3.2.37 \(\int \frac {(e^{\frac {x^3}{2+x}} (48 x^2+16 x^3) \log (\log (2))+(-128-128 x-32 x^2) \log ^2(\log (2))) \log (\log (4))}{e^{\frac {2 x^3}{2+x}} (4+4 x+x^2)+e^{\frac {x^3}{2+x}} (-32 x-32 x^2-8 x^3) \log (\log (2))+(64 x^2+64 x^3+16 x^4) \log ^2(\log (2))} \, dx\)

Optimal. Leaf size=29 \[ \frac {2 \log (\log (4))}{x-\frac {e^{\frac {x^3}{2+x}}}{4 \log (\log (2))}} \]

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Rubi [A]  time = 0.46, antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 3, integrand size = 122, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.025, Rules used = {12, 6688, 6686} \begin {gather*} -\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{x+2}}-4 x \log (\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((E^(x^3/(2 + x))*(48*x^2 + 16*x^3)*Log[Log[2]] + (-128 - 128*x - 32*x^2)*Log[Log[2]]^2)*Log[Log[4]])/(E^(
(2*x^3)/(2 + x))*(4 + 4*x + x^2) + E^(x^3/(2 + x))*(-32*x - 32*x^2 - 8*x^3)*Log[Log[2]] + (64*x^2 + 64*x^3 + 1
6*x^4)*Log[Log[2]]^2),x]

[Out]

(-8*Log[Log[2]]*Log[Log[4]])/(E^(x^3/(2 + x)) - 4*x*Log[Log[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} &=\log (\log (4)) \int \frac {e^{\frac {x^3}{2+x}} \left (48 x^2+16 x^3\right ) \log (\log (2))+\left (-128-128 x-32 x^2\right ) \log ^2(\log (2))}{e^{\frac {2 x^3}{2+x}} \left (4+4 x+x^2\right )+e^{\frac {x^3}{2+x}} \left (-32 x-32 x^2-8 x^3\right ) \log (\log (2))+\left (64 x^2+64 x^3+16 x^4\right ) \log ^2(\log (2))} \, dx\\ &=\log (\log (4)) \int \frac {16 \log (\log (2)) \left (e^{\frac {x^3}{2+x}} x^2 (3+x)-2 (2+x)^2 \log (\log (2))\right )}{(2+x)^2 \left (e^{\frac {x^3}{2+x}}-4 x \log (\log (2))\right )^2} \, dx\\ &=(16 \log (\log (2)) \log (\log (4))) \int \frac {e^{\frac {x^3}{2+x}} x^2 (3+x)-2 (2+x)^2 \log (\log (2))}{(2+x)^2 \left (e^{\frac {x^3}{2+x}}-4 x \log (\log (2))\right )^2} \, dx\\ &=-\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{2+x}}-4 x \log (\log (2))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 28, normalized size = 0.97 \begin {gather*} -\frac {8 \log (\log (2)) \log (\log (4))}{e^{\frac {x^3}{2+x}}-4 x \log (\log (2))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((E^(x^3/(2 + x))*(48*x^2 + 16*x^3)*Log[Log[2]] + (-128 - 128*x - 32*x^2)*Log[Log[2]]^2)*Log[Log[4]]
)/(E^((2*x^3)/(2 + x))*(4 + 4*x + x^2) + E^(x^3/(2 + x))*(-32*x - 32*x^2 - 8*x^3)*Log[Log[2]] + (64*x^2 + 64*x
^3 + 16*x^4)*Log[Log[2]]^2),x]

[Out]

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

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fricas [A]  time = 0.55, size = 35, normalized size = 1.21 \begin {gather*} \frac {8 \, {\left (\log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2}\right )}}{4 \, x \log \left (\log \relax (2)\right ) - e^{\left (\frac {x^{3}}{x + 2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^2-128*x-128)*log(log(2))^2+(16*x^3+48*x^2)*exp(x^3/(2+x))*log(log(2)))*log(2*log(2))/((16*x^
4+64*x^3+64*x^2)*log(log(2))^2+(-8*x^3-32*x^2-32*x)*exp(x^3/(2+x))*log(log(2))+(x^2+4*x+4)*exp(x^3/(2+x))^2),x
, algorithm="fricas")

[Out]

8*(log(2)*log(log(2)) + log(log(2))^2)/(4*x*log(log(2)) - e^(x^3/(x + 2)))

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giac [A]  time = 0.38, size = 31, normalized size = 1.07 \begin {gather*} \frac {8 \, \log \left (2 \, \log \relax (2)\right ) \log \left (\log \relax (2)\right )}{4 \, x \log \left (\log \relax (2)\right ) - e^{\left (\frac {x^{3}}{x + 2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^2-128*x-128)*log(log(2))^2+(16*x^3+48*x^2)*exp(x^3/(2+x))*log(log(2)))*log(2*log(2))/((16*x^
4+64*x^3+64*x^2)*log(log(2))^2+(-8*x^3-32*x^2-32*x)*exp(x^3/(2+x))*log(log(2))+(x^2+4*x+4)*exp(x^3/(2+x))^2),x
, algorithm="giac")

