3.32.62 \(\int \frac {-15 e^{2 x}+e^x (4-4 x)+(-12+90 e^x) \log (\log (4))-135 \log ^2(\log (4))}{e^{2 x}-6 e^x \log (\log (4))+9 \log ^2(\log (4))} \, dx\)

Optimal. Leaf size=22 \[ x \left (-15-\frac {1}{x}+\frac {4}{e^x-3 \log (\log (4))}\right ) \]

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Rubi [B]  time = 0.57, antiderivative size = 129, normalized size of antiderivative = 5.86, number of steps used = 17, number of rules used = 12, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.203, Rules used = {6741, 6742, 2184, 2190, 2279, 2391, 2185, 2191, 2282, 36, 29, 31} \begin {gather*} -\frac {2 x^2}{3 \log (\log (4))}-15 x+\frac {2 (1-x)^2}{3 \log (\log (4))}+\frac {4 (1-x) \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}+\frac {4 x \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}-\frac {4 \log \left (e^x-3 \log (\log (4))\right )}{3 \log (\log (4))}+\frac {4 x}{e^x-3 \log (\log (4))}+\frac {4 x}{3 \log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-15*E^(2*x) + E^x*(4 - 4*x) + (-12 + 90*E^x)*Log[Log[4]] - 135*Log[Log[4]]^2)/(E^(2*x) - 6*E^x*Log[Log[4]
] + 9*Log[Log[4]]^2),x]

[Out]

-15*x + (4*x)/(E^x - 3*Log[Log[4]]) + (2*(1 - x)^2)/(3*Log[Log[4]]) + (4*x)/(3*Log[Log[4]]) - (2*x^2)/(3*Log[L
og[4]]) + (4*(1 - x)*Log[1 - E^x/(3*Log[Log[4]])])/(3*Log[Log[4]]) + (4*x*Log[1 - E^x/(3*Log[Log[4]])])/(3*Log
[Log[4]]) - (4*Log[E^x - 3*Log[Log[4]]])/(3*Log[Log[4]])

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 2184

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2185

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dis
t[1/a, Int[(c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1), x], x] - Dist[b/a, Int[(c + d*x)^m*(F^(g*(e + f*x)
))^n*(a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && ILtQ[p, 0] && IGtQ[m, 0
]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2191

