Integrand size = 169, antiderivative size = 26 \[ \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx=e^{\frac {x}{\log \left (e^x \left (1+4 \left (x+\log ^2(3)\right )\right )+\log (x)\right )}}+x \] Output:
x+exp(x/ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x)))
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx=e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}}+x \] Input:
Integrate[((E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[ 3]^2) + Log[x]]^2 + E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])*(-1 + E ^x*(-5*x - 4*x^2 - 4*x*Log[3]^2) + (E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*L og[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]))/((E^x*(1 + 4*x + 4*Log[3]^2) + L og[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2),x]
Output:
E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]) + 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.
| \(\displaystyle \int \frac {e^{\frac {x}{\log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )}} \left (e^x \left (-4 x^2-5 x-4 x \log ^2(3)\right )+\left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )-1\right )+\left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )}{\left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )} \, dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \int \left (\frac {e^{\frac {x}{\log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )}} \left (-4 e^x x^2+4 e^x x \log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )-5 e^x x \left (1+\frac {4 \log ^2(3)}{5}\right )+\log (x) \log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )+e^x \left (1+4 \log ^2(3)\right ) \log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )-1\right )}{\left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )}+1\right )dx\) |
| \(\Big \downarrow \) 7299 |
| \(\displaystyle \int \left (\frac {e^{\frac {x}{\log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )}} \left (-4 e^x x^2+4 e^x x \log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )-5 e^x x \left (1+\frac {4 \log ^2(3)}{5}\right )+\log (x) \log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )+e^x \left (1+4 \log ^2(3)\right ) \log \left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )-1\right )}{\left (4 e^x x+e^x \left (1+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (4 x+1+4 \log ^2(3)\right )+\log (x)\right )}+1\right )dx\) |
Input:
Int[((E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2 + E^(x/Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]])*(-1 + E^x*(-5 *x - 4*x^2 - 4*x*Log[3]^2) + (E^x*(1 + 4*x + 4*Log[3]^2) + Log[x])*Log[E^x *(1 + 4*x + 4*Log[3]^2) + Log[x]]))/((E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]) *Log[E^x*(1 + 4*x + 4*Log[3]^2) + Log[x]]^2),x]
Output:
$Aborted
Time = 261.68 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00
| method | result | size |
| risch | \(x +{\mathrm e}^{\frac {x}{\ln \left (\ln \left (x \right )+\left (4 \ln \left (3\right )^{2}+4 x +1\right ) {\mathrm e}^{x}\right )}}\) | \(26\) |
| parallelrisch | \(-2 \ln \left (3\right )^{2}-\frac {1}{2}+x +{\mathrm e}^{\frac {x}{\ln \left (\ln \left (x \right )+\left (4 \ln \left (3\right )^{2}+4 x +1\right ) {\mathrm e}^{x}\right )}}\) | \(33\) |
Input:
int((((ln(x)+(4*ln(3)^2+4*x+1)*exp(x))*ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))+ (-4*x*ln(3)^2-4*x^2-5*x)*exp(x)-1)*exp(x/ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x) ))+(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))*ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))^2)/ (ln(x)+(4*ln(3)^2+4*x+1)*exp(x))/ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x))^2,x,me thod=_RETURNVERBOSE)
Output:
x+exp(x/ln(ln(x)+(4*ln(3)^2+4*x+1)*exp(x)))
