Integrand size = 81, antiderivative size = 21 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log \left (x \log ^2\left (e^{2 x}+\frac {1}{5} \left (8+x^2\right )\right )\right ) \] Output:
ln(x*ln(exp(x)^2+1/5*x^2+8/5)^2)
Time = 0.14 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log (x)+2 \log \left (\log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )\right ) \] Input:
Integrate[(20*E^(2*x)*x + 4*x^2 + (8 + 5*E^(2*x) + x^2)*Log[(8 + 5*E^(2*x) + x^2)/5])/((8*x + 5*E^(2*x)*x + x^3)*Log[(8 + 5*E^(2*x) + x^2)/5]),x]
Output:
Log[x] + 2*Log[Log[(8 + 5*E^(2*x) + x^2)/5]]
Time = 0.50 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {7239, 2009}
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.
\(\displaystyle \int \frac {4 x^2+\left (x^2+5 e^{2 x}+8\right ) \log \left (\frac {1}{5} \left (x^2+5 e^{2 x}+8\right )\right )+20 e^{2 x} x}{\left (x^3+5 e^{2 x} x+8 x\right ) \log \left (\frac {1}{5} \left (x^2+5 e^{2 x}+8\right )\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \left (\frac {4 \left (x+5 e^{2 x}\right )}{\left (x^2+5 e^{2 x}+8\right ) \log \left (\frac {1}{5} \left (x^2+5 e^{2 x}+8\right )\right )}+\frac {1}{x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 \log \left (\log \left (\frac {1}{5} \left (x^2+5 e^{2 x}+8\right )\right )\right )+\log (x)\) |
Input:
Int[(20*E^(2*x)*x + 4*x^2 + (8 + 5*E^(2*x) + x^2)*Log[(8 + 5*E^(2*x) + x^2 )/5])/((8*x + 5*E^(2*x)*x + x^3)*Log[(8 + 5*E^(2*x) + x^2)/5]),x]
Output:
Log[x] + 2*Log[Log[(8 + 5*E^(2*x) + x^2)/5]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 0.19 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90
method | result | size |
risch | \(\ln \left (x \right )+2 \ln \left (\ln \left ({\mathrm e}^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right )\right )\) | \(19\) |
parallelrisch | \(\ln \left (x \right )+2 \ln \left (\ln \left ({\mathrm e}^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right )\right )\) | \(19\) |
Input:
int(((5*exp(x)^2+x^2+8)*ln(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^2)/(5*x *exp(x)^2+x^3+8*x)/ln(exp(x)^2+1/5*x^2+8/5),x,method=_RETURNVERBOSE)
Output:
ln(x)+2*ln(ln(exp(2*x)+1/5*x^2+8/5))
Time = 0.07 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log \left (x\right ) + 2 \, \log \left (\log \left (\frac {1}{5} \, x^{2} + e^{\left (2 \, x\right )} + \frac {8}{5}\right )\right ) \] Input:
integrate(((5*exp(x)^2+x^2+8)*log(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^ 2)/(5*x*exp(x)^2+x^3+8*x)/log(exp(x)^2+1/5*x^2+8/5),x, algorithm="fricas")
Output:
log(x) + 2*log(log(1/5*x^2 + e^(2*x) + 8/5))
Time = 0.17 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.95 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log {\left (x \right )} + 2 \log {\left (\log {\left (\frac {x^{2}}{5} + e^{2 x} + \frac {8}{5} \right )} \right )} \] Input:
integrate(((5*exp(x)**2+x**2+8)*ln(exp(x)**2+1/5*x**2+8/5)+20*x*exp(x)**2+ 4*x**2)/(5*x*exp(x)**2+x**3+8*x)/ln(exp(x)**2+1/5*x**2+8/5),x)
Output:
log(x) + 2*log(log(x**2/5 + exp(2*x) + 8/5))
Time = 0.16 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log \left (x\right ) + 2 \, \log \left (-\log \left (5\right ) + \log \left (x^{2} + 5 \, e^{\left (2 \, x\right )} + 8\right )\right ) \] Input:
integrate(((5*exp(x)^2+x^2+8)*log(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^ 2)/(5*x*exp(x)^2+x^3+8*x)/log(exp(x)^2+1/5*x^2+8/5),x, algorithm="maxima")
Output:
log(x) + 2*log(-log(5) + log(x^2 + 5*e^(2*x) + 8))
Time = 0.15 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=\log \left (x\right ) + 2 \, \log \left (\log \left (\frac {1}{5} \, x^{2} + e^{\left (2 \, x\right )} + \frac {8}{5}\right )\right ) \] Input:
integrate(((5*exp(x)^2+x^2+8)*log(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^ 2)/(5*x*exp(x)^2+x^3+8*x)/log(exp(x)^2+1/5*x^2+8/5),x, algorithm="giac")
Output:
log(x) + 2*log(log(1/5*x^2 + e^(2*x) + 8/5))
Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.86 \[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=2\,\ln \left (\ln \left ({\mathrm {e}}^{2\,x}+\frac {x^2}{5}+\frac {8}{5}\right )\right )+\ln \left (x\right ) \] Input:
int((20*x*exp(2*x) + log(exp(2*x) + x^2/5 + 8/5)*(5*exp(2*x) + x^2 + 8) + 4*x^2)/(log(exp(2*x) + x^2/5 + 8/5)*(8*x + 5*x*exp(2*x) + x^3)),x)
Output:
2*log(log(exp(2*x) + x^2/5 + 8/5)) + log(x)
\[ \int \frac {20 e^{2 x} x+4 x^2+\left (8+5 e^{2 x}+x^2\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )}{\left (8 x+5 e^{2 x} x+x^3\right ) \log \left (\frac {1}{5} \left (8+5 e^{2 x}+x^2\right )\right )} \, dx=20 \left (\int \frac {e^{2 x}}{5 e^{2 x} \mathrm {log}\left (e^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right )+\mathrm {log}\left (e^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right ) x^{2}+8 \,\mathrm {log}\left (e^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right )}d x \right )+5 \left (\int \frac {e^{2 x}}{5 e^{2 x} x +x^{3}+8 x}d x \right )+4 \left (\int \frac {x}{5 e^{2 x} \mathrm {log}\left (e^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right )+\mathrm {log}\left (e^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right ) x^{2}+8 \,\mathrm {log}\left (e^{2 x}+\frac {x^{2}}{5}+\frac {8}{5}\right )}d x \right )+\int \frac {x}{5 e^{2 x}+x^{2}+8}d x +8 \left (\int \frac {1}{5 e^{2 x} x +x^{3}+8 x}d x \right ) \] Input:
int(((5*exp(x)^2+x^2+8)*log(exp(x)^2+1/5*x^2+8/5)+20*x*exp(x)^2+4*x^2)/(5* x*exp(x)^2+x^3+8*x)/log(exp(x)^2+1/5*x^2+8/5),x)
Output:
20*int(e**(2*x)/(5*e**(2*x)*log((5*e**(2*x) + x**2 + 8)/5) + log((5*e**(2* x) + x**2 + 8)/5)*x**2 + 8*log((5*e**(2*x) + x**2 + 8)/5)),x) + 5*int(e**( 2*x)/(5*e**(2*x)*x + x**3 + 8*x),x) + 4*int(x/(5*e**(2*x)*log((5*e**(2*x) + x**2 + 8)/5) + log((5*e**(2*x) + x**2 + 8)/5)*x**2 + 8*log((5*e**(2*x) + x**2 + 8)/5)),x) + int(x/(5*e**(2*x) + x**2 + 8),x) + 8*int(1/(5*e**(2*x) *x + x**3 + 8*x),x)