3.4.73 \(\int \frac {(1152 e^3+96 x-1152 \log (4)) \log (-\frac {12}{5+20 x})+(12+48 x) \log ^2(-\frac {12}{5+20 x})}{x^2+4 x^3+e^6 (144+576 x)+e^3 (24 x+96 x^2)+(e^3 (-288-1152 x)-24 x-96 x^2) \log (4)+(144+576 x) \log ^2(4)} \, dx\)

Optimal. Leaf size=34 \[ \frac {\log ^2\left (\frac {3}{-5+5 \left (\frac {3}{4}-x\right )}\right )}{-e^3-\frac {x}{12}+\log (4)} \]

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Rubi [C]  time = 1.25, antiderivative size = 232, normalized size of antiderivative = 6.82, number of steps used = 15, number of rules used = 16, integrand size = 108, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6, 6688, 12, 6742, 36, 31, 2395, 2411, 2343, 2333, 2316, 2315, 2397, 2394, 2393, 2391} \begin {gather*} -\frac {96 \text {Li}_2\left (\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{4 x+1}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \text {Li}_2\left (\frac {4 x+1}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-\frac {12 (4 x+1) \log ^2\left (-\frac {12}{5 (4 x+1)}\right )}{\left (1-48 e^3+48 \log (4)\right ) \left (x+12 \left (e^3-\log (4)\right )\right )}-\frac {96 \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right ) \log \left (-\frac {12}{5 (4 x+1)}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \log \left (-\frac {12}{5 \left (1-48 e^3+48 \log (4)\right )}\right ) \log \left (\frac {x+12 \left (e^3-\log (4)\right )}{4 x+1}\right )}{1-48 e^3+48 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1152*E^3 + 96*x - 1152*Log[4])*Log[-12/(5 + 20*x)] + (12 + 48*x)*Log[-12/(5 + 20*x)]^2)/(x^2 + 4*x^3 + E
^6*(144 + 576*x) + E^3*(24*x + 96*x^2) + (E^3*(-288 - 1152*x) - 24*x - 96*x^2)*Log[4] + (144 + 576*x)*Log[4]^2
),x]

[Out]

(-12*(1 + 4*x)*Log[-12/(5*(1 + 4*x))]^2)/((x + 12*(E^3 - Log[4]))*(1 - 48*E^3 + 48*Log[4])) + (96*Log[(x + 12*
(E^3 - Log[4]))/(1 + 4*x)]*Log[-12/(5*(1 - 48*E^3 + 48*Log[4]))])/(1 - 48*E^3 + 48*Log[4]) - (96*Log[-12/(5*(1
+ 4*x))]*Log[(-4*(x + 12*(E^3 - Log[4])))/(1 - 48*E^3 + 48*Log[4])])/(1 - 48*E^3 + 48*Log[4]) - (96*PolyLog[2
, (4*(x + 12*(E^3 - Log[4])))/(1 + 4*x)])/(1 - 48*E^3 + 48*Log[4]) + (96*PolyLog[2, (1 + 4*x)/(1 - 48*E^3 + 48
*Log[4])])/(1 - 48*E^3 + 48*Log[4])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2316

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

Rule 2333

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)/(x_))^(q_.)*(x_)^(m_.), x_Symbol] :> Int[(e + d*
x)^q*(a + b*Log[c*x^n])^p, x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && EqQ[m, q] && IntegerQ[q]

Rule 2343

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

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 2393

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + (c*e*x)/g])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2394

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

Rule 2395

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

Rule 2397

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)/((f_.) + (g_.)*(x_))^2, x_Symbol] :> Simp[((d +
e*x)*(a + b*Log[c*(d + e*x)^n])^p)/((e*f - d*g)*(f + g*x)), x] - Dist[(b*e*n*p)/(e*f - d*g), Int[(a + b*Log[c*
(d + e*x)^n])^(p - 1)/(f + g*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] && GtQ[p, 0
]