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx=x + e^{\left (\frac {x}{\log \left ({\left (4 \, \log \left (3\right )^{2} + 4 \, x + 1\right )} e^{x} + \log \left (x\right )\right )}\right )} \] Input:
integrate((((log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^2+4*x+ 1)*exp(x))+(-4*x*log(3)^2-4*x^2-5*x)*exp(x)-1)*exp(x/log(log(x)+(4*log(3)^ 2+4*x+1)*exp(x)))+(log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^ 2+4*x+1)*exp(x))^2)/(log(x)+(4*log(3)^2+4*x+1)*exp(x))/log(log(x)+(4*log(3 )^2+4*x+1)*exp(x))^2,x, algorithm="fricas")
Output:
x + e^(x/log((4*log(3)^2 + 4*x + 1)*e^x + log(x)))
Timed out. \[ \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx=\text {Timed out} \] Input:
integrate((((ln(x)+(4*ln(3)**2+4*x+1)*exp(x))*ln(ln(x)+(4*ln(3)**2+4*x+1)* exp(x))+(-4*x*ln(3)**2-4*x**2-5*x)*exp(x)-1)*exp(x/ln(ln(x)+(4*ln(3)**2+4* x+1)*exp(x)))+(ln(x)+(4*ln(3)**2+4*x+1)*exp(x))*ln(ln(x)+(4*ln(3)**2+4*x+1 )*exp(x))**2)/(ln(x)+(4*ln(3)**2+4*x+1)*exp(x))/ln(ln(x)+(4*ln(3)**2+4*x+1 )*exp(x))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((((log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^2+4*x+ 1)*exp(x))+(-4*x*log(3)^2-4*x^2-5*x)*exp(x)-1)*exp(x/log(log(x)+(4*log(3)^ 2+4*x+1)*exp(x)))+(log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^ 2+4*x+1)*exp(x))^2)/(log(x)+(4*log(3)^2+4*x+1)*exp(x))/log(log(x)+(4*log(3 )^2+4*x+1)*exp(x))^2,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is 0which is not of the expected type LIST
Time = 1.64 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx=x + e^{\left (\frac {x}{\log \left (4 \, e^{x} \log \left (3\right )^{2} + 4 \, x e^{x} + e^{x} + \log \left (x\right )\right )}\right )} \] Input:
integrate((((log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^2+4*x+ 1)*exp(x))+(-4*x*log(3)^2-4*x^2-5*x)*exp(x)-1)*exp(x/log(log(x)+(4*log(3)^ 2+4*x+1)*exp(x)))+(log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^ 2+4*x+1)*exp(x))^2)/(log(x)+(4*log(3)^2+4*x+1)*exp(x))/log(log(x)+(4*log(3 )^2+4*x+1)*exp(x))^2,x, algorithm="giac")
Output:
x + e^(x/log(4*e^x*log(3)^2 + 4*x*e^x + e^x + log(x)))
Time = 2.75 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx=x+{\mathrm {e}}^{\frac {x}{\ln \left ({\mathrm {e}}^x+\ln \left (x\right )+4\,{\mathrm {e}}^x\,{\ln \left (3\right )}^2+4\,x\,{\mathrm {e}}^x\right )}} \] Input:
int(-(exp(x/log(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1)))*(exp(x)*(5*x + 4* x*log(3)^2 + 4*x^2) - log(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1))*(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1)) + 1) - log(log(x) + exp(x)*(4*x + 4*log(3 )^2 + 1))^2*(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1)))/(log(log(x) + exp(x) *(4*x + 4*log(3)^2 + 1))^2*(log(x) + exp(x)*(4*x + 4*log(3)^2 + 1))),x)
Output:
x + exp(x/log(exp(x) + log(x) + 4*exp(x)*log(3)^2 + 4*x*exp(x)))
Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )+e^{\frac {x}{\log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )}} \left (-1+e^x \left (-5 x-4 x^2-4 x \log ^2(3)\right )+\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log \left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )\right )}{\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right ) \log ^2\left (e^x \left (1+4 x+4 \log ^2(3)\right )+\log (x)\right )} \, dx=e^{\frac {x}{\mathrm {log}\left (4 e^{x} \mathrm {log}\left (3\right )^{2}+4 e^{x} x +e^{x}+\mathrm {log}\left (x \right )\right )}}+x \] Input:
int((((log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^2+4*x+1)*exp (x))+(-4*x*log(3)^2-4*x^2-5*x)*exp(x)-1)*exp(x/log(log(x)+(4*log(3)^2+4*x+ 1)*exp(x)))+(log(x)+(4*log(3)^2+4*x+1)*exp(x))*log(log(x)+(4*log(3)^2+4*x+ 1)*exp(x))^2)/(log(x)+(4*log(3)^2+4*x+1)*exp(x))/log(log(x)+(4*log(3)^2+4* x+1)*exp(x))^2,x)
Output:
e**(x/log(4*e**x*log(3)**2 + 4*e**x*x + e**x + log(x))) + x