3.80.43 \(\int \frac {40 x-5 x^3+x^4+e (100 x^2-40 x^3+4 x^4)+(-20 x+4 x^2+e (400-160 x+16 x^2)) \log (\frac {x^2+e (-40 x+8 x^2)+e^2 (400-160 x+16 x^2)}{400-160 x+16 x^2})}{-60 x^3+12 x^4+e (1200 x^2-480 x^3+48 x^4)} \, dx\)

Optimal. Leaf size=37 \[ \frac {-x+\frac {x^2}{4}-\log \left (\left (e-\frac {x}{4 (5-x)}\right )^2\right )}{3 x} \]

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Rubi [B]  time = 1.04, antiderivative size = 196, normalized size of antiderivative = 5.30, number of steps used = 16, number of rules used = 9, integrand size = 127, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6688, 12, 6742, 43, 72, 2490, 36, 29, 31} \begin {gather*} \frac {e x}{3 (1+4 e)}+\frac {x}{12 (1+4 e)}+\frac {2}{15} \log (5-x)-\frac {(1+4 e) \log (20 e-(1+4 e) x)}{30 e}+\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)}-\frac {20 e^2 \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}+\frac {\log (20 e-(1+4 e) x)}{30 e}-\frac {(5-x) \log \left (\frac {(20 e-(1+4 e) x)^2}{16 (5-x)^2}\right )}{15 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(40*x - 5*x^3 + x^4 + E*(100*x^2 - 40*x^3 + 4*x^4) + (-20*x + 4*x^2 + E*(400 - 160*x + 16*x^2))*Log[(x^2 +
E*(-40*x + 8*x^2) + E^2*(400 - 160*x + 16*x^2))/(400 - 160*x + 16*x^2)])/(-60*x^3 + 12*x^4 + E*(1200*x^2 - 48
0*x^3 + 48*x^4)),x]

[Out]

x/(12*(1 + 4*E)) + (E*x)/(3*(1 + 4*E)) + (2*Log[5 - x])/15 + Log[20*E - (1 + 4*E)*x]/(30*E) - (5*E*Log[20*E -
(1 + 4*E)*x])/(3*(1 + 4*E)^2) - (20*E^2*Log[20*E - (1 + 4*E)*x])/(3*(1 + 4*E)^2) + (5*E*Log[20*E - (1 + 4*E)*x
])/(3*(1 + 4*E)) - ((1 + 4*E)*Log[20*E - (1 + 4*E)*x])/(30*E) - ((5 - x)*Log[(20*E - (1 + 4*E)*x)^2/(16*(5 - x
)^2)])/(15*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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 2490