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 6688

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

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 {\left (1152 e^3+96 x-1152 \log (4)\right ) \log \left (-\frac {12}{5+20 x}\right )+(12+48 x) \log ^2\left (-\frac {12}{5+20 x}\right )}{x^2+4 x^3+e^3 \left (24 x+96 x^2\right )+\left (e^3 (-288-1152 x)-24 x-96 x^2\right ) \log (4)+(144+576 x) \left (e^6+\log ^2(4)\right )} \, dx\\ &=\int \frac {12 \log \left (-\frac {12}{5+20 x}\right ) \left (8 \left (12 e^3+x-12 \log (4)\right )+(1+4 x) \log \left (-\frac {12}{5+20 x}\right )\right )}{(1+4 x) \left (12 e^3+x-12 \log (4)\right )^2} \, dx\\ &=12 \int \frac {\log \left (-\frac {12}{5+20 x}\right ) \left (8 \left (12 e^3+x-12 \log (4)\right )+(1+4 x) \log \left (-\frac {12}{5+20 x}\right )\right )}{(1+4 x) \left (12 e^3+x-12 \log (4)\right )^2} \, dx\\ &=12 \int \left (\frac {8 \log \left (-\frac {12}{5+20 x}\right )}{(1+4 x) \left (12 e^3+x-12 \log (4)\right )}+\frac {\log ^2\left (-\frac {12}{5+20 x}\right )}{\left (12 e^3+x-12 \log (4)\right )^2}\right ) \, dx\\ &=12 \int \frac {\log ^2\left (-\frac {12}{5+20 x}\right )}{\left (12 e^3+x-12 \log (4)\right )^2} \, dx+96 \int \frac {\log \left (-\frac {12}{5+20 x}\right )}{(1+4 x) \left (12 e^3+x-12 \log (4)\right )} \, dx\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}+\frac {24}{5} \operatorname {Subst}\left (\int \frac {5 \log \left (-\frac {12}{x}\right )}{x \left (\frac {x}{20}+\frac {1}{20} \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )\right )} \, dx,x,5+20 x\right )-\frac {96 \int \frac {\log \left (-\frac {12}{5+20 x}\right )}{12 e^3+x-12 \log (4)} \, dx}{1-48 e^3+48 \log (4)}\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}+24 \operatorname {Subst}\left (\int \frac {\log \left (-\frac {12}{x}\right )}{x \left (\frac {x}{20}+\frac {1}{20} \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )\right )} \, dx,x,5+20 x\right )-\frac {1920 \int \frac {\log \left (\frac {20 \left (12 e^3+x-12 \log (4)\right )}{-5+20 \left (12 e^3-12 \log (4)\right )}\right )}{5+20 x} \, dx}{1-48 e^3+48 \log (4)}\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-24 \operatorname {Subst}\left (\int \frac {\log (-12 x)}{x \left (\frac {1}{20 x}+\frac {1}{20} \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )\right )} \, dx,x,\frac {1}{5+20 x}\right )-\frac {96 \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {x}{-5+20 \left (12 e^3-12 \log (4)\right )}\right )}{x} \, dx,x,5+20 x\right )}{1-48 e^3+48 \log (4)}\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \text {Li}_2\left (\frac {1+4 x}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-24 \operatorname {Subst}\left (\int \frac {\log (-12 x)}{\frac {1}{20}+\frac {1}{20} x \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )} \, dx,x,\frac {1}{5+20 x}\right )\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}+\frac {96 \log \left (\frac {x+12 \left (e^3-\log (4)\right )}{1+4 x}\right ) \log \left (-\frac {12}{5 \left (1-48 e^3+48 \log (4)\right )}\right )}{1-48 e^3+48 \log (4)}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \text {Li}_2\left (\frac {1+4 x}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-24 \operatorname {Subst}\left (\int \frac {\log \left (x \left (5-20 \left (12 e^3-12 \log (4)\right )\right )\right )}{\frac {1}{20}+\frac {1}{20} x \left (-5+20 \left (12 e^3-12 \log (4)\right )\right )} \, dx,x,\frac {1}{5+20 x}\right )\\ &=-\frac {12 (1+4 x) \log ^2\left (-\frac {12}{5 (1+4 x)}\right )}{\left (x+12 \left (e^3-\log (4)\right )\right ) \left (1-48 e^3+48 \log (4)\right )}+\frac {96 \log \left (\frac {x+12 \left (e^3-\log (4)\right )}{1+4 x}\right ) \log \left (-\frac {12}{5 \left (1-48 e^3+48 \log (4)\right )}\right )}{1-48 e^3+48 \log (4)}-\frac {96 \log \left (-\frac {12}{5 (1+4 x)}\right ) \log \left (-\frac {4 \left (x+12 \left (e^3-\log (4)\right )\right )}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}+\frac {96 \text {Li}_2\left (\frac {1+4 x}{1-48 e^3+48 \log (4)}\right )}{1-48 e^3+48 \log (4)}-\frac {96 \text {Li}_2\left (1-\frac {1-48 e^3+48 \log (4)}{1+4 x}\right )}{1-48 e^3+48 \log (4)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 27, normalized size = 0.79 \begin {gather*} -\frac {12 \log ^2\left (-\frac {12}{5+20 x}\right )}{12 e^3+x-12 \log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1152*E^3 + 96*x - 1152*Log[4])*Log[-12/(5 + 20*x)] + (12 + 48*x)*Log[-12/(5 + 20*x)]^2)/(x^2 + 4*x
^3 + E^6*(144 + 576*x) + E^3*(24*x + 96*x^2) + (E^3*(-288 - 1152*x) - 24*x - 96*x^2)*Log[4] + (144 + 576*x)*Lo
g[4]^2),x]

