Integrand size = 77, antiderivative size = 26 \[ \int \frac {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx=\frac {\log \left (2 \left (3-\frac {4}{5} \left (-1+e^3 (2-x)\right )+x\right )\right )}{\log (x)} \] Output:
ln(38/5+2*x-8/5*(2-x)*exp(3))/ln(x)
Time = 0.41 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx=\frac {\log \left (\frac {2}{5} \left (19+4 e^3 (-2+x)+5 x\right )\right )}{\log (x)} \] Input:
Integrate[((5*x + 4*E^3*x)*Log[x] + (-19 + E^3*(8 - 4*x) - 5*x)*Log[(38 + 10*x + E^3*(-16 + 8*x))/5])/((19*x + 5*x^2 + E^3*(-8*x + 4*x^2))*Log[x]^2) ,x]
Output:
Log[(2*(19 + 4*E^3*(-2 + x) + 5*x))/5]/Log[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.
\(\displaystyle \int \frac {\left (4 e^3 x+5 x\right ) \log (x)+\left (e^3 (8-4 x)-5 x-19\right ) \log \left (\frac {1}{5} \left (10 x+e^3 (8 x-16)+38\right )\right )}{\left (5 x^2+e^3 \left (4 x^2-8 x\right )+19 x\right ) \log ^2(x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (4 e^3 x+5 x\right ) \log (x)+\left (e^3 (8-4 x)-5 x-19\right ) \log \left (\frac {1}{5} \left (10 x+e^3 (8 x-16)+38\right )\right )}{x \left (\left (5+4 e^3\right ) x-8 e^3+19\right ) \log ^2(x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {5+4 e^3}{\left (\left (5+4 e^3\right ) x-8 e^3+19\right ) \log (x)}-\frac {\log \left (\frac {2}{5} \left (5+4 e^3\right ) x+\frac {2}{5} \left (19-8 e^3\right )\right )}{x \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \left (5+4 e^3\right ) \int \frac {1}{\left (\left (5+4 e^3\right ) x-8 e^3+19\right ) \log (x)}dx-\int \frac {\log \left (\frac {2}{5} \left (5+4 e^3\right ) x+\frac {2}{5} \left (19-8 e^3\right )\right )}{x \log ^2(x)}dx\) |
Input:
Int[((5*x + 4*E^3*x)*Log[x] + (-19 + E^3*(8 - 4*x) - 5*x)*Log[(38 + 10*x + E^3*(-16 + 8*x))/5])/((19*x + 5*x^2 + E^3*(-8*x + 4*x^2))*Log[x]^2),x]
Output:
$Aborted
Time = 2.24 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {\ln \left (\frac {\left (8 x -16\right ) {\mathrm e}^{3}}{5}+2 x +\frac {38}{5}\right )}{\ln \left (x \right )}\) | \(21\) |
parallelrisch | \(\frac {16 \,{\mathrm e}^{6} \ln \left (\frac {\left (8 x -16\right ) {\mathrm e}^{3}}{5}+2 x +\frac {38}{5}\right )+40 \ln \left (\frac {\left (8 x -16\right ) {\mathrm e}^{3}}{5}+2 x +\frac {38}{5}\right ) {\mathrm e}^{3}+25 \ln \left (\frac {\left (8 x -16\right ) {\mathrm e}^{3}}{5}+2 x +\frac {38}{5}\right )}{\ln \left (x \right ) \left (4 \,{\mathrm e}^{3}+5\right )^{2}}\) | \(72\) |
Input:
int((((-4*x+8)*exp(3)-5*x-19)*ln(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x*exp(3) +5*x)*ln(x))/((4*x^2-8*x)*exp(3)+5*x^2+19*x)/ln(x)^2,x,method=_RETURNVERBO SE)
Output:
1/ln(x)*ln(1/5*(8*x-16)*exp(3)+2*x+38/5)
Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.69 \[ \int \frac {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx=\frac {\log \left (\frac {8}{5} \, {\left (x - 2\right )} e^{3} + 2 \, x + \frac {38}{5}\right )}{\log \left (x\right )} \] Input:
integrate((((-4*x+8)*exp(3)-5*x-19)*log(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x *exp(3)+5*x)*log(x))/((4*x^2-8*x)*exp(3)+5*x^2+19*x)/log(x)^2,x, algorithm ="fricas")
Output:
log(8/5*(x - 2)*e^3 + 2*x + 38/5)/log(x)
Exception generated. \[ \int \frac {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((((-4*x+8)*exp(3)-5*x-19)*ln(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x* exp(3)+5*x)*ln(x))/((4*x**2-8*x)*exp(3)+5*x**2+19*x)/ln(x)**2,x)
Output:
Exception raised: TypeError >> '>' not supported between instances of 'Pol y' and 'int'
Time = 0.16 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx=-\frac {\log \left (5\right ) - \log \left (2\right ) - \log \left (x {\left (4 \, e^{3} + 5\right )} - 8 \, e^{3} + 19\right )}{\log \left (x\right )} \] Input:
integrate((((-4*x+8)*exp(3)-5*x-19)*log(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x *exp(3)+5*x)*log(x))/((4*x^2-8*x)*exp(3)+5*x^2+19*x)/log(x)^2,x, algorithm ="maxima")
Output:
-(log(5) - log(2) - log(x*(4*e^3 + 5) - 8*e^3 + 19))/log(x)
Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx=-\frac {\log \left (5\right ) - \log \left (8 \, x e^{3} + 10 \, x - 16 \, e^{3} + 38\right )}{\log \left (x\right )} \] Input:
integrate((((-4*x+8)*exp(3)-5*x-19)*log(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x *exp(3)+5*x)*log(x))/((4*x^2-8*x)*exp(3)+5*x^2+19*x)/log(x)^2,x, algorithm ="giac")
Output:
-(log(5) - log(8*x*e^3 + 10*x - 16*e^3 + 38))/log(x)
Time = 1.98 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx=\frac {\ln \left (2\,x+\frac {{\mathrm {e}}^3\,\left (8\,x-16\right )}{5}+\frac {38}{5}\right )}{\ln \left (x\right )} \] Input:
int(-(log(2*x + (exp(3)*(8*x - 16))/5 + 38/5)*(5*x + exp(3)*(4*x - 8) + 19 ) - log(x)*(5*x + 4*x*exp(3)))/(log(x)^2*(19*x - exp(3)*(8*x - 4*x^2) + 5* x^2)),x)
Output:
log(2*x + (exp(3)*(8*x - 16))/5 + 38/5)/log(x)
Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {\left (5 x+4 e^3 x\right ) \log (x)+\left (-19+e^3 (8-4 x)-5 x\right ) \log \left (\frac {1}{5} \left (38+10 x+e^3 (-16+8 x)\right )\right )}{\left (19 x+5 x^2+e^3 \left (-8 x+4 x^2\right )\right ) \log ^2(x)} \, dx=\frac {\mathrm {log}\left (\frac {8}{5} e^{3} x -\frac {16}{5} e^{3}+2 x +\frac {38}{5}\right )}{\mathrm {log}\left (x \right )} \] Input:
int((((-4*x+8)*exp(3)-5*x-19)*log(1/5*(8*x-16)*exp(3)+2*x+38/5)+(4*x*exp(3 )+5*x)*log(x))/((4*x^2-8*x)*exp(3)+5*x^2+19*x)/log(x)^2,x)
Output:
log((8*e**3*x - 16*e**3 + 10*x + 38)/5)/log(x)