Integrand size = 156, antiderivative size = 24 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log (x) \log \left (1+\frac {9}{e^{5 e^{2 x}}+x}\right )}{x} \] Output:
ln(1+3/(1/3*x+1/3*exp(5*exp(x)^2)))*ln(x)/x
Time = 0.29 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log (x) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{x} \] Input:
Integrate[(-9*x*Log[x] - 90*E^(5*E^(2*x) + 2*x)*x*Log[x] + (9*x + x^2 + E^ (10*E^(2*x))*(1 - Log[x]) + (-9*x - x^2)*Log[x] + E^(5*E^(2*x))*(9 + 2*x + (-9 - 2*x)*Log[x]))*Log[(9 + E^(5*E^(2*x)) + x)/(E^(5*E^(2*x)) + x)])/(E^ (10*E^(2*x))*x^2 + 9*x^3 + x^4 + E^(5*E^(2*x))*(9*x^2 + 2*x^3)),x]
Output:
(Log[x]*Log[(9 + E^(5*E^(2*x)) + x)/(E^(5*E^(2*x)) + x)])/x
Time = 10.71 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.013, Rules used = {7293, 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.
\(\displaystyle \int \frac {\left (x^2+\left (-x^2-9 x\right ) \log (x)+9 x+e^{10 e^{2 x}} (1-\log (x))+e^{5 e^{2 x}} (2 x+(-2 x-9) \log (x)+9)\right ) \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )-90 e^{2 x+5 e^{2 x}} x \log (x)-9 x \log (x)}{x^4+9 x^3+e^{10 e^{2 x}} x^2+e^{5 e^{2 x}} \left (2 x^3+9 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (-\frac {x^2 \log (x) \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )-x^2 \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )+9 x \log (x)-2 e^{5 e^{2 x}} x \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )+2 e^{5 e^{2 x}} x \log (x) \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )+9 x \log (x) \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )-9 x \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )-9 e^{5 e^{2 x}} \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )-e^{10 e^{2 x}} \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )+9 e^{5 e^{2 x}} \log (x) \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )+e^{10 e^{2 x}} \log (x) \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )}{x^2 \left (x+e^{5 e^{2 x}}\right ) \left (x+e^{5 e^{2 x}}+9\right )}-\frac {90 e^{2 x+5 e^{2 x}} \log (x)}{x \left (x+e^{5 e^{2 x}}\right ) \left (x+e^{5 e^{2 x}}+9\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log (x) \log \left (\frac {x+e^{5 e^{2 x}}+9}{x+e^{5 e^{2 x}}}\right )}{x}\) |
Input:
Int[(-9*x*Log[x] - 90*E^(5*E^(2*x) + 2*x)*x*Log[x] + (9*x + x^2 + E^(10*E^ (2*x))*(1 - Log[x]) + (-9*x - x^2)*Log[x] + E^(5*E^(2*x))*(9 + 2*x + (-9 - 2*x)*Log[x]))*Log[(9 + E^(5*E^(2*x)) + x)/(E^(5*E^(2*x)) + x)])/(E^(10*E^ (2*x))*x^2 + 9*x^3 + x^4 + E^(5*E^(2*x))*(9*x^2 + 2*x^3)),x]
Output:
(Log[x]*Log[(9 + E^(5*E^(2*x)) + x)/(E^(5*E^(2*x)) + x)])/x
Time = 46.79 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.25
method | result | size |
parallelrisch | \(\frac {\ln \left (\frac {{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right ) \ln \left (x \right )}{x}\) | \(30\) |
risch | \(\frac {\ln \left (x \right ) \ln \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{x}-\frac {\ln \left (x \right ) \left (i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right ) \operatorname {csgn}\left (i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right )-i \pi \,\operatorname {csgn}\left (\frac {i}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right )}^{2}+i \pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x +9\right )}{{\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x}\right )}^{3}+2 \ln \left ({\mathrm e}^{5 \,{\mathrm e}^{2 x}}+x \right )\right )}{2 x}\) | \(219\) |
Input:
int((((1-ln(x))*exp(5*exp(x)^2)^2+((-2*x-9)*ln(x)+2*x+9)*exp(5*exp(x)^2)+( -x^2-9*x)*ln(x)+x^2+9*x)*ln((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^2)+x))-90* x*exp(x)^2*ln(x)*exp(5*exp(x)^2)-9*x*ln(x))/(x^2*exp(5*exp(x)^2)^2+(2*x^3+ 9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x,method=_RETURNVERBOSE)
Output:
1/x*ln((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^2)+x))*ln(x)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (22) = 44\).
