3.85.39 \(\int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} (8-16 x-33 x^2-2 x^3)+(32 x^2+4 x^3) \log (2)+(8-x^2) \log ^2(2)+e^x (48 x^2+36 x^3+2 x^4+(16-16 x-34 x^2-2 x^3) \log (2))}{4 x^2} \, dx\)

Optimal. Leaf size=32 \[ -5-x-\left (4+\frac {2}{x}+\frac {x}{4}\right ) \left (-1+\left (e^x-x+\log (2)\right )^2\right ) \]

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Rubi [C]  time = 0.37, antiderivative size = 161, normalized size of antiderivative = 5.03, number of steps used = 24, number of rules used = 7, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.064, Rules used = {12, 14, 2199, 2194, 2177, 2178, 2176} \begin {gather*} \frac {1}{2} \log (256) \text {Ei}(x)-4 \log (2) \text {Ei}(x)-\frac {x^3}{4}+\frac {e^x x^2}{2}-\frac {1}{2} x^2 (8-\log (2))-e^x x-\frac {1}{4} e^{2 x} x+e^x-4 e^{2 x}-\frac {2 e^{2 x}}{x}-\frac {1}{4} x \left (11+\log ^2(2)-32 \log (2)\right )+\frac {2 \left (1-\log ^2(2)\right )}{x}+\frac {1}{2} e^x x (18-\log (2))-\frac {1}{2} e^x (18-\log (2))+\frac {1}{2} e^x (24-17 \log (2))-\frac {e^x \log (256)}{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-8 - 11*x^2 - 32*x^3 - 3*x^4 + E^(2*x)*(8 - 16*x - 33*x^2 - 2*x^3) + (32*x^2 + 4*x^3)*Log[2] + (8 - x^2)*
Log[2]^2 + E^x*(48*x^2 + 36*x^3 + 2*x^4 + (16 - 16*x - 34*x^2 - 2*x^3)*Log[2]))/(4*x^2),x]

[Out]

E^x - 4*E^(2*x) - (2*E^(2*x))/x - E^x*x - (E^(2*x)*x)/4 + (E^x*x^2)/2 - x^3/4 + (E^x*(24 - 17*Log[2]))/2 - (x^
2*(8 - Log[2]))/2 - (E^x*(18 - Log[2]))/2 + (E^x*x*(18 - Log[2]))/2 - 4*ExpIntegralEi[x]*Log[2] + (2*(1 - Log[
2]^2))/x - (x*(11 - 32*Log[2] + Log[2]^2))/4 - (E^x*Log[256])/(2*x) + (ExpIntegralEi[x]*Log[256])/2

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2177

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*(b*F^(g*(e + f*x)))^n)/(d*(m + 1)), x] - Dist[(f*g*n*Log[F])/(d*(m + 1)), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !$UseGamma ===
True

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 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2199

