\(\int \frac {e^{-5+e^{2 e^{-x}}} (e^x (50 e^6-25 x-50 e^3 x)+e^x (-100 e^3+50 x) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} (-100 e^6 x+50 e^3 x^2+(200 e^3 x-50 x^2) \log (x)-100 x \log ^2(x)))}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx\) [6217]

Optimal result

Integrand size = 130, antiderivative size = 34 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=25 e^{-5+e^{2 e^{-x}}} \left (x+x \left (1-\frac {x}{e^3-\log (x)}\right )\right ) \]

[Out]

25*((1-x/(exp(3)-ln(x)))*x+x)*exp(exp(2/exp(x))-5)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(34)=68\).

Time = 0.24 (sec) , antiderivative size = 79, normalized size of antiderivative = 2.32, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {2326} \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\frac {25 e^{x+e^{2 e^{-x}}-5} \left (-e^3 x^2-\left (4 e^3 x-x^2\right ) \log (x)+2 e^6 x+2 x \log ^2(x)\right )}{e^{x+6}+e^x \log ^2(x)-2 e^{x+3} \log (x)} \]

[In]

Int[(E^(-5 + E^(2/E^x))*(E^x*(50*E^6 - 25*x - 50*E^3*x) + E^x*(-100*E^3 + 50*x)*Log[x] + 50*E^x*Log[x]^2 + E^(
2/E^x)*(-100*E^6*x + 50*E^3*x^2 + (200*E^3*x - 50*x^2)*Log[x] - 100*x*Log[x]^2)))/(E^(6 + x) - 2*E^(3 + x)*Log
[x] + E^x*Log[x]^2),x]

[Out]

(25*E^(-5 + E^(2/E^x) + x)*(2*E^6*x - E^3*x^2 - (4*E^3*x - x^2)*Log[x] + 2*x*Log[x]^2))/(E^(6 + x) - 2*E^(3 +
x)*Log[x] + E^x*Log[x]^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 {25 e^{-5+e^{2 e^{-x}}+x} \left (2 e^6 x-e^3 x^2-\left (4 e^3 x-x^2\right ) \log (x)+2 x \log ^2(x)\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=25 e^{-5+e^{2 e^{-x}}} x \left (2+\frac {x}{-e^3+\log (x)}\right ) \]

[In]

Integrate[(E^(-5 + E^(2/E^x))*(E^x*(50*E^6 - 25*x - 50*E^3*x) + E^x*(-100*E^3 + 50*x)*Log[x] + 50*E^x*Log[x]^2
+ E^(2/E^x)*(-100*E^6*x + 50*E^3*x^2 + (200*E^3*x - 50*x^2)*Log[x] - 100*x*Log[x]^2)))/(E^(6 + x) - 2*E^(3 +
x)*Log[x] + E^x*Log[x]^2),x]

[Out]

25*E^(-5 + E^(2/E^x))*x*(2 + x/(-E^3 + Log[x]))

Maple [A] (verified)

Time = 39.14 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03

method	result	size
risch	\(\frac {25 \left (2 \,{\mathrm e}^{3}-x -2 \ln \left (x \right )\right ) x \,{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5}}{{\mathrm e}^{3}-\ln \left (x \right )}\)	\(35\)
parallelrisch	\(-\frac {-50 \,{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5} x +25 \,{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5} x^{2}+50 x \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5}}{{\mathrm e}^{3}-\ln \left (x \right )}\)	\(58\)

[In]

int(((-100*x*ln(x)^2+(200*x*exp(3)-50*x^2)*ln(x)-100*x*exp(3)^2+50*x^2*exp(3))*exp(2/exp(x))+50*exp(x)*ln(x)^2
+(-100*exp(3)+50*x)*exp(x)*ln(x)+(50*exp(3)^2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(x)*ln(x)^2-2
*exp(3)*exp(x)*ln(x)+exp(3)^2*exp(x)),x,method=_RETURNVERBOSE)

[Out]

25*(2*exp(3)-x-2*ln(x))*x/(exp(3)-ln(x))*exp(exp(2*exp(-x))-5)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=-\frac {25 \, {\left (x^{2} - 2 \, x e^{3} + 2 \, x \log \left (x\right )\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )} - 5\right )}}{e^{3} - \log \left (x\right )} \]

[In]

integrate(((-100*x*log(x)^2+(200*x*exp(3)-50*x^2)*log(x)-100*x*exp(3)^2+50*x^2*exp(3))*exp(2/exp(x))+50*exp(x)
*log(x)^2+(-100*exp(3)+50*x)*exp(x)*log(x)+(50*exp(3)^2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(x)
*log(x)^2-2*exp(3)*exp(x)*log(x)+exp(3)^2*exp(x)),x, algorithm="fricas")

