\(\int (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x) \, dx\) [4195]

Optimal result

Integrand size = 30, antiderivative size = 24 \[ \int \left (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x\right ) \, dx=e^{x^2}+e^{-3+5 x^2}+e^x (-1-x) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2207, 2225, 2240} \[ \int \left (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x\right ) \, dx=e^{x^2}+e^{5 x^2-3}-e^x (x+2)+e^x \]

[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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 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]

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \left (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x\right ) \, dx=e^{x^2}+e^{-3+5 x^2}-e^x (1+x) \]

[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)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \left (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x\right ) \, dx=-{\left ({\left (x + 1\right )} e^{\left (x + 3\right )} - e^{\left (5 \, x^{2}\right )} - e^{\left (x^{2} + 3\right )}\right )} e^{\left (-3\right )} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \left (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x\right ) \, dx=\left (- x - 1\right ) e^{x} + \frac {e^{5 x^{2}} + e^{3} e^{x^{2}}}{e^{3}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \left (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x\right ) \, dx=-{\left (x - 1\right )} e^{x} + e^{\left (5 \, x^{2} - 3\right )} + e^{\left (x^{2}\right )} - 2 \, e^{x} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \left (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x\right ) \, dx=-{\left (x + 1\right )} e^{x} + e^{\left (5 \, x^{2} - 3\right )} + e^{\left (x^{2}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \left (e^x (-2-x)+2 e^{x^2} x+10 e^{-3+5 x^2} x\right ) \, dx={\mathrm {e}}^{x^2}-{\mathrm {e}}^x+{\mathrm {e}}^{5\,x^2-3}-x\,{\mathrm {e}}^x \]

[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)