\(\int \frac {e (-1-5 x)+x+x^2-4 x^3+(-4 e-4 x^2) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx\) [1256]

   Optimal result
   Rubi [F]
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 56, antiderivative size = 22 \[ \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx=\frac {(-e+x) (-2 (2+2 x)+\log (x+\log (x)))}{x} \]

[Out]

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

Rubi [F]

\[ \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx=\int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx \]

[In]

Int[(E*(-1 - 5*x) + x + x^2 - 4*x^3 + (-4*E - 4*x^2)*Log[x] + (E*x + E*Log[x])*Log[x + Log[x]])/(x^3 + x^2*Log
[x]),x]

[Out]

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

Rubi steps \begin{align*} \text {integral}& = \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^2 (x+\log (x))} \, dx \\ & = \int \left (\frac {-e+(1-5 e) x+x^2-4 x^3-4 e \log (x)-4 x^2 \log (x)}{x^2 (x+\log (x))}+\frac {e \log (x+\log (x))}{x^2}\right ) \, dx \\ & = e \int \frac {\log (x+\log (x))}{x^2} \, dx+\int \frac {-e+(1-5 e) x+x^2-4 x^3-4 e \log (x)-4 x^2 \log (x)}{x^2 (x+\log (x))} \, dx \\ & = e \int \frac {\log (x+\log (x))}{x^2} \, dx+\int \left (-\frac {4 \left (e+x^2\right )}{x^2}-\frac {(e-x) (1+x)}{x^2 (x+\log (x))}\right ) \, dx \\ & = -\left (4 \int \frac {e+x^2}{x^2} \, dx\right )+e \int \frac {\log (x+\log (x))}{x^2} \, dx-\int \frac {(e-x) (1+x)}{x^2 (x+\log (x))} \, dx \\ & = -\left (4 \int \left (1+\frac {e}{x^2}\right ) \, dx\right )+e \int \frac {\log (x+\log (x))}{x^2} \, dx-\int \left (\frac {1}{-x-\log (x)}+\frac {e}{x^2 (x+\log (x))}+\frac {-1+e}{x (x+\log (x))}\right ) \, dx \\ & = \frac {4 e}{x}-4 x-(-1+e) \int \frac {1}{x (x+\log (x))} \, dx-e \int \frac {1}{x^2 (x+\log (x))} \, dx+e \int \frac {\log (x+\log (x))}{x^2} \, dx-\int \frac {1}{-x-\log (x)} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.18 \[ \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx=\frac {4 e}{x}-4 x+\log (x+\log (x))-\frac {e \log (x+\log (x))}{x} \]

[In]

Integrate[(E*(-1 - 5*x) + x + x^2 - 4*x^3 + (-4*E - 4*x^2)*Log[x] + (E*x + E*Log[x])*Log[x + Log[x]])/(x^3 + x
^2*Log[x]),x]

[Out]

(4*E)/x - 4*x + Log[x + Log[x]] - (E*Log[x + Log[x]])/x

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.45

method result size
parallelrisch \(-\frac {{\mathrm e} \ln \left (x +\ln \left (x \right )\right )-\ln \left (x +\ln \left (x \right )\right ) x +4 x^{2}-4 \,{\mathrm e}}{x}\) \(32\)
risch \(-\frac {{\mathrm e} \ln \left (x +\ln \left (x \right )\right )}{x}+\frac {\ln \left (x +\ln \left (x \right )\right ) x -4 x^{2}+4 \,{\mathrm e}}{x}\) \(35\)

[In]

int(((exp(1)*ln(x)+x*exp(1))*ln(x+ln(x))+(-4*exp(1)-4*x^2)*ln(x)+(-5*x-1)*exp(1)-4*x^3+x^2+x)/(x^2*ln(x)+x^3),
x,method=_RETURNVERBOSE)

[Out]

-(exp(1)*ln(x+ln(x))-ln(x+ln(x))*x+4*x^2-4*exp(1))/x

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx=-\frac {4 \, x^{2} - {\left (x - e\right )} \log \left (x + \log \left (x\right )\right ) - 4 \, e}{x} \]

[In]

integrate(((exp(1)*log(x)+x*exp(1))*log(x+log(x))+(-4*exp(1)-4*x^2)*log(x)+(-5*x-1)*exp(1)-4*x^3+x^2+x)/(x^2*l
og(x)+x^3),x, algorithm="fricas")

[Out]

-(4*x^2 - (x - e)*log(x + log(x)) - 4*e)/x

Sympy [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.23 \[ \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx=- 4 x + \log {\left (x + \log {\left (x \right )} \right )} - \frac {e \log {\left (x + \log {\left (x \right )} \right )}}{x} + \frac {4 e}{x} \]

[In]

integrate(((exp(1)*ln(x)+x*exp(1))*ln(x+ln(x))+(-4*exp(1)-4*x**2)*ln(x)+(-5*x-1)*exp(1)-4*x**3+x**2+x)/(x**2*l
n(x)+x**3),x)

[Out]

-4*x + log(x + log(x)) - E*log(x + log(x))/x + 4*E/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx=-\frac {4 \, x^{2} - {\left (x - e\right )} \log \left (x + \log \left (x\right )\right ) - 4 \, e}{x} \]

[In]

integrate(((exp(1)*log(x)+x*exp(1))*log(x+log(x))+(-4*exp(1)-4*x^2)*log(x)+(-5*x-1)*exp(1)-4*x^3+x^2+x)/(x^2*l
og(x)+x^3),x, algorithm="maxima")

[Out]

-(4*x^2 - (x - e)*log(x + log(x)) - 4*e)/x

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.59 \[ \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx=-\frac {4 \, x^{2} + e \log \left (x + \log \left (x\right )\right ) - x \log \left (-x - \log \left (x\right )\right ) - 4 \, e}{x} \]

[In]

integrate(((exp(1)*log(x)+x*exp(1))*log(x+log(x))+(-4*exp(1)-4*x^2)*log(x)+(-5*x-1)*exp(1)-4*x^3+x^2+x)/(x^2*l
og(x)+x^3),x, algorithm="giac")

[Out]

-(4*x^2 + e*log(x + log(x)) - x*log(-x - log(x)) - 4*e)/x

Mupad [B] (verification not implemented)

Time = 8.75 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.27 \[ \int \frac {e (-1-5 x)+x+x^2-4 x^3+\left (-4 e-4 x^2\right ) \log (x)+(e x+e \log (x)) \log (x+\log (x))}{x^3+x^2 \log (x)} \, dx=\ln \left (x+\ln \left (x\right )\right )-4\,x+\frac {4\,\mathrm {e}}{x}-\frac {\ln \left (x+\ln \left (x\right )\right )\,\mathrm {e}}{x} \]

[In]

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

[Out]

log(x + log(x)) - 4*x + (4*exp(1))/x - (log(x + log(x))*exp(1))/x