Int[((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((a_.) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*
((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m*(a + b*(F^(g*(e + f*x)))^n)^(p + 1))/(b*f*g*n*(p +
1)*Log[F]), x] - Dist[(d*m)/(b*f*g*n*(p + 1)*Log[F]), Int[(c + d*x)^(m - 1)*(a + b*(F^(g*(e + f*x)))^n)^(p + 1
), x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n, p}, x] && NeQ[p, -1]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-15 e^{2 x}+e^x (4-4 x)+\left (-12+90 e^x\right ) \log (\log (4))-135 \log ^2(\log (4))}{\left (e^x-3 \log (\log (4))\right )^2} \, dx\\ &=\int \left (-15-\frac {4 (-1+x)}{e^x-3 \log (\log (4))}-\frac {12 x \log (\log (4))}{\left (e^x-3 \log (\log (4))\right )^2}\right ) \, dx\\ &=-15 x-4 \int \frac {-1+x}{e^x-3 \log (\log (4))} \, dx-(12 \log (\log (4))) \int \frac {x}{\left (e^x-3 \log (\log (4))\right )^2} \, dx\\ &=-15 x+\frac {2 (1-x)^2}{3 \log (\log (4))}-4 \int \frac {e^x x}{\left (e^x-3 \log (\log (4))\right )^2} \, dx+4 \int \frac {x}{e^x-3 \log (\log (4))} \, dx-\frac {4 \int \frac {e^x (-1+x)}{e^x-3 \log (\log (4))} \, dx}{3 \log (\log (4))}\\ &=-15 x+\frac {4 x}{e^x-3 \log (\log (4))}+\frac {2 (1-x)^2}{3 \log (\log (4))}-\frac {2 x^2}{3 \log (\log (4))}+\frac {4 (1-x) \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}-4 \int \frac {1}{e^x-3 \log (\log (4))} \, dx+\frac {4 \int \frac {e^x x}{e^x-3 \log (\log (4))} \, dx}{3 \log (\log (4))}+\frac {4 \int \log \left (1-\frac {e^x}{3 \log (\log (4))}\right ) \, dx}{3 \log (\log (4))}\\ &=-15 x+\frac {4 x}{e^x-3 \log (\log (4))}+\frac {2 (1-x)^2}{3 \log (\log (4))}-\frac {2 x^2}{3 \log (\log (4))}+\frac {4 (1-x) \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}+\frac {4 x \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}-4 \operatorname {Subst}\left (\int \frac {1}{x (x-3 \log (\log (4)))} \, dx,x,e^x\right )-\frac {4 \int \log \left (1-\frac {e^x}{3 \log (\log (4))}\right ) \, dx}{3 \log (\log (4))}+\frac {4 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3 \log (\log (4))}\right )}{x} \, dx,x,e^x\right )}{3 \log (\log (4))}\\ &=-15 x+\frac {4 x}{e^x-3 \log (\log (4))}+\frac {2 (1-x)^2}{3 \log (\log (4))}-\frac {2 x^2}{3 \log (\log (4))}+\frac {4 (1-x) \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}+\frac {4 x \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}-\frac {4 \text {Li}_2\left (\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}+\frac {4 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )}{3 \log (\log (4))}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{x-3 \log (\log (4))} \, dx,x,e^x\right )}{3 \log (\log (4))}-\frac {4 \operatorname {Subst}\left (\int \frac {\log \left (1-\frac {x}{3 \log (\log (4))}\right )}{x} \, dx,x,e^x\right )}{3 \log (\log (4))}\\ &=-15 x+\frac {4 x}{e^x-3 \log (\log (4))}+\frac {2 (1-x)^2}{3 \log (\log (4))}+\frac {4 x}{3 \log (\log (4))}-\frac {2 x^2}{3 \log (\log (4))}+\frac {4 (1-x) \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}+\frac {4 x \log \left (1-\frac {e^x}{3 \log (\log (4))}\right )}{3 \log (\log (4))}-\frac {4 \log \left (e^x-3 \log (\log (4))\right )}{3 \log (\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 18, normalized size = 0.82 \begin {gather*} -15 x+\frac {4 x}{e^x-3 \log (\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-15*E^(2*x) + E^x*(4 - 4*x) + (-12 + 90*E^x)*Log[Log[4]] - 135*Log[Log[4]]^2)/(E^(2*x) - 6*E^x*Log[
Log[4]] + 9*Log[Log[4]]^2),x]

[Out]

-15*x + (4*x)/(E^x - 3*Log[Log[4]])

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fricas [A]  time = 0.54, size = 31, normalized size = 1.41 \begin {gather*} -\frac {15 \, x e^{x} - 45 \, x \log \left (2 \, \log \relax (2)\right ) - 4 \, x}{e^{x} - 3 \, \log \left (2 \, \log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-135*log(2*log(2))^2+(90*exp(x)-12)*log(2*log(2))-15*exp(x)^2+(-4*x+4)*exp(x))/(9*log(2*log(2))^2-6
*exp(x)*log(2*log(2))+exp(x)^2),x, algorithm="fricas")

[Out]

-(15*x*e^x - 45*x*log(2*log(2)) - 4*x)/(e^x - 3*log(2*log(2)))

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giac [A]  time = 0.26, size = 36, normalized size = 1.64 \begin {gather*} -\frac {15 \, x e^{x} - 45 \, x \log \relax (2) - 45 \, x \log \left (\log \relax (2)\right ) - 4 \, x}{e^{x} - 3 \, \log \relax (2) - 3 \, \log \left (\log \relax (2)\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-135*log(2*log(2))^2+(90*exp(x)-12)*log(2*log(2))-15*exp(x)^2+(-4*x+4)*exp(x))/(9*log(2*log(2))^2-6
*exp(x)*log(2*log(2))+exp(x)^2),x, algorithm="giac")

[Out]

-(15*x*e^x - 45*x*log(2) - 45*x*log(log(2)) - 4*x)/(e^x - 3*log(2) - 3*log(log(2)))

