\(\int \frac {2 e^7+e^x (-2 e^7-2 x)+2 x+(2 e^x-2 x) \log (e^x-x)}{(e^x-x) \log ^2(e^x-x)} \, dx\) [5559]

Optimal result

Integrand size = 60, antiderivative size = 17 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \left (e^7+x\right )}{\log \left (e^x-x\right )} \]

[Out]

2*(exp(7)+x)/ln(exp(x)-x)

Rubi [F]

\[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx \]

[In]

Int[(2*E^7 + E^x*(-2*E^7 - 2*x) + 2*x + (2*E^x - 2*x)*Log[E^x - x])/((E^x - x)*Log[E^x - x]^2),x]

[Out]

-2*E^7*Defer[Int][Log[E^x - x]^(-2), x] + 2*E^7*Defer[Int][1/((E^x - x)*Log[E^x - x]^2), x] - 2*Defer[Int][x/L
og[E^x - x]^2, x] + 2*(1 - E^7)*Defer[Int][x/((E^x - x)*Log[E^x - x]^2), x] - 2*Defer[Int][x^2/((E^x - x)*Log[
E^x - x]^2), x] + 2*Defer[Int][Log[E^x - x]^(-1), x]

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {2 (-1+x) \left (e^7+x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )}-\frac {2 \left (e^7+x-\log \left (e^x-x\right )\right )}{\log ^2\left (e^x-x\right )}\right ) \, dx \\ & = -\left (2 \int \frac {(-1+x) \left (e^7+x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx\right )-2 \int \frac {e^7+x-\log \left (e^x-x\right )}{\log ^2\left (e^x-x\right )} \, dx \\ & = -\left (2 \int \left (-\frac {e^7}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )}+\frac {\left (-1+e^7\right ) x}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )}+\frac {x^2}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )}\right ) \, dx\right )-2 \int \left (\frac {e^7+x}{\log ^2\left (e^x-x\right )}-\frac {1}{\log \left (e^x-x\right )}\right ) \, dx \\ & = -\left (2 \int \frac {x^2}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx\right )-2 \int \frac {e^7+x}{\log ^2\left (e^x-x\right )} \, dx+2 \int \frac {1}{\log \left (e^x-x\right )} \, dx+\left (2 e^7\right ) \int \frac {1}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+\left (2 \left (1-e^7\right )\right ) \int \frac {x}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx \\ & = -\left (2 \int \left (\frac {e^7}{\log ^2\left (e^x-x\right )}+\frac {x}{\log ^2\left (e^x-x\right )}\right ) \, dx\right )-2 \int \frac {x^2}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+2 \int \frac {1}{\log \left (e^x-x\right )} \, dx+\left (2 e^7\right ) \int \frac {1}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+\left (2 \left (1-e^7\right )\right ) \int \frac {x}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx \\ & = -\left (2 \int \frac {x}{\log ^2\left (e^x-x\right )} \, dx\right )-2 \int \frac {x^2}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+2 \int \frac {1}{\log \left (e^x-x\right )} \, dx-\left (2 e^7\right ) \int \frac {1}{\log ^2\left (e^x-x\right )} \, dx+\left (2 e^7\right ) \int \frac {1}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx+\left (2 \left (1-e^7\right )\right ) \int \frac {x}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.00 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \left (e^7+x\right )}{\log \left (e^x-x\right )} \]

[In]

Integrate[(2*E^7 + E^x*(-2*E^7 - 2*x) + 2*x + (2*E^x - 2*x)*Log[E^x - x])/((E^x - x)*Log[E^x - x]^2),x]

[Out]

(2*(E^7 + x))/Log[E^x - x]

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{7}+2 x}{\ln \left ({\mathrm e}^{x}-x \right )}\)	\(16\)
norman	\(\frac {2 \,{\mathrm e}^{7}+2 x}{\ln \left ({\mathrm e}^{x}-x \right )}\)	\(19\)
parallelrisch	\(\frac {2 \,{\mathrm e}^{7}+2 x}{\ln \left ({\mathrm e}^{x}-x \right )}\)	\(19\)

[In]

int(((2*exp(x)-2*x)*ln(exp(x)-x)+(-2*exp(7)-2*x)*exp(x)+2*exp(7)+2*x)/(exp(x)-x)/ln(exp(x)-x)^2,x,method=_RETU
RNVERBOSE)

[Out]

2*(exp(7)+x)/ln(exp(x)-x)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \, {\left (x + e^{7}\right )}}{\log \left (-x + e^{x}\right )} \]

[In]

integrate(((2*exp(x)-2*x)*log(exp(x)-x)+(-2*exp(7)-2*x)*exp(x)+2*exp(7)+2*x)/(exp(x)-x)/log(exp(x)-x)^2,x, alg
orithm="fricas")

[Out]

2*(x + e^7)/log(-x + e^x)

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.82 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 x + 2 e^{7}}{\log {\left (- x + e^{x} \right )}} \]

[In]

integrate(((2*exp(x)-2*x)*ln(exp(x)-x)+(-2*exp(7)-2*x)*exp(x)+2*exp(7)+2*x)/(exp(x)-x)/ln(exp(x)-x)**2,x)

[Out]

(2*x + 2*exp(7))/log(-x + exp(x))

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \, {\left (x + e^{7}\right )}}{\log \left (-x + e^{x}\right )} \]

[In]

integrate(((2*exp(x)-2*x)*log(exp(x)-x)+(-2*exp(7)-2*x)*exp(x)+2*exp(7)+2*x)/(exp(x)-x)/log(exp(x)-x)^2,x, alg
orithm="maxima")

[Out]

2*(x + e^7)/log(-x + e^x)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2 \, {\left (x + e^{7}\right )}}{\log \left (-x + e^{x}\right )} \]

[In]

integrate(((2*exp(x)-2*x)*log(exp(x)-x)+(-2*exp(7)-2*x)*exp(x)+2*exp(7)+2*x)/(exp(x)-x)/log(exp(x)-x)^2,x, alg
orithm="giac")

[Out]

2*(x + e^7)/log(-x + e^x)

Mupad [B] (verification not implemented)

Time = 11.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.88 \[ \int \frac {2 e^7+e^x \left (-2 e^7-2 x\right )+2 x+\left (2 e^x-2 x\right ) \log \left (e^x-x\right )}{\left (e^x-x\right ) \log ^2\left (e^x-x\right )} \, dx=\frac {2\,\left (x+{\mathrm {e}}^7\right )}{\ln \left ({\mathrm {e}}^x-x\right )} \]

[In]

int(-(2*x + 2*exp(7) - exp(x)*(2*x + 2*exp(7)) - log(exp(x) - x)*(2*x - 2*exp(x)))/(log(exp(x) - x)^2*(x - exp
(x))),x)

[Out]

(2*(x + exp(7)))/log(exp(x) - x)