\(\int \frac {(-1+x) \log (x)+(-4-x+(-3-x) \log (x)+\log ^2(x)) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx\) [7251]

Optimal result

Integrand size = 53, antiderivative size = 16 \[ \int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx=-1+\frac {x \log (x)}{\log (4+x-\log (x))} \]

[Out]

x/ln(-ln(x)+4+x)*ln(x)-1

Rubi [F]

\[ \int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx=\int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx \]

[In]

Int[((-1 + x)*Log[x] + (-4 - x + (-3 - x)*Log[x] + Log[x]^2)*Log[4 + x - Log[x]])/((-4 - x + Log[x])*Log[4 + x
- Log[x]]^2),x]

[Out]

Defer[Int][Log[x]/((4 + x - Log[x])*Log[4 + x - Log[x]]^2), x] - Defer[Int][(x*Log[x])/((4 + x - Log[x])*Log[4
+ x - Log[x]]^2), x] + Defer[Int][Log[4 + x - Log[x]]^(-1), x] + Defer[Int][Log[x]/Log[4 + x - Log[x]], x]

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

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx=\frac {x \log (x)}{\log (4+x-\log (x))} \]

[In]

Integrate[((-1 + x)*Log[x] + (-4 - x + (-3 - x)*Log[x] + Log[x]^2)*Log[4 + x - Log[x]])/((-4 - x + Log[x])*Log
[4 + x - Log[x]]^2),x]

[Out]

(x*Log[x])/Log[4 + x - Log[x]]

Maple [A] (verified)

Time = 0.81 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {x \ln \left (x \right )}{\ln \left (-\ln \left (x \right )+4+x \right )}\)	\(15\)
parallelrisch	\(\frac {x \ln \left (x \right )}{\ln \left (-\ln \left (x \right )+4+x \right )}\)	\(15\)

[In]

int(((ln(x)^2+(-3-x)*ln(x)-x-4)*ln(-ln(x)+4+x)+(-1+x)*ln(x))/(ln(x)-x-4)/ln(-ln(x)+4+x)^2,x,method=_RETURNVERB
OSE)

[Out]

x/ln(-ln(x)+4+x)*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx=\frac {x \log \left (x\right )}{\log \left (x - \log \left (x\right ) + 4\right )} \]

[In]

integrate(((log(x)^2+(-3-x)*log(x)-x-4)*log(-log(x)+4+x)+(-1+x)*log(x))/(log(x)-x-4)/log(-log(x)+4+x)^2,x, alg
orithm="fricas")

[Out]

x*log(x)/log(x - log(x) + 4)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.75 \[ \int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx=\frac {x \log {\left (x \right )}}{\log {\left (x - \log {\left (x \right )} + 4 \right )}} \]

[In]

integrate(((ln(x)**2+(-3-x)*ln(x)-x-4)*ln(-ln(x)+4+x)+(-1+x)*ln(x))/(ln(x)-x-4)/ln(-ln(x)+4+x)**2,x)

[Out]

x*log(x)/log(x - log(x) + 4)

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx=\frac {x \log \left (x\right )}{\log \left (x - \log \left (x\right ) + 4\right )} \]

[In]

integrate(((log(x)^2+(-3-x)*log(x)-x-4)*log(-log(x)+4+x)+(-1+x)*log(x))/(log(x)-x-4)/log(-log(x)+4+x)^2,x, alg
orithm="maxima")

[Out]

x*log(x)/log(x - log(x) + 4)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx=\frac {x \log \left (x\right )}{\log \left (x - \log \left (x\right ) + 4\right )} \]

[In]

integrate(((log(x)^2+(-3-x)*log(x)-x-4)*log(-log(x)+4+x)+(-1+x)*log(x))/(log(x)-x-4)/log(-log(x)+4+x)^2,x, alg
orithm="giac")

[Out]

x*log(x)/log(x - log(x) + 4)

Mupad [B] (verification not implemented)

Time = 12.52 (sec) , antiderivative size = 82, normalized size of antiderivative = 5.12 \[ \int \frac {(-1+x) \log (x)+\left (-4-x+(-3-x) \log (x)+\log ^2(x)\right ) \log (4+x-\log (x))}{(-4-x+\log (x)) \log ^2(4+x-\log (x))} \, dx=x+3\,\ln \left (x\right )+\frac {5}{x-1}-{\ln \left (x\right )}^2\,\left (\frac {1}{x-1}+1\right )+\frac {x\,\ln \left (x\right )-\frac {x\,\ln \left (x-\ln \left (x\right )+4\right )\,\left (\ln \left (x\right )+1\right )\,\left (x-\ln \left (x\right )+4\right )}{x-1}}{\ln \left (x-\ln \left (x\right )+4\right )}+\frac {\ln \left (x\right )\,\left (x^2+3\right )}{x-1} \]

[In]

int(-(log(x)*(x - 1) - log(x - log(x) + 4)*(x + log(x)*(x + 3) - log(x)^2 + 4))/(log(x - log(x) + 4)^2*(x - lo
g(x) + 4)),x)

[Out]

x + 3*log(x) + 5/(x - 1) - log(x)^2*(1/(x - 1) + 1) + (x*log(x) - (x*log(x - log(x) + 4)*(log(x) + 1)*(x - log
(x) + 4))/(x - 1))/log(x - log(x) + 4) + (log(x)*(x^2 + 3))/(x - 1)