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

Optimal. Leaf size=9 \[ \log \left (\log \left (e^x+\frac {1}{x}\right )\right ) \]

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Rubi [A]  time = 0.09, antiderivative size = 13, normalized size of antiderivative = 1.44, number of steps used = 1, number of rules used = 1, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6684} \begin {gather*} \log \left (\log \left (\frac {e^x x+1}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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 6684

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

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 9, normalized size = 1.00 \begin {gather*} \log \left (\log \left (e^x+\frac {1}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[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)]]

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fricas [A]  time = 0.60, size = 12, normalized size = 1.33 \begin {gather*} \log \left (\log \left (\frac {x e^{x} + 1}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.15, size = 12, normalized size = 1.33 \begin {gather*} \log \left (\log \left (\frac {x e^{x} + 1}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maple [A]  time = 0.08, size = 13, normalized size = 1.44

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

Verification of antiderivative is not currently implemented for this CAS.

[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))

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maxima [A]  time = 0.41, size = 13, normalized size = 1.44 \begin {gather*} \log \left (\log \left (x e^{x} + 1\right ) - \log \relax (x)\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 4.87, size = 8, normalized size = 0.89 \begin {gather*} \ln \left (\ln \left ({\mathrm {e}}^x+\frac {1}{x}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 0.22, size = 10, normalized size = 1.11 \begin {gather*} \log {\left (\log {\left (\frac {x e^{x} + 1}{x} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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