[Out]

-25*(x^2 - 2*x*e^3 + 2*x*log(x))*e^(e^(2*e^(-x)) - 5)/(e^3 - log(x))

Sympy [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\frac {\left (- 25 x^{2} - 50 x \log {\left (x \right )} + 50 x e^{3}\right ) e^{e^{2 e^{- x}} - 5}}{- \log {\left (x \right )} + e^{3}} \]

[In]

integrate(((-100*x*ln(x)**2+(200*x*exp(3)-50*x**2)*ln(x)-100*x*exp(3)**2+50*x**2*exp(3))*exp(2/exp(x))+50*exp(
x)*ln(x)**2+(-100*exp(3)+50*x)*exp(x)*ln(x)+(50*exp(3)**2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(
x)*ln(x)**2-2*exp(3)*exp(x)*ln(x)+exp(3)**2*exp(x)),x)

[Out]

(-25*x**2 - 50*x*log(x) + 50*x*exp(3))*exp(exp(2*exp(-x)) - 5)/(-log(x) + exp(3))

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\frac {25 \, {\left (x^{2} - 2 \, x e^{3} + 2 \, x \log \left (x\right )\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )}\right )}}{e^{5} \log \left (x\right ) - e^{8}} \]

[In]

integrate(((-100*x*log(x)^2+(200*x*exp(3)-50*x^2)*log(x)-100*x*exp(3)^2+50*x^2*exp(3))*exp(2/exp(x))+50*exp(x)
*log(x)^2+(-100*exp(3)+50*x)*exp(x)*log(x)+(50*exp(3)^2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(x)
*log(x)^2-2*exp(3)*exp(x)*log(x)+exp(3)^2*exp(x)),x, algorithm="maxima")

[Out]

25*(x^2 - 2*x*e^3 + 2*x*log(x))*e^(e^(2*e^(-x)))/(e^5*log(x) - e^8)

Giac [F]

\[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\int { \frac {25 \, {\left (2 \, {\left (x - 2 \, e^{3}\right )} e^{x} \log \left (x\right ) + 2 \, e^{x} \log \left (x\right )^{2} - {\left (2 \, x e^{3} + x - 2 \, e^{6}\right )} e^{x} + 2 \, {\left (x^{2} e^{3} - 2 \, x \log \left (x\right )^{2} - 2 \, x e^{6} - {\left (x^{2} - 4 \, x e^{3}\right )} \log \left (x\right )\right )} e^{\left (2 \, e^{\left (-x\right )}\right )}\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )} - 5\right )}}{e^{x} \log \left (x\right )^{2} - 2 \, e^{\left (x + 3\right )} \log \left (x\right ) + e^{\left (x + 6\right )}} \,d x } \]

[In]

integrate(((-100*x*log(x)^2+(200*x*exp(3)-50*x^2)*log(x)-100*x*exp(3)^2+50*x^2*exp(3))*exp(2/exp(x))+50*exp(x)
*log(x)^2+(-100*exp(3)+50*x)*exp(x)*log(x)+(50*exp(3)^2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(x)
*log(x)^2-2*exp(3)*exp(x)*log(x)+exp(3)^2*exp(x)),x, algorithm="giac")

[Out]

integrate(25*(2*(x - 2*e^3)*e^x*log(x) + 2*e^x*log(x)^2 - (2*x*e^3 + x - 2*e^6)*e^x + 2*(x^2*e^3 - 2*x*log(x)^
2 - 2*x*e^6 - (x^2 - 4*x*e^3)*log(x))*e^(2*e^(-x)))*e^(e^(2*e^(-x)) - 5)/(e^x*log(x)^2 - 2*e^(x + 3)*log(x) +
e^(x + 6)), x)

Mupad [B] (verification not implemented)

Time = 12.81 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=50\,x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}-\frac {25\,x^2\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}}{{\mathrm {e}}^3-\ln \left (x\right )} \]

[In]

int((exp(exp(2*exp(-x)) - 5)*(50*exp(x)*log(x)^2 - exp(x)*(25*x - 50*exp(6) + 50*x*exp(3)) + exp(2*exp(-x))*(l
og(x)*(200*x*exp(3) - 50*x^2) - 100*x*log(x)^2 - 100*x*exp(6) + 50*x^2*exp(3)) + exp(x)*log(x)*(50*x - 100*exp
(3))))/(exp(x)*log(x)^2 + exp(6)*exp(x) - 2*exp(3)*exp(x)*log(x)),x)

[Out]

50*x*exp(-5)*exp(exp(2*exp(-x))) - (25*x^2*exp(-5)*exp(exp(2*exp(-x))))/(exp(3) - log(x))