3.13.16 \(\int \frac {-2 x+e^{\log ^2(x)} (-1-2 \log (x))}{e^{4+2 \log ^2(x)} x^2+2 e^{4+\log ^2(x)} x^3+e^4 x^4} \, dx\)

Optimal. Leaf size=24 \[ \frac {x+\frac {4}{e^4 \left (e^{\log ^2(x)}+x\right )}}{4 x} \]

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Rubi [A]  time = 0.30, antiderivative size = 17, normalized size of antiderivative = 0.71, number of steps used = 3, number of rules used = 3, integrand size = 55, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.055, Rules used = {6688, 12, 6687} \begin {gather*} \frac {1}{e^4 x \left (x+e^{\log ^2(x)}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2*x + E^Log[x]^2*(-1 - 2*Log[x]))/(E^(4 + 2*Log[x]^2)*x^2 + 2*E^(4 + Log[x]^2)*x^3 + E^4*x^4),x]

[Out]

1/(E^4*x*(E^Log[x]^2 + 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 6687

Int[(u_)*(y_)^(m_.)*(z_)^(n_.), x_Symbol] :> With[{q = DerivativeDivides[y*z, u*z^(n - m), x]}, Simp[(q*y^(m +
1)*z^(m + 1))/(m + 1), x] /;  !FalseQ[q]] /; FreeQ[{m, n}, 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 {-e^{\log ^2(x)}-2 x-2 e^{\log ^2(x)} \log (x)}{e^4 x^2 \left (e^{\log ^2(x)}+x\right )^2} \, dx\\ &=\frac {\int \frac {-e^{\log ^2(x)}-2 x-2 e^{\log ^2(x)} \log (x)}{x^2 \left (e^{\log ^2(x)}+x\right )^2} \, dx}{e^4}\\ &=\frac {1}{e^4 x \left (e^{\log ^2(x)}+x\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.22, size = 17, normalized size = 0.71 \begin {gather*} \frac {1}{e^4 x \left (e^{\log ^2(x)}+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2*x + E^Log[x]^2*(-1 - 2*Log[x]))/(E^(4 + 2*Log[x]^2)*x^2 + 2*E^(4 + Log[x]^2)*x^3 + E^4*x^4),x]

[Out]

1/(E^4*x*(E^Log[x]^2 + x))

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fricas [A]  time = 0.56, size = 18, normalized size = 0.75 \begin {gather*} \frac {1}{x^{2} e^{4} + x e^{\left (\log \relax (x)^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)-1)*exp(log(x)^2)-2*x)/(x^2*exp(4)*exp(log(x)^2)^2+2*x^3*exp(4)*exp(log(x)^2)+x^4*exp(4))
,x, algorithm="fricas")

[Out]

1/(x^2*e^4 + x*e^(log(x)^2 + 4))

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giac [A]  time = 0.30, size = 18, normalized size = 0.75 \begin {gather*} \frac {1}{x^{2} e^{4} + x e^{\left (\log \relax (x)^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)-1)*exp(log(x)^2)-2*x)/(x^2*exp(4)*exp(log(x)^2)^2+2*x^3*exp(4)*exp(log(x)^2)+x^4*exp(4))
,x, algorithm="giac")

[Out]

1/(x^2*e^4 + x*e^(log(x)^2 + 4))

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maple [A]  time = 0.12, size = 16, normalized size = 0.67

method	result	size
risch	\(\frac {{\mathrm e}^{-4}}{x \left (x +{\mathrm e}^{\ln \relax (x )^{2}}\right )}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*ln(x)-1)*exp(ln(x)^2)-2*x)/(x^2*exp(4)*exp(ln(x)^2)^2+2*x^3*exp(4)*exp(ln(x)^2)+x^4*exp(4)),x,method=
_RETURNVERBOSE)

[Out]

1/x*exp(-4)/(x+exp(ln(x)^2))

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maxima [A]  time = 0.65, size = 18, normalized size = 0.75 \begin {gather*} \frac {1}{x^{2} e^{4} + x e^{\left (\log \relax (x)^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*log(x)-1)*exp(log(x)^2)-2*x)/(x^2*exp(4)*exp(log(x)^2)^2+2*x^3*exp(4)*exp(log(x)^2)+x^4*exp(4))
,x, algorithm="maxima")

[Out]

1/(x^2*e^4 + x*e^(log(x)^2 + 4))

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mupad [B]  time = 1.32, size = 18, normalized size = 0.75 \begin {gather*} \frac {1}{x^2\,{\mathrm {e}}^4+x\,{\mathrm {e}}^4\,{\mathrm {e}}^{{\ln \relax (x)}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + exp(log(x)^2)*(2*log(x) + 1))/(x^4*exp(4) + 2*x^3*exp(4)*exp(log(x)^2) + x^2*exp(2*log(x)^2)*exp(4
)),x)

[Out]

1/(x^2*exp(4) + x*exp(4)*exp(log(x)^2))

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sympy [A]  time = 0.25, size = 19, normalized size = 0.79 \begin {gather*} \frac {1}{x^{2} e^{4} + x e^{4} e^{\log {\relax (x )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*ln(x)-1)*exp(ln(x)**2)-2*x)/(x**2*exp(4)*exp(ln(x)**2)**2+2*x**3*exp(4)*exp(ln(x)**2)+x**4*exp(
4)),x)

[Out]

1/(x**2*exp(4) + x*exp(4)*exp(log(x)**2))

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