Integrand size = 126, antiderivative size = 26 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log ^2\left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right ) \] Output:
ln((ln(x/(2*x*exp(1/x)^2+x)^5)+1)/x)^2
Time = 0.08 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log ^2\left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right ) \] Input:
Integrate[((E^(2/x)*(40 - 20*x) - 10*x + (-2*x - 4*E^(2/x)*x)*Log[x/(x + 2 *E^(2/x)*x)^5])*Log[(1 + Log[x/(x + 2*E^(2/x)*x)^5])/x])/(x^2 + 2*E^(2/x)* x^2 + (x^2 + 2*E^(2/x)*x^2)*Log[x/(x + 2*E^(2/x)*x)^5]),x]
Output:
Log[(1 + Log[x/(x + 2*E^(2/x)*x)^5])/x]^2
Time = 1.92 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.008, Rules used = {7237}
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.
\(\displaystyle \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-4 e^{2/x} x-2 x\right ) \log \left (\frac {x}{\left (2 e^{2/x} x+x\right )^5}\right )\right ) \log \left (\frac {\log \left (\frac {x}{\left (2 e^{2/x} x+x\right )^5}\right )+1}{x}\right )}{2 e^{2/x} x^2+x^2+\left (2 e^{2/x} x^2+x^2\right ) \log \left (\frac {x}{\left (2 e^{2/x} x+x\right )^5}\right )} \, dx\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle \log ^2\left (\frac {\log \left (\frac {x}{\left (2 e^{2/x} x+x\right )^5}\right )+1}{x}\right )\) |
Input:
Int[((E^(2/x)*(40 - 20*x) - 10*x + (-2*x - 4*E^(2/x)*x)*Log[x/(x + 2*E^(2/ x)*x)^5])*Log[(1 + Log[x/(x + 2*E^(2/x)*x)^5])/x])/(x^2 + 2*E^(2/x)*x^2 + (x^2 + 2*E^(2/x)*x^2)*Log[x/(x + 2*E^(2/x)*x)^5]),x]
Output:
Log[(1 + Log[x/(x + 2*E^(2/x)*x)^5])/x]^2
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 13.69 (sec) , antiderivative size = 28682, normalized size of antiderivative = 1103.15
Input:
int(((-4*x*exp(1/x)^2-2*x)*ln(x/(2*x*exp(1/x)^2+x)^5)+(-20*x+40)*exp(1/x)^ 2-10*x)*ln((ln(x/(2*x*exp(1/x)^2+x)^5)+1)/x)/((2*x^2*exp(1/x)^2+x^2)*ln(x/ (2*x*exp(1/x)^2+x)^5)+2*x^2*exp(1/x)^2+x^2),x,method=_RETURNVERBOSE)
Output:
result too large to display
Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (25) = 50\).
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log \left (\frac {\log \left (\frac {1}{32 \, x^{4} e^{\frac {10}{x}} + 80 \, x^{4} e^{\frac {8}{x}} + 80 \, x^{4} e^{\frac {6}{x}} + 40 \, x^{4} e^{\frac {4}{x}} + 10 \, x^{4} e^{\frac {2}{x}} + x^{4}}\right ) + 1}{x}\right )^{2} \] Input:
integrate(((-4*x*exp(1/x)^2-2*x)*log(x/(2*x*exp(1/x)^2+x)^5)+(-20*x+40)*ex p(1/x)^2-10*x)*log((log(x/(2*x*exp(1/x)^2+x)^5)+1)/x)/((2*x^2*exp(1/x)^2+x ^2)*log(x/(2*x*exp(1/x)^2+x)^5)+2*x^2*exp(1/x)^2+x^2),x, algorithm="fricas ")
Output:
log((log(1/(32*x^4*e^(10/x) + 80*x^4*e^(8/x) + 80*x^4*e^(6/x) + 40*x^4*e^( 4/x) + 10*x^4*e^(2/x) + x^4)) + 1)/x)^2
Time = 2.79 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log {\left (\frac {\log {\left (\frac {x}{\left (2 x e^{\frac {2}{x}} + x\right )^{5}} \right )} + 1}{x} \right )}^{2} \] Input:
integrate(((-4*x*exp(1/x)**2-2*x)*ln(x/(2*x*exp(1/x)**2+x)**5)+(-20*x+40)* exp(1/x)**2-10*x)*ln((ln(x/(2*x*exp(1/x)**2+x)**5)+1)/x)/((2*x**2*exp(1/x) **2+x**2)*ln(x/(2*x*exp(1/x)**2+x)**5)+2*x**2*exp(1/x)**2+x**2),x)
Output:
log((log(x/(2*x*exp(2/x) + x)**5) + 1)/x)**2
Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (25) = 50\).
Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 4.31 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=-2 \, \log \left (5\right ) \log \left (x\right ) - \log \left (x\right )^{2} - 2 \, {\left (\log \left (x\right ) - \log \left (\frac {4}{5} \, \log \left (x\right ) + \log \left (2 \, e^{\frac {2}{x}} + 1\right ) - \frac {1}{5}\right )\right )} \log \left (\frac {\log \left (\frac {x}{{\left (2 \, x e^{\frac {2}{x}} + x\right )}^{5}}\right ) + 1}{x}\right ) + 2 \, {\left (\log \left (5\right ) + \log \left (x\right )\right )} \log \left (4 \, \log \left (x\right ) + 5 \, \log \left (2 \, e^{\frac {2}{x}} + 1\right ) - 1\right ) - \log \left (4 \, \log \left (x\right ) + 5 \, \log \left (2 \, e^{\frac {2}{x}} + 1\right ) - 1\right )^{2} \] Input:
integrate(((-4*x*exp(1/x)^2-2*x)*log(x/(2*x*exp(1/x)^2+x)^5)+(-20*x+40)*ex p(1/x)^2-10*x)*log((log(x/(2*x*exp(1/x)^2+x)^5)+1)/x)/((2*x^2*exp(1/x)^2+x ^2)*log(x/(2*x*exp(1/x)^2+x)^5)+2*x^2*exp(1/x)^2+x^2),x, algorithm="maxima ")
Output:
-2*log(5)*log(x) - log(x)^2 - 2*(log(x) - log(4/5*log(x) + log(2*e^(2/x) + 1) - 1/5))*log((log(x/(2*x*e^(2/x) + x)^5) + 1)/x) + 2*(log(5) + log(x))* log(4*log(x) + 5*log(2*e^(2/x) + 1) - 1) - log(4*log(x) + 5*log(2*e^(2/x) + 1) - 1)^2
\[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\int { -\frac {2 \, {\left (10 \, {\left (x - 2\right )} e^{\frac {2}{x}} + {\left (2 \, x e^{\frac {2}{x}} + x\right )} \log \left (\frac {x}{{\left (2 \, x e^{\frac {2}{x}} + x\right )}^{5}}\right ) + 5 \, x\right )} \log \left (\frac {\log \left (\frac {x}{{\left (2 \, x e^{\frac {2}{x}} + x\right )}^{5}}\right ) + 1}{x}\right )}{2 \, x^{2} e^{\frac {2}{x}} + x^{2} + {\left (2 \, x^{2} e^{\frac {2}{x}} + x^{2}\right )} \log \left (\frac {x}{{\left (2 \, x e^{\frac {2}{x}} + x\right )}^{5}}\right )} \,d x } \] Input:
integrate(((-4*x*exp(1/x)^2-2*x)*log(x/(2*x*exp(1/x)^2+x)^5)+(-20*x+40)*ex p(1/x)^2-10*x)*log((log(x/(2*x*exp(1/x)^2+x)^5)+1)/x)/((2*x^2*exp(1/x)^2+x ^2)*log(x/(2*x*exp(1/x)^2+x)^5)+2*x^2*exp(1/x)^2+x^2),x, algorithm="giac")
Output:
integrate(-2*(10*(x - 2)*e^(2/x) + (2*x*e^(2/x) + x)*log(x/(2*x*e^(2/x) + x)^5) + 5*x)*log((log(x/(2*x*e^(2/x) + x)^5) + 1)/x)/(2*x^2*e^(2/x) + x^2 + (2*x^2*e^(2/x) + x^2)*log(x/(2*x*e^(2/x) + x)^5)), x)
Time = 3.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx={\ln \left (\frac {\ln \left (\frac {x}{{\left (x+2\,x\,{\mathrm {e}}^{2/x}\right )}^5}\right )+1}{x}\right )}^2 \] Input:
int(-(log((log(x/(x + 2*x*exp(2/x))^5) + 1)/x)*(10*x + exp(2/x)*(20*x - 40 ) + log(x/(x + 2*x*exp(2/x))^5)*(2*x + 4*x*exp(2/x))))/(log(x/(x + 2*x*exp (2/x))^5)*(2*x^2*exp(2/x) + x^2) + 2*x^2*exp(2/x) + x^2),x)
Output:
log((log(x/(x + 2*x*exp(2/x))^5) + 1)/x)^2
Time = 1.15 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.92 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx={\mathrm {log}\left (\frac {-\mathrm {log}\left (32 e^{\frac {10}{x}} x^{4}+80 e^{\frac {8}{x}} x^{4}+80 e^{\frac {6}{x}} x^{4}+40 e^{\frac {4}{x}} x^{4}+10 e^{\frac {2}{x}} x^{4}+x^{4}\right )+1}{x}\right )}^{2} \] Input:
int(((-4*x*exp(1/x)^2-2*x)*log(x/(2*x*exp(1/x)^2+x)^5)+(-20*x+40)*exp(1/x) ^2-10*x)*log((log(x/(2*x*exp(1/x)^2+x)^5)+1)/x)/((2*x^2*exp(1/x)^2+x^2)*lo g(x/(2*x*exp(1/x)^2+x)^5)+2*x^2*exp(1/x)^2+x^2),x)
Output:
log(( - log(32*e**(10/x)*x**4 + 80*e**(8/x)*x**4 + 80*e**(6/x)*x**4 + 40*e **(4/x)*x**4 + 10*e**(2/x)*x**4 + x**4) + 1)/x)**2