3.99.80 \(\int \frac {-2+e^{1+e^x x} (-6 x+e^x (-6 x^2-6 x^3))}{27 e^{3+3 e^x x} x^4+27 e^{2+2 e^x x} x^3 \log (x)+9 e^{1+e^x x} x^2 \log ^2(x)+x \log ^3(x)} \, dx\)

Optimal. Leaf size=17 \[ \frac {1}{\left (3 e^{1+e^x x} x+\log (x)\right )^2} \]

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Rubi [A]  time = 0.43, antiderivative size = 17, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 91, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {6688, 12, 6686} \begin {gather*} \frac {1}{\left (3 e^{e^x x+1} x+\log (x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-2 + E^(1 + E^x*x)*(-6*x + E^x*(-6*x^2 - 6*x^3)))/(27*E^(3 + 3*E^x*x)*x^4 + 27*E^(2 + 2*E^x*x)*x^3*Log[x]
+ 9*E^(1 + E^x*x)*x^2*Log[x]^2 + x*Log[x]^3),x]

[Out]

(3*E^(1 + E^x*x)*x + Log[x])^(-2)

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 \left (-1-3 e^{1+e^x x} x-3 e^{1+x+e^x x} x^2 (1+x)\right )}{x \left (3 e^{1+e^x x} x+\log (x)\right )^3} \, dx\\ &=2 \int \frac {-1-3 e^{1+e^x x} x-3 e^{1+x+e^x x} x^2 (1+x)}{x \left (3 e^{1+e^x x} x+\log (x)\right )^3} \, dx\\ &=\frac {1}{\left (3 e^{1+e^x x} x+\log (x)\right )^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 17, normalized size = 1.00 \begin {gather*} \frac {1}{\left (3 e^{1+e^x x} x+\log (x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-2 + E^(1 + E^x*x)*(-6*x + E^x*(-6*x^2 - 6*x^3)))/(27*E^(3 + 3*E^x*x)*x^4 + 27*E^(2 + 2*E^x*x)*x^3*
Log[x] + 9*E^(1 + E^x*x)*x^2*Log[x]^2 + x*Log[x]^3),x]

[Out]

(3*E^(1 + E^x*x)*x + Log[x])^(-2)

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fricas [B]  time = 0.70, size = 32, normalized size = 1.88 \begin {gather*} \frac {1}{9 \, x^{2} e^{\left (2 \, x e^{x} + 2\right )} + 6 \, x e^{\left (x e^{x} + 1\right )} \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^3-6*x^2)*exp(x)-6*x)*exp(exp(x)*x+1)-2)/(27*x^4*exp(exp(x)*x+1)^3+27*x^3*log(x)*exp(exp(x)*x
+1)^2+9*x^2*log(x)^2*exp(exp(x)*x+1)+x*log(x)^3),x, algorithm="fricas")

[Out]

1/(9*x^2*e^(2*x*e^x + 2) + 6*x*e^(x*e^x + 1)*log(x) + log(x)^2)

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giac [B]  time = 0.46, size = 32, normalized size = 1.88 \begin {gather*} \frac {1}{9 \, x^{2} e^{\left (2 \, x e^{x} + 2\right )} + 6 \, x e^{\left (x e^{x} + 1\right )} \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^3-6*x^2)*exp(x)-6*x)*exp(exp(x)*x+1)-2)/(27*x^4*exp(exp(x)*x+1)^3+27*x^3*log(x)*exp(exp(x)*x
+1)^2+9*x^2*log(x)^2*exp(exp(x)*x+1)+x*log(x)^3),x, algorithm="giac")

[Out]

1/(9*x^2*e^(2*x*e^x + 2) + 6*x*e^(x*e^x + 1)*log(x) + log(x)^2)

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maple [A]  time = 0.07, size = 16, normalized size = 0.94

method	result	size
risch	\(\frac {1}{\left (3 \,{\mathrm e}^{{\mathrm e}^{x} x +1} x +\ln \relax (x )\right )^{2}}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-6*x^3-6*x^2)*exp(x)-6*x)*exp(exp(x)*x+1)-2)/(27*x^4*exp(exp(x)*x+1)^3+27*x^3*ln(x)*exp(exp(x)*x+1)^2+9
*x^2*ln(x)^2*exp(exp(x)*x+1)+x*ln(x)^3),x,method=_RETURNVERBOSE)

[Out]

1/(3*exp(exp(x)*x+1)*x+ln(x))^2

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maxima [B]  time = 0.45, size = 32, normalized size = 1.88 \begin {gather*} \frac {1}{9 \, x^{2} e^{\left (2 \, x e^{x} + 2\right )} + 6 \, x e^{\left (x e^{x} + 1\right )} \log \relax (x) + \log \relax (x)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x^3-6*x^2)*exp(x)-6*x)*exp(exp(x)*x+1)-2)/(27*x^4*exp(exp(x)*x+1)^3+27*x^3*log(x)*exp(exp(x)*x
+1)^2+9*x^2*log(x)^2*exp(exp(x)*x+1)+x*log(x)^3),x, algorithm="maxima")

[Out]

1/(9*x^2*e^(2*x*e^x + 2) + 6*x*e^(x*e^x + 1)*log(x) + log(x)^2)

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mupad [B]  time = 5.81, size = 32, normalized size = 1.88 \begin {gather*} \frac {1}{{\ln \relax (x)}^2+9\,x^2\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^x}\,{\mathrm {e}}^2+6\,x\,{\mathrm {e}}^{x\,{\mathrm {e}}^x}\,\mathrm {e}\,\ln \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x*exp(x) + 1)*(6*x + exp(x)*(6*x^2 + 6*x^3)) + 2)/(x*log(x)^3 + 27*x^4*exp(3*x*exp(x) + 3) + 27*x^3*
exp(2*x*exp(x) + 2)*log(x) + 9*x^2*exp(x*exp(x) + 1)*log(x)^2),x)

[Out]

1/(log(x)^2 + 9*x^2*exp(2*x*exp(x))*exp(2) + 6*x*exp(x*exp(x))*exp(1)*log(x))

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sympy [B]  time = 0.42, size = 36, normalized size = 2.12 \begin {gather*} \frac {1}{9 x^{2} e^{2 x e^{x} + 2} + 6 x e^{x e^{x} + 1} \log {\relax (x )} + \log {\relax (x )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x**3-6*x**2)*exp(x)-6*x)*exp(exp(x)*x+1)-2)/(27*x**4*exp(exp(x)*x+1)**3+27*x**3*ln(x)*exp(exp(
x)*x+1)**2+9*x**2*ln(x)**2*exp(exp(x)*x+1)+x*ln(x)**3),x)

[Out]

1/(9*x**2*exp(2*x*exp(x) + 2) + 6*x*exp(x*exp(x) + 1)*log(x) + log(x)**2)

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