\(\int \frac {e^{-x-8 x^2} (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9))}{(25+10 x+x^2) \log ^4(9)} \, dx\) [7488]

Optimal result

Integrand size = 60, antiderivative size = 29 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {-x+\frac {e^{-x-8 x^2} x^2}{\log ^4(9)}}{5+x} \]

[Out]

(1/16*x^2/ln(3)^4/exp(2*x^2)^4/exp(x)-x)/(5+x)

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(64\) vs. \(2(29)=58\).

Time = 1.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21, number of steps used = 42, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {12, 27, 6874, 2274, 2266, 2236, 2272, 2273} \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {e^{-8 x^2-x} x}{\log ^4(9)}-\frac {5 e^{-8 x^2-x}}{\log ^4(9)}+\frac {25 e^{-8 x^2-x}}{(x+5) \log ^4(9)}+\frac {5}{x+5} \]

[In]

Int[(E^(-x - 8*x^2)*(10*x - 4*x^2 - 81*x^3 - 16*x^4 - 5*E^(x + 8*x^2)*Log[9]^4))/((25 + 10*x + x^2)*Log[9]^4),
x]

[Out]

5/(5 + x) - (5*E^(-x - 8*x^2))/Log[9]^4 + (E^(-x - 8*x^2)*x)/Log[9]^4 + (25*E^(-x - 8*x^2))/((5 + x)*Log[9]^4)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 2236

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

Rule 2266

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[F^(a - b^2/(4*c)), Int[F^((b + 2*c*x)^2/(4*c))
, x], x] /; FreeQ[{F, a, b, c}, x]

Rule 2272

Int[(F_)^((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)*((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[e*(F^(a + b*x + c*x^2)/(2
*c*Log[F])), x] - Dist[(b*e - 2*c*d)/(2*c), Int[F^(a + b*x + c*x^2), x], x] /; FreeQ[{F, a, b, c, d, e}, x] &&
NeQ[b*e - 2*c*d, 0]