Int[(F_)^((c_.)*(v_))*(u_)^(m_.)*(w_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), w*NormalizePo
werOfLinear[u, x]^m, x], x] /; FreeQ[{F, c}, x] && PolynomialQ[w, x] && LinearQ[v, x] && PowerOfLinearQ[u, x]
&& IntegerQ[m] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {-8-11 x^2-32 x^3-3 x^4+e^{2 x} \left (8-16 x-33 x^2-2 x^3\right )+\left (32 x^2+4 x^3\right ) \log (2)+\left (8-x^2\right ) \log ^2(2)+e^x \left (48 x^2+36 x^3+2 x^4+\left (16-16 x-34 x^2-2 x^3\right ) \log (2)\right )}{x^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {e^{2 x} \left (-8+16 x+33 x^2+2 x^3\right )}{x^2}+\frac {-3 x^4-4 x^3 (8-\log (2))-8 \left (1-\log ^2(2)\right )-x^2 \left (11-32 \log (2)+\log ^2(2)\right )}{x^2}+\frac {2 e^x \left (x^4+x^2 (24-17 \log (2))+x^3 (18-\log (2))-8 x \log (2)+\log (256)\right )}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{4} \int \frac {e^{2 x} \left (-8+16 x+33 x^2+2 x^3\right )}{x^2} \, dx\right )+\frac {1}{4} \int \frac {-3 x^4-4 x^3 (8-\log (2))-8 \left (1-\log ^2(2)\right )-x^2 \left (11-32 \log (2)+\log ^2(2)\right )}{x^2} \, dx+\frac {1}{2} \int \frac {e^x \left (x^4+x^2 (24-17 \log (2))+x^3 (18-\log (2))-8 x \log (2)+\log (256)\right )}{x^2} \, dx\\ &=-\left (\frac {1}{4} \int \left (33 e^{2 x}-\frac {8 e^{2 x}}{x^2}+\frac {16 e^{2 x}}{x}+2 e^{2 x} x\right ) \, dx\right )+\frac {1}{4} \int \left (-3 x^2+4 x (-8+\log (2))-11 \left (1+\frac {1}{11} (-32+\log (2)) \log (2)\right )+\frac {8 \left (-1+\log ^2(2)\right )}{x^2}\right ) \, dx+\frac {1}{2} \int \left (e^x x^2+24 e^x \left (1-\frac {17 \log (2)}{24}\right )-e^x x (-18+\log (2))-\frac {8 e^x \log (2)}{x}+\frac {e^x \log (256)}{x^2}\right ) \, dx\\ &=-\frac {x^3}{4}-\frac {1}{2} x^2 (8-\log (2))+\frac {2 \left (1-\log ^2(2)\right )}{x}-\frac {1}{4} x \left (11-32 \log (2)+\log ^2(2)\right )-\frac {1}{2} \int e^{2 x} x \, dx+\frac {1}{2} \int e^x x^2 \, dx+2 \int \frac {e^{2 x}}{x^2} \, dx-4 \int \frac {e^{2 x}}{x} \, dx-\frac {33}{4} \int e^{2 x} \, dx+\frac {1}{2} (24-17 \log (2)) \int e^x \, dx+\frac {1}{2} (18-\log (2)) \int e^x x \, dx-(4 \log (2)) \int \frac {e^x}{x} \, dx+\frac {1}{2} \log (256) \int \frac {e^x}{x^2} \, dx\\ &=-\frac {33 e^{2 x}}{8}-\frac {2 e^{2 x}}{x}-\frac {1}{4} e^{2 x} x+\frac {e^x x^2}{2}-\frac {x^3}{4}-4 \text {Ei}(2 x)+\frac {1}{2} e^x (24-17 \log (2))-\frac {1}{2} x^2 (8-\log (2))+\frac {1}{2} e^x x (18-\log (2))-4 \text {Ei}(x) \log (2)+\frac {2 \left (1-\log ^2(2)\right )}{x}-\frac {1}{4} x \left (11-32 \log (2)+\log ^2(2)\right )-\frac {e^x \log (256)}{2 x}+\frac {1}{4} \int e^{2 x} \, dx+4 \int \frac {e^{2 x}}{x} \, dx+\frac {1}{2} (-18+\log (2)) \int e^x \, dx+\frac {1}{2} \log (256) \int \frac {e^x}{x} \, dx-\int e^x x \, dx\\ &=-4 e^{2 x}-\frac {2 e^{2 x}}{x}-e^x x-\frac {1}{4} e^{2 x} x+\frac {e^x x^2}{2}-\frac {x^3}{4}+\frac {1}{2} e^x (24-17 \log (2))-\frac {1}{2} x^2 (8-\log (2))-\frac {1}{2} e^x (18-\log (2))+\frac {1}{2} e^x x (18-\log (2))-4 \text {Ei}(x) \log (2)+\frac {2 \left (1-\log ^2(2)\right )}{x}-\frac {1}{4} x \left (11-32 \log (2)+\log ^2(2)\right )-\frac {e^x \log (256)}{2 x}+\frac {1}{2} \text {Ei}(x) \log (256)+\int e^x \, dx\\ &=e^x-4 e^{2 x}-\frac {2 e^{2 x}}{x}-e^x x-\frac {1}{4} e^{2 x} x+\frac {e^x x^2}{2}-\frac {x^3}{4}+\frac {1}{2} e^x (24-17 \log (2))-\frac {1}{2} x^2 (8-\log (2))-\frac {1}{2} e^x (18-\log (2))+\frac {1}{2} e^x x (18-\log (2))-4 \text {Ei}(x) \log (2)+\frac {2 \left (1-\log ^2(2)\right )}{x}-\frac {1}{4} x \left (11-32 \log (2)+\log ^2(2)\right )-\frac {e^x \log (256)}{2 x}+\frac {1}{2} \text {Ei}(x) \log (256)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.14, size = 82, normalized size = 2.56 \begin {gather*} -\frac {x^4+e^{2 x} \left (8+16 x+x^2\right )+8 \left (-1+\log ^2(2)\right )-x^3 (-16+\log (4))+e^x \left (-2 x^3+x^2 (-32+\log (4))+\log (65536)+x (-16+\log (4294967296))\right )+x^2 \left (11+\log ^2(2)-\log (4294967296)\right )}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-8 - 11*x^2 - 32*x^3 - 3*x^4 + E^(2*x)*(8 - 16*x - 33*x^2 - 2*x^3) + (32*x^2 + 4*x^3)*Log[2] + (8 -
x^2)*Log[2]^2 + E^x*(48*x^2 + 36*x^3 + 2*x^4 + (16 - 16*x - 34*x^2 - 2*x^3)*Log[2]))/(4*x^2),x]