[Out]

(-12*Log[-12/(5 + 20*x)]^2)/(12*E^3 + x - 12*Log[4])

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fricas [A]  time = 0.59, size = 26, normalized size = 0.76 \begin {gather*} -\frac {12 \, \log \left (-\frac {12}{5 \, {\left (4 \, x + 1\right )}}\right )^{2}}{x + 12 \, e^{3} - 24 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x+12)*log(-12/(20*x+5))^2+(-2304*log(2)+1152*exp(3)+96*x)*log(-12/(20*x+5)))/(4*(576*x+144)*log
(2)^2+2*((-1152*x-288)*exp(3)-96*x^2-24*x)*log(2)+(576*x+144)*exp(3)^2+(96*x^2+24*x)*exp(3)+4*x^3+x^2),x, algo
rithm="fricas")

[Out]

-12*log(-12/5/(4*x + 1))^2/(x + 12*e^3 - 24*log(2))

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giac [B]  time = 0.59, size = 56, normalized size = 1.65 \begin {gather*} -\frac {48 \, \log \left (-\frac {12}{5 \, {\left (4 \, x + 1\right )}}\right )^{2}}{{\left (4 \, x + 1\right )} {\left (\frac {48 \, e^{3}}{4 \, x + 1} - \frac {96 \, \log \relax (2)}{4 \, x + 1} - \frac {1}{4 \, x + 1} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x+12)*log(-12/(20*x+5))^2+(-2304*log(2)+1152*exp(3)+96*x)*log(-12/(20*x+5)))/(4*(576*x+144)*log
(2)^2+2*((-1152*x-288)*exp(3)-96*x^2-24*x)*log(2)+(576*x+144)*exp(3)^2+(96*x^2+24*x)*exp(3)+4*x^3+x^2),x, algo
rithm="giac")

[Out]

-48*log(-12/5/(4*x + 1))^2/((4*x + 1)*(48*e^3/(4*x + 1) - 96*log(2)/(4*x + 1) - 1/(4*x + 1) + 1))

