3.39.44 \(\int e^{-x^2} (10 x+e^{x^2} (1+2 x+3 x^2)) \, dx\)

Optimal. Leaf size=18 \[ -5 e^{-x^2}+x+x \left (x+x^2\right ) \]

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Rubi [A]  time = 0.06, antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 3, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6688, 14, 2209} \begin {gather*} x^3+x^2-5 e^{-x^2}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x + E^x^2*(1 + 2*x + 3*x^2))/E^x^2,x]

[Out]

-5/E^x^2 + x + x^2 + x^3

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 6688

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

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 17, normalized size = 0.94 \begin {gather*} -5 e^{-x^2}+x+x^2+x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x + E^x^2*(1 + 2*x + 3*x^2))/E^x^2,x]

[Out]

-5/E^x^2 + x + x^2 + x^3

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fricas [A]  time = 0.52, size = 22, normalized size = 1.22 \begin {gather*} {\left ({\left (x^{3} + x^{2} + x\right )} e^{\left (x^{2}\right )} - 5\right )} e^{\left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+2*x+1)*exp(x^2)+10*x)/exp(x^2),x, algorithm="fricas")

[Out]

((x^3 + x^2 + x)*e^(x^2) - 5)*e^(-x^2)

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giac [A]  time = 0.15, size = 16, normalized size = 0.89 \begin {gather*} x^{3} + x^{2} + x - 5 \, e^{\left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+2*x+1)*exp(x^2)+10*x)/exp(x^2),x, algorithm="giac")

[Out]

x^3 + x^2 + x - 5*e^(-x^2)

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maple [A]  time = 0.02, size = 17, normalized size = 0.94

method	result	size
default	\(x^{3}+x^{2}+x -5 \,{\mathrm e}^{-x^{2}}\)	\(17\)
risch	\(x^{3}+x^{2}+x -5 \,{\mathrm e}^{-x^{2}}\)	\(17\)
norman	\(\left (-5+x^{2} {\mathrm e}^{x^{2}}+x^{3} {\mathrm e}^{x^{2}}+{\mathrm e}^{x^{2}} x \right ) {\mathrm e}^{-x^{2}}\)	\(32\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2+2*x+1)*exp(x^2)+10*x)/exp(x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.35, size = 16, normalized size = 0.89 \begin {gather*} x^{3} + x^{2} + x - 5 \, e^{\left (-x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+2*x+1)*exp(x^2)+10*x)/exp(x^2),x, algorithm="maxima")

[Out]

x^3 + x^2 + x - 5*e^(-x^2)

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mupad [B]  time = 0.07, size = 16, normalized size = 0.89 \begin {gather*} x-5\,{\mathrm {e}}^{-x^2}+x^2+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-x^2)*(10*x + exp(x^2)*(2*x + 3*x^2 + 1)),x)

[Out]

x - 5*exp(-x^2) + x^2 + x^3

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sympy [A]  time = 0.09, size = 14, normalized size = 0.78 \begin {gather*} x^{3} + x^{2} + x - 5 e^{- x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2+2*x+1)*exp(x**2)+10*x)/exp(x**2),x)

[Out]

x**3 + x**2 + x - 5*exp(-x**2)

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