3.72.31 \(\int \frac {-10 x-2 x^3+e^2 (10+2 x^2)+e^5 (10+2 x^2)+(-10 x^2+8 x^3+2 x^4) \log ^2(-1+x)+(10+2 x^2) \log (\frac {-5-x}{x})+\log (-1+x) (10 x+10 x^2-6 x^3-2 x^4+e^2 (-10 x+8 x^2+2 x^3)+e^5 (-10 x+8 x^2+2 x^3)+(-10 x+8 x^2+2 x^3) \log (\frac {-5-x}{x}))}{-5 x+4 x^2+x^3} \, dx\)

Optimal. Leaf size=26 \[ \left (e^2+e^5-x+\log \left (-1-\frac {5}{x}\right )+x \log (-1+x)\right )^2 \]

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Rubi [A]  time = 0.51, antiderivative size = 30, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 5, integrand size = 172, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.029, Rules used = {6, 1594, 6688, 12, 6686} \begin {gather*} \left (-x+x \log (x-1)+\log \left (-\frac {x+5}{x}\right )+e^2 \left (1+e^3\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-10*x - 2*x^3 + E^2*(10 + 2*x^2) + E^5*(10 + 2*x^2) + (-10*x^2 + 8*x^3 + 2*x^4)*Log[-1 + x]^2 + (10 + 2*x
^2)*Log[(-5 - x)/x] + Log[-1 + x]*(10*x + 10*x^2 - 6*x^3 - 2*x^4 + E^2*(-10*x + 8*x^2 + 2*x^3) + E^5*(-10*x +
8*x^2 + 2*x^3) + (-10*x + 8*x^2 + 2*x^3)*Log[(-5 - x)/x]))/(-5*x + 4*x^2 + x^3),x]

[Out]

(E^2*(1 + E^3) - x + x*Log[-1 + x] + Log[-((5 + x)/x)])^2

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 1594

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^
(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] && PosQ[r - p]

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 x-2 x^3+\left (e^2+e^5\right ) \left (10+2 x^2\right )+\left (-10 x^2+8 x^3+2 x^4\right ) \log ^2(-1+x)+\left (10+2 x^2\right ) \log \left (\frac {-5-x}{x}\right )+\log (-1+x) \left (10 x+10 x^2-6 x^3-2 x^4+e^2 \left (-10 x+8 x^2+2 x^3\right )+e^5 \left (-10 x+8 x^2+2 x^3\right )+\left (-10 x+8 x^2+2 x^3\right ) \log \left (\frac {-5-x}{x}\right )\right )}{-5 x+4 x^2+x^3} \, dx\\ &=\int \frac {-10 x-2 x^3+\left (e^2+e^5\right ) \left (10+2 x^2\right )+\left (-10 x^2+8 x^3+2 x^4\right ) \log ^2(-1+x)+\left (10+2 x^2\right ) \log \left (\frac {-5-x}{x}\right )+\log (-1+x) \left (10 x+10 x^2-6 x^3-2 x^4+e^2 \left (-10 x+8 x^2+2 x^3\right )+e^5 \left (-10 x+8 x^2+2 x^3\right )+\left (-10 x+8 x^2+2 x^3\right ) \log \left (\frac {-5-x}{x}\right )\right )}{x \left (-5+4 x+x^2\right )} \, dx\\ &=\int \frac {2 \left (5+x^2+x \left (-5+4 x+x^2\right ) \log (-1+x)\right ) \left (-e^2 \left (1+e^3\right )+x-x \log (-1+x)-\log \left (-\frac {5+x}{x}\right )\right )}{x \left (5-4 x-x^2\right )} \, dx\\ &=2 \int \frac {\left (5+x^2+x \left (-5+4 x+x^2\right ) \log (-1+x)\right ) \left (-e^2 \left (1+e^3\right )+x-x \log (-1+x)-\log \left (-\frac {5+x}{x}\right )\right )}{x \left (5-4 x-x^2\right )} \, dx\\ &=\left (e^2 \left (1+e^3\right )-x+x \log (-1+x)+\log \left (-\frac {5+x}{x}\right )\right )^2\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.04, size = 27, normalized size = 1.04 \begin {gather*} \left (e^2+e^5-x+x \log (-1+x)+\log \left (-\frac {5+x}{x}\right )\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10*x - 2*x^3 + E^2*(10 + 2*x^2) + E^5*(10 + 2*x^2) + (-10*x^2 + 8*x^3 + 2*x^4)*Log[-1 + x]^2 + (10
+ 2*x^2)*Log[(-5 - x)/x] + Log[-1 + x]*(10*x + 10*x^2 - 6*x^3 - 2*x^4 + E^2*(-10*x + 8*x^2 + 2*x^3) + E^5*(-1
0*x + 8*x^2 + 2*x^3) + (-10*x + 8*x^2 + 2*x^3)*Log[(-5 - x)/x]))/(-5*x + 4*x^2 + x^3),x]