Time = 0.10 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log \left (x\right ) \log \left (\frac {{\left (x + 9\right )} e^{\left (2 \, x\right )} + e^{\left (2 \, x + 5 \, e^{\left (2 \, x\right )}\right )}}{x e^{\left (2 \, x\right )} + e^{\left (2 \, x + 5 \, e^{\left (2 \, x\right )}\right )}}\right )}{x} \] Input:
integrate((((1-log(x))*exp(5*exp(x)^2)^2+((-2*x-9)*log(x)+2*x+9)*exp(5*exp (x)^2)+(-x^2-9*x)*log(x)+x^2+9*x)*log((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^ 2)+x))-90*x*exp(x)^2*log(x)*exp(5*exp(x)^2)-9*x*log(x))/(x^2*exp(5*exp(x)^ 2)^2+(2*x^3+9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x, algorithm="fricas")
Output:
log(x)*log(((x + 9)*e^(2*x) + e^(2*x + 5*e^(2*x)))/(x*e^(2*x) + e^(2*x + 5 *e^(2*x))))/x
Time = 2.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log {\left (x \right )} \log {\left (\frac {x + e^{5 e^{2 x}} + 9}{x + e^{5 e^{2 x}}} \right )}}{x} \] Input:
integrate((((1-ln(x))*exp(5*exp(x)**2)**2+((-2*x-9)*ln(x)+2*x+9)*exp(5*exp (x)**2)+(-x**2-9*x)*ln(x)+x**2+9*x)*ln((exp(5*exp(x)**2)+x+9)/(exp(5*exp(x )**2)+x))-90*x*exp(x)**2*ln(x)*exp(5*exp(x)**2)-9*x*ln(x))/(x**2*exp(5*exp (x)**2)**2+(2*x**3+9*x**2)*exp(5*exp(x)**2)+x**4+9*x**3),x)
Output:
log(x)*log((x + exp(5*exp(2*x)) + 9)/(x + exp(5*exp(2*x))))/x
Time = 0.11 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log \left (x + e^{\left (5 \, e^{\left (2 \, x\right )}\right )} + 9\right ) \log \left (x\right ) - \log \left (x + e^{\left (5 \, e^{\left (2 \, x\right )}\right )}\right ) \log \left (x\right )}{x} \] Input:
integrate((((1-log(x))*exp(5*exp(x)^2)^2+((-2*x-9)*log(x)+2*x+9)*exp(5*exp (x)^2)+(-x^2-9*x)*log(x)+x^2+9*x)*log((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^ 2)+x))-90*x*exp(x)^2*log(x)*exp(5*exp(x)^2)-9*x*log(x))/(x^2*exp(5*exp(x)^ 2)^2+(2*x^3+9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x, algorithm="maxima")
Output:
(log(x + e^(5*e^(2*x)) + 9)*log(x) - log(x + e^(5*e^(2*x)))*log(x))/x
Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\log \left (x + e^{\left (5 \, e^{\left (2 \, x\right )}\right )} + 9\right ) \log \left (x\right ) - \log \left (x + e^{\left (5 \, e^{\left (2 \, x\right )}\right )}\right ) \log \left (x\right )}{x} \] Input:
integrate((((1-log(x))*exp(5*exp(x)^2)^2+((-2*x-9)*log(x)+2*x+9)*exp(5*exp (x)^2)+(-x^2-9*x)*log(x)+x^2+9*x)*log((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^ 2)+x))-90*x*exp(x)^2*log(x)*exp(5*exp(x)^2)-9*x*log(x))/(x^2*exp(5*exp(x)^ 2)^2+(2*x^3+9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x, algorithm="giac")
Output:
(log(x + e^(5*e^(2*x)) + 9)*log(x) - log(x + e^(5*e^(2*x)))*log(x))/x
Time = 3.41 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\ln \left (\frac {x+{\mathrm {e}}^{5\,{\mathrm {e}}^{2\,x}}+9}{x+{\mathrm {e}}^{5\,{\mathrm {e}}^{2\,x}}}\right )\,\ln \left (x\right )}{x} \] Input:
int(-(9*x*log(x) - log((x + exp(5*exp(2*x)) + 9)/(x + exp(5*exp(2*x))))*(9 *x + exp(5*exp(2*x))*(2*x - log(x)*(2*x + 9) + 9) - log(x)*(9*x + x^2) - e xp(10*exp(2*x))*(log(x) - 1) + x^2) + 90*x*exp(5*exp(2*x))*exp(2*x)*log(x) )/(exp(5*exp(2*x))*(9*x^2 + 2*x^3) + x^2*exp(10*exp(2*x)) + 9*x^3 + x^4),x )
Output:
(log((x + exp(5*exp(2*x)) + 9)/(x + exp(5*exp(2*x))))*log(x))/x
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38 \[ \int \frac {-9 x \log (x)-90 e^{5 e^{2 x}+2 x} x \log (x)+\left (9 x+x^2+e^{10 e^{2 x}} (1-\log (x))+\left (-9 x-x^2\right ) \log (x)+e^{5 e^{2 x}} (9+2 x+(-9-2 x) \log (x))\right ) \log \left (\frac {9+e^{5 e^{2 x}}+x}{e^{5 e^{2 x}}+x}\right )}{e^{10 e^{2 x}} x^2+9 x^3+x^4+e^{5 e^{2 x}} \left (9 x^2+2 x^3\right )} \, dx=\frac {\mathrm {log}\left (\frac {e^{5 e^{2 x}}+x +9}{e^{5 e^{2 x}}+x}\right ) \mathrm {log}\left (x \right )}{x} \] Input:
int((((1-log(x))*exp(5*exp(x)^2)^2+((-2*x-9)*log(x)+2*x+9)*exp(5*exp(x)^2) +(-x^2-9*x)*log(x)+x^2+9*x)*log((exp(5*exp(x)^2)+x+9)/(exp(5*exp(x)^2)+x)) -90*x*exp(x)^2*log(x)*exp(5*exp(x)^2)-9*x*log(x))/(x^2*exp(5*exp(x)^2)^2+( 2*x^3+9*x^2)*exp(5*exp(x)^2)+x^4+9*x^3),x)
Output:
(log((e**(5*e**(2*x)) + x + 9)/(e**(5*e**(2*x)) + x))*log(x))/x