3.44.21 \(\int \frac {-36+e^{-6+2 x} (-9+18 x)}{(16 x+4 e^{-6+2 x} x) \log (\frac {x}{4+e^{-6+2 x}})+(-16 x-4 e^{-6+2 x} x) \log (\frac {x}{4+e^{-6+2 x}}) \log (\frac {1}{5} \log (\frac {x}{4+e^{-6+2 x}}))+(4 x+e^{-6+2 x} x) \log (\frac {x}{4+e^{-6+2 x}}) \log ^2(\frac {1}{5} \log (\frac {x}{4+e^{-6+2 x}}))} \, dx\)

Optimal. Leaf size=25 \[ \frac {9}{-2+\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \]

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Rubi [A]  time = 0.60, antiderivative size = 27, normalized size of antiderivative = 1.08, number of steps used = 3, number of rules used = 3, integrand size = 145, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.021, Rules used = {6688, 12, 6686} \begin {gather*} -\frac {9}{2-\log \left (\frac {1}{5} \log \left (\frac {x}{e^{2 x-6}+4}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-36 + E^(-6 + 2*x)*(-9 + 18*x))/((16*x + 4*E^(-6 + 2*x)*x)*Log[x/(4 + E^(-6 + 2*x))] + (-16*x - 4*E^(-6 +
2*x)*x)*Log[x/(4 + E^(-6 + 2*x))]*Log[Log[x/(4 + E^(-6 + 2*x))]/5] + (4*x + E^(-6 + 2*x)*x)*Log[x/(4 + E^(-6
+ 2*x))]*Log[Log[x/(4 + E^(-6 + 2*x))]/5]^2),x]

[Out]

-9/(2 - Log[Log[x/(4 + E^(-6 + 2*x))]/5])

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 {9 \left (-4 e^6+e^{2 x} (-1+2 x)\right )}{\left (4 e^6+e^{2 x}\right ) x \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \left (2-\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )\right )^2} \, dx\\ &=9 \int \frac {-4 e^6+e^{2 x} (-1+2 x)}{\left (4 e^6+e^{2 x}\right ) x \log \left (\frac {x}{4+e^{-6+2 x}}\right ) \left (2-\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )\right )^2} \, dx\\ &=-\frac {9}{2-\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 25, normalized size = 1.00 \begin {gather*} \frac {9}{-2+\log \left (\frac {1}{5} \log \left (\frac {x}{4+e^{-6+2 x}}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-36 + E^(-6 + 2*x)*(-9 + 18*x))/((16*x + 4*E^(-6 + 2*x)*x)*Log[x/(4 + E^(-6 + 2*x))] + (-16*x - 4*E
^(-6 + 2*x)*x)*Log[x/(4 + E^(-6 + 2*x))]*Log[Log[x/(4 + E^(-6 + 2*x))]/5] + (4*x + E^(-6 + 2*x)*x)*Log[x/(4 +
E^(-6 + 2*x))]*Log[Log[x/(4 + E^(-6 + 2*x))]/5]^2),x]

[Out]

9/(-2 + Log[Log[x/(4 + E^(-6 + 2*x))]/5])

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fricas [A]  time = 0.71, size = 22, normalized size = 0.88 \begin {gather*} \frac {9}{\log \left (\frac {1}{5} \, \log \left (\frac {x}{e^{\left (2 \, x - 6\right )} + 4}\right )\right ) - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x-9)*exp(x-3)^2-36)/((x*exp(x-3)^2+4*x)*log(x/(exp(x-3)^2+4))*log(1/5*log(x/(exp(x-3)^2+4)))^2+
(-4*x*exp(x-3)^2-16*x)*log(x/(exp(x-3)^2+4))*log(1/5*log(x/(exp(x-3)^2+4)))+(4*x*exp(x-3)^2+16*x)*log(x/(exp(x
-3)^2+4))),x, algorithm="fricas")

[Out]

9/(log(1/5*log(x/(e^(2*x - 6) + 4))) - 2)

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giac [B]  time = 6.83, size = 129, normalized size = 5.16 \begin {gather*} -\frac {9 \, {\left (\log \left (\frac {x}{4 \, e^{6} + e^{\left (2 \, x\right )}}\right ) + 6\right )}}{\log \relax (5) \log \relax (x) - \log \relax (5) \log \left (4 \, e^{6} + e^{\left (2 \, x\right )}\right ) - \log \relax (x) \log \left (\log \left (\frac {x}{4 \, e^{6} + e^{\left (2 \, x\right )}}\right ) + 6\right ) + \log \left (4 \, e^{6} + e^{\left (2 \, x\right )}\right ) \log \left (\log \left (\frac {x}{4 \, e^{6} + e^{\left (2 \, x\right )}}\right ) + 6\right ) + 6 \, \log \relax (5) + 2 \, \log \relax (x) - 2 \, \log \left (4 \, e^{6} + e^{\left (2 \, x\right )}\right ) - 6 \, \log \left (\log \left (\frac {x}{4 \, e^{6} + e^{\left (2 \, x\right )}}\right ) + 6\right ) + 12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x-9)*exp(x-3)^2-36)/((x*exp(x-3)^2+4*x)*log(x/(exp(x-3)^2+4))*log(1/5*log(x/(exp(x-3)^2+4)))^2+
(-4*x*exp(x-3)^2-16*x)*log(x/(exp(x-3)^2+4))*log(1/5*log(x/(exp(x-3)^2+4)))+(4*x*exp(x-3)^2+16*x)*log(x/(exp(x
-3)^2+4))),x, algorithm="giac")