[Out]

(E^2 + E^5 - x + x*Log[-1 + x] + Log[-((5 + x)/x)])^2

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fricas [B]  time = 0.72, size = 88, normalized size = 3.38 \begin {gather*} x^{2} \log \left (x - 1\right )^{2} + x^{2} - 2 \, x e^{5} - 2 \, x e^{2} - 2 \, {\left (x^{2} - x e^{5} - x e^{2} - x \log \left (-\frac {x + 5}{x}\right )\right )} \log \left (x - 1\right ) - 2 \, {\left (x - e^{5} - e^{2}\right )} \log \left (-\frac {x + 5}{x}\right ) + \log \left (-\frac {x + 5}{x}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4+8*x^3-10*x^2)*log(x-1)^2+((2*x^3+8*x^2-10*x)*log((-x-5)/x)+(2*x^3+8*x^2-10*x)*exp(5)+(2*x^3+
8*x^2-10*x)*exp(2)-2*x^4-6*x^3+10*x^2+10*x)*log(x-1)+(2*x^2+10)*log((-x-5)/x)+(2*x^2+10)*exp(5)+(2*x^2+10)*exp
(2)-2*x^3-10*x)/(x^3+4*x^2-5*x),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.20, size = 149, normalized size = 5.73 \begin {gather*} x^{2} \log \left (x - 1\right )^{2} - 2 \, x^{2} \log \left (x - 1\right ) + 2 \, x e^{5} \log \left (x - 1\right ) + 2 \, x e^{2} \log \left (x - 1\right ) - 2 \, x \log \left (x - 1\right ) \log \relax (x) + 2 \, x \log \left (x - 1\right ) \log \left (-x - 5\right ) + x^{2} - 2 \, x e^{5} - 2 \, x e^{2} + 2 \, e^{5} \log \left (x + 5\right ) + 2 \, e^{2} \log \left (x + 5\right ) - \log \left (x + 5\right )^{2} + 2 \, x \log \relax (x) - 2 \, e^{5} \log \relax (x) - 2 \, e^{2} \log \relax (x) + \log \relax (x)^{2} - 2 \, x \log \left (-x - 5\right ) + 2 \, \log \left (x + 5\right ) \log \left (-x - 5\right ) - 2 \, \log \relax (x) \log \left (-x - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4+8*x^3-10*x^2)*log(x-1)^2+((2*x^3+8*x^2-10*x)*log((-x-5)/x)+(2*x^3+8*x^2-10*x)*exp(5)+(2*x^3+
8*x^2-10*x)*exp(2)-2*x^4-6*x^3+10*x^2+10*x)*log(x-1)+(2*x^2+10)*log((-x-5)/x)+(2*x^2+10)*exp(5)+(2*x^2+10)*exp
(2)-2*x^3-10*x)/(x^3+4*x^2-5*x),x, algorithm="giac")

