3.63.82 \(\int \frac {10 x^3+(45 x^2-15 x^3) \log (-3+x)+(-75+25 x+e^{x^2} (-15+5 x-30 x^2+10 x^3)+(-30 x+10 x^2) \log (3)) \log ^3(-3+x)}{(-3+x) \log ^3(-3+x)} \, dx\)

Optimal. Leaf size=25 \[ 5 x \left (5+e^{x^2}+x \log (3)-\frac {x^2}{\log ^2(-3+x)}\right ) \]

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Rubi [B]  time = 1.15, antiderivative size = 131, normalized size of antiderivative = 5.24, number of steps used = 43, number of rules used = 18, integrand size = 79, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.228, Rules used = {6742, 2226, 2204, 2212, 6688, 2411, 2353, 2297, 2298, 2302, 30, 2306, 2309, 2178, 2400, 2399, 2389, 2390} \begin {gather*} 5 e^{x^2} x-\frac {15 x^2 (3-x)}{\log (x-3)}+\frac {5}{2} x^2 \log (9)+25 x+\frac {5 (3-x)^3}{\log ^2(x-3)}-\frac {45 (3-x)^2}{\log ^2(x-3)}+\frac {135 (3-x)}{\log ^2(x-3)}-\frac {135}{\log ^2(x-3)}+\frac {15 (3-x)^3}{\log (x-3)}-\frac {90 (3-x)^2}{\log (x-3)}+\frac {135 (3-x)}{\log (x-3)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(10*x^3 + (45*x^2 - 15*x^3)*Log[-3 + x] + (-75 + 25*x + E^x^2*(-15 + 5*x - 30*x^2 + 10*x^3) + (-30*x + 10*
x^2)*Log[3])*Log[-3 + x]^3)/((-3 + x)*Log[-3 + x]^3),x]

[Out]

25*x + 5*E^x^2*x + (5*x^2*Log[9])/2 - 135/Log[-3 + x]^2 + (135*(3 - x))/Log[-3 + x]^2 - (45*(3 - x)^2)/Log[-3
+ x]^2 + (5*(3 - x)^3)/Log[-3 + x]^2 + (135*(3 - x))/Log[-3 + x] - (90*(3 - x)^2)/Log[-3 + x] + (15*(3 - x)^3)
/Log[-3 + x] - (15*(3 - x)*x^2)/Log[-3 + x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2204

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[(F^a*Sqrt[Pi]*Erfi[(c + d*x)*Rt[b*Log[F], 2
]])/(2*d*Rt[b*Log[F], 2]), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2226

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*(u_), x_Symbol] :> Int[ExpandLinearProduct[F^(a + b*(c + d*
x)^n), u, c, d, x], x] /; FreeQ[{F, a, b, c, d, n}, x] && PolynomialQ[u, x]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2302

