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

Optimal. Leaf size=15 \[ \frac {5}{2} e x \left (x+e^{e^x} x\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 23, normalized size of antiderivative = 1.53, number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2288} \begin {gather*} \frac {5}{2} e^{e^x+1} x^2+\frac {5 e x^2}{2} \end {gather*}

Antiderivative was successfully verified.

[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 2288

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 {gather*} \begin {aligned} \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 {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 18, normalized size = 1.20 \begin {gather*} \frac {5}{2} e \left (x^2+e^{e^x} x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[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

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fricas [A]  time = 0.58, size = 24, normalized size = 1.60 \begin {gather*} \frac {5}{2} \, x^{2} e + 5 \, x e^{\left (e^{x} - \log \left (\frac {2}{x}\right ) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 0.18, size = 18, normalized size = 1.20 \begin {gather*} \frac {5}{2} \, x^{2} e + \frac {5}{2} \, x^{2} e^{\left (e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.24, size = 19, normalized size = 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	\(\frac {5 x^{2} {\mathrm e} \,{\mathrm e}^{{\mathrm e}^{x}}}{2}+5 x^{2} {\mathrm e}^{1-\ln \left (\frac {2}{x}\right )-\ln \relax (x )}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.46, size = 18, normalized size = 1.20 \begin {gather*} \frac {5}{2} \, x^{2} e + \frac {5}{2} \, x^{2} e^{\left (e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 4.19, size = 12, normalized size = 0.80 \begin {gather*} \frac {5\,x^2\,\mathrm {e}\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}+1\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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sympy [A]  time = 0.15, size = 24, normalized size = 1.60 \begin {gather*} \frac {5 e x^{2} e^{e^{x}}}{2} + \frac {5 e x^{2}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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