[Out]

x^2*log(x - 1)^2 - 2*x^2*log(x - 1) + 2*x*e^5*log(x - 1) + 2*x*e^2*log(x - 1) - 2*x*log(x - 1)*log(x) + 2*x*lo
g(x - 1)*log(-x - 5) + x^2 - 2*x*e^5 - 2*x*e^2 + 2*e^5*log(x + 5) + 2*e^2*log(x + 5) - log(x + 5)^2 + 2*x*log(
x) - 2*e^5*log(x) - 2*e^2*log(x) + log(x)^2 - 2*x*log(-x - 5) + 2*log(x + 5)*log(-x - 5) - 2*log(x)*log(-x - 5
)

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maple [C]  time = 0.40, size = 604, normalized size = 23.23

method	result	size
risch	\(-2 \,{\mathrm e}^{2} x -2 x \,{\mathrm e}^{5}+\ln \left (5+x \right )^{2}+\ln \relax (x )^{2}+x^{2}+x^{2} \ln \left (x -1\right )^{2}-2 \,{\mathrm e}^{2} \ln \relax (x )-2 \,{\mathrm e}^{5} \ln \relax (x )+2 x \ln \relax (x )+2 i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}+2 i \pi x \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}-2 i \ln \left (5+x \right ) \pi \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}-i \pi x \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{3}-i \ln \relax (x ) \pi \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{3}+i \ln \left (5+x \right ) \pi \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{3}-2 x \ln \left (5+x \right )-2 \ln \relax (x ) \ln \left (5+x \right )+2 \ln \left (5+x \right ) {\mathrm e}^{5}+2 \ln \left (5+x \right ) {\mathrm e}^{2}+\left (2 x \ln \left (5+x \right )-2 x \ln \relax (x )-i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )+i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}-2 i \pi x \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}+i \pi x \,\mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}+i \pi x \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{3}+2 x \,{\mathrm e}^{5}+2 i \pi x +2 \,{\mathrm e}^{2} x -2 x^{2}\right ) \ln \left (x -1\right )+2 i \ln \left (5+x \right ) \pi -2 i \ln \relax (x ) \pi -2 i \pi x +i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )+i \ln \relax (x ) \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )-i \ln \left (5+x \right ) \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )-i \pi x \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}-i \pi x \,\mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}-i \ln \relax (x ) \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}+i \ln \left (5+x \right ) \pi \,\mathrm {csgn}\left (\frac {i}{x}\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}+i \ln \left (5+x \right ) \pi \,\mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}-i \ln \relax (x ) \pi \,\mathrm {csgn}\left (i \left (5+x \right )\right ) \mathrm {csgn}\left (\frac {i \left (5+x \right )}{x}\right )^{2}\)	\(604\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^4+8*x^3-10*x^2)*ln(x-1)^2+((2*x^3+8*x^2-10*x)*ln((-x-5)/x)+(2*x^3+8*x^2-10*x)*exp(5)+(2*x^3+8*x^2-10
*x)*exp(2)-2*x^4-6*x^3+10*x^2+10*x)*ln(x-1)+(2*x^2+10)*ln((-x-5)/x)+(2*x^2+10)*exp(5)+(2*x^2+10)*exp(2)-2*x^3-
10*x)/(x^3+4*x^2-5*x),x,method=_RETURNVERBOSE)

[Out]

