\(\int (5 e x+\frac {1}{2} e^{1+e^x} x (10+5 e^x x)) \, dx\) [6616]

Optimal result

Integrand size = 25, antiderivative size = 15 \[ \int \left (5 e x+\frac {1}{2} e^{1+e^x} x \left (10+5 e^x x\right )\right ) \, dx=\frac {5}{2} e x \left (x+e^{e^x} x\right ) \]

[Out]

5*(x*exp(exp(x))+x)*exp(-ln(2/x)+1)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.53, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2326} \[ \int \left (5 e x+\frac {1}{2} e^{1+e^x} x \left (10+5 e^x x\right )\right ) \, dx=\frac {5}{2} e^{e^x+1} x^2+\frac {5 e x^2}{2} \]

[In]

Int[5*E*x + (E^(1 + E^x)*x*(10 + 5*E^x*x))/2,x]

[Out]

(5*E*x^2)/2 + (5*E^(1 + E^x)*x^2)/2

Rule 2326

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps \begin{align*} \text {integral}& = \frac {5 e x^2}{2}+\frac {1}{2} \int e^{1+e^x} x \left (10+5 e^x x\right ) \, dx \\ & = \frac {5 e x^2}{2}+\frac {5}{2} e^{1+e^x} x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \left (5 e x+\frac {1}{2} e^{1+e^x} x \left (10+5 e^x x\right )\right ) \, dx=\frac {5}{2} e \left (x^2+e^{e^x} x^2\right ) \]

[In]

Integrate[5*E*x + (E^(1 + E^x)*x*(10 + 5*E^x*x))/2,x]

[Out]

(5*E*(x^2 + E^E^x*x^2))/2

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.27

method	result	size
norman	\(\frac {5 x^{2} {\mathrm e}}{2}+\frac {5 x^{2} {\mathrm e} \,{\mathrm e}^{{\mathrm e}^{x}}}{2}\)	\(19\)
risch	\(\frac {5 x^{2} {\mathrm e}^{{\mathrm e}^{x}+1}}{2}+\frac {5 x^{2} {\mathrm e}}{2}\)	\(19\)
default	\(\frac {5 x^{2} {\mathrm e} \,{\mathrm e}^{{\mathrm e}^{x}}}{2}+5 x \,{\mathrm e}^{-\ln \left (\frac {2}{x}\right )+1}\)	\(38\)
parallelrisch	\(\frac {5 \,{\mathrm e}^{-\ln \left (\frac {2}{x}\right )+1} x^{2} {\mathrm e}^{{\mathrm e}^{x}}+5 \,{\mathrm e}^{-\ln \left (\frac {2}{x}\right )+1} x^{2}}{x}\)	\(41\)

[In]

int((5*exp(x)*x+10)*exp(-ln(2/x)+1)*exp(exp(x))+10*exp(-ln(2/x)+1),x,method=_RETURNVERBOSE)

[Out]

5/2*x^2*exp(1)+5/2*x^2*exp(1)*exp(exp(x))

Fricas [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.60 \[ \int \left (5 e x+\frac {1}{2} e^{1+e^x} x \left (10+5 e^x x\right )\right ) \, dx=\frac {5}{2} \, x^{2} e + 5 \, x e^{\left (e^{x} - \log \left (\frac {2}{x}\right ) + 1\right )} \]

[In]

integrate((5*exp(x)*x+10)*exp(-log(2/x)+1)*exp(exp(x))+10*exp(-log(2/x)+1),x, algorithm="fricas")

[Out]

5/2*x^2*e + 5*x*e^(e^x - log(2/x) + 1)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.60 \[ \int \left (5 e x+\frac {1}{2} e^{1+e^x} x \left (10+5 e^x x\right )\right ) \, dx=\frac {5 e x^{2} e^{e^{x}}}{2} + \frac {5 e x^{2}}{2} \]

[In]

integrate((5*exp(x)*x+10)*exp(-ln(2/x)+1)*exp(exp(x))+10*exp(-ln(2/x)+1),x)

[Out]

5*E*x**2*exp(exp(x))/2 + 5*E*x**2/2

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \left (5 e x+\frac {1}{2} e^{1+e^x} x \left (10+5 e^x x\right )\right ) \, dx=\frac {5}{2} \, x^{2} e + \frac {5}{2} \, x^{2} e^{\left (e^{x} + 1\right )} \]

[In]

integrate((5*exp(x)*x+10)*exp(-log(2/x)+1)*exp(exp(x))+10*exp(-log(2/x)+1),x, algorithm="maxima")

[Out]

5/2*x^2*e + 5/2*x^2*e^(e^x + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.20 \[ \int \left (5 e x+\frac {1}{2} e^{1+e^x} x \left (10+5 e^x x\right )\right ) \, dx=\frac {5}{2} \, x^{2} e + \frac {5}{2} \, x^{2} e^{\left (e^{x} + 1\right )} \]

[In]

integrate((5*exp(x)*x+10)*exp(-log(2/x)+1)*exp(exp(x))+10*exp(-log(2/x)+1),x, algorithm="giac")

[Out]

5/2*x^2*e + 5/2*x^2*e^(e^x + 1)

Mupad [B] (verification not implemented)

Time = 12.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.80 \[ \int \left (5 e x+\frac {1}{2} e^{1+e^x} x \left (10+5 e^x x\right )\right ) \, dx=\frac {5\,x^2\,\mathrm {e}\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}+1\right )}{2} \]

[In]

int(10*exp(1 - log(2/x)) + exp(exp(x))*exp(1 - log(2/x))*(5*x*exp(x) + 10),x)

[Out]

(5*x^2*exp(1)*(exp(exp(x)) + 1))/2