Integrand size = 78, antiderivative size = 25 \[ \int \frac {e^{\frac {2+x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )}{x}} \left (2 x+\left (2+x-2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (-2-x+2 x^3\right ) \log (x)\right )}{-x^2 \log \left (\frac {1}{x}\right )+x^2 \log (x)} \, dx=\frac {5 e^{\frac {2}{x}+x^2} \left (\log \left (\frac {1}{x}\right )-\log (x)\right )}{x} \] Output:
exp(2/x+ln(5*(ln(1/x)-ln(x))/x)+x^2)
Time = 0.17 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {2+x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )}{x}} \left (2 x+\left (2+x-2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (-2-x+2 x^3\right ) \log (x)\right )}{-x^2 \log \left (\frac {1}{x}\right )+x^2 \log (x)} \, dx=\frac {5 e^{\frac {2}{x}+x^2} \left (\log \left (\frac {1}{x}\right )-\log (x)\right )}{x} \] Input:
Integrate[(E^((2 + x^3 + x*Log[(5*Log[x^(-1)] - 5*Log[x])/x])/x)*(2*x + (2 + x - 2*x^3)*Log[x^(-1)] + (-2 - x + 2*x^3)*Log[x]))/(-(x^2*Log[x^(-1)]) + x^2*Log[x]),x]
Output:
(5*E^(2/x + x^2)*(Log[x^(-1)] - Log[x]))/x
Leaf count is larger than twice the leaf count of optimal. \(72\) vs. \(2(25)=50\).
Time = 1.10 (sec) , antiderivative size = 72, normalized size of antiderivative = 2.88, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {7239, 27, 25, 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.
\(\displaystyle \int \frac {e^{\frac {x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )+2}{x}} \left (\left (-2 x^3+x+2\right ) \log \left (\frac {1}{x}\right )+\left (2 x^3-x-2\right ) \log (x)+2 x\right )}{x^2 \log (x)-x^2 \log \left (\frac {1}{x}\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {5 e^{x^2+\frac {2}{x}} \left (\left (2 x^3-x-2\right ) \log \left (\frac {1}{x}\right )+\left (-2 x^3+x+2\right ) \log (x)-2 x\right )}{x^3}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 5 \int -\frac {e^{x^2+\frac {2}{x}} \left (2 x+\left (-2 x^3+x+2\right ) \log \left (\frac {1}{x}\right )-\left (-2 x^3+x+2\right ) \log (x)\right )}{x^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -5 \int \frac {e^{x^2+\frac {2}{x}} \left (2 x+\left (-2 x^3+x+2\right ) \log \left (\frac {1}{x}\right )-\left (-2 x^3+x+2\right ) \log (x)\right )}{x^3}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -5 \int \left (\frac {e^{x^2+\frac {2}{x}} \left (-2 \log \left (\frac {1}{x}\right ) x^3+\log \left (\frac {1}{x}\right ) x+2 x+2 \log \left (\frac {1}{x}\right )\right )}{x^3}+\frac {e^{x^2+\frac {2}{x}} \left (2 x^3-x-2\right ) \log (x)}{x^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 \left (\frac {e^{x^2+\frac {2}{x}} \left (1-x^3\right ) \log (x)}{\left (\frac {1}{x^2}-x\right ) x^3}-\frac {e^{x^2+\frac {2}{x}} \left (1-x^3\right ) \log \left (\frac {1}{x}\right )}{\left (\frac {1}{x^2}-x\right ) x^3}\right )\) |
Input:
Int[(E^((2 + x^3 + x*Log[(5*Log[x^(-1)] - 5*Log[x])/x])/x)*(2*x + (2 + x - 2*x^3)*Log[x^(-1)] + (-2 - x + 2*x^3)*Log[x]))/(-(x^2*Log[x^(-1)]) + x^2* Log[x]),x]
Output:
-5*(-((E^(2/x + x^2)*(1 - x^3)*Log[x^(-1)])/((x^(-2) - x)*x^3)) + (E^(2/x + x^2)*(1 - x^3)*Log[x])/((x^(-2) - x)*x^3))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 3.54 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12
method | result | size |
parallelrisch | \({\mathrm e}^{\frac {x \ln \left (-\frac {5 \left (\ln \left (x \right )-\ln \left (\frac {1}{x}\right )\right )}{x}\right )+x^{3}+2}{x}}\) | \(28\) |
risch | \(-\frac {10 \,{\mathrm e}^{\frac {i \pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{2} \operatorname {csgn}\left (\frac {i}{x}\right )}{2}+\frac {i \pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{2} \operatorname {csgn}\left (i \ln \left (x \right )\right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right ) \operatorname {csgn}\left (i \ln \left (x \right )\right )}{2}-i \pi \operatorname {csgn}\left (\frac {i \ln \left (x \right )}{x}\right )^{2}+x^{2}+\frac {2}{x}} \ln \left (x \right )}{x}\) | \(117\) |
Input:
int(((2*x^3-x-2)*ln(x)+(-2*x^3+x+2)*ln(1/x)+2*x)*exp((x*ln((-5*ln(x)+5*ln( 1/x))/x)+x^3+2)/x)/(x^2*ln(x)-x^2*ln(1/x)),x,method=_RETURNVERBOSE)