-2*exp(2)*x-2*x*exp(5)+ln(5+x)^2+ln(x)^2+x^2+x^2*ln(x-1)^2-2*exp(2)*ln(x)-2*exp(5)*ln(x)+2*x*ln(x)+2*I*ln(x)*P
i*csgn(I/x*(5+x))^2+2*I*Pi*x*csgn(I/x*(5+x))^2-2*I*ln(5+x)*Pi*csgn(I/x*(5+x))^2+2*I*ln(5+x)*Pi-2*I*ln(x)*Pi-2*
x*ln(5+x)-2*ln(x)*ln(5+x)+2*ln(5+x)*exp(5)+2*ln(5+x)*exp(2)+(2*x*ln(5+x)-2*x*ln(x)-I*Pi*x*csgn(I/x)*csgn(I*(5+
x))*csgn(I/x*(5+x))+I*Pi*x*csgn(I/x)*csgn(I/x*(5+x))^2-2*I*Pi*x*csgn(I/x*(5+x))^2+I*Pi*x*csgn(I*(5+x))*csgn(I/
x*(5+x))^2+I*Pi*x*csgn(I/x*(5+x))^3+2*x*exp(5)+2*I*Pi*x+2*exp(2)*x-2*x^2)*ln(x-1)-2*I*Pi*x-I*ln(x)*Pi*csgn(I/x
*(5+x))^3-I*Pi*x*csgn(I/x)*csgn(I/x*(5+x))^2-I*Pi*x*csgn(I*(5+x))*csgn(I/x*(5+x))^2-I*ln(x)*Pi*csgn(I/x)*csgn(
I/x*(5+x))^2+I*ln(5+x)*Pi*csgn(I/x*(5+x))^3+I*ln(5+x)*Pi*csgn(I/x)*csgn(I/x*(5+x))^2+I*ln(x)*Pi*csgn(I/x)*csgn
(I*(5+x))*csgn(I/x*(5+x))+I*Pi*x*csgn(I/x)*csgn(I*(5+x))*csgn(I/x*(5+x))-I*ln(5+x)*Pi*csgn(I/x)*csgn(I*(5+x))*
csgn(I/x*(5+x))+I*ln(5+x)*Pi*csgn(I*(5+x))*csgn(I/x*(5+x))^2-I*ln(x)*Pi*csgn(I*(5+x))*csgn(I/x*(5+x))^2-I*Pi*x
*csgn(I/x*(5+x))^3

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maxima [B]  time = 0.42, size = 141, normalized size = 5.42 \begin {gather*} x^{2} \log \left (x - 1\right )^{2} + x^{2} - 2 \, x {\left (e^{5} + e^{2}\right )} + \frac {1}{3} \, {\left (\log \left (x + 5\right ) + 5 \, \log \left (x - 1\right ) - 6 \, \log \relax (x)\right )} e^{5} + \frac {1}{3} \, {\left (\log \left (x + 5\right ) + 5 \, \log \left (x - 1\right ) - 6 \, \log \relax (x)\right )} e^{2} - \frac {1}{3} \, {\left (6 \, x^{2} - 6 \, x {\left (e^{5} + e^{2}\right )} + 6 \, x \log \relax (x) + 5 \, e^{5} + 5 \, e^{2}\right )} \log \left (x - 1\right ) + 2 \, x \log \relax (x) + \log \relax (x)^{2} + \frac {1}{3} \, {\left (6 \, x \log \left (x - 1\right ) - 6 \, x + 5 \, e^{5} + 5 \, e^{2} - 6 \, \log \relax (x)\right )} \log \left (-x - 5\right ) + \log \left (-x - 5\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^4+8*x^3-10*x^2)*log(x-1)^2+((2*x^3+8*x^2-10*x)*log((-x-5)/x)+(2*x^3+8*x^2-10*x)*exp(5)+(2*x^3+
8*x^2-10*x)*exp(2)-2*x^4-6*x^3+10*x^2+10*x)*log(x-1)+(2*x^2+10)*log((-x-5)/x)+(2*x^2+10)*exp(5)+(2*x^2+10)*exp
(2)-2*x^3-10*x)/(x^3+4*x^2-5*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.42, size = 97, normalized size = 3.73 \begin {gather*} x^2-\ln \left (x-1\right )\,\left (2\,x^2-2\,x\,{\mathrm {e}}^2\,\left ({\mathrm {e}}^3+1\right )\right )-x\,\left (2\,{\mathrm {e}}^2+2\,{\mathrm {e}}^5\right )-\ln \left (-\frac {x+5}{x}\right )\,\left (2\,x-2\,x\,\ln \left (x-1\right )\right )+x^2\,{\ln \left (x-1\right )}^2+{\ln \left (-\frac {x+5}{x}\right )}^2-{\mathrm {e}}^2\,\mathrm {atan}\left (\frac {x\,2{}\mathrm {i}}{5}+1{}\mathrm {i}\right )\,\left ({\mathrm {e}}^3+1\right )\,4{}\mathrm {i} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x - 1)^2*(8*x^3 - 10*x^2 + 2*x^4) - 10*x + exp(2)*(2*x^2 + 10) + exp(5)*(2*x^2 + 10) + log(-(x + 5)/x
)*(2*x^2 + 10) + log(x - 1)*(10*x + exp(2)*(8*x^2 - 10*x + 2*x^3) + exp(5)*(8*x^2 - 10*x + 2*x^3) + log(-(x +
5)/x)*(8*x^2 - 10*x + 2*x^3) + 10*x^2 - 6*x^3 - 2*x^4) - 2*x^3)/(4*x^2 - 5*x + x^3),x)

