3.21.30 \(\int \frac {-\log (4)+(x \log (4)+e^{e^8+2 e^4 x+x^2} (2 e^4 x+2 x^2) \log (4)) \log (x)}{(16 x+e^{2 e^8+4 e^4 x+2 x^2} x-8 x^2+x^3+e^{e^8+2 e^4 x+x^2} (-8 x+2 x^2)) \log (x)+(8 x-2 e^{e^8+2 e^4 x+x^2} x-2 x^2) \log (x) \log (\log (x))+x \log (x) \log ^2(\log (x))} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{e^2}-\frac {\log (4)}{-4+e^{\left (e^4+x\right )^2}+x-\log (\log (x))} \]

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Rubi [A]  time = 1.23, antiderivative size = 24, normalized size of antiderivative = 0.89, number of steps used = 3, number of rules used = 3, integrand size = 150, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6688, 12, 6686} \begin {gather*} \frac {\log (4)}{-x-e^{\left (x+e^4\right )^2}+\log (\log (x))+4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-Log[4] + (x*Log[4] + E^(E^8 + 2*E^4*x + x^2)*(2*E^4*x + 2*x^2)*Log[4])*Log[x])/((16*x + E^(2*E^8 + 4*E^4
*x + 2*x^2)*x - 8*x^2 + x^3 + E^(E^8 + 2*E^4*x + x^2)*(-8*x + 2*x^2))*Log[x] + (8*x - 2*E^(E^8 + 2*E^4*x + x^2
)*x - 2*x^2)*Log[x]*Log[Log[x]] + x*Log[x]*Log[Log[x]]^2),x]

[Out]

Log[4]/(4 - E^(E^4 + x)^2 - x + Log[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 {\log (4) \left (-1+x \left (1+2 e^{4+e^8+2 e^4 x+x^2}+2 e^{\left (e^4+x\right )^2} x\right ) \log (x)\right )}{x \log (x) \left (4-e^{\left (e^4+x\right )^2}-x+\log (\log (x))\right )^2} \, dx\\ &=\log (4) \int \frac {-1+x \left (1+2 e^{4+e^8+2 e^4 x+x^2}+2 e^{\left (e^4+x\right )^2} x\right ) \log (x)}{x \log (x) \left (4-e^{\left (e^4+x\right )^2}-x+\log (\log (x))\right )^2} \, dx\\ &=\frac {\log (4)}{4-e^{\left (e^4+x\right )^2}-x+\log (\log (x))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.16, size = 23, normalized size = 0.85 \begin {gather*} -\frac {\log (4)}{-4+e^{\left (e^4+x\right )^2}+x-\log (\log (x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-Log[4] + (x*Log[4] + E^(E^8 + 2*E^4*x + x^2)*(2*E^4*x + 2*x^2)*Log[4])*Log[x])/((16*x + E^(2*E^8 +
4*E^4*x + 2*x^2)*x - 8*x^2 + x^3 + E^(E^8 + 2*E^4*x + x^2)*(-8*x + 2*x^2))*Log[x] + (8*x - 2*E^(E^8 + 2*E^4*x
+ x^2)*x - 2*x^2)*Log[x]*Log[Log[x]] + x*Log[x]*Log[Log[x]]^2),x]

[Out]

-(Log[4]/(-4 + E^(E^4 + x)^2 + x - Log[Log[x]]))

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fricas [A]  time = 0.89, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, \log \relax (2)}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \relax (x)\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x*exp(4)+2*x^2)*log(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*log(2))*log(x)-2*log(2))/(x*log(x)*lo
g(log(x))^2+(-2*x*exp(exp(4)^2+2*x*exp(4)+x^2)-2*x^2+8*x)*log(x)*log(log(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2
+(2*x^2-8*x)*exp(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*log(x)),x, algorithm="fricas")

[Out]

-2*log(2)/(x + e^(x^2 + 2*x*e^4 + e^8) - log(log(x)) - 4)

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giac [A]  time = 0.73, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, \log \relax (2)}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \relax (x)\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x*exp(4)+2*x^2)*log(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*log(2))*log(x)-2*log(2))/(x*log(x)*lo
g(log(x))^2+(-2*x*exp(exp(4)^2+2*x*exp(4)+x^2)-2*x^2+8*x)*log(x)*log(log(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2
+(2*x^2-8*x)*exp(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*log(x)),x, algorithm="giac")

