\(\int \frac {(e^{2/x} (40-20 x)-10 x+(-2 x-4 e^{2/x} x) \log (\frac {x}{(x+2 e^{2/x} x)^5})) \log (\frac {1+\log (\frac {x}{(x+2 e^{2/x} x)^5})}{x})}{x^2+2 e^{2/x} x^2+(x^2+2 e^{2/x} x^2) \log (\frac {x}{(x+2 e^{2/x} x)^5})} \, dx\) [3295]

Optimal result

Integrand size = 126, antiderivative size = 26 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log ^2\left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right ) \]

[Out]

ln((ln(x/(2*x*exp(1/x)^2+x)^5)+1)/x)^2

Rubi [A] (verified)

Time = 1.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6873, 6874, 6818} \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log ^2\left (\frac {\log \left (\frac {x}{\left (2 e^{2/x} x+x\right )^5}\right )+1}{x}\right ) \]

[In]

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]

[Out]

Log[(1 + Log[x/(x + 2*E^(2/x)*x)^5])/x]^2

Rule 6818

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*(y^(m + 1)/(m + 1)), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6873

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps \begin{align*} \text {integral}& = \log ^2\left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.00 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log ^2\left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right ) \]

[In]

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]

[Out]

Log[(1 + Log[x/(x + 2*E^(2/x)*x)^5])/x]^2

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 33.45 (sec) , antiderivative size = 28642, normalized size of antiderivative = 1101.62

method	result	size
risch	\(\text {Expression too large to display}\)	\(28642\)

[In]

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)

[Out]

result too large to display

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (25) = 50\).

Time = 0.25 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.73 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log \left (\frac {\log \left (\frac {1}{32 \, x^{4} e^{\frac {10}{x}} + 80 \, x^{4} e^{\frac {8}{x}} + 80 \, x^{4} e^{\frac {6}{x}} + 40 \, x^{4} e^{\frac {4}{x}} + 10 \, x^{4} e^{\frac {2}{x}} + x^{4}}\right ) + 1}{x}\right )^{2} \]

[In]

integrate(((-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)*log(x/(2*x*exp(1/x)^2+x)^5)+2*x^2*exp(1/x)^2+x^2),x, algorithm="fric
as")

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.93 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\log {\left (\frac {\log {\left (\frac {x}{\left (2 x e^{\frac {2}{x}} + x\right )^{5}} \right )} + 1}{x} \right )}^{2} \]

[In]

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)

[Out]

log((log(x/(2*x*exp(2/x) + x)**5) + 1)/x)**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 112 vs. \(2 (25) = 50\).

Time = 0.33 (sec) , antiderivative size = 112, normalized size of antiderivative = 4.31 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=-2 \, \log \left (5\right ) \log \left (x\right ) - \log \left (x\right )^{2} - 2 \, {\left (\log \left (x\right ) - \log \left (\frac {4}{5} \, \log \left (x\right ) + \log \left (2 \, e^{\frac {2}{x}} + 1\right ) - \frac {1}{5}\right )\right )} \log \left (\frac {\log \left (\frac {x}{{\left (2 \, x e^{\frac {2}{x}} + x\right )}^{5}}\right ) + 1}{x}\right ) + 2 \, {\left (\log \left (5\right ) + \log \left (x\right )\right )} \log \left (4 \, \log \left (x\right ) + 5 \, \log \left (2 \, e^{\frac {2}{x}} + 1\right ) - 1\right ) - \log \left (4 \, \log \left (x\right ) + 5 \, \log \left (2 \, e^{\frac {2}{x}} + 1\right ) - 1\right )^{2} \]

[In]

integrate(((-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)*log(x/(2*x*exp(1/x)^2+x)^5)+2*x^2*exp(1/x)^2+x^2),x, algorithm="maxi
ma")

[Out]

-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

Giac [F]

\[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx=\int { -\frac {2 \, {\left (10 \, {\left (x - 2\right )} e^{\frac {2}{x}} + {\left (2 \, x e^{\frac {2}{x}} + x\right )} \log \left (\frac {x}{{\left (2 \, x e^{\frac {2}{x}} + x\right )}^{5}}\right ) + 5 \, x\right )} \log \left (\frac {\log \left (\frac {x}{{\left (2 \, x e^{\frac {2}{x}} + x\right )}^{5}}\right ) + 1}{x}\right )}{2 \, x^{2} e^{\frac {2}{x}} + x^{2} + {\left (2 \, x^{2} e^{\frac {2}{x}} + x^{2}\right )} \log \left (\frac {x}{{\left (2 \, x e^{\frac {2}{x}} + x\right )}^{5}}\right )} \,d x } \]

[In]

integrate(((-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)*log(x/(2*x*exp(1/x)^2+x)^5)+2*x^2*exp(1/x)^2+x^2),x, algorithm="giac
")

[Out]

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)

Mupad [B] (verification not implemented)

Time = 11.03 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {\left (e^{2/x} (40-20 x)-10 x+\left (-2 x-4 e^{2/x} x\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )\right ) \log \left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right )}{x^2+2 e^{2/x} x^2+\left (x^2+2 e^{2/x} x^2\right ) \log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )} \, dx={\ln \left (\frac {\ln \left (\frac {x}{{\left (x+2\,x\,{\mathrm {e}}^{2/x}\right )}^5}\right )+1}{x}\right )}^2 \]

[In]

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)

[Out]

log((log(x/(x + 2*x*exp(2/x))^5) + 1)/x)^2