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2306

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

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2353

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2389

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

Rule 2390

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

Rule 2399

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

Rule 2400

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
d + e*x)*(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1))/(b*e*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), I
nt[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Dist[(q*(e*f - d*g))/(b*e*n*(p + 1)), Int[(f + g*x
)^(q - 1)*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x]) /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && LtQ[p, -1] && GtQ[q, 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 \left (5 e^{x^2} \left (1+2 x^2\right )+\frac {5 \left (-2 x^3-9 x^2 \log (-3+x)+3 x^3 \log (-3+x)+15 \log ^3(-3+x)-5 x \left (1-\frac {3 \log (9)}{5}\right ) \log ^3(-3+x)-x^2 \log (9) \log ^3(-3+x)\right )}{(3-x) \log ^3(-3+x)}\right ) \, dx\\ &=5 \int e^{x^2} \left (1+2 x^2\right ) \, dx+5 \int \frac {-2 x^3-9 x^2 \log (-3+x)+3 x^3 \log (-3+x)+15 \log ^3(-3+x)-5 x \left (1-\frac {3 \log (9)}{5}\right ) \log ^3(-3+x)-x^2 \log (9) \log ^3(-3+x)}{(3-x) \log ^3(-3+x)} \, dx\\ &=5 \int \left (e^{x^2}+2 e^{x^2} x^2\right ) \, dx+5 \int \left (5+x \log (9)+\frac {2 x^3}{(-3+x) \log ^3(-3+x)}-\frac {3 x^2}{\log ^2(-3+x)}\right ) \, dx\\ &=25 x+\frac {5}{2} x^2 \log (9)+5 \int e^{x^2} \, dx+10 \int e^{x^2} x^2 \, dx+10 \int \frac {x^3}{(-3+x) \log ^3(-3+x)} \, dx-15 \int \frac {x^2}{\log ^2(-3+x)} \, dx\\ &=25 x+5 e^{x^2} x+\frac {5}{2} \sqrt {\pi } \text {erfi}(x)+\frac {5}{2} x^2 \log (9)-\frac {15 (3-x) x^2}{\log (-3+x)}-5 \int e^{x^2} \, dx+10 \operatorname {Subst}\left (\int \frac {(3+x)^3}{x \log ^3(x)} \, dx,x,-3+x\right )-45 \int \frac {x^2}{\log (-3+x)} \, dx+90 \int \frac {x}{\log (-3+x)} \, dx\\ &=25 x+5 e^{x^2} x+\frac {5}{2} x^2 \log (9)-\frac {15 (3-x) x^2}{\log (-3+x)}+10 \operatorname {Subst}\left (\int \left (\frac {27}{\log ^3(x)}+\frac {27}{x \log ^3(x)}+\frac {9 x}{\log ^3(x)}+\frac {x^2}{\log ^3(x)}\right ) \, dx,x,-3+x\right )-45 \int \left (\frac {9}{\log (-3+x)}+\frac {6 (-3+x)}{\log (-3+x)}+\frac {(-3+x)^2}{\log (-3+x)}\right ) \, dx+90 \int \left (\frac {3}{\log (-3+x)}+\frac {-3+x}{\log (-3+x)}\right ) \, dx\\ &=25 x+5 e^{x^2} x+\frac {5}{2} x^2 \log (9)-\frac {15 (3-x) x^2}{\log (-3+x)}+10 \operatorname {Subst}\left (\int \frac {x^2}{\log ^3(x)} \, dx,x,-3+x\right )-45 \int \frac {(-3+x)^2}{\log (-3+x)} \, dx+90 \int \frac {-3+x}{\log (-3+x)} \, dx+90 \operatorname {Subst}\left (\int \frac {x}{\log ^3(x)} \, dx,x,-3+x\right )+270 \int \frac {1}{\log (-3+x)} \, dx-270 \int \frac {-3+x}{\log (-3+x)} \, dx+270 \operatorname {Subst}\left (\int \frac {1}{\log ^3(x)} \, dx,x,-3+x\right )+270 \operatorname {Subst}\left (\int \frac {1}{x \log ^3(x)} \, dx,x,-3+x\right )-405 \int \frac {1}{\log (-3+x)} \, dx\\ &=25 x+5 e^{x^2} x+\frac {5}{2} x^2 \log (9)+\frac {135 (3-x)}{\log ^2(-3+x)}-\frac {45 (3-x)^2}{\log ^2(-3+x)}+\frac {5 (3-x)^3}{\log ^2(-3+x)}-\frac {15 (3-x) x^2}{\log (-3+x)}+15 \operatorname {Subst}\left (\int \frac {x^2}{\log ^2(x)} \, dx,x,-3+x\right )-45 \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-3+x\right )+90 \operatorname {Subst}\left (\int \frac {x}{\log ^2(x)} \, dx,x,-3+x\right )+90 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-3+x\right )+135 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-3+x\right )+270 \operatorname {Subst}\left (\int \frac {1}{x^3} \, dx,x,\log (-3+x)\right )+270 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+x\right )-270 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-3+x\right )-405 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+x\right )\\ &=25 x+5 e^{x^2} x+\frac {5}{2} x^2 \log (9)-\frac {135}{\log ^2(-3+x)}+\frac {135 (3-x)}{\log ^2(-3+x)}-\frac {45 (3-x)^2}{\log ^2(-3+x)}+\frac {5 (3-x)^3}{\log ^2(-3+x)}+\frac {135 (3-x)}{\log (-3+x)}-\frac {90 (3-x)^2}{\log (-3+x)}+\frac {15 (3-x)^3}{\log (-3+x)}-\frac {15 (3-x) x^2}{\log (-3+x)}-135 \text {li}(-3+x)-45 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-3+x)\right )+45 \operatorname {Subst}\left (\int \frac {x^2}{\log (x)} \, dx,x,-3+x\right )+90 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-3+x)\right )+135 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-3+x\right )+180 \operatorname {Subst}\left (\int \frac {x}{\log (x)} \, dx,x,-3+x\right )-270 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-3+x)\right )\\ &=25 x+5 e^{x^2} x-180 \text {Ei}(2 \log (-3+x))-45 \text {Ei}(3 \log (-3+x))+\frac {5}{2} x^2 \log (9)-\frac {135}{\log ^2(-3+x)}+\frac {135 (3-x)}{\log ^2(-3+x)}-\frac {45 (3-x)^2}{\log ^2(-3+x)}+\frac {5 (3-x)^3}{\log ^2(-3+x)}+\frac {135 (3-x)}{\log (-3+x)}-\frac {90 (3-x)^2}{\log (-3+x)}+\frac {15 (3-x)^3}{\log (-3+x)}-\frac {15 (3-x) x^2}{\log (-3+x)}+45 \operatorname {Subst}\left (\int \frac {e^{3 x}}{x} \, dx,x,\log (-3+x)\right )+180 \operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log (-3+x)\right )\\ &=25 x+5 e^{x^2} x+\frac {5}{2} x^2 \log (9)-\frac {135}{\log ^2(-3+x)}+\frac {135 (3-x)}{\log ^2(-3+x)}-\frac {45 (3-x)^2}{\log ^2(-3+x)}+\frac {5 (3-x)^3}{\log ^2(-3+x)}+\frac {135 (3-x)}{\log (-3+x)}-\frac {90 (3-x)^2}{\log (-3+x)}+\frac {15 (3-x)^3}{\log (-3+x)}-\frac {15 (3-x) x^2}{\log (-3+x)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 34, normalized size = 1.36 \begin {gather*} 5 \left (-15+\left (5+e^{x^2}\right ) x+x^2 \log (3)-\log (19683)-\frac {x^3}{\log ^2(-3+x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(10*x^3 + (45*x^2 - 15*x^3)*Log[-3 + x] + (-75 + 25*x + E^x^2*(-15 + 5*x - 30*x^2 + 10*x^3) + (-30*x
+ 10*x^2)*Log[3])*Log[-3 + x]^3)/((-3 + x)*Log[-3 + x]^3),x]

