3.64.4 \(\int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} (-6 e^5 x^4-18 e^5 x^2 \log (5))} \, dx\)

Optimal. Leaf size=28 \[ 5-\frac {3}{-x^2+3 \left (e^{\frac {1}{e^5 x}}-\log (5)\right )} \]

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Rubi [A]  time = 0.36, antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, integrand size = 99, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {3}{-x^2+3 e^{\frac {1}{e^5 x}}-\log (125)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-9*E^(1/(E^5*x)) - 6*E^5*x^3)/(9*E^(5 + 2/(E^5*x))*x^2 + E^5*x^6 + 6*E^5*x^4*Log[5] + 9*E^5*x^2*Log[5]^2
+ E^(1/(E^5*x))*(-6*E^5*x^4 - 18*E^5*x^2*Log[5])),x]

[Out]

-3/(3*E^(1/(E^5*x)) - x^2 - Log[125])

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 {3 \left (-3 e^{\frac {1}{e^5 x}}-2 e^5 x^3\right )}{e^5 x^2 \left (3 e^{\frac {1}{e^5 x}}-x^2-\log (125)\right )^2} \, dx\\ &=\frac {3 \int \frac {-3 e^{\frac {1}{e^5 x}}-2 e^5 x^3}{x^2 \left (3 e^{\frac {1}{e^5 x}}-x^2-\log (125)\right )^2} \, dx}{e^5}\\ &=-\frac {3}{3 e^{\frac {1}{e^5 x}}-x^2-\log (125)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 21, normalized size = 0.75 \begin {gather*} \frac {3}{-3 e^{\frac {1}{e^5 x}}+x^2+\log (125)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-9*E^(1/(E^5*x)) - 6*E^5*x^3)/(9*E^(5 + 2/(E^5*x))*x^2 + E^5*x^6 + 6*E^5*x^4*Log[5] + 9*E^5*x^2*Log
[5]^2 + E^(1/(E^5*x))*(-6*E^5*x^4 - 18*E^5*x^2*Log[5])),x]

[Out]

3/(-3*E^(1/(E^5*x)) + x^2 + Log[125])

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fricas [A]  time = 0.57, size = 21, normalized size = 0.75 \begin {gather*} \frac {3}{x^{2} - 3 \, e^{\left (\frac {e^{\left (-5\right )}}{x}\right )} + 3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^2+(-18*x^2*exp(5)*log(5)-6*x^4*exp(5
))*exp(1/x/exp(5))+9*x^2*exp(5)*log(5)^2+6*x^4*exp(5)*log(5)+x^6*exp(5)),x, algorithm="fricas")

[Out]

3/(x^2 - 3*e^(e^(-5)/x) + 3*log(5))

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giac [A]  time = 0.18, size = 28, normalized size = 1.00 \begin {gather*} -\frac {3}{x^{2} {\left (\frac {3 \, e^{\left (\frac {e^{\left (-5\right )}}{x}\right )}}{x^{2}} - \frac {3 \, \log \relax (5)}{x^{2}} - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^2+(-18*x^2*exp(5)*log(5)-6*x^4*exp(5
))*exp(1/x/exp(5))+9*x^2*exp(5)*log(5)^2+6*x^4*exp(5)*log(5)+x^6*exp(5)),x, algorithm="giac")

[Out]

-3/(x^2*(3*e^(e^(-5)/x)/x^2 - 3*log(5)/x^2 - 1))

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maple [A]  time = 1.86, size = 22, normalized size = 0.79

method	result	size
risch	\(\frac {3}{x^{2}+3 \ln \relax (5)-3 \,{\mathrm e}^{\frac {{\mathrm e}^{-5}}{x}}}\)	\(22\)
norman	\(\frac {3}{x^{2}+3 \ln \relax (5)-3 \,{\mathrm e}^{\frac {{\mathrm e}^{-5}}{x}}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^2+(-18*x^2*exp(5)*ln(5)-6*x^4*exp(5))*exp(
1/x/exp(5))+9*x^2*exp(5)*ln(5)^2+6*x^4*exp(5)*ln(5)+x^6*exp(5)),x,method=_RETURNVERBOSE)

[Out]

3/(x^2+3*ln(5)-3*exp(1/x*exp(-5)))

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maxima [A]  time = 0.49, size = 21, normalized size = 0.75 \begin {gather*} \frac {3}{x^{2} - 3 \, e^{\left (\frac {e^{\left (-5\right )}}{x}\right )} + 3 \, \log \relax (5)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^2+(-18*x^2*exp(5)*log(5)-6*x^4*exp(5
))*exp(1/x/exp(5))+9*x^2*exp(5)*log(5)^2+6*x^4*exp(5)*log(5)+x^6*exp(5)),x, algorithm="maxima")

[Out]

3/(x^2 - 3*e^(e^(-5)/x) + 3*log(5))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {9\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-5}}{x}}+6\,x^3\,{\mathrm {e}}^5}{x^6\,{\mathrm {e}}^5-{\mathrm {e}}^{\frac {{\mathrm {e}}^{-5}}{x}}\,\left (6\,{\mathrm {e}}^5\,x^4+18\,{\mathrm {e}}^5\,\ln \relax (5)\,x^2\right )+6\,x^4\,{\mathrm {e}}^5\,\ln \relax (5)+9\,x^2\,{\mathrm {e}}^5\,{\ln \relax (5)}^2+9\,x^2\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-5}}{x}}\,{\mathrm {e}}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(9*exp(exp(-5)/x) + 6*x^3*exp(5))/(x^6*exp(5) - exp(exp(-5)/x)*(6*x^4*exp(5) + 18*x^2*exp(5)*log(5)) + 6*
x^4*exp(5)*log(5) + 9*x^2*exp(5)*log(5)^2 + 9*x^2*exp((2*exp(-5))/x)*exp(5)),x)

[Out]

int(-(9*exp(exp(-5)/x) + 6*x^3*exp(5))/(x^6*exp(5) - exp(exp(-5)/x)*(6*x^4*exp(5) + 18*x^2*exp(5)*log(5)) + 6*
x^4*exp(5)*log(5) + 9*x^2*exp(5)*log(5)^2 + 9*x^2*exp((2*exp(-5))/x)*exp(5)), x)

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sympy [A]  time = 0.15, size = 20, normalized size = 0.71 \begin {gather*} - \frac {3}{- x^{2} + 3 e^{\frac {1}{x e^{5}}} - 3 \log {\relax (5 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-9*exp(1/x/exp(5))-6*x**3*exp(5))/(9*x**2*exp(5)*exp(1/x/exp(5))**2+(-18*x**2*exp(5)*ln(5)-6*x**4*e
xp(5))*exp(1/x/exp(5))+9*x**2*exp(5)*ln(5)**2+6*x**4*exp(5)*ln(5)+x**6*exp(5)),x)

[Out]

-3/(-x**2 + 3*exp(exp(-5)/x) - 3*log(5))

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