3.23.5 \(\int \frac {e^{\frac {4+x}{5+3 x}} (-138-111 x-27 x^2)+e^{\frac {4+x}{5+3 x}} (-46-37 x-9 x^2) \log ^2(\log (4))}{225+420 x+286 x^2+84 x^3+9 x^4} \, dx\)

Optimal. Leaf size=28 \[ 3+\frac {e^{\frac {4+x}{5+3 x}} \left (3+\log ^2(\log (4))\right )}{3+x} \]

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Rubi [A]  time = 1.18, antiderivative size = 26, normalized size of antiderivative = 0.93, number of steps used = 25, number of rules used = 10, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6688, 12, 6742, 2232, 2231, 2230, 2210, 2233, 2178, 2209} \begin {gather*} \frac {e^{\frac {x+4}{3 x+5}} \left (3+\log ^2(\log (4))\right )}{x+3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((4 + x)/(5 + 3*x))*(-138 - 111*x - 27*x^2) + E^((4 + x)/(5 + 3*x))*(-46 - 37*x - 9*x^2)*Log[Log[4]]^2)
/(225 + 420*x + 286*x^2 + 84*x^3 + 9*x^4),x]

[Out]

(E^((4 + x)/(5 + 3*x))*(3 + Log[Log[4]]^2))/(3 + 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 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 2209

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

Rule 2210

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

Rule 2230

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

Rule 2231

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

Rule 2232

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))*((g_.) + (h_.)*(x_))^(m_), x_Symbol] :> S
imp[((g + h*x)^(m + 1)*F^(e + (f*(a + b*x))/(c + d*x)))/(h*(m + 1)), x] - Dist[(f*(b*c - a*d)*Log[F])/(h*(m +
1)), Int[((g + h*x)^(m + 1)*F^(e + (f*(a + b*x))/(c + d*x)))/(c + d*x)^2, x], x] /; FreeQ[{F, a, b, c, d, e, f
, g, h}, x] && NeQ[b*c - a*d, 0] && NeQ[d*g - c*h, 0] && ILtQ[m, -1]

Rule 2233

Int[(F_)^((e_.) + ((f_.)*((a_.) + (b_.)*(x_)))/((c_.) + (d_.)*(x_)))/(((g_.) + (h_.)*(x_))*((i_.) + (j_.)*(x_)
)), x_Symbol] :> -Dist[d/(h*(d*i - c*j)), Subst[Int[F^(e + (f*(b*i - a*j))/(d*i - c*j) - ((b*c - a*d)*f*x)/(d*
i - c*j))/x, x], x, (i + j*x)/(c + d*x)], x] /; FreeQ[{F, a, b, c, d, e, f, g, h}, x] && EqQ[d*g - c*h, 0]