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)/((g_.) + (h_.)*(x_))^
2, x_Symbol] :> Simp[((a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/((b*g - a*h)*(g + h*x)), x] - Dist[(p*
r*s*(b*c - a*d))/(b*g - a*h), Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1)/((c + d*x)*(g + h*x)), x], x] /
; FreeQ[{a, b, c, d, e, f, g, h, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && EqQ[p + q, 0] && NeQ[b*g - a*h, 0] &&
IGtQ[s, 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 {40 x+4 e (-5+x)^2 x^2-5 x^3+x^4+4 (-5+x) (4 e (-5+x)+x) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )}{12 (5-x) x^2 (20 e-(1+4 e) x)} \, dx\\ &=\frac {1}{12} \int \frac {40 x+4 e (-5+x)^2 x^2-5 x^3+x^4+4 (-5+x) (4 e (-5+x)+x) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )}{(5-x) x^2 (20 e-(1+4 e) x)} \, dx\\ &=\frac {1}{12} \int \left (\frac {4 e (5-x)}{20 e-(1+4 e) x}+\frac {40}{(5-x) x (20 e-(1+4 e) x)}+\frac {5 x}{(-5+x) (20 e-(1+4 e) x)}+\frac {x^2}{(5-x) (20 e-(1+4 e) x)}+\frac {4 \log \left (\frac {(-20 e+(1+4 e) x)^2}{16 (-5+x)^2}\right )}{x^2}\right ) \, dx\\ &=\frac {1}{12} \int \frac {x^2}{(5-x) (20 e-(1+4 e) x)} \, dx+\frac {1}{3} \int \frac {\log \left (\frac {(-20 e+(1+4 e) x)^2}{16 (-5+x)^2}\right )}{x^2} \, dx+\frac {5}{12} \int \frac {x}{(-5+x) (20 e-(1+4 e) x)} \, dx+\frac {10}{3} \int \frac {1}{(5-x) x (20 e-(1+4 e) x)} \, dx+\frac {1}{3} e \int \frac {5-x}{20 e-(1+4 e) x} \, dx\\ &=-\frac {(5-x) \log \left (\frac {(20 e-(1+4 e) x)^2}{16 (5-x)^2}\right )}{15 x}+\frac {1}{12} \int \left (\frac {1}{1+4 e}+\frac {5}{-5+x}+\frac {80 e^2}{(1+4 e) (20 e-(1+4 e) x)}\right ) \, dx+\frac {5}{12} \int \left (\frac {1}{5-x}+\frac {4 e}{-20 e+(1+4 e) x}\right ) \, dx+\frac {2}{3} \int \frac {1}{x (-20 e+(1+4 e) x)} \, dx+\frac {10}{3} \int \left (\frac {1}{25 (-5+x)}+\frac {1}{100 e x}+\frac {(1+4 e)^2}{100 e (20 e-(1+4 e) x)}\right ) \, dx+\frac {1}{3} e \int \left (\frac {1}{1+4 e}+\frac {5}{(1+4 e) (20 e-(1+4 e) x)}\right ) \, dx\\ &=\frac {x}{12 (1+4 e)}+\frac {e x}{3 (1+4 e)}+\frac {2}{15} \log (5-x)+\frac {\log (x)}{30 e}-\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {20 e^2 \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}+\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)}-\frac {(1+4 e) \log (20 e-(1+4 e) x)}{30 e}-\frac {(5-x) \log \left (\frac {(20 e-(1+4 e) x)^2}{16 (5-x)^2}\right )}{15 x}-\frac {\int \frac {1}{x} \, dx}{30 e}+\frac {(1+4 e) \int \frac {1}{-20 e+(1+4 e) x} \, dx}{30 e}\\ &=\frac {x}{12 (1+4 e)}+\frac {e x}{3 (1+4 e)}+\frac {2}{15} \log (5-x)+\frac {\log (20 e-(1+4 e) x)}{30 e}-\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {20 e^2 \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}+\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)}-\frac {(1+4 e) \log (20 e-(1+4 e) x)}{30 e}-\frac {(5-x) \log \left (\frac {(20 e-(1+4 e) x)^2}{16 (5-x)^2}\right )}{15 x}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.39, size = 323, normalized size = 8.73 \begin {gather*} \frac {5 e x^2-(1+4 e)^2 x \log ^2(-4 e (-5+x)-x)+2 (1+4 e)^2 x \log \left (-\frac {1}{5} (1+4 e) (-5+x)\right ) \log \left (\frac {5}{4 e (-5+x)+x}\right )+x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )+8 e x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )+16 e^2 x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )-20 e \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+8 e x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+16 e^2 x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+(1+4 e)^2 x \log (-4 e (-5+x)-x) \left (2 \log \left (-\frac {1}{5} (1+4 e) (-5+x)\right )+\log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )\right )}{60 e x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(40*x - 5*x^3 + x^4 + E*(100*x^2 - 40*x^3 + 4*x^4) + (-20*x + 4*x^2 + E*(400 - 160*x + 16*x^2))*Log[
(x^2 + E*(-40*x + 8*x^2) + E^2*(400 - 160*x + 16*x^2))/(400 - 160*x + 16*x^2)])/(-60*x^3 + 12*x^4 + E*(1200*x^
2 - 480*x^3 + 48*x^4)),x]