[Out]

-2*log(2)/(x + e^(x^2 + 2*x*e^4 + e^8) - log(log(x)) - 4)

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maple [A]  time = 0.06, size = 27, normalized size = 1.00

method	result	size
risch	\(-\frac {2 \ln \relax (2)}{x +{\mathrm e}^{{\mathrm e}^{8}+2 x \,{\mathrm e}^{4}+x^{2}}-\ln \left (\ln \relax (x )\right )-4}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*(2*x*exp(4)+2*x^2)*ln(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*ln(2))*ln(x)-2*ln(2))/(x*ln(x)*ln(ln(x))^2+(
-2*x*exp(exp(4)^2+2*x*exp(4)+x^2)-2*x^2+8*x)*ln(x)*ln(ln(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2+(2*x^2-8*x)*exp
(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*ln(x)),x,method=_RETURNVERBOSE)

[Out]

-2*ln(2)/(x+exp(exp(8)+2*x*exp(4)+x^2)-ln(ln(x))-4)

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maxima [A]  time = 0.65, size = 26, normalized size = 0.96 \begin {gather*} -\frac {2 \, \log \relax (2)}{x + e^{\left (x^{2} + 2 \, x e^{4} + e^{8}\right )} - \log \left (\log \relax (x)\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x*exp(4)+2*x^2)*log(2)*exp(exp(4)^2+2*x*exp(4)+x^2)+2*x*log(2))*log(x)-2*log(2))/(x*log(x)*lo
g(log(x))^2+(-2*x*exp(exp(4)^2+2*x*exp(4)+x^2)-2*x^2+8*x)*log(x)*log(log(x))+(x*exp(exp(4)^2+2*x*exp(4)+x^2)^2
+(2*x^2-8*x)*exp(exp(4)^2+2*x*exp(4)+x^2)+x^3-8*x^2+16*x)*log(x)),x, algorithm="maxima")

[Out]

-2*log(2)/(x + e^(x^2 + 2*x*e^4 + e^8) - log(log(x)) - 4)

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mupad [B]  time = 1.57, size = 28, normalized size = 1.04 \begin {gather*} -\frac {2\,\ln \relax (2)}{x-\ln \left (\ln \relax (x)\right )+{\mathrm {e}}^{x^2}\,{\mathrm {e}}^{2\,x\,{\mathrm {e}}^4}\,{\mathrm {e}}^{{\mathrm {e}}^8}-4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*log(2) - log(x)*(2*x*log(2) + 2*exp(exp(8) + 2*x*exp(4) + x^2)*log(2)*(2*x*exp(4) + 2*x^2)))/(log(x)*(
16*x - exp(exp(8) + 2*x*exp(4) + x^2)*(8*x - 2*x^2) + x*exp(2*exp(8) + 4*x*exp(4) + 2*x^2) - 8*x^2 + x^3) - lo
g(log(x))*log(x)*(2*x*exp(exp(8) + 2*x*exp(4) + x^2) - 8*x + 2*x^2) + x*log(log(x))^2*log(x)),x)

[Out]

-(2*log(2))/(x - log(log(x)) + exp(x^2)*exp(2*x*exp(4))*exp(exp(8)) - 4)

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sympy [A]  time = 0.39, size = 29, normalized size = 1.07 \begin {gather*} - \frac {2 \log {\relax (2 )}}{x + e^{x^{2} + 2 x e^{4} + e^{8}} - \log {\left (\log {\relax (x )} \right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*(2*x*exp(4)+2*x**2)*ln(2)*exp(exp(4)**2+2*x*exp(4)+x**2)+2*x*ln(2))*ln(x)-2*ln(2))/(x*ln(x)*ln(l
n(x))**2+(-2*x*exp(exp(4)**2+2*x*exp(4)+x**2)-2*x**2+8*x)*ln(x)*ln(ln(x))+(x*exp(exp(4)**2+2*x*exp(4)+x**2)**2
+(2*x**2-8*x)*exp(exp(4)**2+2*x*exp(4)+x**2)+x**3-8*x**2+16*x)*ln(x)),x)

[Out]

-2*log(2)/(x + exp(x**2 + 2*x*exp(4) + exp(8)) - log(log(x)) - 4)

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