[Out]

5*(-15 + (5 + E^x^2)*x + x^2*Log[3] - Log[19683] - x^3/Log[-3 + x]^2)

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fricas [A]  time = 0.67, size = 36, normalized size = 1.44 \begin {gather*} -\frac {5 \, {\left (x^{3} - {\left (x^{2} \log \relax (3) + x e^{\left (x^{2}\right )} + 5 \, x\right )} \log \left (x - 3\right )^{2}\right )}}{\log \left (x - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3-30*x^2+5*x-15)*exp(x^2)+(10*x^2-30*x)*log(3)+25*x-75)*log(x-3)^3+(-15*x^3+45*x^2)*log(x-3)
+10*x^3)/(x-3)/log(x-3)^3,x, algorithm="fricas")

[Out]

-5*(x^3 - (x^2*log(3) + x*e^(x^2) + 5*x)*log(x - 3)^2)/log(x - 3)^2

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giac [A]  time = 0.21, size = 47, normalized size = 1.88 \begin {gather*} \frac {5 \, {\left (x^{2} \log \relax (3) \log \left (x - 3\right )^{2} + x e^{\left (x^{2}\right )} \log \left (x - 3\right )^{2} - x^{3} + 5 \, x \log \left (x - 3\right )^{2}\right )}}{\log \left (x - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3-30*x^2+5*x-15)*exp(x^2)+(10*x^2-30*x)*log(3)+25*x-75)*log(x-3)^3+(-15*x^3+45*x^2)*log(x-3)
+10*x^3)/(x-3)/log(x-3)^3,x, algorithm="giac")

