\(\int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log (\frac {e^{679} x-\log (x)}{x})+(e^{679} x \log (2)-\log (2) \log (x)) \log ^2(\frac {e^{679} x-\log (x)}{x})}{-e^{679} x+\log (x)} \, dx\) [4750]

Optimal result

Integrand size = 89, antiderivative size = 24 \[ \int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx=-3+x \log (2) \left (x-\log ^2\left (e^{679}-\frac {\log (x)}{x}\right )\right ) \]

[Out]

x*ln(2)*(x-ln(exp(679)-ln(x)/x)^2)-3

Rubi [F]

\[ \int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx=\int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx \]

[In]

Int[(-2*E^679*x^2*Log[2] + 2*x*Log[2]*Log[x] + (-2*Log[2] + 2*Log[2]*Log[x])*Log[(E^679*x - Log[x])/x] + (E^67
9*x*Log[2] - Log[2]*Log[x])*Log[(E^679*x - Log[x])/x]^2)/(-(E^679*x) + Log[x]),x]

[Out]

(x^2*Log[4])/2 + 2*Log[2]*Defer[Int][Log[E^679 - Log[x]/x]/(E^679*x - Log[x]), x] - 2*Log[2]*Defer[Int][(Log[x
]*Log[E^679 - Log[x]/x])/(E^679*x - Log[x]), x] - Log[2]*Defer[Int][Log[E^679 - Log[x]/x]^2, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (x \log (4)-\frac {2 \log (2) (-1+\log (x)) \log \left (e^{679}-\frac {\log (x)}{x}\right )}{e^{679} x-\log (x)}-\log (2) \log ^2\left (e^{679}-\frac {\log (x)}{x}\right )\right ) \, dx \\ & = \frac {1}{2} x^2 \log (4)-\log (2) \int \log ^2\left (e^{679}-\frac {\log (x)}{x}\right ) \, dx-(2 \log (2)) \int \frac {(-1+\log (x)) \log \left (e^{679}-\frac {\log (x)}{x}\right )}{e^{679} x-\log (x)} \, dx \\ & = \frac {1}{2} x^2 \log (4)-\log (2) \int \log ^2\left (e^{679}-\frac {\log (x)}{x}\right ) \, dx-(2 \log (2)) \int \left (-\frac {\log \left (e^{679}-\frac {\log (x)}{x}\right )}{e^{679} x-\log (x)}+\frac {\log (x) \log \left (e^{679}-\frac {\log (x)}{x}\right )}{e^{679} x-\log (x)}\right ) \, dx \\ & = \frac {1}{2} x^2 \log (4)-\log (2) \int \log ^2\left (e^{679}-\frac {\log (x)}{x}\right ) \, dx+(2 \log (2)) \int \frac {\log \left (e^{679}-\frac {\log (x)}{x}\right )}{e^{679} x-\log (x)} \, dx-(2 \log (2)) \int \frac {\log (x) \log \left (e^{679}-\frac {\log (x)}{x}\right )}{e^{679} x-\log (x)} \, dx \\ \end{align*}

Mathematica [F]

\[ \int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx=\int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx \]

[In]

Integrate[(-2*E^679*x^2*Log[2] + 2*x*Log[2]*Log[x] + (-2*Log[2] + 2*Log[2]*Log[x])*Log[(E^679*x - Log[x])/x] +
(E^679*x*Log[2] - Log[2]*Log[x])*Log[(E^679*x - Log[x])/x]^2)/(-(E^679*x) + Log[x]),x]

[Out]

Integrate[(-2*E^679*x^2*Log[2] + 2*x*Log[2]*Log[x] + (-2*Log[2] + 2*Log[2]*Log[x])*Log[(E^679*x - Log[x])/x] +
(E^679*x*Log[2] - Log[2]*Log[x])*Log[(E^679*x - Log[x])/x]^2)/(-(E^679*x) + Log[x]), x]

Maple [A] (verified)

Time = 1.49 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21

method	result	size
parallelrisch	\(-\ln \left (2\right ) x \ln \left (\frac {-\ln \left (x \right )+x \,{\mathrm e}^{679}}{x}\right )^{2}+x^{2} \ln \left (2\right )\)	\(29\)
risch	\(\text {Expression too large to display}\)	\(676\)

[In]

int(((-ln(2)*ln(x)+x*exp(679)*ln(2))*ln((-ln(x)+x*exp(679))/x)^2+(2*ln(2)*ln(x)-2*ln(2))*ln((-ln(x)+x*exp(679)
)/x)+2*x*ln(2)*ln(x)-2*x^2*exp(679)*ln(2))/(ln(x)-x*exp(679)),x,method=_RETURNVERBOSE)

