Integrand size = 109, antiderivative size = 31 \[ \int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{25-10 x+x^2} \, dx=\left (-2+e^{\frac {e^{4 e^x}}{-5+x}} \left (-e^5+e^x+x\right )\right ) \log (3) \] Output:
((exp(x)-exp(5)+x)*exp(exp(exp(x))^4/(-5+x))-2)*ln(3)
Time = 0.07 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{25-10 x+x^2} \, dx=e^{\frac {e^{4 e^x}}{-5+x}} \left (-e^5+e^x+x\right ) \log (3) \] Input:
Integrate[(E^(E^(4*E^x)/(-5 + x))*((25 - 10*x + x^2)*Log[3] + E^x*(25 - 10 *x + x^2)*Log[3] + E^(4*E^x)*((E^5 - x)*Log[3] + E^(2*x)*(-20 + 4*x)*Log[3 ] + E^x*(-1 + E^5*(20 - 4*x) - 20*x + 4*x^2)*Log[3])))/(25 - 10*x + x^2),x ]
Output:
E^(E^(4*E^x)/(-5 + x))*(-E^5 + E^x + x)*Log[3]
Leaf count is larger than twice the leaf count of optimal. \(123\) vs. \(2(31)=62\).
Time = 0.66 (sec) , antiderivative size = 123, normalized size of antiderivative = 3.97, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.009, Rules used = {2726}
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.
\(\displaystyle \int \frac {e^{\frac {e^{4 e^x}}{x-5}} \left (e^x \left (x^2-10 x+25\right ) \log (3)+\left (x^2-10 x+25\right ) \log (3)+e^{4 e^x} \left (e^x \left (4 x^2-20 x+e^5 (20-4 x)-1\right ) \log (3)+\left (e^5-x\right ) \log (3)+e^{2 x} (4 x-20) \log (3)\right )\right )}{x^2-10 x+25} \, dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle \frac {e^{4 e^x-\frac {e^{4 e^x}}{5-x}} \left (e^x \left (-4 x^2+20 x-4 e^5 (5-x)+1\right ) \log (3)+4 e^{2 x} (5-x) \log (3)-\left (e^5-x\right ) \log (3)\right )}{\left (\frac {4 e^{x+4 e^x}}{5-x}+\frac {e^{4 e^x}}{(5-x)^2}\right ) \left (x^2-10 x+25\right )}\) |
Input:
Int[(E^(E^(4*E^x)/(-5 + x))*((25 - 10*x + x^2)*Log[3] + E^x*(25 - 10*x + x ^2)*Log[3] + E^(4*E^x)*((E^5 - x)*Log[3] + E^(2*x)*(-20 + 4*x)*Log[3] + E^ x*(-1 + E^5*(20 - 4*x) - 20*x + 4*x^2)*Log[3])))/(25 - 10*x + x^2),x]
Output:
(E^(4*E^x - E^(4*E^x)/(5 - x))*(4*E^(2*x)*(5 - x)*Log[3] - (E^5 - x)*Log[3 ] + E^x*(1 - 4*E^5*(5 - x) + 20*x - 4*x^2)*Log[3]))/((E^(4*E^x)/(5 - x)^2 + (4*E^(4*E^x + x))/(5 - x))*(25 - 10*x + x^2))
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 39.47 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87
method | result | size |
risch | \(-\left ({\mathrm e}^{5}-x -{\mathrm e}^{x}\right ) \ln \left (3\right ) {\mathrm e}^{\frac {{\mathrm e}^{4 \,{\mathrm e}^{x}}}{-5+x}}\) | \(27\) |
parallelrisch | \(-{\mathrm e}^{5} \ln \left (3\right ) {\mathrm e}^{\frac {{\mathrm e}^{4 \,{\mathrm e}^{x}}}{-5+x}}+\ln \left (3\right ) x \,{\mathrm e}^{\frac {{\mathrm e}^{4 \,{\mathrm e}^{x}}}{-5+x}}+\ln \left (3\right ) {\mathrm e}^{x} {\mathrm e}^{\frac {{\mathrm e}^{4 \,{\mathrm e}^{x}}}{-5+x}}\) | \(53\) |
Input:
int((((4*x-20)*ln(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*ln(3)*exp(x) +(exp(5)-x)*ln(3))*exp(exp(x))^4+(x^2-10*x+25)*ln(3)*exp(x)+(x^2-10*x+25)* ln(3))*exp(exp(exp(x))^4/(-5+x))/(x^2-10*x+25),x,method=_RETURNVERBOSE)
Output:
-(exp(5)-x-exp(x))*ln(3)*exp(exp(4*exp(x))/(-5+x))
Time = 0.09 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{25-10 x+x^2} \, dx={\left ({\left (x - e^{5}\right )} \log \left (3\right ) + e^{x} \log \left (3\right )\right )} e^{\left (\frac {e^{\left (4 \, e^{x}\right )}}{x - 5}\right )} \] Input:
integrate((((4*x-20)*log(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*log(3 )*exp(x)+(exp(5)-x)*log(3))*exp(exp(x))^4+(x^2-10*x+25)*log(3)*exp(x)+(x^2 -10*x+25)*log(3))*exp(exp(exp(x))^4/(-5+x))/(x^2-10*x+25),x, algorithm="fr icas")
Output:
((x - e^5)*log(3) + e^x*log(3))*e^(e^(4*e^x)/(x - 5))
Time = 11.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{25-10 x+x^2} \, dx=\left (x \log {\left (3 \right )} + e^{x} \log {\left (3 \right )} - e^{5} \log {\left (3 \right )}\right ) e^{\frac {e^{4 e^{x}}}{x - 5}} \] Input:
integrate((((4*x-20)*ln(3)*exp(x)**2+((-4*x+20)*exp(5)+4*x**2-20*x-1)*ln(3 )*exp(x)+(exp(5)-x)*ln(3))*exp(exp(x))**4+(x**2-10*x+25)*ln(3)*exp(x)+(x** 2-10*x+25)*ln(3))*exp(exp(exp(x))**4/(-5+x))/(x**2-10*x+25),x)
Output:
(x*log(3) + exp(x)*log(3) - exp(5)*log(3))*exp(exp(4*exp(x))/(x - 5))
Time = 0.19 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{25-10 x+x^2} \, dx={\left (x \log \left (3\right ) - e^{5} \log \left (3\right ) + e^{x} \log \left (3\right )\right )} e^{\left (\frac {e^{\left (4 \, e^{x}\right )}}{x - 5}\right )} \] Input:
integrate((((4*x-20)*log(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*log(3 )*exp(x)+(exp(5)-x)*log(3))*exp(exp(x))^4+(x^2-10*x+25)*log(3)*exp(x)+(x^2 -10*x+25)*log(3))*exp(exp(exp(x))^4/(-5+x))/(x^2-10*x+25),x, algorithm="ma xima")
Output:
(x*log(3) - e^5*log(3) + e^x*log(3))*e^(e^(4*e^x)/(x - 5))
\[ \int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{25-10 x+x^2} \, dx=\int { \frac {{\left ({\left (x^{2} - 10 \, x + 25\right )} e^{x} \log \left (3\right ) + {\left (4 \, {\left (x - 5\right )} e^{\left (2 \, x\right )} \log \left (3\right ) + {\left (4 \, x^{2} - 4 \, {\left (x - 5\right )} e^{5} - 20 \, x - 1\right )} e^{x} \log \left (3\right ) - {\left (x - e^{5}\right )} \log \left (3\right )\right )} e^{\left (4 \, e^{x}\right )} + {\left (x^{2} - 10 \, x + 25\right )} \log \left (3\right )\right )} e^{\left (\frac {e^{\left (4 \, e^{x}\right )}}{x - 5}\right )}}{x^{2} - 10 \, x + 25} \,d x } \] Input:
integrate((((4*x-20)*log(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*log(3 )*exp(x)+(exp(5)-x)*log(3))*exp(exp(x))^4+(x^2-10*x+25)*log(3)*exp(x)+(x^2 -10*x+25)*log(3))*exp(exp(exp(x))^4/(-5+x))/(x^2-10*x+25),x, algorithm="gi ac")
Output:
integrate(((x^2 - 10*x + 25)*e^x*log(3) + (4*(x - 5)*e^(2*x)*log(3) + (4*x ^2 - 4*(x - 5)*e^5 - 20*x - 1)*e^x*log(3) - (x - e^5)*log(3))*e^(4*e^x) + (x^2 - 10*x + 25)*log(3))*e^(e^(4*e^x)/(x - 5))/(x^2 - 10*x + 25), x)
Time = 3.12 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{25-10 x+x^2} \, dx={\mathrm {e}}^{\frac {{\mathrm {e}}^{4\,{\mathrm {e}}^x}}{x-5}}\,\left (x\,\ln \left (3\right )-{\mathrm {e}}^5\,\ln \left (3\right )+{\mathrm {e}}^x\,\ln \left (3\right )\right ) \] Input:
int((exp(exp(4*exp(x))/(x - 5))*(log(3)*(x^2 - 10*x + 25) - exp(4*exp(x))* (log(3)*(x - exp(5)) + exp(x)*log(3)*(20*x - 4*x^2 + exp(5)*(4*x - 20) + 1 ) - exp(2*x)*log(3)*(4*x - 20)) + exp(x)*log(3)*(x^2 - 10*x + 25)))/(x^2 - 10*x + 25),x)
Output:
exp(exp(4*exp(x))/(x - 5))*(x*log(3) - exp(5)*log(3) + exp(x)*log(3))
Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.90 \[ \int \frac {e^{\frac {e^{4 e^x}}{-5+x}} \left (\left (25-10 x+x^2\right ) \log (3)+e^x \left (25-10 x+x^2\right ) \log (3)+e^{4 e^x} \left (\left (e^5-x\right ) \log (3)+e^{2 x} (-20+4 x) \log (3)+e^x \left (-1+e^5 (20-4 x)-20 x+4 x^2\right ) \log (3)\right )\right )}{25-10 x+x^2} \, dx=e^{\frac {e^{4 e^{x}}}{-5+x}} \mathrm {log}\left (3\right ) \left (e^{x}-e^{5}+x \right ) \] Input:
int((((4*x-20)*log(3)*exp(x)^2+((-4*x+20)*exp(5)+4*x^2-20*x-1)*log(3)*exp( x)+(exp(5)-x)*log(3))*exp(exp(x))^4+(x^2-10*x+25)*log(3)*exp(x)+(x^2-10*x+ 25)*log(3))*exp(exp(exp(x))^4/(-5+x))/(x^2-10*x+25),x)
Output:
e**(e**(4*e**x)/(x - 5))*log(3)*(e**x - e**5 + x)