[Out]

-1/4*(x^4 + E^(2*x)*(8 + 16*x + x^2) + 8*(-1 + Log[2]^2) - x^3*(-16 + Log[4]) + E^x*(-2*x^3 + x^2*(-32 + Log[4
]) + Log[65536] + x*(-16 + Log[4294967296])) + x^2*(11 + Log[2]^2 - Log[4294967296]))/x

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fricas [B]  time = 0.58, size = 84, normalized size = 2.62 \begin {gather*} -\frac {x^{4} + 16 \, x^{3} + {\left (x^{2} + 8\right )} \log \relax (2)^{2} + 11 \, x^{2} + {\left (x^{2} + 16 \, x + 8\right )} e^{\left (2 \, x\right )} - 2 \, {\left (x^{3} + 16 \, x^{2} - {\left (x^{2} + 16 \, x + 8\right )} \log \relax (2) + 8 \, x\right )} e^{x} - 2 \, {\left (x^{3} + 16 \, x^{2}\right )} \log \relax (2) - 8}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*x^3-33*x^2-16*x+8)*exp(x)^2+((-2*x^3-34*x^2-16*x+16)*log(2)+2*x^4+36*x^3+48*x^2)*exp(x)+(-x
^2+8)*log(2)^2+(4*x^3+32*x^2)*log(2)-3*x^4-32*x^3-11*x^2-8)/x^2,x, algorithm="fricas")

[Out]

-1/4*(x^4 + 16*x^3 + (x^2 + 8)*log(2)^2 + 11*x^2 + (x^2 + 16*x + 8)*e^(2*x) - 2*(x^3 + 16*x^2 - (x^2 + 16*x +
8)*log(2) + 8*x)*e^x - 2*(x^3 + 16*x^2)*log(2) - 8)/x

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giac [B]  time = 0.14, size = 110, normalized size = 3.44 \begin {gather*} -\frac {x^{4} - 2 \, x^{3} e^{x} - 2 \, x^{3} \log \relax (2) + 2 \, x^{2} e^{x} \log \relax (2) + x^{2} \log \relax (2)^{2} + 16 \, x^{3} + x^{2} e^{\left (2 \, x\right )} - 32 \, x^{2} e^{x} - 32 \, x^{2} \log \relax (2) + 32 \, x e^{x} \log \relax (2) + 11 \, x^{2} + 16 \, x e^{\left (2 \, x\right )} - 16 \, x e^{x} + 16 \, e^{x} \log \relax (2) + 8 \, \log \relax (2)^{2} + 8 \, e^{\left (2 \, x\right )} - 8}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*x^3-33*x^2-16*x+8)*exp(x)^2+((-2*x^3-34*x^2-16*x+16)*log(2)+2*x^4+36*x^3+48*x^2)*exp(x)+(-x
^2+8)*log(2)^2+(4*x^3+32*x^2)*log(2)-3*x^4-32*x^3-11*x^2-8)/x^2,x, algorithm="giac")

[Out]

-1/4*(x^4 - 2*x^3*e^x - 2*x^3*log(2) + 2*x^2*e^x*log(2) + x^2*log(2)^2 + 16*x^3 + x^2*e^(2*x) - 32*x^2*e^x - 3
2*x^2*log(2) + 32*x*e^x*log(2) + 11*x^2 + 16*x*e^(2*x) - 16*x*e^x + 16*e^x*log(2) + 8*log(2)^2 + 8*e^(2*x) - 8
)/x

