3.91.51 \(\int \frac {e^{e^x} (-1+e^x (-17+e^2-x))}{\log (3)} \, dx\)

Optimal. Leaf size=18 \[ \frac {e^{e^x} \left (-17+e^2-x\right )}{\log (3)} \]

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 1.06, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 2288} \begin {gather*} -\frac {e^{e^x} \left (x-e^2+17\right )}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^E^x*(-1 + E^x*(-17 + E^2 - x)))/Log[3],x]

[Out]

-((E^E^x*(17 - E^2 + x))/Log[3])

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 {\int e^{e^x} \left (-1+e^x \left (-17+e^2-x\right )\right ) \, dx}{\log (3)}\\ &=-\frac {e^{e^x} \left (17-e^2+x\right )}{\log (3)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 1.00 \begin {gather*} \frac {e^{e^x} \left (-17+e^2-x\right )}{\log (3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^E^x*(-1 + E^x*(-17 + E^2 - x)))/Log[3],x]

[Out]

(E^E^x*(-17 + E^2 - x))/Log[3]

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fricas [A]  time = 0.45, size = 16, normalized size = 0.89 \begin {gather*} -\frac {{\left (x - e^{2} + 17\right )} e^{\left (e^{x}\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)-x-17)*exp(x)-1)*exp(exp(x))/log(3),x, algorithm="fricas")

[Out]

-(x - e^2 + 17)*e^(e^x)/log(3)

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giac [B]  time = 0.20, size = 33, normalized size = 1.83 \begin {gather*} -\frac {{\left (x e^{\left (x + e^{x}\right )} - e^{\left (x + e^{x} + 2\right )} + 17 \, e^{\left (x + e^{x}\right )}\right )} e^{\left (-x\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)-x-17)*exp(x)-1)*exp(exp(x))/log(3),x, algorithm="giac")

[Out]

-(x*e^(x + e^x) - e^(x + e^x + 2) + 17*e^(x + e^x))*e^(-x)/log(3)

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maple [A]  time = 0.06, size = 16, normalized size = 0.89

method	result	size
risch	\(\frac {\left ({\mathrm e}^{2}-x -17\right ) {\mathrm e}^{{\mathrm e}^{x}}}{\ln \relax (3)}\)	\(16\)
norman	\(\frac {\left ({\mathrm e}^{2}-17\right ) {\mathrm e}^{{\mathrm e}^{x}}}{\ln \relax (3)}-\frac {x \,{\mathrm e}^{{\mathrm e}^{x}}}{\ln \relax (3)}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((exp(2)-x-17)*exp(x)-1)*exp(exp(x))/ln(3),x,method=_RETURNVERBOSE)

[Out]

(exp(2)-x-17)/ln(3)*exp(exp(x))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\frac {{\left (x - e^{2}\right )} e^{\left (e^{x}\right )} + 17 \, e^{\left (e^{x}\right )}}{\log \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)-x-17)*exp(x)-1)*exp(exp(x))/log(3),x, algorithm="maxima")

[Out]

-((x - e^2)*e^(e^x) + Ei(e^x) + 17*e^(e^x) - integrate(e^(e^x), x))/log(3)

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mupad [B]  time = 0.07, size = 16, normalized size = 0.89 \begin {gather*} -\frac {{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (x-{\mathrm {e}}^2+17\right )}{\ln \relax (3)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(exp(x))*(exp(x)*(x - exp(2) + 17) + 1))/log(3),x)

[Out]

-(exp(exp(x))*(x - exp(2) + 17))/log(3)

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sympy [A]  time = 0.18, size = 14, normalized size = 0.78 \begin {gather*} \frac {\left (- x - 17 + e^{2}\right ) e^{e^{x}}}{\log {\relax (3 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((exp(2)-x-17)*exp(x)-1)*exp(exp(x))/ln(3),x)

[Out]

(-x - 17 + exp(2))*exp(exp(x))/log(3)

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