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 {e^{\frac {4+x}{5+3 x}} \left (46+37 x+9 x^2\right ) \left (-3-\log ^2(\log (4))\right )}{\left (15+14 x+3 x^2\right )^2} \, dx\\ &=\left (-3-\log ^2(\log (4))\right ) \int \frac {e^{\frac {4+x}{5+3 x}} \left (46+37 x+9 x^2\right )}{\left (15+14 x+3 x^2\right )^2} \, dx\\ &=\left (-3-\log ^2(\log (4))\right ) \int \left (\frac {e^{\frac {4+x}{5+3 x}}}{(3+x)^2}+\frac {7 e^{\frac {4+x}{5+3 x}}}{16 (3+x)}+\frac {21 e^{\frac {4+x}{5+3 x}}}{4 (5+3 x)^2}-\frac {21 e^{\frac {4+x}{5+3 x}}}{16 (5+3 x)}\right ) \, dx\\ &=\left (-3-\log ^2(\log (4))\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{(3+x)^2} \, dx-\frac {1}{16} \left (7 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{3+x} \, dx+\frac {1}{16} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{5+3 x} \, dx-\frac {1}{4} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{(5+3 x)^2} \, dx\\ &=\frac {e^{\frac {4+x}{5+3 x}} \left (3+\log ^2(\log (4))\right )}{3+x}-\frac {1}{16} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{5+3 x} \, dx+\frac {1}{16} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}}}{5+3 x} \, dx+\frac {1}{4} \left (7 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{(3+x) (5+3 x)} \, dx-\frac {1}{4} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}}}{(5+3 x)^2} \, dx+\left (7 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{(3+x) (5+3 x)^2} \, dx\\ &=\frac {3}{4} e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}} \left (3+\log ^2(\log (4))\right )+\frac {e^{\frac {4+x}{5+3 x}} \left (3+\log ^2(\log (4))\right )}{3+x}-\frac {7}{16} \sqrt [3]{e} \text {Ei}\left (\frac {7}{3 (5+3 x)}\right ) \left (3+\log ^2(\log (4))\right )-\frac {1}{16} \left (7 \left (3+\log ^2(\log (4))\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-\frac {1}{4}+\frac {7 x}{4}}}{x} \, dx,x,\frac {3+x}{5+3 x}\right )-\frac {1}{16} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}}}{5+3 x} \, dx+\left (7 \left (3+\log ^2(\log (4))\right )\right ) \int \left (\frac {e^{\frac {4+x}{5+3 x}}}{16 (3+x)}+\frac {3 e^{\frac {4+x}{5+3 x}}}{4 (5+3 x)^2}-\frac {3 e^{\frac {4+x}{5+3 x}}}{16 (5+3 x)}\right ) \, dx\\ &=\frac {3}{4} e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}} \left (3+\log ^2(\log (4))\right )+\frac {e^{\frac {4+x}{5+3 x}} \left (3+\log ^2(\log (4))\right )}{3+x}-\frac {7 \text {Ei}\left (\frac {7 (3+x)}{4 (5+3 x)}\right ) \left (3+\log ^2(\log (4))\right )}{16 \sqrt [4]{e}}+\frac {1}{16} \left (7 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{3+x} \, dx-\frac {1}{16} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{5+3 x} \, dx+\frac {1}{4} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{(5+3 x)^2} \, dx\\ &=\frac {3}{4} e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}} \left (3+\log ^2(\log (4))\right )+\frac {e^{\frac {4+x}{5+3 x}} \left (3+\log ^2(\log (4))\right )}{3+x}-\frac {7 \text {Ei}\left (\frac {7 (3+x)}{4 (5+3 x)}\right ) \left (3+\log ^2(\log (4))\right )}{16 \sqrt [4]{e}}+\frac {1}{16} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{5+3 x} \, dx-\frac {1}{16} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}}}{5+3 x} \, dx-\frac {1}{4} \left (7 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {4+x}{5+3 x}}}{(3+x) (5+3 x)} \, dx+\frac {1}{4} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}}}{(5+3 x)^2} \, dx\\ &=\frac {e^{\frac {4+x}{5+3 x}} \left (3+\log ^2(\log (4))\right )}{3+x}+\frac {7}{16} \sqrt [3]{e} \text {Ei}\left (\frac {7}{3 (5+3 x)}\right ) \left (3+\log ^2(\log (4))\right )-\frac {7 \text {Ei}\left (\frac {7 (3+x)}{4 (5+3 x)}\right ) \left (3+\log ^2(\log (4))\right )}{16 \sqrt [4]{e}}+\frac {1}{16} \left (7 \left (3+\log ^2(\log (4))\right )\right ) \operatorname {Subst}\left (\int \frac {e^{-\frac {1}{4}+\frac {7 x}{4}}}{x} \, dx,x,\frac {3+x}{5+3 x}\right )+\frac {1}{16} \left (21 \left (3+\log ^2(\log (4))\right )\right ) \int \frac {e^{\frac {1}{3}+\frac {7}{3 (5+3 x)}}}{5+3 x} \, dx\\ &=\frac {e^{\frac {4+x}{5+3 x}} \left (3+\log ^2(\log (4))\right )}{3+x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.13, size = 26, normalized size = 0.93 \begin {gather*} \frac {e^{\frac {4+x}{5+3 x}} \left (3+\log ^2(\log (4))\right )}{3+x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((4 + x)/(5 + 3*x))*(-138 - 111*x - 27*x^2) + E^((4 + x)/(5 + 3*x))*(-46 - 37*x - 9*x^2)*Log[Log[
4]]^2)/(225 + 420*x + 286*x^2 + 84*x^3 + 9*x^4),x]

[Out]

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

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fricas [A]  time = 0.54, size = 41, normalized size = 1.46 \begin {gather*} \frac {e^{\left (\frac {x + 4}{3 \, x + 5}\right )} \log \left (2 \, \log \relax (2)\right )^{2} + 3 \, e^{\left (\frac {x + 4}{3 \, x + 5}\right )}}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-37*x-46)*exp((4+x)/(3*x+5))*log(2*log(2))^2+(-27*x^2-111*x-138)*exp((4+x)/(3*x+5)))/(9*x^4+
84*x^3+286*x^2+420*x+225),x, algorithm="fricas")

