\(\int e^{-x^2} (10 x+e^{x^2} (1+2 x+3 x^2)) \, dx\) [3844]

Optimal result

Integrand size = 28, antiderivative size = 18 \[ \int e^{-x^2} \left (10 x+e^{x^2} \left (1+2 x+3 x^2\right )\right ) \, dx=-5 e^{-x^2}+x+x \left (x+x^2\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6820, 14, 2240} \[ \int e^{-x^2} \left (10 x+e^{x^2} \left (1+2 x+3 x^2\right )\right ) \, dx=x^3+x^2-5 e^{-x^2}+x \]

[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 2240

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 6820

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

Rubi steps \begin{align*} \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{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.33 \[ \int e^{-x^2} \left (10 x+e^{x^2} \left (1+2 x+3 x^2\right )\right ) \, dx=-5 e^{-x^2}+x+x^3-\log \left (e^{-x^2}\right ) \]

[In]

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

[Out]

-5/E^x^2 + x + x^3 - Log[E^(-x^2)]

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.94

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

[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)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.22 \[ \int e^{-x^2} \left (10 x+e^{x^2} \left (1+2 x+3 x^2\right )\right ) \, dx={\left ({\left (x^{3} + x^{2} + x\right )} e^{\left (x^{2}\right )} - 5\right )} e^{\left (-x^{2}\right )} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int e^{-x^2} \left (10 x+e^{x^2} \left (1+2 x+3 x^2\right )\right ) \, dx=x^{3} + x^{2} + x - 5 e^{- x^{2}} \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^{-x^2} \left (10 x+e^{x^2} \left (1+2 x+3 x^2\right )\right ) \, dx=x^{3} + x^{2} + x - 5 \, e^{\left (-x^{2}\right )} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^{-x^2} \left (10 x+e^{x^2} \left (1+2 x+3 x^2\right )\right ) \, dx=x^{3} + x^{2} + x - 5 \, e^{\left (-x^{2}\right )} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.89 \[ \int e^{-x^2} \left (10 x+e^{x^2} \left (1+2 x+3 x^2\right )\right ) \, dx=x-5\,{\mathrm {e}}^{-x^2}+x^2+x^3 \]

[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