\(\int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx\) [8790]

Optimal result

Integrand size = 34, antiderivative size = 12 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=\log (x)+x (-2+\log (1+\log (x))) \]

[Out]

ln(x)+(-2+ln(ln(x)+1))*x

Rubi [F]

\[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=\int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx \]

[In]

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

[Out]

-2*x + ExpIntegralEi[1 + Log[x]]/E + Log[x] + Defer[Int][Log[1 + Log[x]], x]

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

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=-2 x+\log (x)+x \log (1+\log (x)) \]

[In]

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

[Out]

-2*x + Log[x] + x*Log[1 + Log[x]]

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17

method	result	size
default	\(\ln \left (x \right )-2 x +\ln \left (\ln \left (x \right )+1\right ) x\)	\(14\)
norman	\(\ln \left (x \right )-2 x +\ln \left (\ln \left (x \right )+1\right ) x\)	\(14\)
risch	\(\ln \left (x \right )-2 x +\ln \left (\ln \left (x \right )+1\right ) x\)	\(14\)
parallelrisch	\(\ln \left (\ln \left (x \right )+1\right ) x -2 x +\ln \left (x \right )-\frac {1}{2}\)	\(15\)

[In]

int(((x*ln(x)+x)*ln(ln(x)+1)+(1-2*x)*ln(x)-x+1)/(x*ln(x)+x),x,method=_RETURNVERBOSE)

[Out]

ln(x)-2*x+ln(ln(x)+1)*x

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=x \log \left (\log \left (x\right ) + 1\right ) - 2 \, x + \log \left (x\right ) \]

[In]

integrate(((x*log(x)+x)*log(log(x)+1)+(1-2*x)*log(x)-x+1)/(x*log(x)+x),x, algorithm="fricas")

[Out]

x*log(log(x) + 1) - 2*x + log(x)

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.17 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=x \log {\left (\log {\left (x \right )} + 1 \right )} - 2 x + \log {\left (x \right )} \]

[In]

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

[Out]

x*log(log(x) + 1) - 2*x + log(x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.67 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx={\left (x - 1\right )} \log \left (\log \left (x\right ) + 1\right ) - 2 \, x + \log \left (x\right ) + \log \left (\log \left (x\right ) + 1\right ) \]

[In]

integrate(((x*log(x)+x)*log(log(x)+1)+(1-2*x)*log(x)-x+1)/(x*log(x)+x),x, algorithm="maxima")

[Out]

(x - 1)*log(log(x) + 1) - 2*x + log(x) + log(log(x) + 1)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=x \log \left (\log \left (x\right ) + 1\right ) - 2 \, x + \log \left (x\right ) \]

[In]

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

[Out]

x*log(log(x) + 1) - 2*x + log(x)

Mupad [B] (verification not implemented)

Time = 15.41 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.08 \[ \int \frac {1-x+(1-2 x) \log (x)+(x+x \log (x)) \log (1+\log (x))}{x+x \log (x)} \, dx=\ln \left (x\right )-2\,x+x\,\ln \left (\ln \left (x\right )+1\right ) \]

[In]

int(-(x + log(x)*(2*x - 1) - log(log(x) + 1)*(x + x*log(x)) - 1)/(x + x*log(x)),x)

[Out]

log(x) - 2*x + x*log(log(x) + 1)