[Out]

-ln(2)*x*ln((-ln(x)+x*exp(679))/x)^2+x^2*ln(2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx=-x \log \left (2\right ) \log \left (\frac {x e^{679} - \log \left (x\right )}{x}\right )^{2} + x^{2} \log \left (2\right ) \]

[In]

integrate(((-log(2)*log(x)+x*exp(679)*log(2))*log((-log(x)+x*exp(679))/x)^2+(2*log(2)*log(x)-2*log(2))*log((-l
og(x)+x*exp(679))/x)+2*x*log(2)*log(x)-2*x^2*exp(679)*log(2))/(log(x)-x*exp(679)),x, algorithm="fricas")

[Out]

-x*log(2)*log((x*e^679 - log(x))/x)^2 + x^2*log(2)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx=x^{2} \log {\left (2 \right )} - x \log {\left (2 \right )} \log {\left (\frac {x e^{679} - \log {\left (x \right )}}{x} \right )}^{2} \]

[In]

integrate(((-ln(2)*ln(x)+x*exp(679)*ln(2))*ln((-ln(x)+x*exp(679))/x)**2+(2*ln(2)*ln(x)-2*ln(2))*ln((-ln(x)+x*e
xp(679))/x)+2*x*ln(2)*ln(x)-2*x**2*exp(679)*ln(2))/(ln(x)-x*exp(679)),x)

[Out]

x**2*log(2) - x*log(2)*log((x*exp(679) - log(x))/x)**2

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).

Time = 0.29 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx=-x \log \left (2\right ) \log \left (x e^{679} - \log \left (x\right )\right )^{2} + 2 \, x \log \left (2\right ) \log \left (x e^{679} - \log \left (x\right )\right ) \log \left (x\right ) - x \log \left (2\right ) \log \left (x\right )^{2} + x^{2} \log \left (2\right ) \]

[In]

integrate(((-log(2)*log(x)+x*exp(679)*log(2))*log((-log(x)+x*exp(679))/x)^2+(2*log(2)*log(x)-2*log(2))*log((-l
og(x)+x*exp(679))/x)+2*x*log(2)*log(x)-2*x^2*exp(679)*log(2))/(log(x)-x*exp(679)),x, algorithm="maxima")

[Out]

-x*log(2)*log(x*e^679 - log(x))^2 + 2*x*log(2)*log(x*e^679 - log(x))*log(x) - x*log(2)*log(x)^2 + x^2*log(2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).

Time = 0.31 (sec) , antiderivative size = 50, normalized size of antiderivative = 2.08 \[ \int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx=-x \log \left (2\right ) \log \left (x e^{679} - \log \left (x\right )\right )^{2} + 2 \, x \log \left (2\right ) \log \left (x e^{679} - \log \left (x\right )\right ) \log \left (x\right ) - x \log \left (2\right ) \log \left (x\right )^{2} + x^{2} \log \left (2\right ) \]

[In]

integrate(((-log(2)*log(x)+x*exp(679)*log(2))*log((-log(x)+x*exp(679))/x)^2+(2*log(2)*log(x)-2*log(2))*log((-l
og(x)+x*exp(679))/x)+2*x*log(2)*log(x)-2*x^2*exp(679)*log(2))/(log(x)-x*exp(679)),x, algorithm="giac")

[Out]

-x*log(2)*log(x*e^679 - log(x))^2 + 2*x*log(2)*log(x*e^679 - log(x))*log(x) - x*log(2)*log(x)^2 + x^2*log(2)

Mupad [B] (verification not implemented)

Time = 11.54 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {-2 e^{679} x^2 \log (2)+2 x \log (2) \log (x)+(-2 \log (2)+2 \log (2) \log (x)) \log \left (\frac {e^{679} x-\log (x)}{x}\right )+\left (e^{679} x \log (2)-\log (2) \log (x)\right ) \log ^2\left (\frac {e^{679} x-\log (x)}{x}\right )}{-e^{679} x+\log (x)} \, dx=x\,\ln \left (2\right )\,\left (x-{\ln \left (-\frac {\ln \left (x\right )-x\,{\mathrm {e}}^{679}}{x}\right )}^2\right ) \]

[In]

int(-(log(-(log(x) - x*exp(679))/x)*(2*log(2) - 2*log(2)*log(x)) + log(-(log(x) - x*exp(679))/x)^2*(log(2)*log
(x) - x*exp(679)*log(2)) + 2*x^2*exp(679)*log(2) - 2*x*log(2)*log(x))/(log(x) - x*exp(679)),x)

[Out]

x*log(2)*(x - log(-(log(x) - x*exp(679))/x)^2)