Integrand size = 73, antiderivative size = 22 \[ \int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx=\frac {2}{-5+e^{\frac {1}{12} e^{-5+2 x}} \log (x)} \] Output:
2/(ln(x)*exp(exp(-ln(12)+2*x-5))-5)
Time = 0.15 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05 \[ \int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx=-\frac {2}{5-e^{\frac {1}{12} e^{-5+2 x}} \log (x)} \] Input:
Integrate[(E^(E^(-5 + 2*x)/12)*(-2 - (E^(-5 + 2*x)*x*Log[x])/3))/(25*x - 1 0*E^(E^(-5 + 2*x)/12)*x*Log[x] + E^(E^(-5 + 2*x)/6)*x*Log[x]^2),x]
Output:
-2/(5 - E^(E^(-5 + 2*x)/12)*Log[x])
Time = 1.52 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {7292, 7259, 17}
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.
\(\displaystyle \int \frac {e^{\frac {1}{12} e^{2 x-5}} \left (-\frac {1}{3} e^{2 x-5} x \log (x)-2\right )}{25 x+e^{\frac {1}{6} e^{2 x-5}} x \log ^2(x)-10 e^{\frac {1}{12} e^{2 x-5}} x \log (x)} \, dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {e^{\frac {1}{12} e^{2 x-5}} \left (-\frac {1}{3} e^{2 x-5} x \log (x)-2\right )}{x \left (5-e^{\frac {1}{12} e^{2 x-5}} \log (x)\right )^2}dx\) |
\(\Big \downarrow \) 7259 |
\(\displaystyle -2 \int \frac {1}{\left (5-e^{\frac {1}{12} e^{2 x-5}} \log (x)\right )^2}d\left (e^{\frac {1}{12} e^{2 x-5}} \log (x)\right )\) |
\(\Big \downarrow \) 17 |
\(\displaystyle -\frac {2}{5-e^{\frac {1}{12} e^{2 x-5}} \log (x)}\) |
Input:
Int[(E^(E^(-5 + 2*x)/12)*(-2 - (E^(-5 + 2*x)*x*Log[x])/3))/(25*x - 10*E^(E ^(-5 + 2*x)/12)*x*Log[x] + E^(E^(-5 + 2*x)/6)*x*Log[x]^2),x]
Output:
-2/(5 - E^(E^(-5 + 2*x)/12)*Log[x])
Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 )/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
Int[(u_)*((a_) + (b_.)*(v_)^(p_.)*(w_)^(p_.))^(m_.), x_Symbol] :> With[{c = Simplify[u/(w*D[v, x] + v*D[w, x])]}, Simp[c Subst[Int[(a + b*x^p)^m, x] , x, v*w], x] /; FreeQ[c, x]] /; FreeQ[{a, b, m, p}, x] && IntegerQ[p]
Time = 0.42 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {2}{\ln \left (x \right ) {\mathrm e}^{\frac {{\mathrm e}^{-5+2 x}}{12}}-5}\) | \(19\) |
parallelrisch | \(\frac {2 \ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{-\ln \left (12\right )+2 x -5}}}{5 \left (\ln \left (x \right ) {\mathrm e}^{{\mathrm e}^{-\ln \left (12\right )+2 x -5}}-5\right )}\) | \(34\) |
Input:
int((-4*x*exp(-ln(12)+2*x-5)*ln(x)-2)*exp(exp(-ln(12)+2*x-5))/(x*ln(x)^2*e xp(exp(-ln(12)+2*x-5))^2-10*x*ln(x)*exp(exp(-ln(12)+2*x-5))+25*x),x,method =_RETURNVERBOSE)
Output:
2/(ln(x)*exp(1/12*exp(-5+2*x))-5)
Time = 0.09 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx=\frac {2}{e^{\left (e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )} \log \left (x\right ) - 5} \] Input:
integrate((-4*x*exp(-log(12)+2*x-5)*log(x)-2)*exp(exp(-log(12)+2*x-5))/(x* log(x)^2*exp(exp(-log(12)+2*x-5))^2-10*x*log(x)*exp(exp(-log(12)+2*x-5))+2 5*x),x, algorithm="fricas")
Output:
2/(e^(e^(2*x - log(12) - 5))*log(x) - 5)
Time = 0.11 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx=\frac {2}{e^{\frac {e^{2 x - 5}}{12}} \log {\left (x \right )} - 5} \] Input:
integrate((-4*x*exp(-ln(12)+2*x-5)*ln(x)-2)*exp(exp(-ln(12)+2*x-5))/(x*ln( x)**2*exp(exp(-ln(12)+2*x-5))**2-10*x*ln(x)*exp(exp(-ln(12)+2*x-5))+25*x), x)
Output:
2/(exp(exp(2*x - 5)/12)*log(x) - 5)
\[ \int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx=\int { -\frac {2 \, {\left (2 \, x e^{\left (2 \, x - \log \left (12\right ) - 5\right )} \log \left (x\right ) + 1\right )} e^{\left (e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )}}{x e^{\left (2 \, e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )} \log \left (x\right )^{2} - 10 \, x e^{\left (e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )} \log \left (x\right ) + 25 \, x} \,d x } \] Input:
integrate((-4*x*exp(-log(12)+2*x-5)*log(x)-2)*exp(exp(-log(12)+2*x-5))/(x* log(x)^2*exp(exp(-log(12)+2*x-5))^2-10*x*log(x)*exp(exp(-log(12)+2*x-5))+2 5*x),x, algorithm="maxima")
Output:
-2*integrate(1/6*(x*e^(2*x - 5)*log(x) + 6)*e^(1/12*e^(2*x - 5))/(x*e^(1/6 *e^(2*x - 5))*log(x)^2 - 10*x*e^(1/12*e^(2*x - 5))*log(x) + 25*x), x)
\[ \int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx=\int { -\frac {2 \, {\left (2 \, x e^{\left (2 \, x - \log \left (12\right ) - 5\right )} \log \left (x\right ) + 1\right )} e^{\left (e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )}}{x e^{\left (2 \, e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )} \log \left (x\right )^{2} - 10 \, x e^{\left (e^{\left (2 \, x - \log \left (12\right ) - 5\right )}\right )} \log \left (x\right ) + 25 \, x} \,d x } \] Input:
integrate((-4*x*exp(-log(12)+2*x-5)*log(x)-2)*exp(exp(-log(12)+2*x-5))/(x* log(x)^2*exp(exp(-log(12)+2*x-5))^2-10*x*log(x)*exp(exp(-log(12)+2*x-5))+2 5*x),x, algorithm="giac")
Output:
integrate(-2*(2*x*e^(2*x - log(12) - 5)*log(x) + 1)*e^(e^(2*x - log(12) - 5))/(x*e^(2*e^(2*x - log(12) - 5))*log(x)^2 - 10*x*e^(e^(2*x - log(12) - 5 ))*log(x) + 25*x), x)
Time = 3.34 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.82 \[ \int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx=\frac {2}{{\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}\,{\mathrm {e}}^{-5}}{12}}\,\ln \left (x\right )-5} \] Input:
int(-(exp(exp(2*x - log(12) - 5))*(4*x*exp(2*x - log(12) - 5)*log(x) + 2)) /(25*x + x*exp(2*exp(2*x - log(12) - 5))*log(x)^2 - 10*x*exp(exp(2*x - log (12) - 5))*log(x)),x)
Output:
2/(exp((exp(2*x)*exp(-5))/12)*log(x) - 5)
Time = 0.21 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.64 \[ \int \frac {e^{\frac {1}{12} e^{-5+2 x}} \left (-2-\frac {1}{3} e^{-5+2 x} x \log (x)\right )}{25 x-10 e^{\frac {1}{12} e^{-5+2 x}} x \log (x)+e^{\frac {1}{6} e^{-5+2 x}} x \log ^2(x)} \, dx=\frac {2 e^{\frac {e^{2 x}}{12 e^{5}}} \mathrm {log}\left (x \right )}{5 e^{\frac {e^{2 x}}{12 e^{5}}} \mathrm {log}\left (x \right )-25} \] Input:
int((-4*x*exp(-log(12)+2*x-5)*log(x)-2)*exp(exp(-log(12)+2*x-5))/(x*log(x) ^2*exp(exp(-log(12)+2*x-5))^2-10*x*log(x)*exp(exp(-log(12)+2*x-5))+25*x),x )
Output:
(2*e**(e**(2*x)/(12*e**5))*log(x))/(5*(e**(e**(2*x)/(12*e**5))*log(x) - 5) )