Integrand size = 78, antiderivative size = 38 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {\frac {x}{5}+\frac {e^3}{\frac {4 x}{5}-\frac {5 e^{-e^x}}{\log (2)}}}{2 x} \] Output:
1/2*(1/5*x+exp(3)/(4/5*x-5/ln(2)/exp(exp(x))))/x
Time = 1.43 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.82 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=-\frac {5 e^{3+e^x} \log (2)}{2 \left (25 x-e^{e^x} x^2 \log (16)\right )} \] Input:
Integrate[(-40*E^(3 + 2*E^x)*x*Log[2]^2 + E^E^x*(125*E^3*Log[2] - 125*E^(3 + x)*x*Log[2]))/(1250*x^2 - 400*E^E^x*x^3*Log[2] + 32*E^(2*E^x)*x^4*Log[2 ]^2),x]
Output:
(-5*E^(3 + E^x)*Log[2])/(2*(25*x - E^E^x*x^2*Log[16]))
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.
\(\displaystyle \int \frac {e^{e^x} \left (125 e^3 \log (2)-125 e^{x+3} x \log (2)\right )-40 e^{2 e^x+3} x \log ^2(2)}{32 e^{2 e^x} x^4 \log ^2(2)-400 e^{e^x} x^3 \log (2)+1250 x^2} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {5 e^{e^x+3} \log (2) \left (-25 e^x x-8 e^{e^x} x \log (2)+25\right )}{2 x^2 \left (25-e^{e^x} x \log (16)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5}{2} \log (2) \int \frac {e^{3+e^x} \left (-25 e^x x-8 e^{e^x} \log (2) x+25\right )}{x^2 \left (25-e^{e^x} x \log (16)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {5}{2} \log (2) \int \left (-\frac {e^{3+e^x} \left (8 e^{e^x} x \log (2)-25\right )}{x^2 \left (e^{e^x} x \log (16)-25\right )^2}-\frac {25 e^{x+e^x+3}}{x \left (e^{e^x} x \log (16)-25\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5}{2} \log (2) \left (-25 \int \frac {e^{3+e^x}}{x^2 \left (e^{e^x} x \log (16)-25\right )^2}dx-\frac {8 \log (2) \int \frac {e^{3+e^x}}{x^2 \left (e^{e^x} x \log (16)-25\right )}dx}{\log (16)}-25 \int \frac {e^{x+e^x+3}}{x \left (e^{e^x} x \log (16)-25\right )^2}dx\right )\) |
Input:
Int[(-40*E^(3 + 2*E^x)*x*Log[2]^2 + E^E^x*(125*E^3*Log[2] - 125*E^(3 + x)* x*Log[2]))/(1250*x^2 - 400*E^E^x*x^3*Log[2] + 32*E^(2*E^x)*x^4*Log[2]^2),x ]
Output:
$Aborted
Time = 0.67 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66
method | result | size |
norman | \(\frac {5 \,{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{3} \ln \left (2\right )}{2 x \left (4 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (2\right ) x -25\right )}\) | \(25\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{{\mathrm e}^{x}} {\mathrm e}^{3} \ln \left (2\right )}{2 x \left (4 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (2\right ) x -25\right )}\) | \(25\) |
risch | \(\frac {5 \,{\mathrm e}^{3}}{8 x^{2}}+\frac {125 \,{\mathrm e}^{3}}{8 x^{2} \left (4 \,{\mathrm e}^{{\mathrm e}^{x}} \ln \left (2\right ) x -25\right )}\) | \(28\) |
Input:
int((-40*x*exp(3)*ln(2)^2*exp(exp(x))^2+(-125*x*exp(3)*ln(2)*exp(x)+125*ex p(3)*ln(2))*exp(exp(x)))/(32*x^4*ln(2)^2*exp(exp(x))^2-400*x^3*ln(2)*exp(e xp(x))+1250*x^2),x,method=_RETURNVERBOSE)
Output:
5/2*exp(exp(x))*exp(3)*ln(2)/x/(4*exp(exp(x))*ln(2)*x-25)
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5 \, e^{\left (e^{x} + 3\right )} \log \left (2\right )}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \left (2\right ) - 25 \, x\right )}} \] Input:
integrate((-40*x*exp(3)*log(2)^2*exp(exp(x))^2+(-125*x*exp(3)*log(2)*exp(x )+125*exp(3)*log(2))*exp(exp(x)))/(32*x^4*log(2)^2*exp(exp(x))^2-400*x^3*l og(2)*exp(exp(x))+1250*x^2),x, algorithm="fricas")
Output:
5/2*e^(e^x + 3)*log(2)/(4*x^2*e^(e^x)*log(2) - 25*x)
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.84 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {125 e^{3}}{32 x^{3} e^{e^{x}} \log {\left (2 \right )} - 200 x^{2}} + \frac {5 e^{3}}{8 x^{2}} \] Input:
integrate((-40*x*exp(3)*ln(2)**2*exp(exp(x))**2+(-125*x*exp(3)*ln(2)*exp(x )+125*exp(3)*ln(2))*exp(exp(x)))/(32*x**4*ln(2)**2*exp(exp(x))**2-400*x**3 *ln(2)*exp(exp(x))+1250*x**2),x)
Output:
125*exp(3)/(32*x**3*exp(exp(x))*log(2) - 200*x**2) + 5*exp(3)/(8*x**2)
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5 \, e^{\left (e^{x} + 3\right )} \log \left (2\right )}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \left (2\right ) - 25 \, x\right )}} \] Input:
integrate((-40*x*exp(3)*log(2)^2*exp(exp(x))^2+(-125*x*exp(3)*log(2)*exp(x )+125*exp(3)*log(2))*exp(exp(x)))/(32*x^4*log(2)^2*exp(exp(x))^2-400*x^3*l og(2)*exp(exp(x))+1250*x^2),x, algorithm="maxima")
Output:
5/2*e^(e^x + 3)*log(2)/(4*x^2*e^(e^x)*log(2) - 25*x)
Time = 0.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.66 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5 \, e^{\left (e^{x} + 3\right )} \log \left (2\right )}{2 \, {\left (4 \, x^{2} e^{\left (e^{x}\right )} \log \left (2\right ) - 25 \, x\right )}} \] Input:
integrate((-40*x*exp(3)*log(2)^2*exp(exp(x))^2+(-125*x*exp(3)*log(2)*exp(x )+125*exp(3)*log(2))*exp(exp(x)))/(32*x^4*log(2)^2*exp(exp(x))^2-400*x^3*l og(2)*exp(exp(x))+1250*x^2),x, algorithm="giac")
Output:
5/2*e^(e^x + 3)*log(2)/(4*x^2*e^(e^x)*log(2) - 25*x)
Time = 7.93 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.42 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5\,{\mathrm {e}}^3}{8\,x^2}+\frac {125\,\left ({\mathrm {e}}^3+x\,{\mathrm {e}}^{x+3}\right )}{32\,x\,\left (x^2\,\ln \left (2\right )+x^3\,{\mathrm {e}}^x\,\ln \left (2\right )\right )\,\left ({\mathrm {e}}^{{\mathrm {e}}^x}-\frac {25}{4\,x\,\ln \left (2\right )}\right )} \] Input:
int((exp(exp(x))*(125*exp(3)*log(2) - 125*x*exp(3)*exp(x)*log(2)) - 40*x*e xp(3)*exp(2*exp(x))*log(2)^2)/(1250*x^2 + 32*x^4*exp(2*exp(x))*log(2)^2 - 400*x^3*exp(exp(x))*log(2)),x)
Output:
(5*exp(3))/(8*x^2) + (125*(exp(3) + x*exp(x + 3)))/(32*x*(x^2*log(2) + x^3 *exp(x)*log(2))*(exp(exp(x)) - 25/(4*x*log(2))))
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.76 \[ \int \frac {-40 e^{3+2 e^x} x \log ^2(2)+e^{e^x} \left (125 e^3 \log (2)-125 e^{3+x} x \log (2)\right )}{1250 x^2-400 e^{e^x} x^3 \log (2)+32 e^{2 e^x} x^4 \log ^2(2)} \, dx=\frac {5 e^{e^{x}} \mathrm {log}\left (2\right ) e^{3}}{2 x \left (4 e^{e^{x}} \mathrm {log}\left (2\right ) x -25\right )} \] Input:
int((-40*x*exp(3)*log(2)^2*exp(exp(x))^2+(-125*x*exp(3)*log(2)*exp(x)+125* exp(3)*log(2))*exp(exp(x)))/(32*x^4*log(2)^2*exp(exp(x))^2-400*x^3*log(2)* exp(exp(x))+1250*x^2),x)
Output:
(5*e**(e**x)*log(2)*e**3)/(2*x*(4*e**(e**x)*log(2)*x - 25))