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maple [A]  time = 8.10, size = 27, normalized size = 0.79

method	result	size
norman	\(-\frac {12 \ln \left (-\frac {12}{20 x +5}\right )^{2}}{x +12 \,{\mathrm e}^{3}-24 \ln \relax (2)}\)	\(27\)
risch	\(-\frac {12 \ln \left (-\frac {12}{20 x +5}\right )^{2}}{x +12 \,{\mathrm e}^{3}-24 \ln \relax (2)}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((48*x+12)*ln(-12/(20*x+5))^2+(-2304*ln(2)+1152*exp(3)+96*x)*ln(-12/(20*x+5)))/(4*(576*x+144)*ln(2)^2+2*((
-1152*x-288)*exp(3)-96*x^2-24*x)*ln(2)+(576*x+144)*exp(3)^2+(96*x^2+24*x)*exp(3)+4*x^3+x^2),x,method=_RETURNVE
RBOSE)

[Out]

-12*ln(-12/(20*x+5))^2/(x+12*exp(3)-24*ln(2))

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maxima [B]  time = 0.65, size = 73, normalized size = 2.15 \begin {gather*} -\frac {12 \, {\left (\log \relax (5)^{2} - 2 \, \log \relax (5) \log \relax (3) + \log \relax (3)^{2} - 4 \, {\left (\log \relax (5) - \log \relax (3)\right )} \log \relax (2) + 4 \, \log \relax (2)^{2} + 2 \, {\left (\log \relax (5) - \log \relax (3) - 2 \, \log \relax (2)\right )} \log \left (-4 \, x - 1\right ) + \log \left (-4 \, x - 1\right )^{2}\right )}}{x + 12 \, e^{3} - 24 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x+12)*log(-12/(20*x+5))^2+(-2304*log(2)+1152*exp(3)+96*x)*log(-12/(20*x+5)))/(4*(576*x+144)*log
(2)^2+2*((-1152*x-288)*exp(3)-96*x^2-24*x)*log(2)+(576*x+144)*exp(3)^2+(96*x^2+24*x)*exp(3)+4*x^3+x^2),x, algo
rithm="maxima")

[Out]

-12*(log(5)^2 - 2*log(5)*log(3) + log(3)^2 - 4*(log(5) - log(3))*log(2) + 4*log(2)^2 + 2*(log(5) - log(3) - 2*
log(2))*log(-4*x - 1) + log(-4*x - 1)^2)/(x + 12*e^3 - 24*log(2))

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mupad [B]  time = 0.66, size = 26, normalized size = 0.76 \begin {gather*} -\frac {12\,{\ln \left (-\frac {12}{20\,x+5}\right )}^2}{x+12\,{\mathrm {e}}^3-24\,\ln \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(-12/(20*x + 5))^2*(48*x + 12) + log(-12/(20*x + 5))*(96*x + 1152*exp(3) - 2304*log(2)))/(exp(3)*(24*x
+ 96*x^2) - 2*log(2)*(24*x + 96*x^2 + exp(3)*(1152*x + 288)) + 4*log(2)^2*(576*x + 144) + x^2 + 4*x^3 + exp(6
)*(576*x + 144)),x)

[Out]

-(12*log(-12/(20*x + 5))^2)/(x + 12*exp(3) - 24*log(2))

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sympy [A]  time = 0.25, size = 26, normalized size = 0.76 \begin {gather*} - \frac {12 \log {\left (- \frac {12}{20 x + 5} \right )}^{2}}{x - 24 \log {\relax (2 )} + 12 e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((48*x+12)*ln(-12/(20*x+5))**2+(-2304*ln(2)+1152*exp(3)+96*x)*ln(-12/(20*x+5)))/(4*(576*x+144)*ln(2)
**2+2*((-1152*x-288)*exp(3)-96*x**2-24*x)*ln(2)+(576*x+144)*exp(3)**2+(96*x**2+24*x)*exp(3)+4*x**3+x**2),x)

[Out]

-12*log(-12/(20*x + 5))**2/(x - 24*log(2) + 12*exp(3))

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