[Out]

(5*E*x^2 - (1 + 4*E)^2*x*Log[-4*E*(-5 + x) - x]^2 + 2*(1 + 4*E)^2*x*Log[-1/5*((1 + 4*E)*(-5 + x))]*Log[5/(4*E*
(-5 + x) + x)] + x*Log[5/(4*E*(-5 + x) + x)]^2 + 8*E*x*Log[5/(4*E*(-5 + x) + x)]^2 + 16*E^2*x*Log[5/(4*E*(-5 +
x) + x)]^2 - 20*E*Log[(4*E*(-5 + x) + x)^2/(16*(-5 + x)^2)] + x*Log[5/(4*E*(-5 + x) + x)]*Log[(4*E*(-5 + x) +
x)^2/(16*(-5 + x)^2)] + 8*E*x*Log[5/(4*E*(-5 + x) + x)]*Log[(4*E*(-5 + x) + x)^2/(16*(-5 + x)^2)] + 16*E^2*x*
Log[5/(4*E*(-5 + x) + x)]*Log[(4*E*(-5 + x) + x)^2/(16*(-5 + x)^2)] + (1 + 4*E)^2*x*Log[-4*E*(-5 + x) - x]*(2*
Log[-1/5*((1 + 4*E)*(-5 + x))] + Log[(4*E*(-5 + x) + x)^2/(16*(-5 + x)^2)]))/(60*E*x)

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fricas [A]  time = 0.59, size = 51, normalized size = 1.38 \begin {gather*} \frac {x^{2} - 4 \, \log \left (\frac {x^{2} + 16 \, {\left (x^{2} - 10 \, x + 25\right )} e^{2} + 8 \, {\left (x^{2} - 5 \, x\right )} e}{16 \, {\left (x^{2} - 10 \, x + 25\right )}}\right )}{12 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2-160*x+400)*exp(1)+4*x^2-20*x)*log(((16*x^2-160*x+400)*exp(1)^2+(8*x^2-40*x)*exp(1)+x^2)/(1
6*x^2-160*x+400))+(4*x^4-40*x^3+100*x^2)*exp(1)+x^4-5*x^3+40*x)/((48*x^4-480*x^3+1200*x^2)*exp(1)+12*x^4-60*x^
3),x, algorithm="fricas")

[Out]

1/12*(x^2 - 4*log(1/16*(x^2 + 16*(x^2 - 10*x + 25)*e^2 + 8*(x^2 - 5*x)*e)/(x^2 - 10*x + 25)))/x

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giac [A]  time = 0.41, size = 56, normalized size = 1.51 \begin {gather*} \frac {x^{2} - 4 \, \log \left (\frac {16 \, x^{2} e^{2} + 8 \, x^{2} e + x^{2} - 160 \, x e^{2} - 40 \, x e + 400 \, e^{2}}{16 \, {\left (x^{2} - 10 \, x + 25\right )}}\right )}{12 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2-160*x+400)*exp(1)+4*x^2-20*x)*log(((16*x^2-160*x+400)*exp(1)^2+(8*x^2-40*x)*exp(1)+x^2)/(1
6*x^2-160*x+400))+(4*x^4-40*x^3+100*x^2)*exp(1)+x^4-5*x^3+40*x)/((48*x^4-480*x^3+1200*x^2)*exp(1)+12*x^4-60*x^
3),x, algorithm="giac")

[Out]

1/12*(x^2 - 4*log(1/16*(16*x^2*e^2 + 8*x^2*e + x^2 - 160*x*e^2 - 40*x*e + 400*e^2)/(x^2 - 10*x + 25)))/x