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maple [B]  time = 0.07, size = 101, normalized size = 3.16

method	result	size
norman	\(\frac {\left (\frac {\ln \relax (2)}{2}-4\right ) x^{3}+\left (-\frac {\ln \relax (2)^{2}}{4}+8 \ln \relax (2)-\frac {11}{4}\right ) x^{2}+\left (4-8 \ln \relax (2)\right ) x \,{\mathrm e}^{x}+\left (8-\frac {\ln \relax (2)}{2}\right ) x^{2} {\mathrm e}^{x}-\frac {x^{4}}{4}-2 \,{\mathrm e}^{2 x}-4 x \,{\mathrm e}^{2 x}+\frac {{\mathrm e}^{x} x^{3}}{2}-4 \,{\mathrm e}^{x} \ln \relax (2)-\frac {{\mathrm e}^{2 x} x^{2}}{4}-2 \ln \relax (2)^{2}+2}{x}\)	\(101\)
risch	\(-\frac {x \ln \relax (2)^{2}}{4}+\frac {x^{2} \ln \relax (2)}{2}-\frac {x^{3}}{4}+8 x \ln \relax (2)-4 x^{2}-\frac {11 x}{4}-\frac {2 \ln \relax (2)^{2}}{x}+\frac {2}{x}-\frac {\left (x^{2}+16 x +8\right ) {\mathrm e}^{2 x}}{4 x}-\frac {\left (x^{2} \ln \relax (2)-x^{3}+16 x \ln \relax (2)-16 x^{2}+8 \ln \relax (2)-8 x \right ) {\mathrm e}^{x}}{2 x}\)	\(101\)
default	\(-4 x^{2}-\frac {11 x}{4}+\frac {2}{x}-\frac {x^{3}}{4}-4 \,{\mathrm e}^{2 x}-\frac {x \ln \relax (2)^{2}}{4}-\frac {x \,{\mathrm e}^{2 x}}{4}+\frac {x^{2} \ln \relax (2)}{2}+8 \,{\mathrm e}^{x} x +4 \,{\mathrm e}^{x}+\frac {{\mathrm e}^{x} x^{2}}{2}-8 \,{\mathrm e}^{x} \ln \relax (2)-\frac {2 \,{\mathrm e}^{2 x}}{x}-\frac {2 \ln \relax (2)^{2}}{x}-\frac {4 \ln \relax (2) {\mathrm e}^{x}}{x}-\frac {x \ln \relax (2) {\mathrm e}^{x}}{2}+8 x \ln \relax (2)\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((-2*x^3-33*x^2-16*x+8)*exp(x)^2+((-2*x^3-34*x^2-16*x+16)*ln(2)+2*x^4+36*x^3+48*x^2)*exp(x)+(-x^2+8)*l
n(2)^2+(4*x^3+32*x^2)*ln(2)-3*x^4-32*x^3-11*x^2-8)/x^2,x,method=_RETURNVERBOSE)

[Out]

((1/2*ln(2)-4)*x^3+(-1/4*ln(2)^2+8*ln(2)-11/4)*x^2+(4-8*ln(2))*x*exp(x)+(8-1/2*ln(2))*x^2*exp(x)-1/4*x^4-2*exp
(x)^2-4*x*exp(x)^2+1/2*exp(x)*x^3-4*exp(x)*ln(2)-1/4*exp(x)^2*x^2-2*ln(2)^2+2)/x

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maxima [C]  time = 0.39, size = 130, normalized size = 4.06 \begin {gather*} -\frac {1}{4} \, x^{3} + \frac {1}{2} \, x^{2} \log \relax (2) - \frac {1}{2} \, {\left (x - 1\right )} e^{x} \log \relax (2) - \frac {1}{4} \, x \log \relax (2)^{2} - 4 \, x^{2} - \frac {1}{8} \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} + \frac {1}{2} \, {\left (x^{2} - 2 \, x + 2\right )} e^{x} + 9 \, {\left (x - 1\right )} e^{x} + 8 \, x \log \relax (2) - 4 \, {\rm Ei}\relax (x) \log \relax (2) - \frac {17}{2} \, e^{x} \log \relax (2) + 4 \, \Gamma \left (-1, -x\right ) \log \relax (2) - \frac {11}{4} \, x - \frac {2 \, \log \relax (2)^{2}}{x} + \frac {2}{x} - 4 \, {\rm Ei}\left (2 \, x\right ) - \frac {33}{8} \, e^{\left (2 \, x\right )} + 12 \, e^{x} + 4 \, \Gamma \left (-1, -2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*x^3-33*x^2-16*x+8)*exp(x)^2+((-2*x^3-34*x^2-16*x+16)*log(2)+2*x^4+36*x^3+48*x^2)*exp(x)+(-x
^2+8)*log(2)^2+(4*x^3+32*x^2)*log(2)-3*x^4-32*x^3-11*x^2-8)/x^2,x, algorithm="maxima")