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maple [A]  time = 0.16, size = 24, normalized size = 1.09

method	result	size
risch	\(-15 x -\frac {4 x}{3 \ln \relax (2)+3 \ln \left (\ln \relax (2)\right )-{\mathrm e}^{x}}\)	\(24\)
norman	\(\frac {\left (-4-45 \ln \left (\ln \relax (2)\right )-45 \ln \relax (2)\right ) x +15 \,{\mathrm e}^{x} x}{3 \ln \left (2 \ln \relax (2)\right )-{\mathrm e}^{x}}\)	\(35\)
default	\(-\frac {4}{-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}}-\frac {45 \ln \relax (2)}{-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}}-\frac {45 \ln \left (\ln \relax (2)\right )}{-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}}-15 \ln \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )-\frac {4 \ln \relax (2) \ln \left ({\mathrm e}^{x}\right )}{3 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}+\frac {12 \ln \relax (2)}{\left (3 \ln \relax (2)+3 \ln \left (\ln \relax (2)\right )\right ) \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}+\frac {4 \ln \relax (2) \ln \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}{3 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}-\frac {15 \ln \relax (2)^{2} \ln \left ({\mathrm e}^{x}\right )}{\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}+\frac {135 \ln \relax (2)^{2}}{\left (3 \ln \relax (2)+3 \ln \left (\ln \relax (2)\right )\right ) \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}+\frac {15 \ln \relax (2)^{2} \ln \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}{\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}-\frac {4 \ln \left (\ln \relax (2)\right ) \ln \left ({\mathrm e}^{x}\right )}{3 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}+\frac {12 \ln \left (\ln \relax (2)\right )}{\left (3 \ln \relax (2)+3 \ln \left (\ln \relax (2)\right )\right ) \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}+\frac {4 \ln \left (\ln \relax (2)\right ) \ln \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}{3 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}-\frac {15 \ln \left (\ln \relax (2)\right )^{2} \ln \left ({\mathrm e}^{x}\right )}{\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}+\frac {135 \ln \left (\ln \relax (2)\right )^{2}}{\left (3 \ln \relax (2)+3 \ln \left (\ln \relax (2)\right )\right ) \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}+\frac {15 \ln \left (\ln \relax (2)\right )^{2} \ln \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}{\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}-\frac {4 \ln \left (3 \ln \relax (2)+3 \ln \left (\ln \relax (2)\right )-{\mathrm e}^{x}\right )}{3 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )}-\frac {4 x \,{\mathrm e}^{x}}{3 \left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right ) \left (3 \ln \relax (2)+3 \ln \left (\ln \relax (2)\right )-{\mathrm e}^{x}\right )}-\frac {30 \ln \left (\ln \relax (2)\right ) \ln \relax (2) \ln \left ({\mathrm e}^{x}\right )}{\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}+\frac {270 \ln \left (\ln \relax (2)\right ) \ln \relax (2)}{\left (3 \ln \relax (2)+3 \ln \left (\ln \relax (2)\right )\right ) \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}+\frac {30 \ln \left (\ln \relax (2)\right ) \ln \relax (2) \ln \left (-3 \ln \relax (2)-3 \ln \left (\ln \relax (2)\right )+{\mathrm e}^{x}\right )}{\left (\ln \relax (2)+\ln \left (\ln \relax (2)\right )\right )^{2}}\)	\(501\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-135*ln(2*ln(2))^2+(90*exp(x)-12)*ln(2*ln(2))-15*exp(x)^2+(-4*x+4)*exp(x))/(9*ln(2*ln(2))^2-6*exp(x)*ln(2
*ln(2))+exp(x)^2),x,method=_RETURNVERBOSE)

[Out]

-15*x-4*x/(3*ln(2)+3*ln(ln(2))-exp(x))

