3.100.50 \(\int \frac {1}{4} e^{1-e^x+x} x (8+4 x-4 e^x x) \, dx\)

Optimal. Leaf size=14 \[ e^{1-e^x+x} x^2 \]

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Rubi [B]  time = 0.07, antiderivative size = 29, normalized size of antiderivative = 2.07, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {12, 2288} \begin {gather*} \frac {e^{x-e^x+1} x \left (x-e^x x\right )}{1-e^x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(1 - E^x + x)*x*(8 + 4*x - 4*E^x*x))/4,x]

[Out]

(E^(1 - E^x + x)*x*(x - E^x*x))/(1 - E^x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 {1}{4} \int e^{1-e^x+x} x \left (8+4 x-4 e^x x\right ) \, dx\\ &=\frac {e^{1-e^x+x} x \left (x-e^x x\right )}{1-e^x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 14, normalized size = 1.00 \begin {gather*} e^{1-e^x+x} x^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(1 - E^x + x)*x*(8 + 4*x - 4*E^x*x))/4,x]

[Out]

E^(1 - E^x + x)*x^2

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fricas [A]  time = 0.59, size = 16, normalized size = 1.14 \begin {gather*} 4 \, e^{\left (x - e^{x} + \log \left (\frac {1}{4} \, x^{2}\right ) + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(x)*x+4*x+8)*exp(log(1/4*x^2)+1-exp(x)+x)/x,x, algorithm="fricas")

[Out]

4*e^(x - e^x + log(1/4*x^2) + 1)

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giac [A]  time = 0.16, size = 12, normalized size = 0.86 \begin {gather*} x^{2} e^{\left (x - e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(x)*x+4*x+8)*exp(log(1/4*x^2)+1-exp(x)+x)/x,x, algorithm="giac")

[Out]

x^2*e^(x - e^x + 1)

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maple [A]  time = 0.16, size = 17, normalized size = 1.21

method	result	size
default	\(4 \,{\mathrm e}^{\ln \left (\frac {x^{2}}{4}\right )+1-{\mathrm e}^{x}+x}\)	\(17\)
norman	\(4 \,{\mathrm e}^{\ln \left (\frac {x^{2}}{4}\right )+1-{\mathrm e}^{x}+x}\)	\(17\)
risch	\(x^{2} {\mathrm e}^{1-\frac {i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}}{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )-\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}}{2}-{\mathrm e}^{x}+x}\)	\(62\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*exp(x)*x+4*x+8)*exp(ln(1/4*x^2)+1-exp(x)+x)/x,x,method=_RETURNVERBOSE)

[Out]

4*exp(ln(1/4*x^2)+1-exp(x)+x)

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maxima [A]  time = 0.40, size = 12, normalized size = 0.86 \begin {gather*} x^{2} e^{\left (x - e^{x} + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(x)*x+4*x+8)*exp(log(1/4*x^2)+1-exp(x)+x)/x,x, algorithm="maxima")

[Out]

x^2*e^(x - e^x + 1)

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mupad [B]  time = 8.89, size = 13, normalized size = 0.93 \begin {gather*} x^2\,\mathrm {e}\,{\mathrm {e}}^{-{\mathrm {e}}^x}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + log(x^2/4) - exp(x) + 1)*(4*x - 4*x*exp(x) + 8))/x,x)

[Out]

x^2*exp(1)*exp(-exp(x))*exp(x)

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sympy [A]  time = 0.17, size = 10, normalized size = 0.71 \begin {gather*} x^{2} e^{x - e^{x} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(x)*x+4*x+8)*exp(ln(1/4*x**2)+1-exp(x)+x)/x,x)

[Out]

x**2*exp(x - exp(x) + 1)

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