Output:
exp((x*ln(-5*(ln(x)-ln(1/x))/x)+x^3+2)/x)
Time = 0.10 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {e^{\frac {2+x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )}{x}} \left (2 x+\left (2+x-2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (-2-x+2 x^3\right ) \log (x)\right )}{-x^2 \log \left (\frac {1}{x}\right )+x^2 \log (x)} \, dx=e^{\left (\frac {x^{3} + x \log \left (\frac {10 \, \log \left (\frac {1}{x}\right )}{x}\right ) + 2}{x}\right )} \] Input:
integrate(((2*x^3-x-2)*log(x)+(-2*x^3+x+2)*log(1/x)+2*x)*exp((x*log((-5*lo g(x)+5*log(1/x))/x)+x^3+2)/x)/(x^2*log(x)-x^2*log(1/x)),x, algorithm="fric as")
Output:
e^((x^3 + x*log(10*log(1/x)/x) + 2)/x)
Time = 0.25 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {e^{\frac {2+x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )}{x}} \left (2 x+\left (2+x-2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (-2-x+2 x^3\right ) \log (x)\right )}{-x^2 \log \left (\frac {1}{x}\right )+x^2 \log (x)} \, dx=e^{\frac {x^{3} + x \log {\left (- \frac {10 \log {\left (x \right )}}{x} \right )} + 2}{x}} \] Input:
integrate(((2*x**3-x-2)*ln(x)+(-2*x**3+x+2)*ln(1/x)+2*x)*exp((x*ln((-5*ln( x)+5*ln(1/x))/x)+x**3+2)/x)/(x**2*ln(x)-x**2*ln(1/x)),x)
Output:
exp((x**3 + x*log(-10*log(x)/x) + 2)/x)
Time = 0.09 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.68 \[ \int \frac {e^{\frac {2+x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )}{x}} \left (2 x+\left (2+x-2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (-2-x+2 x^3\right ) \log (x)\right )}{-x^2 \log \left (\frac {1}{x}\right )+x^2 \log (x)} \, dx=-\frac {10 \, e^{\left (x^{2} + \frac {2}{x}\right )} \log \left (x\right )}{x} \] Input:
integrate(((2*x^3-x-2)*log(x)+(-2*x^3+x+2)*log(1/x)+2*x)*exp((x*log((-5*lo g(x)+5*log(1/x))/x)+x^3+2)/x)/(x^2*log(x)-x^2*log(1/x)),x, algorithm="maxi ma")
Output:
-10*e^(x^2 + 2/x)*log(x)/x
Time = 0.13 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {2+x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )}{x}} \left (2 x+\left (2+x-2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (-2-x+2 x^3\right ) \log (x)\right )}{-x^2 \log \left (\frac {1}{x}\right )+x^2 \log (x)} \, dx=e^{\left (x^{2} + \frac {2}{x} + \log \left (-\frac {10 \, \log \left (x\right )}{x}\right )\right )} \] Input:
integrate(((2*x^3-x-2)*log(x)+(-2*x^3+x+2)*log(1/x)+2*x)*exp((x*log((-5*lo g(x)+5*log(1/x))/x)+x^3+2)/x)/(x^2*log(x)-x^2*log(1/x)),x, algorithm="giac ")
Output:
e^(x^2 + 2/x + log(-10*log(x)/x))
Timed out. \[ \int \frac {e^{\frac {2+x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )}{x}} \left (2 x+\left (2+x-2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (-2-x+2 x^3\right ) \log (x)\right )}{-x^2 \log \left (\frac {1}{x}\right )+x^2 \log (x)} \, dx=\int \frac {{\mathrm {e}}^{\frac {x\,\ln \left (\frac {5\,\ln \left (\frac {1}{x}\right )-5\,\ln \left (x\right )}{x}\right )+x^3+2}{x}}\,\left (2\,x-\ln \left (x\right )\,\left (-2\,x^3+x+2\right )+\ln \left (\frac {1}{x}\right )\,\left (-2\,x^3+x+2\right )\right )}{x^2\,\ln \left (x\right )-x^2\,\ln \left (\frac {1}{x}\right )} \,d x \] Input:
int((exp((x*log((5*log(1/x) - 5*log(x))/x) + x^3 + 2)/x)*(2*x - log(x)*(x - 2*x^3 + 2) + log(1/x)*(x - 2*x^3 + 2)))/(x^2*log(x) - x^2*log(1/x)),x)
Output:
int((exp((x*log((5*log(1/x) - 5*log(x))/x) + x^3 + 2)/x)*(2*x - log(x)*(x - 2*x^3 + 2) + log(1/x)*(x - 2*x^3 + 2)))/(x^2*log(x) - x^2*log(1/x)), x)
Time = 0.17 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.72 \[ \int \frac {e^{\frac {2+x^3+x \log \left (\frac {5 \log \left (\frac {1}{x}\right )-5 \log (x)}{x}\right )}{x}} \left (2 x+\left (2+x-2 x^3\right ) \log \left (\frac {1}{x}\right )+\left (-2-x+2 x^3\right ) \log (x)\right )}{-x^2 \log \left (\frac {1}{x}\right )+x^2 \log (x)} \, dx=-\frac {10 e^{\frac {x^{3}+2}{x}} \mathrm {log}\left (x \right )}{x} \] Input:
int(((2*x^3-x-2)*log(x)+(-2*x^3+x+2)*log(1/x)+2*x)*exp((x*log((-5*log(x)+5 *log(1/x))/x)+x^3+2)/x)/(x^2*log(x)-x^2*log(1/x)),x)
Output:
( - 10*e**((x**3 + 2)/x)*log(x))/x