Integrand size = 72, antiderivative size = 23 \[ \int \frac {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx=\frac {e^x}{\frac {64 x}{25}-\frac {16 x}{\log \left (3 e^x\right )}} \] Output:
exp(x)/(64/25*x-16*x/ln(3*exp(x)))
Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx=\frac {25 e^x \log \left (3 e^x\right )}{16 x \left (-25+4 \log \left (3 e^x\right )\right )} \] Input:
Integrate[(-625*E^x*x + E^x*(625 - 625*x)*Log[3*E^x] + E^x*(-100 + 100*x)* Log[3*E^x]^2)/(10000*x^2 - 3200*x^2*Log[3*E^x] + 256*x^2*Log[3*E^x]^2),x]
Output:
(25*E^x*Log[3*E^x])/(16*x*(-25 + 4*Log[3*E^x]))
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.
\(\displaystyle \int \frac {-625 e^x x+e^x (100 x-100) \log ^2\left (3 e^x\right )+e^x (625-625 x) \log \left (3 e^x\right )}{10000 x^2+256 x^2 \log ^2\left (3 e^x\right )-3200 x^2 \log \left (3 e^x\right )} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-625 e^x x+e^x (100 x-100) \log ^2\left (3 e^x\right )+e^x (625-625 x) \log \left (3 e^x\right )}{16 x^2 \left (25-4 \log \left (3 e^x\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{16} \int -\frac {25 \left (4 e^x (1-x) \log ^2\left (3 e^x\right )-25 e^x (1-x) \log \left (3 e^x\right )+25 e^x x\right )}{x^2 \left (25-4 \log \left (3 e^x\right )\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {25}{16} \int \frac {4 e^x (1-x) \log ^2\left (3 e^x\right )-25 e^x (1-x) \log \left (3 e^x\right )+25 e^x x}{x^2 \left (25-4 \log \left (3 e^x\right )\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {25}{16} \int \left (-\frac {4 e^x \log ^2\left (3 e^x\right )}{x \left (4 \log \left (3 e^x\right )-25\right )^2}+\frac {4 e^x \log ^2\left (3 e^x\right )}{x^2 \left (4 \log \left (3 e^x\right )-25\right )^2}+\frac {25 e^x \log \left (3 e^x\right )}{x \left (4 \log \left (3 e^x\right )-25\right )^2}-\frac {25 e^x \log \left (3 e^x\right )}{x^2 \left (4 \log \left (3 e^x\right )-25\right )^2}+\frac {25 e^x}{x \left (4 \log \left (3 e^x\right )-25\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {25}{16} \left (\frac {25}{4} \int \frac {e^x}{x^2 \left (4 \log \left (3 e^x\right )-25\right )}dx+25 \int \frac {e^x}{x \left (4 \log \left (3 e^x\right )-25\right )^2}dx-\frac {25}{4} \int \frac {e^x}{x \left (4 \log \left (3 e^x\right )-25\right )}dx-\frac {e^x}{4 x}\right )\) |
Input:
Int[(-625*E^x*x + E^x*(625 - 625*x)*Log[3*E^x] + E^x*(-100 + 100*x)*Log[3* E^x]^2)/(10000*x^2 - 3200*x^2*Log[3*E^x] + 256*x^2*Log[3*E^x]^2),x]
Output:
$Aborted
Time = 0.23 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04
method | result | size |
norman | \(\frac {25 \,{\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}\right )}{16 x \left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-25\right )}\) | \(24\) |
parallelrisch | \(\frac {25 \,{\mathrm e}^{x} \ln \left (3 \,{\mathrm e}^{x}\right )}{16 x \left (4 \ln \left (3 \,{\mathrm e}^{x}\right )-25\right )}\) | \(24\) |
risch | \(\frac {25 \,{\mathrm e}^{x}}{64 x}+\frac {625 \,{\mathrm e}^{x}}{64 x \left (-25+4 \ln \left (3\right )+4 \ln \left ({\mathrm e}^{x}\right )\right )}\) | \(29\) |
default | \(\text {Expression too large to display}\) | \(702\) |
