Integrand size = 130, antiderivative size = 34 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=25 e^{-5+e^{2 e^{-x}}} \left (x+x \left (1-\frac {x}{e^3-\log (x)}\right )\right ) \]
Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=25 e^{-5+e^{2 e^{-x}}} x \left (2+\frac {x}{-e^3+\log (x)}\right ) \]
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]
Leaf count is larger than twice the leaf count of optimal. \(79\) vs. \(2(34)=68\).
Time = 0.47 (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 = {2726}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {e^{e^{2 e^{-x}}-5} \left (e^{2 e^{-x}} \left (50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 e^6 x-100 x \log ^2(x)\right )+e^x \left (-50 e^3 x-25 x+50 e^6\right )+50 e^x \log ^2(x)+e^x \left (50 x-100 e^3\right ) \log (x)\right )}{e^{x+6}+e^x \log ^2(x)-2 e^{x+3} \log (x)} \, dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {25 e^{x+e^{2 e^{-x}}-5} \left (-e^3 x^2-\left (4 e^3 x-x^2\right ) \log (x)+2 e^6 x+2 x \log ^2(x)\right )}{e^{x+6}+e^x \log ^2(x)-2 e^{x+3} \log (x)}\) |
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]
(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)
3.3.17.3.1 Defintions of rubi rules used
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]
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 | \(-\frac {-50 \,{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5} x +25 \,{\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5} x^{2}+50 x \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{2 \,{\mathrm e}^{-x}}-5}}{{\mathrm e}^{3}-\ln \left (x \right )}\) | \(58\) |
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*e xp(3)^2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(x)*ln(x)^2-2*e xp(3)*exp(x)*ln(x)+exp(3)^2*exp(x)),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.03 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=-\frac {25 \, {\left (x^{2} - 2 \, x e^{3} + 2 \, x \log \left (x\right )\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )} - 5\right )}}{e^{3} - \log \left (x\right )} \]
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)*lo g(x)+(50*exp(3)^2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(x)*l og(x)^2-2*exp(3)*exp(x)*log(x)+exp(3)^2*exp(x)),x, algorithm=\
Time = 0.35 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\frac {\left (- 25 x^{2} - 50 x \log {\left (x \right )} + 50 x e^{3}\right ) e^{e^{2 e^{- x}} - 5}}{- \log {\left (x \right )} + e^{3}} \]
integrate(((-100*x*ln(x)**2+(200*x*exp(3)-50*x**2)*ln(x)-100*x*exp(3)**2+5 0*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)
Time = 0.27 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.06 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\frac {25 \, {\left (x^{2} - 2 \, x e^{3} + 2 \, x \log \left (x\right )\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )}\right )}}{e^{5} \log \left (x\right ) - e^{8}} \]
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)*lo g(x)+(50*exp(3)^2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(x)*l og(x)^2-2*exp(3)*exp(x)*log(x)+exp(3)^2*exp(x)),x, algorithm=\
\[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=\int { \frac {25 \, {\left (2 \, {\left (x - 2 \, e^{3}\right )} e^{x} \log \left (x\right ) + 2 \, e^{x} \log \left (x\right )^{2} - {\left (2 \, x e^{3} + x - 2 \, e^{6}\right )} e^{x} + 2 \, {\left (x^{2} e^{3} - 2 \, x \log \left (x\right )^{2} - 2 \, x e^{6} - {\left (x^{2} - 4 \, x e^{3}\right )} \log \left (x\right )\right )} e^{\left (2 \, e^{\left (-x\right )}\right )}\right )} e^{\left (e^{\left (2 \, e^{\left (-x\right )}\right )} - 5\right )}}{e^{x} \log \left (x\right )^{2} - 2 \, e^{\left (x + 3\right )} \log \left (x\right ) + e^{\left (x + 6\right )}} \,d x } \]
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)*lo g(x)+(50*exp(3)^2-50*x*exp(3)-25*x)*exp(x))*exp(exp(2/exp(x))-5)/(exp(x)*l og(x)^2-2*exp(3)*exp(x)*log(x)+exp(3)^2*exp(x)),x, algorithm=\
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)
Time = 12.81 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-5+e^{2 e^{-x}}} \left (e^x \left (50 e^6-25 x-50 e^3 x\right )+e^x \left (-100 e^3+50 x\right ) \log (x)+50 e^x \log ^2(x)+e^{2 e^{-x}} \left (-100 e^6 x+50 e^3 x^2+\left (200 e^3 x-50 x^2\right ) \log (x)-100 x \log ^2(x)\right )\right )}{e^{6+x}-2 e^{3+x} \log (x)+e^x \log ^2(x)} \, dx=50\,x\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}-\frac {25\,x^2\,{\mathrm {e}}^{-5}\,{\mathrm {e}}^{{\mathrm {e}}^{2\,{\mathrm {e}}^{-x}}}}{{\mathrm {e}}^3-\ln \left (x\right )} \]