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

Optimal. Leaf size=26 \[ \log ^2\left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 2.07, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 3, integrand size = 126, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {6741, 6742, 6686} \begin {gather*} \log ^2\left (\frac {\log \left (\frac {x}{\left (2 e^{2/x} x+x\right )^5}\right )+1}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[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 6686

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 6741

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

Rule 6742

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 26, normalized size = 1.00 \begin {gather*} \log ^2\left (\frac {1+\log \left (\frac {x}{\left (x+2 e^{2/x} x\right )^5}\right )}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[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

________________________________________________________________________________________

fricas [B]  time = 0.60, size = 71, normalized size = 2.73 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [C]  time = 3.47, size = 28486, normalized size = 1095.62

method	result	size
risch	\(\text {Expression too large to display}\)	\(28486\)

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

maxima [B]  time = 0.70, size = 112, normalized size = 4.31 \begin {gather*} -2 \, \log \relax (5) \log \relax (x) - \log \relax (x)^{2} - 2 \, {\left (\log \relax (x) - \log \left (\frac {4}{5} \, \log \relax (x) + \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 \relax (5) + \log \relax (x)\right )} \log \left (4 \, \log \relax (x) + 5 \, \log \left (2 \, e^{\frac {2}{x}} + 1\right ) - 1\right ) - \log \left (4 \, \log \relax (x) + 5 \, \log \left (2 \, e^{\frac {2}{x}} + 1\right ) - 1\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

mupad [B]  time = 3.69, size = 25, normalized size = 0.96 \begin {gather*} {\ln \left (\frac {\ln \left (\frac {x}{{\left (x+2\,x\,{\mathrm {e}}^{2/x}\right )}^5}\right )+1}{x}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

sympy [A]  time = 7.24, size = 20, normalized size = 0.77 \begin {gather*} \log {\left (\frac {\log {\left (\frac {x}{\left (2 x e^{\frac {2}{x}} + x\right )^{5}} \right )} + 1}{x} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________