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maple [A]  time = 0.29, size = 53, normalized size = 1.43

method	result	size
risch	\(-\frac {\ln \left (\frac {\left (16 x^{2}-160 x +400\right ) {\mathrm e}^{2}+\left (8 x^{2}-40 x \right ) {\mathrm e}+x^{2}}{16 x^{2}-160 x +400}\right )}{3 x}+\frac {x}{12}\)	\(53\)
norman	\(\frac {\frac {x^{2}}{12}-\frac {\ln \left (\frac {\left (16 x^{2}-160 x +400\right ) {\mathrm e}^{2}+\left (8 x^{2}-40 x \right ) {\mathrm e}+x^{2}}{16 x^{2}-160 x +400}\right )}{3}}{x}\)	\(58\)
derivativedivides	\(-\frac {4 \ln \relax (2)}{15 \left (\frac {5}{x -5}+1\right )}-\frac {-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{x -5}+\frac {25}{\left (x -5\right )^{2}}+8 \,{\mathrm e}+\frac {10}{x -5}+1\right )}{60}-\frac {{\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{x -5}+\frac {25}{\left (x -5\right )^{2}}+8 \,{\mathrm e}+\frac {10}{x -5}+1\right )}{12 \left (x -5\right )}}{\frac {5}{x -5}+1}+\frac {x}{12}-\frac {5}{12}-\frac {2 \ln \left (4 \,{\mathrm e}+\frac {5}{x -5}+1\right )}{15}-\frac {{\mathrm e}^{-1} \ln \left (4 \,{\mathrm e}+\frac {5}{x -5}+1\right )}{30}\)	\(166\)
default	\(-\frac {4 \ln \relax (2)}{15 \left (\frac {5}{x -5}+1\right )}-\frac {-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{x -5}+\frac {25}{\left (x -5\right )^{2}}+8 \,{\mathrm e}+\frac {10}{x -5}+1\right )}{60}-\frac {{\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{x -5}+\frac {25}{\left (x -5\right )^{2}}+8 \,{\mathrm e}+\frac {10}{x -5}+1\right )}{12 \left (x -5\right )}}{\frac {5}{x -5}+1}+\frac {x}{12}-\frac {5}{12}-\frac {2 \ln \left (4 \,{\mathrm e}+\frac {5}{x -5}+1\right )}{15}-\frac {{\mathrm e}^{-1} \ln \left (4 \,{\mathrm e}+\frac {5}{x -5}+1\right )}{30}\)	\(166\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x^2-160*x+400)*exp(1)+4*x^2-20*x)*ln(((16*x^2-160*x+400)*exp(1)^2+(8*x^2-40*x)*exp(1)+x^2)/(16*x^2-1
60*x+400))+(4*x^4-40*x^3+100*x^2)*exp(1)+x^4-5*x^3+40*x)/((48*x^4-480*x^3+1200*x^2)*exp(1)+12*x^4-60*x^3),x,me
thod=_RETURNVERBOSE)

[Out]

-1/3/x*ln(((16*x^2-160*x+400)*exp(2)+(8*x^2-40*x)*exp(1)+x^2)/(16*x^2-160*x+400))+1/12*x

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maxima [B]  time = 0.54, size = 269, normalized size = 7.27 \begin {gather*} -\frac {1}{30} \, {\left (4 \, e + 1\right )} e^{\left (-1\right )} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - \frac {1}{3} \, {\left (\frac {80 \, e^{2} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{16 \, e^{2} + 8 \, e + 1} - \frac {x}{4 \, e + 1} - 5 \, \log \left (x - 5\right )\right )} e + \frac {10}{3} \, {\left (\frac {4 \, e \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{4 \, e + 1} - \log \left (x - 5\right )\right )} e - \frac {5}{3} \, {\left (\log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - \log \left (x - 5\right )\right )} e + \frac {{\left (40 \, e \log \relax (2) + {\left (x {\left (4 \, e + 1\right )} - 20 \, e\right )} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - 4 \, {\left (x e - 5 \, e\right )} \log \left (x - 5\right )\right )} e^{\left (-1\right )}}{30 \, x} - \frac {20 \, e^{2} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{3 \, {\left (16 \, e^{2} + 8 \, e + 1\right )}} + \frac {5 \, e \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{3 \, {\left (4 \, e + 1\right )}} + \frac {x}{12 \, {\left (4 \, e + 1\right )}} + \frac {2}{15} \, \log \left (x - 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x^2-160*x+400)*exp(1)+4*x^2-20*x)*log(((16*x^2-160*x+400)*exp(1)^2+(8*x^2-40*x)*exp(1)+x^2)/(1
6*x^2-160*x+400))+(4*x^4-40*x^3+100*x^2)*exp(1)+x^4-5*x^3+40*x)/((48*x^4-480*x^3+1200*x^2)*exp(1)+12*x^4-60*x^
3),x, algorithm="maxima")