[Out]

5*(x^2*log(3)*log(x - 3)^2 + x*e^(x^2)*log(x - 3)^2 - x^3 + 5*x*log(x - 3)^2)/log(x - 3)^2

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maple [A]  time = 0.36, size = 30, normalized size = 1.20

method	result	size
risch	\(5 x^{2} \ln \relax (3)+5 \,{\mathrm e}^{x^{2}} x +25 x -\frac {5 x^{3}}{\ln \left (x -3\right )^{2}}\)	\(30\)
default	\(25 x +5 \,{\mathrm e}^{x^{2}} x +5 x^{2} \ln \relax (3)-\frac {5 \left (x -3\right )^{3}}{\ln \left (x -3\right )^{2}}-\frac {45 \left (x -3\right )^{2}}{\ln \left (x -3\right )^{2}}-\frac {135 \left (x -3\right )}{\ln \left (x -3\right )^{2}}-\frac {135}{\ln \left (x -3\right )^{2}}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((10*x^3-30*x^2+5*x-15)*exp(x^2)+(10*x^2-30*x)*ln(3)+25*x-75)*ln(x-3)^3+(-15*x^3+45*x^2)*ln(x-3)+10*x^3)/
(x-3)/ln(x-3)^3,x,method=_RETURNVERBOSE)

[Out]

5*x^2*ln(3)+5*exp(x^2)*x+25*x-5*x^3/ln(x-3)^2

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maxima [B]  time = 0.52, size = 69, normalized size = 2.76 \begin {gather*} -30 \, {\left (x + 3 \, \log \left (x - 3\right )\right )} \log \relax (3) + 90 \, \log \relax (3) \log \left (x - 3\right ) + 25 \, x + \frac {5 \, {\left (x e^{\left (x^{2}\right )} \log \left (x - 3\right )^{2} - x^{3} + {\left (x^{2} \log \relax (3) + 6 \, x \log \relax (3)\right )} \log \left (x - 3\right )^{2}\right )}}{\log \left (x - 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x^3-30*x^2+5*x-15)*exp(x^2)+(10*x^2-30*x)*log(3)+25*x-75)*log(x-3)^3+(-15*x^3+45*x^2)*log(x-3)
+10*x^3)/(x-3)/log(x-3)^3,x, algorithm="maxima")

[Out]

-30*(x + 3*log(x - 3))*log(3) + 90*log(3)*log(x - 3) + 25*x + 5*(x*e^(x^2)*log(x - 3)^2 - x^3 + (x^2*log(3) +
6*x*log(3))*log(x - 3)^2)/log(x - 3)^2

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mupad [B]  time = 0.23, size = 29, normalized size = 1.16 \begin {gather*} 25\,x+5\,x\,{\mathrm {e}}^{x^2}+5\,x^2\,\ln \relax (3)-\frac {5\,x^3}{{\ln \left (x-3\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - 3)*(45*x^2 - 15*x^3) + log(x - 3)^3*(25*x - log(3)*(30*x - 10*x^2) + exp(x^2)*(5*x - 30*x^2 + 10*
x^3 - 15) - 75) + 10*x^3)/(log(x - 3)^3*(x - 3)),x)

[Out]

25*x + 5*x*exp(x^2) + 5*x^2*log(3) - (5*x^3)/log(x - 3)^2

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sympy [A]  time = 0.39, size = 31, normalized size = 1.24 \begin {gather*} - \frac {5 x^{3}}{\log {\left (x - 3 \right )}^{2}} + 5 x^{2} \log {\relax (3 )} + 5 x e^{x^{2}} + 25 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((10*x**3-30*x**2+5*x-15)*exp(x**2)+(10*x**2-30*x)*ln(3)+25*x-75)*ln(x-3)**3+(-15*x**3+45*x**2)*ln(
x-3)+10*x**3)/(x-3)/ln(x-3)**3,x)

[Out]

-5*x**3/log(x - 3)**2 + 5*x**2*log(3) + 5*x*exp(x**2) + 25*x

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