Rule 2273

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

Rule 2274

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{25+10 x+x^2} \, dx}{\log ^4(9)} \\ & = \frac {\int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{(5+x)^2} \, dx}{\log ^4(9)} \\ & = \frac {\int \left (\frac {10 e^{-x-8 x^2} x}{(5+x)^2}-\frac {4 e^{-x-8 x^2} x^2}{(5+x)^2}-\frac {81 e^{-x-8 x^2} x^3}{(5+x)^2}-\frac {16 e^{-x-8 x^2} x^4}{(5+x)^2}-\frac {5 \log ^4(9)}{(5+x)^2}\right ) \, dx}{\log ^4(9)} \\ & = \frac {5}{5+x}-\frac {4 \int \frac {e^{-x-8 x^2} x^2}{(5+x)^2} \, dx}{\log ^4(9)}+\frac {10 \int \frac {e^{-x-8 x^2} x}{(5+x)^2} \, dx}{\log ^4(9)}-\frac {16 \int \frac {e^{-x-8 x^2} x^4}{(5+x)^2} \, dx}{\log ^4(9)}-\frac {81 \int \frac {e^{-x-8 x^2} x^3}{(5+x)^2} \, dx}{\log ^4(9)} \\ & = \frac {5}{5+x}-\frac {4 \int \left (e^{-x-8 x^2}+\frac {25 e^{-x-8 x^2}}{(5+x)^2}-\frac {10 e^{-x-8 x^2}}{5+x}\right ) \, dx}{\log ^4(9)}+\frac {10 \int \left (-\frac {5 e^{-x-8 x^2}}{(5+x)^2}+\frac {e^{-x-8 x^2}}{5+x}\right ) \, dx}{\log ^4(9)}-\frac {16 \int \left (75 e^{-x-8 x^2}-10 e^{-x-8 x^2} x+e^{-x-8 x^2} x^2+\frac {625 e^{-x-8 x^2}}{(5+x)^2}-\frac {500 e^{-x-8 x^2}}{5+x}\right ) \, dx}{\log ^4(9)}-\frac {81 \int \left (-10 e^{-x-8 x^2}+e^{-x-8 x^2} x-\frac {125 e^{-x-8 x^2}}{(5+x)^2}+\frac {75 e^{-x-8 x^2}}{5+x}\right ) \, dx}{\log ^4(9)} \\ & = \frac {5}{5+x}-\frac {4 \int e^{-x-8 x^2} \, dx}{\log ^4(9)}+\frac {10 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {16 \int e^{-x-8 x^2} x^2 \, dx}{\log ^4(9)}+\frac {40 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {50 \int \frac {e^{-x-8 x^2}}{(5+x)^2} \, dx}{\log ^4(9)}-\frac {81 \int e^{-x-8 x^2} x \, dx}{\log ^4(9)}-\frac {100 \int \frac {e^{-x-8 x^2}}{(5+x)^2} \, dx}{\log ^4(9)}+\frac {160 \int e^{-x-8 x^2} x \, dx}{\log ^4(9)}+\frac {810 \int e^{-x-8 x^2} \, dx}{\log ^4(9)}-\frac {1200 \int e^{-x-8 x^2} \, dx}{\log ^4(9)}-\frac {6075 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {8000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {10000 \int \frac {e^{-x-8 x^2}}{(5+x)^2} \, dx}{\log ^4(9)}+\frac {10125 \int \frac {e^{-x-8 x^2}}{(5+x)^2} \, dx}{\log ^4(9)} \\ & = \frac {5}{5+x}-\frac {79 e^{-x-8 x^2}}{16 \log ^4(9)}+\frac {e^{-x-8 x^2} x}{\log ^4(9)}+\frac {25 e^{-x-8 x^2}}{(5+x) \log ^4(9)}-\frac {\int e^{-x-8 x^2} \, dx}{\log ^4(9)}+\frac {\int e^{-x-8 x^2} x \, dx}{\log ^4(9)}+\frac {81 \int e^{-x-8 x^2} \, dx}{16 \log ^4(9)}-\frac {10 \int e^{-x-8 x^2} \, dx}{\log ^4(9)}+\frac {10 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {40 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {800 \int e^{-x-8 x^2} \, dx}{\log ^4(9)}+\frac {1600 \int e^{-x-8 x^2} \, dx}{\log ^4(9)}-\frac {3950 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {6075 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {7900 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {8000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {160000 \int e^{-x-8 x^2} \, dx}{\log ^4(9)}-\frac {162000 \int e^{-x-8 x^2} \, dx}{\log ^4(9)}-\frac {790000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {799875 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {\left (4 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)}+\frac {\left (810 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)}-\frac {\left (1200 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)} \\ & = \frac {5}{5+x}-\frac {5 e^{-x-8 x^2}}{\log ^4(9)}+\frac {e^{-x-8 x^2} x}{\log ^4(9)}+\frac {25 e^{-x-8 x^2}}{(5+x) \log ^4(9)}+\frac {403 \sqrt [32]{e} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {1+16 x}{4 \sqrt {2}}\right )}{2 \log ^4(9)}-\frac {150 \sqrt [32]{e} \sqrt {2 \pi } \text {erf}\left (\frac {1+16 x}{4 \sqrt {2}}\right )}{\log ^4(9)}-\frac {\int e^{-x-8 x^2} \, dx}{16 \log ^4(9)}+\frac {10 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {40 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {3950 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {6075 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {7900 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {8000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {790000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {799875 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {\sqrt [32]{e} \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)}+\frac {\left (81 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{16 \log ^4(9)}-\frac {\left (10 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)}+\frac {\left (800 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)}+\frac {\left (1600 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)}+\frac {\left (160000 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)}-\frac {\left (162000 \sqrt [32]{e}\right ) \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{\log ^4(9)} \\ & = \frac {5}{5+x}-\frac {5 e^{-x-8 x^2}}{\log ^4(9)}+\frac {e^{-x-8 x^2} x}{\log ^4(9)}+\frac {25 e^{-x-8 x^2}}{(5+x) \log ^4(9)}+\frac {12801 \sqrt [32]{e} \sqrt {\frac {\pi }{2}} \text {erf}\left (\frac {1+16 x}{4 \sqrt {2}}\right )}{64 \log ^4(9)}-\frac {100 \sqrt [32]{e} \sqrt {2 \pi } \text {erf}\left (\frac {1+16 x}{4 \sqrt {2}}\right )}{\log ^4(9)}+\frac {10 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {40 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {3950 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {6075 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {7900 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {8000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {790000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {799875 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {\sqrt [32]{e} \int e^{-\frac {1}{32} (-1-16 x)^2} \, dx}{16 \log ^4(9)} \\ & = \frac {5}{5+x}-\frac {5 e^{-x-8 x^2}}{\log ^4(9)}+\frac {e^{-x-8 x^2} x}{\log ^4(9)}+\frac {25 e^{-x-8 x^2}}{(5+x) \log ^4(9)}+\frac {10 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {40 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {3950 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {6075 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {7900 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {8000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}-\frac {790000 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)}+\frac {799875 \int \frac {e^{-x-8 x^2}}{5+x} \, dx}{\log ^4(9)} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.61 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.07 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {e^{-x (1+8 x)} x^2+5 \log ^4(9)}{(5+x) \log ^4(9)} \]

[In]

Integrate[(E^(-x - 8*x^2)*(10*x - 4*x^2 - 81*x^3 - 16*x^4 - 5*E^(x + 8*x^2)*Log[9]^4))/((25 + 10*x + x^2)*Log[
9]^4),x]