[Out]

-1/30*(4*e + 1)*e^(-1)*log(x*(4*e + 1) - 20*e) - 1/3*(80*e^2*log(x*(4*e + 1) - 20*e)/(16*e^2 + 8*e + 1) - x/(4
*e + 1) - 5*log(x - 5))*e + 10/3*(4*e*log(x*(4*e + 1) - 20*e)/(4*e + 1) - log(x - 5))*e - 5/3*(log(x*(4*e + 1)
- 20*e) - log(x - 5))*e + 1/30*(40*e*log(2) + (x*(4*e + 1) - 20*e)*log(x*(4*e + 1) - 20*e) - 4*(x*e - 5*e)*lo
g(x - 5))*e^(-1)/x - 20/3*e^2*log(x*(4*e + 1) - 20*e)/(16*e^2 + 8*e + 1) + 5/3*e*log(x*(4*e + 1) - 20*e)/(4*e
+ 1) + 1/12*x/(4*e + 1) + 2/15*log(x - 5)

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mupad [B]  time = 3.85, size = 53, normalized size = 1.43 \begin {gather*} \frac {x}{12}-\frac {\ln \left (\frac {{\mathrm {e}}^2\,\left (16\,x^2-160\,x+400\right )-\mathrm {e}\,\left (40\,x-8\,x^2\right )+x^2}{16\,x^2-160\,x+400}\right )}{3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((40*x + log((exp(2)*(16*x^2 - 160*x + 400) - exp(1)*(40*x - 8*x^2) + x^2)/(16*x^2 - 160*x + 400))*(exp(1)*
(16*x^2 - 160*x + 400) - 20*x + 4*x^2) + exp(1)*(100*x^2 - 40*x^3 + 4*x^4) - 5*x^3 + x^4)/(exp(1)*(1200*x^2 -
480*x^3 + 48*x^4) - 60*x^3 + 12*x^4),x)

[Out]

x/12 - log((exp(2)*(16*x^2 - 160*x + 400) - exp(1)*(40*x - 8*x^2) + x^2)/(16*x^2 - 160*x + 400))/(3*x)

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sympy [A]  time = 0.30, size = 46, normalized size = 1.24 \begin {gather*} \frac {x}{12} - \frac {\log {\left (\frac {x^{2} + e \left (8 x^{2} - 40 x\right ) + \left (16 x^{2} - 160 x + 400\right ) e^{2}}{16 x^{2} - 160 x + 400} \right )}}{3 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x**2-160*x+400)*exp(1)+4*x**2-20*x)*ln(((16*x**2-160*x+400)*exp(1)**2+(8*x**2-40*x)*exp(1)+x**
2)/(16*x**2-160*x+400))+(4*x**4-40*x**3+100*x**2)*exp(1)+x**4-5*x**3+40*x)/((48*x**4-480*x**3+1200*x**2)*exp(1
)+12*x**4-60*x**3),x)

[Out]

x/12 - log((x**2 + E*(8*x**2 - 40*x) + (16*x**2 - 160*x + 400)*exp(2))/(16*x**2 - 160*x + 400))/(3*x)

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