[Out]

(e^((x + 4)/(3*x + 5))*log(2*log(2))^2 + 3*e^((x + 4)/(3*x + 5)))/(x + 3)

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giac [B]  time = 2.22, size = 198, normalized size = 7.07 \begin {gather*} \frac {\frac {3 \, {\left (x + 4\right )} e^{\left (\frac {x + 4}{3 \, x + 5}\right )} \log \relax (2)^{2}}{3 \, x + 5} - e^{\left (\frac {x + 4}{3 \, x + 5}\right )} \log \relax (2)^{2} + \frac {6 \, {\left (x + 4\right )} e^{\left (\frac {x + 4}{3 \, x + 5}\right )} \log \relax (2) \log \left (\log \relax (2)\right )}{3 \, x + 5} - 2 \, e^{\left (\frac {x + 4}{3 \, x + 5}\right )} \log \relax (2) \log \left (\log \relax (2)\right ) + \frac {3 \, {\left (x + 4\right )} e^{\left (\frac {x + 4}{3 \, x + 5}\right )} \log \left (\log \relax (2)\right )^{2}}{3 \, x + 5} - e^{\left (\frac {x + 4}{3 \, x + 5}\right )} \log \left (\log \relax (2)\right )^{2} + \frac {9 \, {\left (x + 4\right )} e^{\left (\frac {x + 4}{3 \, x + 5}\right )}}{3 \, x + 5} - 3 \, e^{\left (\frac {x + 4}{3 \, x + 5}\right )}}{\frac {4 \, {\left (x + 4\right )}}{3 \, x + 5} + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-37*x-46)*exp((4+x)/(3*x+5))*log(2*log(2))^2+(-27*x^2-111*x-138)*exp((4+x)/(3*x+5)))/(9*x^4+
84*x^3+286*x^2+420*x+225),x, algorithm="giac")

[Out]

(3*(x + 4)*e^((x + 4)/(3*x + 5))*log(2)^2/(3*x + 5) - e^((x + 4)/(3*x + 5))*log(2)^2 + 6*(x + 4)*e^((x + 4)/(3
*x + 5))*log(2)*log(log(2))/(3*x + 5) - 2*e^((x + 4)/(3*x + 5))*log(2)*log(log(2)) + 3*(x + 4)*e^((x + 4)/(3*x
+ 5))*log(log(2))^2/(3*x + 5) - e^((x + 4)/(3*x + 5))*log(log(2))^2 + 9*(x + 4)*e^((x + 4)/(3*x + 5))/(3*x +
5) - 3*e^((x + 4)/(3*x + 5)))/(4*(x + 4)/(3*x + 5) + 1)

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maple [A]  time = 0.16, size = 28, normalized size = 1.00

