Integrand size = 74, antiderivative size = 26 \begin {dmath*} \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=-4-\frac {4 x}{5}+\log ^2\left (1-e^3+e^{e^x} x^4\right ) \end {dmath*}
Time = 0.62 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \begin {dmath*} \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=\frac {1}{5} \left (-4 x+5 \log ^2\left (1-e^3+e^{e^x} x^4\right )\right ) \end {dmath*}
Integrate[(-4 + 4*E^3 - 4*E^E^x*x^4 + E^E^x*(40*x^3 + 10*E^x*x^4)*Log[1 - E^3 + E^E^x*x^4])/(5 - 5*E^3 + 5*E^E^x*x^4),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.
\(\displaystyle \int \frac {-4 e^{e^x} x^4+e^{e^x} \left (10 e^x x^4+40 x^3\right ) \log \left (e^{e^x} x^4-e^3+1\right )+4 e^3-4}{5 e^{e^x} x^4-5 e^3+5} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {-4 e^{e^x} x^4+e^{e^x} \left (10 e^x x^4+40 x^3\right ) \log \left (e^{e^x} x^4-e^3+1\right )-4 \left (1-e^3\right )}{5 \left (e^{e^x} x^4-e^3+1\right )}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{5} \int -\frac {2 \left (2 e^{e^x} x^4-5 e^{e^x} \left (e^x x^4+4 x^3\right ) \log \left (e^{e^x} x^4-e^3+1\right )+2 \left (1-e^3\right )\right )}{e^{e^x} x^4-e^3+1}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {2}{5} \int \frac {2 e^{e^x} x^4-5 e^{e^x} \left (e^x x^4+4 x^3\right ) \log \left (e^{e^x} x^4-e^3+1\right )+2 \left (1-e^3\right )}{e^{e^x} x^4-e^3+1}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {2}{5} \int \left (\frac {2 \left (-e^{e^x} x^4+10 e^{e^x} \log \left (e^{e^x} x^4-e^3+1\right ) x^3+e^3-1\right )}{-e^{e^x} x^4+e^3-1}-\frac {5 e^{x+e^x} x^4 \log \left (e^{e^x} x^4-e^3+1\right )}{e^{e^x} x^4-e^3+1}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2}{5} \left (5 \int \frac {e^{x+e^x} x^4 \int \frac {e^{x+e^x} x^4}{e^{e^x} x^4-e^3+1}dx}{e^{e^x} x^4-e^3+1}dx-20 \left (1-e^3\right ) \log \left (e^{e^x} x^4-e^3+1\right ) \int \frac {1}{x \left (-e^{e^x} x^4+e^3-1\right )}dx-5 \log \left (e^{e^x} x^4-e^3+1\right ) \int \frac {e^{x+e^x} x^4}{e^{e^x} x^4-e^3+1}dx-20 \int \frac {\log \left (e^{e^x} x^4-e^3+1\right )}{x}dx+20 \left (1-e^3\right ) \int \frac {e^{x+e^x} x^4 \int -\frac {1}{e^{e^x} x^5+\left (1-e^3\right ) x}dx}{e^{e^x} x^4-e^3+1}dx+20 \int \frac {e^{e^x} x^3 \int \frac {e^{x+e^x} x^4}{e^{e^x} x^4-e^3+1}dx}{e^{e^x} x^4-e^3+1}dx+80 \left (1-e^3\right ) \int \frac {e^{e^x} x^3 \int -\frac {1}{e^{e^x} x^5+\left (1-e^3\right ) x}dx}{e^{e^x} x^4-e^3+1}dx+2 x\right )\) |
Int[(-4 + 4*E^3 - 4*E^E^x*x^4 + E^E^x*(40*x^3 + 10*E^x*x^4)*Log[1 - E^3 + E^E^x*x^4])/(5 - 5*E^3 + 5*E^E^x*x^4),x]
3.10.61.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Time = 222.63 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.81
method | result | size |
risch | \(-\frac {4 x}{5}+\ln \left (x^{4} {\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{3}+1\right )^{2}\) | \(21\) |
parallelrisch | \(-\frac {4 x}{5}+\ln \left (x^{4} {\mathrm e}^{{\mathrm e}^{x}}-{\mathrm e}^{3}+1\right )^{2}\) | \(21\) |
int(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*ln(x^4*exp(exp(x))-exp(3)+1)-4*x^4 *exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x,method=_RETURNVE RBOSE)
Time = 0.28 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \begin {dmath*} \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=\log \left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right )^{2} - \frac {4}{5} \, x \end {dmath*}
integrate(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*log(x^4*exp(exp(x))-exp(3)+1 )-4*x^4*exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x, algorith m=\
Time = 0.37 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \begin {dmath*} \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=- \frac {4 x}{5} + \log {\left (x^{4} e^{e^{x}} - e^{3} + 1 \right )}^{2} \end {dmath*}
integrate(((10*exp(x)*x**4+40*x**3)*exp(exp(x))*ln(x**4*exp(exp(x))-exp(3) +1)-4*x**4*exp(exp(x))+4*exp(3)-4)/(5*x**4*exp(exp(x))-5*exp(3)+5),x)
\begin {dmath*} \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=\int { -\frac {2 \, {\left (2 \, x^{4} e^{\left (e^{x}\right )} - 5 \, {\left (x^{4} e^{x} + 4 \, x^{3}\right )} e^{\left (e^{x}\right )} \log \left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right ) - 2 \, e^{3} + 2\right )}}{5 \, {\left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right )}} \,d x } \end {dmath*}
integrate(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*log(x^4*exp(exp(x))-exp(3)+1 )-4*x^4*exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x, algorith m=\
-2/5*integrate((2*x^4*e^(e^x) - 5*(x^4*e^x + 4*x^3)*e^(e^x)*log(x^4*e^(e^x ) - e^3 + 1) - 2*e^3 + 2)/(x^4*e^(e^x) - e^3 + 1), x)
\begin {dmath*} \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx=\int { -\frac {2 \, {\left (2 \, x^{4} e^{\left (e^{x}\right )} - 5 \, {\left (x^{4} e^{x} + 4 \, x^{3}\right )} e^{\left (e^{x}\right )} \log \left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right ) - 2 \, e^{3} + 2\right )}}{5 \, {\left (x^{4} e^{\left (e^{x}\right )} - e^{3} + 1\right )}} \,d x } \end {dmath*}
integrate(((10*exp(x)*x^4+40*x^3)*exp(exp(x))*log(x^4*exp(exp(x))-exp(3)+1 )-4*x^4*exp(exp(x))+4*exp(3)-4)/(5*x^4*exp(exp(x))-5*exp(3)+5),x, algorith m=\
integrate(-2/5*(2*x^4*e^(e^x) - 5*(x^4*e^x + 4*x^3)*e^(e^x)*log(x^4*e^(e^x ) - e^3 + 1) - 2*e^3 + 2)/(x^4*e^(e^x) - e^3 + 1), x)
Time = 0.43 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \begin {dmath*} \int \frac {-4+4 e^3-4 e^{e^x} x^4+e^{e^x} \left (40 x^3+10 e^x x^4\right ) \log \left (1-e^3+e^{e^x} x^4\right )}{5-5 e^3+5 e^{e^x} x^4} \, dx={\ln \left (x^4\,{\mathrm {e}}^{{\mathrm {e}}^x}-{\mathrm {e}}^3+1\right )}^2-\frac {4\,x}{5} \end {dmath*}