3.43.47 \(\int \frac {-32-2 x}{e^{3 e^2} (e^4 x+2 e^2 x^2+x^3)+e^{3 e^2} (32 e^2 x+32 x^2) \log (x)+256 e^{3 e^2} x \log ^2(x)} \, dx\)

Optimal. Leaf size=20 \[ \frac {2 e^{-3 e^2}}{e^2+x+16 \log (x)} \]

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Rubi [A]  time = 0.17, antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6688, 12, 6686} \begin {gather*} \frac {2 e^{-3 e^2}}{x+16 \log (x)+e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-32 - 2*x)/(E^(3*E^2)*(E^4*x + 2*E^2*x^2 + x^3) + E^(3*E^2)*(32*E^2*x + 32*x^2)*Log[x] + 256*E^(3*E^2)*x*
Log[x]^2),x]

[Out]

2/(E^(3*E^2)*(E^2 + x + 16*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 {2 e^{-3 e^2} (-16-x)}{x \left (e^2+x+16 \log (x)\right )^2} \, dx\\ &=\left (2 e^{-3 e^2}\right ) \int \frac {-16-x}{x \left (e^2+x+16 \log (x)\right )^2} \, dx\\ &=\frac {2 e^{-3 e^2}}{e^2+x+16 \log (x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 20, normalized size = 1.00 \begin {gather*} \frac {2 e^{-3 e^2}}{e^2+x+16 \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-32 - 2*x)/(E^(3*E^2)*(E^4*x + 2*E^2*x^2 + x^3) + E^(3*E^2)*(32*E^2*x + 32*x^2)*Log[x] + 256*E^(3*E
^2)*x*Log[x]^2),x]

[Out]

2/(E^(3*E^2)*(E^2 + x + 16*Log[x]))

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fricas [A]  time = 2.32, size = 24, normalized size = 1.20 \begin {gather*} \frac {2}{{\left (x + e^{2}\right )} e^{\left (3 \, e^{2}\right )} + 16 \, e^{\left (3 \, e^{2}\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-32)/(256*x*exp(3*exp(2))*log(x)^2+(32*exp(2)*x+32*x^2)*exp(3*exp(2))*log(x)+(x*exp(2)^2+2*x^2*
exp(2)+x^3)*exp(3*exp(2))),x, algorithm="fricas")

[Out]

2/((x + e^2)*e^(3*e^2) + 16*e^(3*e^2)*log(x))

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giac [A]  time = 0.22, size = 28, normalized size = 1.40 \begin {gather*} \frac {2}{x e^{\left (3 \, e^{2}\right )} + 16 \, e^{\left (3 \, e^{2}\right )} \log \relax (x) + e^{\left (3 \, e^{2} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-32)/(256*x*exp(3*exp(2))*log(x)^2+(32*exp(2)*x+32*x^2)*exp(3*exp(2))*log(x)+(x*exp(2)^2+2*x^2*
exp(2)+x^3)*exp(3*exp(2))),x, algorithm="giac")

[Out]

2/(x*e^(3*e^2) + 16*e^(3*e^2)*log(x) + e^(3*e^2 + 2))

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maple [A]  time = 0.12, size = 18, normalized size = 0.90

method	result	size
norman	\(\frac {2 \,{\mathrm e}^{-3 \,{\mathrm e}^{2}}}{16 \ln \relax (x )+x +{\mathrm e}^{2}}\)	\(18\)
risch	\(\frac {2 \,{\mathrm e}^{-3 \,{\mathrm e}^{2}}}{16 \ln \relax (x )+x +{\mathrm e}^{2}}\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*x-32)/(256*x*exp(3*exp(2))*ln(x)^2+(32*exp(2)*x+32*x^2)*exp(3*exp(2))*ln(x)+(x*exp(2)^2+2*x^2*exp(2)+x
^3)*exp(3*exp(2))),x,method=_RETURNVERBOSE)

[Out]

2/exp(exp(2))^3/(16*ln(x)+x+exp(2))

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maxima [A]  time = 0.38, size = 28, normalized size = 1.40 \begin {gather*} \frac {2}{x e^{\left (3 \, e^{2}\right )} + 16 \, e^{\left (3 \, e^{2}\right )} \log \relax (x) + e^{\left (3 \, e^{2} + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-32)/(256*x*exp(3*exp(2))*log(x)^2+(32*exp(2)*x+32*x^2)*exp(3*exp(2))*log(x)+(x*exp(2)^2+2*x^2*
exp(2)+x^3)*exp(3*exp(2))),x, algorithm="maxima")

[Out]

2/(x*e^(3*e^2) + 16*e^(3*e^2)*log(x) + e^(3*e^2 + 2))

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mupad [B]  time = 3.78, size = 40, normalized size = 2.00 \begin {gather*} \frac {2\,x^2}{x^3\,{\mathrm {e}}^{3\,{\mathrm {e}}^2}+16\,x^2\,{\mathrm {e}}^{3\,{\mathrm {e}}^2}\,\ln \relax (x)+x^2\,{\mathrm {e}}^{3\,{\mathrm {e}}^2}\,{\mathrm {e}}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + 32)/(exp(3*exp(2))*(x*exp(4) + 2*x^2*exp(2) + x^3) + exp(3*exp(2))*log(x)*(32*x*exp(2) + 32*x^2) +
256*x*exp(3*exp(2))*log(x)^2),x)

[Out]

(2*x^2)/(x^3*exp(3*exp(2)) + 16*x^2*exp(3*exp(2))*log(x) + x^2*exp(3*exp(2))*exp(2))

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sympy [A]  time = 0.13, size = 31, normalized size = 1.55 \begin {gather*} \frac {2}{x e^{3 e^{2}} + 16 e^{3 e^{2}} \log {\relax (x )} + e^{2} e^{3 e^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*x-32)/(256*x*exp(3*exp(2))*ln(x)**2+(32*exp(2)*x+32*x**2)*exp(3*exp(2))*ln(x)+(x*exp(2)**2+2*x**
2*exp(2)+x**3)*exp(3*exp(2))),x)

[Out]

2/(x*exp(3*exp(2)) + 16*exp(3*exp(2))*log(x) + exp(2)*exp(3*exp(2)))

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