3.21.6 \(\int \frac {-8+8 x}{e^{2 e^2} x+2 e^{e^2} x^2+x^3+(-2 e^{e^2} x-2 x^2) \log (x)+x \log ^2(x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {8}{-e^{e^2}-x+\log (x)} \]

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Rubi [A]  time = 0.30, antiderivative size = 15, normalized size of antiderivative = 0.88, number of steps used = 3, number of rules used = 3, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {8}{x-\log (x)+e^{e^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 + 8*x)/(E^(2*E^2)*x + 2*E^E^2*x^2 + x^3 + (-2*E^E^2*x - 2*x^2)*Log[x] + x*Log[x]^2),x]

[Out]

-8/(E^E^2 + x - Log[x])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {8 (-1+x)}{x \left (e^{e^2}+x-\log (x)\right )^2} \, dx\\ &=8 \int \frac {-1+x}{x \left (e^{e^2}+x-\log (x)\right )^2} \, dx\\ &=-\frac {8}{e^{e^2}+x-\log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 0.88 \begin {gather*} -\frac {8}{e^{e^2}+x-\log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 + 8*x)/(E^(2*E^2)*x + 2*E^E^2*x^2 + x^3 + (-2*E^E^2*x - 2*x^2)*Log[x] + x*Log[x]^2),x]

[Out]

-8/(E^E^2 + x - Log[x])

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fricas [A]  time = 0.73, size = 13, normalized size = 0.76 \begin {gather*} -\frac {8}{x + e^{\left (e^{2}\right )} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-8)/(x*log(x)^2+(-2*x*exp(exp(2))-2*x^2)*log(x)+x*exp(exp(2))^2+2*x^2*exp(exp(2))+x^3),x, algori
thm="fricas")

[Out]

-8/(x + e^(e^2) - log(x))

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giac [A]  time = 0.18, size = 13, normalized size = 0.76 \begin {gather*} -\frac {8}{x + e^{\left (e^{2}\right )} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-8)/(x*log(x)^2+(-2*x*exp(exp(2))-2*x^2)*log(x)+x*exp(exp(2))^2+2*x^2*exp(exp(2))+x^3),x, algori
thm="giac")

[Out]

-8/(x + e^(e^2) - log(x))

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maple [A]  time = 0.09, size = 14, normalized size = 0.82

method	result	size
norman	\(-\frac {8}{{\mathrm e}^{{\mathrm e}^{2}}+x -\ln \relax (x )}\)	\(14\)
risch	\(-\frac {8}{{\mathrm e}^{{\mathrm e}^{2}}+x -\ln \relax (x )}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x-8)/(x*ln(x)^2+(-2*x*exp(exp(2))-2*x^2)*ln(x)+x*exp(exp(2))^2+2*x^2*exp(exp(2))+x^3),x,method=_RETURNV
ERBOSE)

[Out]

-8/(exp(exp(2))+x-ln(x))

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maxima [A]  time = 0.70, size = 13, normalized size = 0.76 \begin {gather*} -\frac {8}{x + e^{\left (e^{2}\right )} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-8)/(x*log(x)^2+(-2*x*exp(exp(2))-2*x^2)*log(x)+x*exp(exp(2))^2+2*x^2*exp(exp(2))+x^3),x, algori
thm="maxima")

[Out]

-8/(x + e^(e^2) - log(x))

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mupad [B]  time = 1.50, size = 13, normalized size = 0.76 \begin {gather*} -\frac {8}{x+{\mathrm {e}}^{{\mathrm {e}}^2}-\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x - 8)/(x*exp(2*exp(2)) + 2*x^2*exp(exp(2)) + x*log(x)^2 - log(x)*(2*x*exp(exp(2)) + 2*x^2) + x^3),x)

[Out]

-8/(x + exp(exp(2)) - log(x))

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sympy [A]  time = 0.11, size = 10, normalized size = 0.59 \begin {gather*} \frac {8}{- x + \log {\relax (x )} - e^{e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*x-8)/(x*ln(x)**2+(-2*x*exp(exp(2))-2*x**2)*ln(x)+x*exp(exp(2))**2+2*x**2*exp(exp(2))+x**3),x)

[Out]

8/(-x + log(x) - exp(exp(2)))

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