[Out]

-9*(log(x/(4*e^6 + e^(2*x))) + 6)/(log(5)*log(x) - log(5)*log(4*e^6 + e^(2*x)) - log(x)*log(log(x/(4*e^6 + e^(
2*x))) + 6) + log(4*e^6 + e^(2*x))*log(log(x/(4*e^6 + e^(2*x))) + 6) + 6*log(5) + 2*log(x) - 2*log(4*e^6 + e^(
2*x)) - 6*log(log(x/(4*e^6 + e^(2*x))) + 6) + 12)

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maple [C]  time = 0.15, size = 98, normalized size = 3.92

method	result	size
risch	\(\frac {9}{\ln \left (\frac {\ln \relax (x )}{5}-\frac {\ln \left ({\mathrm e}^{2 x -6}+4\right )}{5}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{2 x -6}+4}\right ) \left (-\mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{2 x -6}+4}\right )+\mathrm {csgn}\left (i x \right )\right ) \left (-\mathrm {csgn}\left (\frac {i x}{{\mathrm e}^{2 x -6}+4}\right )+\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 x -6}+4}\right )\right )}{10}\right )-2}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((18*x-9)*exp(x-3)^2-36)/((x*exp(x-3)^2+4*x)*ln(x/(exp(x-3)^2+4))*ln(1/5*ln(x/(exp(x-3)^2+4)))^2+(-4*x*exp
(x-3)^2-16*x)*ln(x/(exp(x-3)^2+4))*ln(1/5*ln(x/(exp(x-3)^2+4)))+(4*x*exp(x-3)^2+16*x)*ln(x/(exp(x-3)^2+4))),x,
method=_RETURNVERBOSE)

[Out]

9/(ln(1/5*ln(x)-1/5*ln(exp(2*x-6)+4)-1/10*I*Pi*csgn(I*x/(exp(2*x-6)+4))*(-csgn(I*x/(exp(2*x-6)+4))+csgn(I*x))*
(-csgn(I*x/(exp(2*x-6)+4))+csgn(I/(exp(2*x-6)+4))))-2)

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maxima [A]  time = 0.59, size = 27, normalized size = 1.08 \begin {gather*} -\frac {9}{\log \relax (5) - \log \left (\log \relax (x) - \log \left (4 \, e^{6} + e^{\left (2 \, x\right )}\right ) + 6\right ) + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x-9)*exp(x-3)^2-36)/((x*exp(x-3)^2+4*x)*log(x/(exp(x-3)^2+4))*log(1/5*log(x/(exp(x-3)^2+4)))^2+
(-4*x*exp(x-3)^2-16*x)*log(x/(exp(x-3)^2+4))*log(1/5*log(x/(exp(x-3)^2+4)))+(4*x*exp(x-3)^2+16*x)*log(x/(exp(x
-3)^2+4))),x, algorithm="maxima")

[Out]

-9/(log(5) - log(log(x) - log(4*e^6 + e^(2*x)) + 6) + 2)

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mupad [B]  time = 4.09, size = 25, normalized size = 1.00 \begin {gather*} \frac {9}{\ln \left (\frac {\ln \relax (x)}{5}-\frac {\ln \left ({\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^6\right )}{5}+\frac {6}{5}\right )-2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(2*x - 6)*(18*x - 9) - 36)/(log(x/(exp(2*x - 6) + 4))*(16*x + 4*x*exp(2*x - 6)) - log(log(x/(exp(2*x -
6) + 4))/5)*log(x/(exp(2*x - 6) + 4))*(16*x + 4*x*exp(2*x - 6)) + log(log(x/(exp(2*x - 6) + 4))/5)^2*log(x/(e
xp(2*x - 6) + 4))*(4*x + x*exp(2*x - 6))),x)

[Out]

9/(log(log(x)/5 - log(exp(2*x) + 4*exp(6))/5 + 6/5) - 2)

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sympy [A]  time = 0.81, size = 17, normalized size = 0.68 \begin {gather*} \frac {9}{\log {\left (\frac {\log {\left (\frac {x}{e^{2 x - 6} + 4} \right )}}{5} \right )} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((18*x-9)*exp(x-3)**2-36)/((x*exp(x-3)**2+4*x)*ln(x/(exp(x-3)**2+4))*ln(1/5*ln(x/(exp(x-3)**2+4)))**
2+(-4*x*exp(x-3)**2-16*x)*ln(x/(exp(x-3)**2+4))*ln(1/5*ln(x/(exp(x-3)**2+4)))+(4*x*exp(x-3)**2+16*x)*ln(x/(exp
(x-3)**2+4))),x)

[Out]

9/(log(log(x/(exp(2*x - 6) + 4))/5) - 2)

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