[Out]

(x^2/E^(x*(1 + 8*x)) + 5*Log[9]^4)/((5 + x)*Log[9]^4)

Maple [A] (verified)

Time = 0.52 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10

method	result	size
risch	\(\frac {5}{5+x}+\frac {x^{2} {\mathrm e}^{-x \left (8 x +1\right )}}{16 \left (5+x \right ) \ln \left (3\right )^{4}}\)	\(32\)
parallelrisch	\(\frac {\left (80 \ln \left (3\right )^{4} {\mathrm e}^{x} {\mathrm e}^{8 x^{2}}+x^{2}\right ) {\mathrm e}^{-x} {\mathrm e}^{-8 x^{2}}}{16 \ln \left (3\right )^{4} \left (5+x \right )}\)	\(44\)

[In]

int(1/16*(-80*ln(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x)/(x^2+10*x+25)/ln(3)^4/exp(x)/exp(2*x^2)^4,
x,method=_RETURNVERBOSE)

[Out]

5/(5+x)+1/16/(5+x)*x^2/ln(3)^4*exp(-x*(8*x+1))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.34 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {{\left (80 \, e^{\left (8 \, x^{2} + x\right )} \log \left (3\right )^{4} + x^{2}\right )} e^{\left (-8 \, x^{2} - x\right )}}{16 \, {\left (x + 5\right )} \log \left (3\right )^{4}} \]

[In]

integrate(1/16*(-80*log(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x)/(x^2+10*x+25)/log(3)^4/exp(x)/exp(2
*x^2)^4,x, algorithm="fricas")

[Out]

1/16*(80*e^(8*x^2 + x)*log(3)^4 + x^2)*e^(-8*x^2 - x)/((x + 5)*log(3)^4)

Sympy [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {x^{2} e^{- 8 x^{2}}}{16 x e^{x} \log {\left (3 \right )}^{4} + 80 e^{x} \log {\left (3 \right )}^{4}} + \frac {5}{x + 5} \]

[In]

integrate(1/16*(-80*ln(3)**4*exp(x)*exp(2*x**2)**4-16*x**4-81*x**3-4*x**2+10*x)/(x**2+10*x+25)/ln(3)**4/exp(x)
/exp(2*x**2)**4,x)

[Out]

x**2*exp(-8*x**2)/(16*x*exp(x)*log(3)**4 + 80*exp(x)*log(3)**4) + 5/(x + 5)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.28 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {\frac {80 \, \log \left (3\right )^{4}}{x + 5} + \frac {x^{2} e^{\left (-8 \, x^{2} - x\right )}}{x + 5}}{16 \, \log \left (3\right )^{4}} \]

[In]

integrate(1/16*(-80*log(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x)/(x^2+10*x+25)/log(3)^4/exp(x)/exp(2
*x^2)^4,x, algorithm="maxima")

[Out]

1/16*(80*log(3)^4/(x + 5) + x^2*e^(-8*x^2 - x)/(x + 5))/log(3)^4

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {80 \, \log \left (3\right )^{4} + x^{2} e^{\left (-8 \, x^{2} - x\right )}}{16 \, {\left (x + 5\right )} \log \left (3\right )^{4}} \]

[In]

integrate(1/16*(-80*log(3)^4*exp(x)*exp(2*x^2)^4-16*x^4-81*x^3-4*x^2+10*x)/(x^2+10*x+25)/log(3)^4/exp(x)/exp(2
*x^2)^4,x, algorithm="giac")

[Out]

1/16*(80*log(3)^4 + x^2*e^(-8*x^2 - x))/((x + 5)*log(3)^4)

Mupad [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.79 \[ \int \frac {e^{-x-8 x^2} \left (10 x-4 x^2-81 x^3-16 x^4-5 e^{x+8 x^2} \log ^4(9)\right )}{\left (25+10 x+x^2\right ) \log ^4(9)} \, dx=\frac {x^2-16\,x\,{\mathrm {e}}^{8\,x^2+x}\,{\ln \left (3\right )}^4}{80\,{\mathrm {e}}^{8\,x^2+x}\,{\ln \left (3\right )}^4+16\,x\,{\mathrm {e}}^{8\,x^2+x}\,{\ln \left (3\right )}^4} \]

[In]

int(-(exp(-x)*exp(-8*x^2)*(x^2/4 - (5*x)/8 + (81*x^3)/16 + x^4 + 5*exp(8*x^2)*exp(x)*log(3)^4))/(log(3)^4*(10*
x + x^2 + 25)),x)

[Out]

(x^2 - 16*x*exp(x + 8*x^2)*log(3)^4)/(80*exp(x + 8*x^2)*log(3)^4 + 16*x*exp(x + 8*x^2)*log(3)^4)