[Out]

8*log(2*log(2))*log(log(2))/(4*x*log(log(2)) - e^(x^3/(x + 2)))

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maple [B]  time = 0.31, size = 58, normalized size = 2.00

method	result	size
risch	\(\frac {8 \ln \left (\ln \relax (2)\right ) \ln \relax (2)}{4 x \ln \left (\ln \relax (2)\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}+\frac {8 \ln \left (\ln \relax (2)\right )^{2}}{4 x \ln \left (\ln \relax (2)\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}}\)	\(58\)
norman	\(\frac {\left (8 \ln \left (\ln \relax (2)\right ) \ln \relax (2)+8 \ln \left (\ln \relax (2)\right )^{2}\right ) x +16 \ln \left (\ln \relax (2)\right ) \ln \relax (2)+16 \ln \left (\ln \relax (2)\right )^{2}}{\left (2+x \right ) \left (4 x \ln \left (\ln \relax (2)\right )-{\mathrm e}^{\frac {x^{3}}{2+x}}\right )}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-32*x^2-128*x-128)*ln(ln(2))^2+(16*x^3+48*x^2)*exp(x^3/(2+x))*ln(ln(2)))*ln(2*ln(2))/((16*x^4+64*x^3+64*
x^2)*ln(ln(2))^2+(-8*x^3-32*x^2-32*x)*exp(x^3/(2+x))*ln(ln(2))+(x^2+4*x+4)*exp(x^3/(2+x))^2),x,method=_RETURNV
ERBOSE)

[Out]

8*ln(ln(2))/(4*x*ln(ln(2))-exp(x^3/(2+x)))*ln(2)+8*ln(ln(2))^2/(4*x*ln(ln(2))-exp(x^3/(2+x)))

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maxima [A]  time = 0.64, size = 51, normalized size = 1.76 \begin {gather*} \frac {8 \, e^{\left (2 \, x + \frac {8}{x + 2}\right )} \log \left (2 \, \log \relax (2)\right ) \log \left (\log \relax (2)\right )}{4 \, x e^{\left (2 \, x + \frac {8}{x + 2}\right )} \log \left (\log \relax (2)\right ) - e^{\left (x^{2} + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x^2-128*x-128)*log(log(2))^2+(16*x^3+48*x^2)*exp(x^3/(2+x))*log(log(2)))*log(2*log(2))/((16*x^
4+64*x^3+64*x^2)*log(log(2))^2+(-8*x^3-32*x^2-32*x)*exp(x^3/(2+x))*log(log(2))+(x^2+4*x+4)*exp(x^3/(2+x))^2),x
, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.44, size = 29, normalized size = 1.00 \begin {gather*} -\frac {8\,\ln \left (2\,\ln \relax (2)\right )\,\ln \left (\ln \relax (2)\right )}{{\mathrm {e}}^{\frac {x^3}{x+2}}-4\,x\,\ln \left (\ln \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(2*log(2))*(log(log(2))^2*(128*x + 32*x^2 + 128) - exp(x^3/(x + 2))*log(log(2))*(48*x^2 + 16*x^3)))/(
exp((2*x^3)/(x + 2))*(4*x + x^2 + 4) + log(log(2))^2*(64*x^2 + 64*x^3 + 16*x^4) - exp(x^3/(x + 2))*log(log(2))
*(32*x + 32*x^2 + 8*x^3)),x)

[Out]

-(8*log(2*log(2))*log(log(2)))/(exp(x^3/(x + 2)) - 4*x*log(log(2)))

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sympy [A]  time = 0.25, size = 36, normalized size = 1.24 \begin {gather*} \frac {- 8 \log {\left (\log {\relax (2 )} \right )}^{2} - 8 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )}}{- 4 x \log {\left (\log {\relax (2 )} \right )} + e^{\frac {x^{3}}{x + 2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-32*x**2-128*x-128)*ln(ln(2))**2+(16*x**3+48*x**2)*exp(x**3/(2+x))*ln(ln(2)))*ln(2*ln(2))/((16*x**
4+64*x**3+64*x**2)*ln(ln(2))**2+(-8*x**3-32*x**2-32*x)*exp(x**3/(2+x))*ln(ln(2))+(x**2+4*x+4)*exp(x**3/(2+x))*
*2),x)

[Out]

(-8*log(log(2))**2 - 8*log(2)*log(log(2)))/(-4*x*log(log(2)) + exp(x**3/(x + 2)))

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