3.42.95 \(\int (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x) \, dx\)

Optimal. Leaf size=24 \[ e^{x^2}+e^{-3+5 x^2}+e^x (-1-x) \]

________________________________________________________________________________________

Rubi [A]  time = 0.03, antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2176, 2194, 2209} \begin {gather*} e^{x^2}+e^{5 x^2-3}-e^x (x+2)+e^x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^x*(-2 - x) + 2*E^x^2*x + 10*E^(-3 + 5*x^2)*x,x]

[Out]

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

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 23, normalized size = 0.96 \begin {gather*} e^{x^2}+e^{-3+5 x^2}-e^x (1+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^x*(-2 - x) + 2*E^x^2*x + 10*E^(-3 + 5*x^2)*x,x]

[Out]

E^x^2 + E^(-3 + 5*x^2) - E^x*(1 + x)

________________________________________________________________________________________

fricas [A]  time = 0.64, size = 29, normalized size = 1.21 \begin {gather*} -{\left ({\left (x + 1\right )} e^{\left (x + 3\right )} - e^{\left (5 \, x^{2}\right )} - e^{\left (x^{2} + 3\right )}\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*x*exp(5*x^2-3)+2*exp(x^2)*x+(-x-2)*exp(x),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.23, size = 20, normalized size = 0.83 \begin {gather*} -{\left (x + 1\right )} e^{x} + e^{\left (5 \, x^{2} - 3\right )} + e^{\left (x^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*x*exp(5*x^2-3)+2*exp(x^2)*x+(-x-2)*exp(x),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.07, size = 22, normalized size = 0.92

method	result	size
risch	\({\mathrm e}^{x^{2}}+{\mathrm e}^{5 x^{2}-3}+\left (-x -1\right ) {\mathrm e}^{x}\)	\(22\)
default	\(-{\mathrm e}^{x} x -{\mathrm e}^{x}+{\mathrm e}^{5 x^{2}-3}+{\mathrm e}^{x^{2}}\)	\(23\)
norman	\({\mathrm e}^{-3} {\mathrm e}^{5 x^{2}}-{\mathrm e}^{x} x -{\mathrm e}^{x}+{\mathrm e}^{x^{2}}\)	\(26\)
meijerg	\(-{\mathrm e}^{-3} \left (1-{\mathrm e}^{5 x^{2}}\right )+{\mathrm e}^{x^{2}}+\frac {\left (-2 x +2\right ) {\mathrm e}^{x}}{2}-2 \,{\mathrm e}^{x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(10*x*exp(5*x^2-3)+2*exp(x^2)*x+(-x-2)*exp(x),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 24, normalized size = 1.00 \begin {gather*} -{\left (x - 1\right )} e^{x} + e^{\left (5 \, x^{2} - 3\right )} + e^{\left (x^{2}\right )} - 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*x*exp(5*x^2-3)+2*exp(x^2)*x+(-x-2)*exp(x),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 22, normalized size = 0.92 \begin {gather*} {\mathrm {e}}^{x^2}-{\mathrm {e}}^x+{\mathrm {e}}^{5\,x^2-3}-x\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*exp(x^2) - exp(x)*(x + 2) + 10*x*exp(5*x^2 - 3),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.24, size = 26, normalized size = 1.08 \begin {gather*} \left (- x - 1\right ) e^{x} + \frac {e^{5 x^{2}} + e^{3} e^{x^{2}}}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(10*x*exp(5*x**2-3)+2*exp(x**2)*x+(-x-2)*exp(x),x)

[Out]

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

________________________________________________________________________________________