method	result	size
gosper	\(\frac {\left (\ln \left (2 \ln \relax (2)\right )^{2}+3\right ) {\mathrm e}^{\frac {4+x}{3 x +5}}}{3+x}\)	\(28\)
risch	\(\frac {\left (\ln \relax (2)^{2}+2 \ln \relax (2) \ln \left (\ln \relax (2)\right )+\ln \left (\ln \relax (2)\right )^{2}+3\right ) {\mathrm e}^{\frac {4+x}{3 x +5}}}{3+x}\)	\(37\)
norman	\(\frac {\left (15+10 \ln \relax (2) \ln \left (\ln \relax (2)\right )+5 \ln \relax (2)^{2}+5 \ln \left (\ln \relax (2)\right )^{2}\right ) {\mathrm e}^{\frac {4+x}{3 x +5}}+\left (9+6 \ln \relax (2) \ln \left (\ln \relax (2)\right )+3 \ln \relax (2)^{2}+3 \ln \left (\ln \relax (2)\right )^{2}\right ) x \,{\mathrm e}^{\frac {4+x}{3 x +5}}}{3 x^{2}+14 x +15}\)	\(86\)
derivativedivides	\(\frac {9 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{4}-\frac {21 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{16 \left (\frac {7}{3 \left (3 x +5\right )}+\frac {7}{12}\right )}+\frac {3 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}} \ln \relax (2)^{2}}{4}-\frac {7 \ln \relax (2)^{2} {\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{16 \left (\frac {7}{3 \left (3 x +5\right )}+\frac {7}{12}\right )}+\frac {3 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}} \ln \left (\ln \relax (2)\right )^{2}}{4}-\frac {7 \ln \left (\ln \relax (2)\right )^{2} {\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{16 \left (\frac {7}{3 \left (3 x +5\right )}+\frac {7}{12}\right )}+\frac {3 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}} \ln \relax (2) \ln \left (\ln \relax (2)\right )}{2}-\frac {7 \ln \relax (2) \ln \left (\ln \relax (2)\right ) {\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{8 \left (\frac {7}{3 \left (3 x +5\right )}+\frac {7}{12}\right )}\)	\(194\)
default	\(\frac {9 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{4}-\frac {21 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{16 \left (\frac {7}{3 \left (3 x +5\right )}+\frac {7}{12}\right )}+\frac {3 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}} \ln \relax (2)^{2}}{4}-\frac {7 \ln \relax (2)^{2} {\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{16 \left (\frac {7}{3 \left (3 x +5\right )}+\frac {7}{12}\right )}+\frac {3 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}} \ln \left (\ln \relax (2)\right )^{2}}{4}-\frac {7 \ln \left (\ln \relax (2)\right )^{2} {\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{16 \left (\frac {7}{3 \left (3 x +5\right )}+\frac {7}{12}\right )}+\frac {3 \,{\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}} \ln \relax (2) \ln \left (\ln \relax (2)\right )}{2}-\frac {7 \ln \relax (2) \ln \left (\ln \relax (2)\right ) {\mathrm e}^{\frac {1}{3}+\frac {7}{3 \left (3 x +5\right )}}}{8 \left (\frac {7}{3 \left (3 x +5\right )}+\frac {7}{12}\right )}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-9*x^2-37*x-46)*exp((4+x)/(3*x+5))*ln(2*ln(2))^2+(-27*x^2-111*x-138)*exp((4+x)/(3*x+5)))/(9*x^4+84*x^3+2
86*x^2+420*x+225),x,method=_RETURNVERBOSE)

[Out]

(ln(2*ln(2))^2+3)/(3+x)*exp((4+x)/(3*x+5))

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maxima [A]  time = 0.69, size = 36, normalized size = 1.29 \begin {gather*} \frac {{\left (\log \relax (2)^{2} + 2 \, \log \relax (2) \log \left (\log \relax (2)\right ) + \log \left (\log \relax (2)\right )^{2} + 3\right )} e^{\left (\frac {7}{3 \, {\left (3 \, x + 5\right )}} + \frac {1}{3}\right )}}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x^2-37*x-46)*exp((4+x)/(3*x+5))*log(2*log(2))^2+(-27*x^2-111*x-138)*exp((4+x)/(3*x+5)))/(9*x^4+
84*x^3+286*x^2+420*x+225),x, algorithm="maxima")

[Out]

(log(2)^2 + 2*log(2)*log(log(2)) + log(log(2))^2 + 3)*e^(7/3/(3*x + 5) + 1/3)/(x + 3)

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mupad [B]  time = 0.19, size = 25, normalized size = 0.89 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x+4}{3\,x+5}}\,\left ({\ln \left (\ln \relax (4)\right )}^2+3\right )}{x+3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((x + 4)/(3*x + 5))*(111*x + 27*x^2 + 138) + log(2*log(2))^2*exp((x + 4)/(3*x + 5))*(37*x + 9*x^2 + 4
6))/(420*x + 286*x^2 + 84*x^3 + 9*x^4 + 225),x)

[Out]

(exp((x + 4)/(3*x + 5))*(log(log(4))^2 + 3))/(x + 3)

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sympy [A]  time = 0.24, size = 36, normalized size = 1.29 \begin {gather*} \frac {\left (2 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )} + \log {\left (\log {\relax (2 )} \right )}^{2} + \log {\relax (2 )}^{2} + 3\right ) e^{\frac {x + 4}{3 x + 5}}}{x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-9*x**2-37*x-46)*exp((4+x)/(3*x+5))*ln(2*ln(2))**2+(-27*x**2-111*x-138)*exp((4+x)/(3*x+5)))/(9*x**
4+84*x**3+286*x**2+420*x+225),x)

[Out]

(2*log(2)*log(log(2)) + log(log(2))**2 + log(2)**2 + 3)*exp((x + 4)/(3*x + 5))/(x + 3)

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