Integrand size = 63, antiderivative size = 24 \[ \int \frac {-6+10 e^{x+\frac {1}{2} (20+11 x)}+e^{x+\frac {1}{2} (20+11 x)} (-10+65 x) \log (x)}{6 x+10 e^{x+\frac {1}{2} (20+11 x)} x \log (x)} \, dx=\log \left (\frac {3+5 e^{4 x+\frac {5 (4+x)}{2}} \log (x)}{x}\right ) \] Output:
ln((3+5*exp(x)*exp(11/2*x+10)*ln(x))/x)
Time = 0.14 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-6+10 e^{x+\frac {1}{2} (20+11 x)}+e^{x+\frac {1}{2} (20+11 x)} (-10+65 x) \log (x)}{6 x+10 e^{x+\frac {1}{2} (20+11 x)} x \log (x)} \, dx=\frac {1}{2} \left (-2 \log (x)+2 \log \left (3+5 e^{10+\frac {13 x}{2}} \log (x)\right )\right ) \] Input:
Integrate[(-6 + 10*E^(x + (20 + 11*x)/2) + E^(x + (20 + 11*x)/2)*(-10 + 65 *x)*Log[x])/(6*x + 10*E^(x + (20 + 11*x)/2)*x*Log[x]),x]
Output:
(-2*Log[x] + 2*Log[3 + 5*E^(10 + (13*x)/2)*Log[x]])/2
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.
\(\displaystyle \int \frac {10 e^{x+\frac {1}{2} (11 x+20)}+e^{x+\frac {1}{2} (11 x+20)} (65 x-10) \log (x)-6}{6 x+10 e^{x+\frac {1}{2} (11 x+20)} x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {10 e^{x+\frac {1}{2} (11 x+20)}+e^{x+\frac {1}{2} (11 x+20)} (65 x-10) \log (x)-6}{2 x \left (5 e^{\frac {13 x}{2}+10} \log (x)+3\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int -\frac {5 e^{x+\frac {1}{2} (11 x+20)} (2-13 x) \log (x)-10 e^{x+\frac {1}{2} (11 x+20)}+6}{x \left (5 e^{\frac {13 x}{2}+10} \log (x)+3\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {5 e^{x+\frac {1}{2} (11 x+20)} (2-13 x) \log (x)-10 e^{x+\frac {1}{2} (11 x+20)}+6}{x \left (5 e^{\frac {13 x}{2}+10} \log (x)+3\right )}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {1}{2} \int \left (\frac {-13 x \log (x)+2 \log (x)-2}{x \log (x)}+\frac {3 (13 x \log (x)+2)}{x \log (x) \left (5 e^{\frac {13 x}{2}+10} \log (x)+3\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-78 \text {Subst}\left (\int \frac {1}{5 e^{13 x+10} \log (2 x)+3}dx,x,\frac {x}{2}\right )-6 \int \frac {1}{x \log (x) \left (5 e^{\frac {13 x}{2}+10} \log (x)+3\right )}dx+13 x-2 \log (x)+2 \log (\log (x))\right )\) |
Input:
Int[(-6 + 10*E^(x + (20 + 11*x)/2) + E^(x + (20 + 11*x)/2)*(-10 + 65*x)*Lo g[x])/(6*x + 10*E^(x + (20 + 11*x)/2)*x*Log[x]),x]
Output:
$Aborted
Time = 0.18 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83
method | result | size |
parallelrisch | \(-\ln \left (x \right )+\ln \left ({\mathrm e}^{x} {\mathrm e}^{\frac {11 x}{2}+10} \ln \left (x \right )+\frac {3}{5}\right )\) | \(20\) |
risch | \(\frac {13 x}{2}-\ln \left (x \right )+\ln \left (\ln \left (x \right )+\frac {3 \,{\mathrm e}^{-\frac {13 x}{2}-10}}{5}\right )\) | \(21\) |
Input:
int(((65*x-10)*exp(x)*exp(11/2*x+10)*ln(x)+10*exp(x)*exp(11/2*x+10)-6)/(10 *x*exp(x)*exp(11/2*x+10)*ln(x)+6*x),x,method=_RETURNVERBOSE)
Output:
-ln(x)+ln(exp(x)*exp(11/2*x+10)*ln(x)+3/5)
Time = 0.08 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-6+10 e^{x+\frac {1}{2} (20+11 x)}+e^{x+\frac {1}{2} (20+11 x)} (-10+65 x) \log (x)}{6 x+10 e^{x+\frac {1}{2} (20+11 x)} x \log (x)} \, dx=\frac {13}{2} \, x + \log \left ({\left (5 \, e^{\left (\frac {13}{2} \, x + 10\right )} \log \left (x\right ) + 3\right )} e^{\left (-\frac {13}{2} \, x - 10\right )}\right ) - \log \left (x\right ) \] Input:
integrate(((65*x-10)*exp(x)*exp(11/2*x+10)*log(x)+10*exp(x)*exp(11/2*x+10) -6)/(10*x*exp(x)*exp(11/2*x+10)*log(x)+6*x),x, algorithm="fricas")
Output:
13/2*x + log((5*e^(13/2*x + 10)*log(x) + 3)*e^(-13/2*x - 10)) - log(x)
Time = 0.18 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {-6+10 e^{x+\frac {1}{2} (20+11 x)}+e^{x+\frac {1}{2} (20+11 x)} (-10+65 x) \log (x)}{6 x+10 e^{x+\frac {1}{2} (20+11 x)} x \log (x)} \, dx=\frac {13 x}{2} - \log {\left (x \right )} + \log {\left (\log {\left (x \right )} + \frac {3}{5 e^{10} \left (e^{x}\right )^{\frac {13}{2}}} \right )} \] Input:
integrate(((65*x-10)*exp(x)*exp(11/2*x+10)*ln(x)+10*exp(x)*exp(11/2*x+10)- 6)/(10*x*exp(x)*exp(11/2*x+10)*ln(x)+6*x),x)
Output:
13*x/2 - log(x) + log(log(x) + 3*exp(-10)/(5*exp(x)**(13/2)))
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-6+10 e^{x+\frac {1}{2} (20+11 x)}+e^{x+\frac {1}{2} (20+11 x)} (-10+65 x) \log (x)}{6 x+10 e^{x+\frac {1}{2} (20+11 x)} x \log (x)} \, dx=-\log \left (x\right ) + \log \left (\frac {{\left (5 \, e^{\left (\frac {13}{2} \, x + 10\right )} \log \left (x\right ) + 3\right )} e^{\left (-10\right )}}{5 \, \log \left (x\right )}\right ) + \log \left (\log \left (x\right )\right ) \] Input:
integrate(((65*x-10)*exp(x)*exp(11/2*x+10)*log(x)+10*exp(x)*exp(11/2*x+10) -6)/(10*x*exp(x)*exp(11/2*x+10)*log(x)+6*x),x, algorithm="maxima")
Output:
-log(x) + log(1/5*(5*e^(13/2*x + 10)*log(x) + 3)*e^(-10)/log(x)) + log(log (x))
Time = 0.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-6+10 e^{x+\frac {1}{2} (20+11 x)}+e^{x+\frac {1}{2} (20+11 x)} (-10+65 x) \log (x)}{6 x+10 e^{x+\frac {1}{2} (20+11 x)} x \log (x)} \, dx=\log \left (5 \, e^{\left (\frac {13}{2} \, x + 10\right )} \log \left (x\right ) + 3\right ) - \log \left (\frac {1}{2} \, x\right ) \] Input:
integrate(((65*x-10)*exp(x)*exp(11/2*x+10)*log(x)+10*exp(x)*exp(11/2*x+10) -6)/(10*x*exp(x)*exp(11/2*x+10)*log(x)+6*x),x, algorithm="giac")
Output:
log(5*e^(13/2*x + 10)*log(x) + 3) - log(1/2*x)
Time = 2.35 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.79 \[ \int \frac {-6+10 e^{x+\frac {1}{2} (20+11 x)}+e^{x+\frac {1}{2} (20+11 x)} (-10+65 x) \log (x)}{6 x+10 e^{x+\frac {1}{2} (20+11 x)} x \log (x)} \, dx=\ln \left (3\,{\mathrm {e}}^{-10}+5\,{\left ({\mathrm {e}}^x\right )}^{13/2}\,\ln \left (x\right )\right )-\ln \left (x\right ) \] Input:
int((10*exp((11*x)/2 + 10)*exp(x) + exp((11*x)/2 + 10)*exp(x)*log(x)*(65*x - 10) - 6)/(6*x + 10*x*exp((11*x)/2 + 10)*exp(x)*log(x)),x)
Output:
log(3*exp(-10) + 5*exp(x)^(13/2)*log(x)) - log(x)
Time = 0.20 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.83 \[ \int \frac {-6+10 e^{x+\frac {1}{2} (20+11 x)}+e^{x+\frac {1}{2} (20+11 x)} (-10+65 x) \log (x)}{6 x+10 e^{x+\frac {1}{2} (20+11 x)} x \log (x)} \, dx=\mathrm {log}\left (5 e^{\frac {13 x}{2}} \mathrm {log}\left (x \right ) e^{10}+3\right )-\mathrm {log}\left (x \right ) \] Input:
int(((65*x-10)*exp(x)*exp(11/2*x+10)*log(x)+10*exp(x)*exp(11/2*x+10)-6)/(1 0*x*exp(x)*exp(11/2*x+10)*log(x)+6*x),x)
Output:
log(5*e**((13*x)/2)*log(x)*e**10 + 3) - log(x)