[Out]

-1/4*x^3 + 1/2*x^2*log(2) - 1/2*(x - 1)*e^x*log(2) - 1/4*x*log(2)^2 - 4*x^2 - 1/8*(2*x - 1)*e^(2*x) + 1/2*(x^2
- 2*x + 2)*e^x + 9*(x - 1)*e^x + 8*x*log(2) - 4*Ei(x)*log(2) - 17/2*e^x*log(2) + 4*gamma(-1, -x)*log(2) - 11/
4*x - 2*log(2)^2/x + 2/x - 4*Ei(2*x) - 33/8*e^(2*x) + 12*e^x + 4*gamma(-1, -2*x)

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mupad [B]  time = 5.22, size = 90, normalized size = 2.81 \begin {gather*} x^2\,\left (\frac {\ln \relax (4)}{4}+\frac {{\mathrm {e}}^x}{2}-4\right )-\frac {2\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^x\,\ln \relax (2)+2\,{\ln \relax (2)}^2-2}{x}-{\mathrm {e}}^x\,\left (8\,\ln \relax (2)-4\right )-4\,{\mathrm {e}}^{2\,x}-x\,\left (\frac {{\mathrm {e}}^{2\,x}}{4}-8\,\ln \relax (2)+\frac {{\mathrm {e}}^x\,\left (\ln \relax (4)-32\right )}{4}+\frac {{\ln \relax (2)}^2}{4}+\frac {11}{4}\right )-\frac {x^3}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((exp(2*x)*(16*x + 33*x^2 + 2*x^3 - 8))/4 + (log(2)^2*(x^2 - 8))/4 - (log(2)*(32*x^2 + 4*x^3))/4 - (exp(x
)*(48*x^2 - log(2)*(16*x + 34*x^2 + 2*x^3 - 16) + 36*x^3 + 2*x^4))/4 + (11*x^2)/4 + 8*x^3 + (3*x^4)/4 + 2)/x^2
,x)

[Out]

x^2*(log(4)/4 + exp(x)/2 - 4) - (2*exp(2*x) + 4*exp(x)*log(2) + 2*log(2)^2 - 2)/x - exp(x)*(8*log(2) - 4) - 4*
exp(2*x) - x*(exp(2*x)/4 - 8*log(2) + (exp(x)*(log(4) - 32))/4 + log(2)^2/4 + 11/4) - x^3/4

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sympy [B]  time = 0.26, size = 110, normalized size = 3.44 \begin {gather*} - \frac {x^{3}}{4} - \frac {x^{2} \left (16 - 2 \log {\relax (2 )}\right )}{4} - \frac {x \left (- 32 \log {\relax (2 )} + \log {\relax (2 )}^{2} + 11\right )}{4} - \frac {-8 + 8 \log {\relax (2 )}^{2}}{4 x} + \frac {\left (- 2 x^{3} - 32 x^{2} - 16 x\right ) e^{2 x} + \left (4 x^{4} - 4 x^{3} \log {\relax (2 )} + 64 x^{3} - 64 x^{2} \log {\relax (2 )} + 32 x^{2} - 32 x \log {\relax (2 )}\right ) e^{x}}{8 x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((-2*x**3-33*x**2-16*x+8)*exp(x)**2+((-2*x**3-34*x**2-16*x+16)*ln(2)+2*x**4+36*x**3+48*x**2)*exp
(x)+(-x**2+8)*ln(2)**2+(4*x**3+32*x**2)*ln(2)-3*x**4-32*x**3-11*x**2-8)/x**2,x)

[Out]

-x**3/4 - x**2*(16 - 2*log(2))/4 - x*(-32*log(2) + log(2)**2 + 11)/4 - (-8 + 8*log(2)**2)/(4*x) + ((-2*x**3 -
32*x**2 - 16*x)*exp(2*x) + (4*x**4 - 4*x**3*log(2) + 64*x**3 - 64*x**2*log(2) + 32*x**2 - 32*x*log(2))*exp(x))
/(8*x**2)

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