Input:
int(((100*x-100)*exp(x)*ln(3*exp(x))^2+(-625*x+625)*exp(x)*ln(3*exp(x))-62 5*exp(x)*x)/(256*x^2*ln(3*exp(x))^2-3200*x^2*ln(3*exp(x))+10000*x^2),x,met hod=_RETURNVERBOSE)
Output:
25/16*exp(x)*ln(3*exp(x))/x/(4*ln(3*exp(x))-25)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx=\frac {25 \, {\left (x + \log \left (3\right )\right )} e^{x}}{16 \, {\left (4 \, x^{2} + 4 \, x \log \left (3\right ) - 25 \, x\right )}} \] Input:
integrate(((100*x-100)*exp(x)*log(3*exp(x))^2+(-625*x+625)*exp(x)*log(3*ex p(x))-625*exp(x)*x)/(256*x^2*log(3*exp(x))^2-3200*x^2*log(3*exp(x))+10000* x^2),x, algorithm="fricas")
Output:
25/16*(x + log(3))*e^x/(4*x^2 + 4*x*log(3) - 25*x)
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx=\frac {\left (25 x + 25 \log {\left (3 \right )}\right ) e^{x}}{64 x^{2} - 400 x + 64 x \log {\left (3 \right )}} \] Input:
integrate(((100*x-100)*exp(x)*ln(3*exp(x))**2+(-625*x+625)*exp(x)*ln(3*exp (x))-625*exp(x)*x)/(256*x**2*ln(3*exp(x))**2-3200*x**2*ln(3*exp(x))+10000* x**2),x)
Output:
(25*x + 25*log(3))*exp(x)/(64*x**2 - 400*x + 64*x*log(3))
Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx=\frac {25 \, {\left (x + \log \left (3\right )\right )} e^{x}}{16 \, {\left (4 \, x^{2} + x {\left (4 \, \log \left (3\right ) - 25\right )}\right )}} \] Input:
integrate(((100*x-100)*exp(x)*log(3*exp(x))^2+(-625*x+625)*exp(x)*log(3*ex p(x))-625*exp(x)*x)/(256*x^2*log(3*exp(x))^2-3200*x^2*log(3*exp(x))+10000* x^2),x, algorithm="maxima")
Output:
25/16*(x + log(3))*e^x/(4*x^2 + x*(4*log(3) - 25))
Time = 0.12 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.22 \[ \int \frac {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx=\frac {25 \, {\left (x e^{x} + e^{x} \log \left (3\right )\right )}}{16 \, {\left (4 \, x^{2} + 4 \, x \log \left (3\right ) - 25 \, x\right )}} \] Input:
integrate(((100*x-100)*exp(x)*log(3*exp(x))^2+(-625*x+625)*exp(x)*log(3*ex p(x))-625*exp(x)*x)/(256*x^2*log(3*exp(x))^2-3200*x^2*log(3*exp(x))+10000* x^2),x, algorithm="giac")
Output:
25/16*(x*e^x + e^x*log(3))/(4*x^2 + 4*x*log(3) - 25*x)
Time = 3.64 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx=\frac {{\mathrm {e}}^x\,\left (\frac {25\,x}{64}+\frac {25\,\ln \left (3\right )}{64}\right )}{x^2+\left (\ln \left (3\right )-\frac {25}{4}\right )\,x} \] Input:
int(-(625*x*exp(x) + exp(x)*log(3*exp(x))*(625*x - 625) - exp(x)*log(3*exp (x))^2*(100*x - 100))/(256*x^2*log(3*exp(x))^2 + 10000*x^2 - 3200*x^2*log( 3*exp(x))),x)
Output:
(exp(x)*((25*x)/64 + (25*log(3))/64))/(x*(log(3) - 25/4) + x^2)
Time = 0.19 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-625 e^x x+e^x (625-625 x) \log \left (3 e^x\right )+e^x (-100+100 x) \log ^2\left (3 e^x\right )}{10000 x^2-3200 x^2 \log \left (3 e^x\right )+256 x^2 \log ^2\left (3 e^x\right )} \, dx=\frac {25 e^{x} \mathrm {log}\left (3 e^{x}\right )}{16 x \left (4 \,\mathrm {log}\left (3 e^{x}\right )-25\right )} \] Input:
int(((100*x-100)*exp(x)*log(3*exp(x))^2+(-625*x+625)*exp(x)*log(3*exp(x))- 625*exp(x)*x)/(256*x^2*log(3*exp(x))^2-3200*x^2*log(3*exp(x))+10000*x^2),x )
Output:
(25*e**x*log(3*e**x))/(16*x*(4*log(3*e**x) - 25))