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maxima [B]  time = 0.97, size = 227, normalized size = 10.32 \begin {gather*} 15 \, {\left (\frac {3}{e^{x} \log \left (2 \, \log \relax (2)\right ) - 3 \, \log \left (2 \, \log \relax (2)\right )^{2}} - \frac {x}{\log \left (2 \, \log \relax (2)\right )^{2}} + \frac {\log \left (e^{x} - 3 \, \log \left (2 \, \log \relax (2)\right )\right )}{\log \left (2 \, \log \relax (2)\right )^{2}}\right )} \log \left (2 \, \log \relax (2)\right )^{2} + \frac {4}{3} \, {\left (\frac {3}{e^{x} \log \left (2 \, \log \relax (2)\right ) - 3 \, \log \left (2 \, \log \relax (2)\right )^{2}} - \frac {x}{\log \left (2 \, \log \relax (2)\right )^{2}} + \frac {\log \left (e^{x} - 3 \, \log \left (2 \, \log \relax (2)\right )\right )}{\log \left (2 \, \log \relax (2)\right )^{2}}\right )} \log \left (2 \, \log \relax (2)\right ) + \frac {4 \, x e^{x}}{3 \, {\left ({\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} e^{x} - 3 \, \log \relax (2)^{2} - 6 \, \log \relax (2) \log \left (\log \relax (2)\right ) - 3 \, \log \left (\log \relax (2)\right )^{2}\right )}} - \frac {45 \, \log \left (2 \, \log \relax (2)\right )}{e^{x} - 3 \, \log \left (2 \, \log \relax (2)\right )} - \frac {4 \, \log \left (e^{x} - 3 \, \log \relax (2) - 3 \, \log \left (\log \relax (2)\right )\right )}{3 \, {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )}} - \frac {4}{e^{x} - 3 \, \log \left (2 \, \log \relax (2)\right )} - 15 \, \log \left (e^{x} - 3 \, \log \left (2 \, \log \relax (2)\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-135*log(2*log(2))^2+(90*exp(x)-12)*log(2*log(2))-15*exp(x)^2+(-4*x+4)*exp(x))/(9*log(2*log(2))^2-6
*exp(x)*log(2*log(2))+exp(x)^2),x, algorithm="maxima")

[Out]

15*(3/(e^x*log(2*log(2)) - 3*log(2*log(2))^2) - x/log(2*log(2))^2 + log(e^x - 3*log(2*log(2)))/log(2*log(2))^2
)*log(2*log(2))^2 + 4/3*(3/(e^x*log(2*log(2)) - 3*log(2*log(2))^2) - x/log(2*log(2))^2 + log(e^x - 3*log(2*log
(2)))/log(2*log(2))^2)*log(2*log(2)) + 4/3*x*e^x/((log(2) + log(log(2)))*e^x - 3*log(2)^2 - 6*log(2)*log(log(2
)) - 3*log(log(2))^2) - 45*log(2*log(2))/(e^x - 3*log(2*log(2))) - 4/3*log(e^x - 3*log(2) - 3*log(log(2)))/(lo
g(2) + log(log(2))) - 4/(e^x - 3*log(2*log(2))) - 15*log(e^x - 3*log(2*log(2)))

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mupad [B]  time = 0.14, size = 29, normalized size = 1.32 \begin {gather*} -\frac {x\,\left (\ln \left ({\ln \relax (4)}^{45}\right )+4\right )-15\,x\,{\mathrm {e}}^x}{\ln \left ({\ln \relax (4)}^3\right )-{\mathrm {e}}^x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(15*exp(2*x) - log(2*log(2))*(90*exp(x) - 12) + exp(x)*(4*x - 4) + 135*log(2*log(2))^2)/(exp(2*x) - 6*log
(2*log(2))*exp(x) + 9*log(2*log(2))^2),x)

[Out]

-(x*(log(log(4)^45) + 4) - 15*x*exp(x))/(log(log(4)^3) - exp(x))

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sympy [A]  time = 0.13, size = 20, normalized size = 0.91 \begin {gather*} - 15 x + \frac {4 x}{e^{x} - 3 \log {\relax (2 )} - 3 \log {\left (\log {\relax (2 )} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-135*ln(2*ln(2))**2+(90*exp(x)-12)*ln(2*ln(2))-15*exp(x)**2+(-4*x+4)*exp(x))/(9*ln(2*ln(2))**2-6*ex
p(x)*ln(2*ln(2))+exp(x)**2),x)

[Out]

-15*x + 4*x/(exp(x) - 3*log(2) - 3*log(log(2)))

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