[Out]

x^2 - log(x - 1)*(2*x^2 - 2*x*exp(2)*(exp(3) + 1)) - x*(2*exp(2) + 2*exp(5)) - log(-(x + 5)/x)*(2*x - 2*x*log(
x - 1)) + x^2*log(x - 1)^2 + log(-(x + 5)/x)^2 - exp(2)*atan((x*2i)/5 + 1i)*(exp(3) + 1)*4i

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sympy [B]  time = 0.94, size = 202, normalized size = 7.77 \begin {gather*} x^{2} \log {\left (x - 1 \right )}^{2} + x^{2} + x \left (- 2 e^{5} - 2 e^{2}\right ) + \left (2 x \log {\left (x - 1 \right )} - 2 x\right ) \log {\left (\frac {- x - 5}{x} \right )} + \left (- 2 x^{2} + 2 x e^{2} + 2 x e^{5}\right ) \log {\left (x - 1 \right )} + \log {\left (\frac {- x - 5}{x} \right )}^{2} - 2 \left (1 + e\right ) \left (- e + 1 + e^{2}\right ) e^{2} \log {\left (x + \frac {- 10 \left (1 + e\right ) \left (- e + 1 + e^{2}\right ) e^{2} + 10 e^{2} + 10 e^{5}}{4 e^{2} + 4 e^{5}} \right )} + 2 \left (1 + e\right ) \left (- e + 1 + e^{2}\right ) e^{2} \log {\left (x + \frac {10 e^{2} + 10 e^{5} + 10 \left (1 + e\right ) \left (- e + 1 + e^{2}\right ) e^{2}}{4 e^{2} + 4 e^{5}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**4+8*x**3-10*x**2)*ln(x-1)**2+((2*x**3+8*x**2-10*x)*ln((-x-5)/x)+(2*x**3+8*x**2-10*x)*exp(5)+(
2*x**3+8*x**2-10*x)*exp(2)-2*x**4-6*x**3+10*x**2+10*x)*ln(x-1)+(2*x**2+10)*ln((-x-5)/x)+(2*x**2+10)*exp(5)+(2*
x**2+10)*exp(2)-2*x**3-10*x)/(x**3+4*x**2-5*x),x)

[Out]

x**2*log(x - 1)**2 + x**2 + x*(-2*exp(5) - 2*exp(2)) + (2*x*log(x - 1) - 2*x)*log((-x - 5)/x) + (-2*x**2 + 2*x
*exp(2) + 2*x*exp(5))*log(x - 1) + log((-x - 5)/x)**2 - 2*(1 + E)*(-E + 1 + exp(2))*exp(2)*log(x + (-10*(1 + E
)*(-E + 1 + exp(2))*exp(2) + 10*exp(2) + 10*exp(5))/(4*exp(2) + 4*exp(5))) + 2*(1 + E)*(-E + 1 + exp(2))*exp(2
)*log(x + (10*exp(2) + 10*exp(5) + 10*(1 + E)*(-E + 1 + exp(2))*exp(2))/(4*exp(2) + 4*exp(5)))

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