\mathrm {e}}^{x^2}\,\left (x+2\,x\,{\mathrm {e}}^8+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(log(x + 2*x*exp(8) + 1) + x^2)*(2*x + exp(8)*(4*x^2 + 2) + 2*x^2 + 1))/(x + 2*x*exp(8) + 1),x)

[Out]

exp(x^2)*(x + 2*x*exp(8) + 1)

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sympy [A]  time = 0.12, size = 14, normalized size = 0.93 \begin {gather*} \left (x + 2 x e^{8} + 1\right ) e^{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**2+2)*exp(4)**2+2*x**2+2*x+1)*exp(ln(2*x*exp(4)**2+x+1)+x**2)/(2*x*exp(4)**2+x+1),x)

[Out]

(x + 2*x*exp(8) + 1)*exp(x**2)

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