\(\int \frac {-1+e^x x^2}{(x+e^x x^2) \log (\frac {1+e^x x}{x})} \, dx\) [7532]

Optimal result

Integrand size = 35, antiderivative size = 9 \[ \int \frac {-1+e^x x^2}{\left (x+e^x x^2\right ) \log \left (\frac {1+e^x x}{x}\right )} \, dx=\log \left (\log \left (e^x+\frac {1}{x}\right )\right ) \]

[Out]

ln(ln(exp(x)+1/x))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.44, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6816} \[ \int \frac {-1+e^x x^2}{\left (x+e^x x^2\right ) \log \left (\frac {1+e^x x}{x}\right )} \, dx=\log \left (\log \left (\frac {e^x x+1}{x}\right )\right ) \]

[In]

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

[Out]

Log[Log[(1 + E^x*x)/x]]

Rule 6816

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 9, normalized size of antiderivative = 1.00 \[ \int \frac {-1+e^x x^2}{\left (x+e^x x^2\right ) \log \left (\frac {1+e^x x}{x}\right )} \, dx=\log \left (\log \left (e^x+\frac {1}{x}\right )\right ) \]

[In]

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

[Out]

Log[Log[E^x + x^(-1)]]

Maple [A] (verified)

Time = 0.15 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.44

method	result	size
norman	\(\ln \left (\ln \left (\frac {{\mathrm e}^{x} x +1}{x}\right )\right )\)	\(13\)
parallelrisch	\(\ln \left (\ln \left (\frac {{\mathrm e}^{x} x +1}{x}\right )\right )\)	\(13\)
risch	\(\ln \left (\ln \left ({\mathrm e}^{x} x +1\right )+\frac {i \left (\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x +1\right )\right ) {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +1\right )}{x}\right )}^{2}-\pi \,\operatorname {csgn}\left (i \left ({\mathrm e}^{x} x +1\right )\right ) \operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +1\right )}{x}\right ) \operatorname {csgn}\left (\frac {i}{x}\right )-\pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +1\right )}{x}\right )}^{3}+\pi {\operatorname {csgn}\left (\frac {i \left ({\mathrm e}^{x} x +1\right )}{x}\right )}^{2} \operatorname {csgn}\left (\frac {i}{x}\right )+2 i \ln \left (x \right )\right )}{2}\right )\)	\(121\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.33 \[ \int \frac {-1+e^x x^2}{\left (x+e^x x^2\right ) \log \left (\frac {1+e^x x}{x}\right )} \, dx=\log \left (\log \left (\frac {x e^{x} + 1}{x}\right )\right ) \]

[In]

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

[Out]

log(log((x*e^x + 1)/x))

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 10, normalized size of antiderivative = 1.11 \[ \int \frac {-1+e^x x^2}{\left (x+e^x x^2\right ) \log \left (\frac {1+e^x x}{x}\right )} \, dx=\log {\left (\log {\left (\frac {x e^{x} + 1}{x} \right )} \right )} \]

[In]

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

[Out]

log(log((x*exp(x) + 1)/x))

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.44 \[ \int \frac {-1+e^x x^2}{\left (x+e^x x^2\right ) \log \left (\frac {1+e^x x}{x}\right )} \, dx=\log \left (\log \left (x e^{x} + 1\right ) - \log \left (x\right )\right ) \]

[In]

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

[Out]

log(log(x*e^x + 1) - log(x))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 12, normalized size of antiderivative = 1.33 \[ \int \frac {-1+e^x x^2}{\left (x+e^x x^2\right ) \log \left (\frac {1+e^x x}{x}\right )} \, dx=\log \left (\log \left (\frac {x e^{x} + 1}{x}\right )\right ) \]

[In]

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

[Out]

log(log((x*e^x + 1)/x))

Mupad [B] (verification not implemented)

Time = 13.06 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.89 \[ \int \frac {-1+e^x x^2}{\left (x+e^x x^2\right ) \log \left (\frac {1+e^x x}{x}\right )} \, dx=\ln \left (\ln \left ({\mathrm {e}}^x+\frac {1}{x}\right )\right ) \]

[In]

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

[Out]

log(log(exp(x) + 1/x))