Integrand size = 127, antiderivative size = 37 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {-x+\frac {x^2}{4}-\log \left (\left (e-\frac {x}{4 (5-x)}\right )^2\right )}{3 x} \] Output:
1/3*(1/4*x^2-x-ln((exp(1)-x/(-4*x+20))^2))/x
Leaf count is larger than twice the leaf count of optimal. \(323\) vs. \(2(37)=74\).
Time = 0.22 (sec) , antiderivative size = 323, normalized size of antiderivative = 8.73 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {5 e x^2-(1+4 e)^2 x \log ^2(-4 e (-5+x)-x)+2 (1+4 e)^2 x \log \left (-\frac {1}{5} (1+4 e) (-5+x)\right ) \log \left (\frac {5}{4 e (-5+x)+x}\right )+x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )+8 e x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )+16 e^2 x \log ^2\left (\frac {5}{4 e (-5+x)+x}\right )-20 e \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+8 e x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+16 e^2 x \log \left (\frac {5}{4 e (-5+x)+x}\right ) \log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )+(1+4 e)^2 x \log (-4 e (-5+x)-x) \left (2 \log \left (-\frac {1}{5} (1+4 e) (-5+x)\right )+\log \left (\frac {(4 e (-5+x)+x)^2}{16 (-5+x)^2}\right )\right )}{60 e x} \] Input:
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]
Output:
(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)
Leaf count is larger than twice the leaf count of optimal. \(213\) vs. \(2(37)=74\).
Time = 1.18 (sec) , antiderivative size = 213, normalized size of antiderivative = 5.76, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.024, Rules used = {2026, 7279, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^4-5 x^3+\left (4 x^2+e \left (16 x^2-160 x+400\right )-20 x\right ) \log \left (\frac {x^2+e \left (8 x^2-40 x\right )+e^2 \left (16 x^2-160 x+400\right )}{16 x^2-160 x+400}\right )+e \left (4 x^4-40 x^3+100 x^2\right )+40 x}{12 x^4-60 x^3+e \left (48 x^4-480 x^3+1200 x^2\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {x^4-5 x^3+\left (4 x^2+e \left (16 x^2-160 x+400\right )-20 x\right ) \log \left (\frac {x^2+e \left (8 x^2-40 x\right )+e^2 \left (16 x^2-160 x+400\right )}{16 x^2-160 x+400}\right )+e \left (4 x^4-40 x^3+100 x^2\right )+40 x}{x^2 \left (12 (1+4 e) x^2-60 (1+8 e) x+1200 e\right )}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {x^2}{12 (5-x) (20 e-(1+4 e) x)}+\frac {\log \left (\frac {((1+4 e) x-20 e)^2}{16 (x-5)^2}\right )}{3 x^2}+\frac {5 x}{12 (x-5) (20 e-(1+4 e) x)}+\frac {e (5-x)}{3 (20 e-(1+4 e) x)}+\frac {10}{3 (5-x) (20 e-(1+4 e) x) x}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e x}{3 (1+4 e)}+\frac {x}{12 (1+4 e)}+\frac {2}{15} \log (5-x)+\frac {\log (x)}{30 e}-\frac {\log \left (-\frac {x}{5-x}\right )}{30 e}-\frac {(1+4 e) \log (20 e-(1+4 e) x)}{30 e}+\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)}-\frac {20 e^2 \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {5 e \log (20 e-(1+4 e) x)}{3 (1+4 e)^2}-\frac {(20 e-(1+4 e) x) \log \left (\frac {(20 e-(1+4 e) x)^2}{16 (5-x)^2}\right )}{60 e x}\) |
Input:
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 - 480* x^3 + 48*x^4)),x]
Output:
x/(12*(1 + 4*E)) + (E*x)/(3*(1 + 4*E)) + (2*Log[5 - x])/15 + Log[x]/(30*E) - Log[-(x/(5 - 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) - (( 20*E - (1 + 4*E)*x)*Log[(20*E - (1 + 4*E)*x)^2/(16*(5 - x)^2)])/(60*E*x)
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 2.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 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 | \(\frac {\frac {x^{2}}{12}-\frac {\ln \left (\frac {\left (16 x^{2}-160 x +400\right ) {\mathrm e}^{2}+\left (8 x^{2}-40 x \right ) {\mathrm e}+x^{2}}{16 x^{2}-160 x +400}\right )}{3}}{x}\) | \(58\) |
| parallelrisch | \(-\frac {\left (-10000 x^{2} {\mathrm e}^{2}+40000 \ln \left (\frac {\left (16 x^{2}-160 x +400\right ) {\mathrm e}^{2}+\left (8 x^{2}-40 x \right ) {\mathrm e}+x^{2}}{16 x^{2}-160 x +400}\right ) {\mathrm e}^{2}\right ) {\mathrm e}^{-2}}{120000 x}\) | \(70\) |
| derivativedivides | \(-\frac {4 \ln \left (2\right )}{15 \left (1+\frac {5}{-5+x}\right )}-\frac {-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{60}-\frac {{\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{12 \left (-5+x \right )}}{1+\frac {5}{-5+x}}-\frac {5}{12}+\frac {x}{12}-\frac {\left (8 \,{\mathrm e}+2\right ) {\mathrm e}^{-1} \ln \left (4 \,{\mathrm e}+1+\frac {5}{-5+x}\right )}{60}\) | \(156\) |
| default | \(-\frac {4 \ln \left (2\right )}{15 \left (1+\frac {5}{-5+x}\right )}-\frac {-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{60}-\frac {{\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{12 \left (-5+x \right )}}{1+\frac {5}{-5+x}}-\frac {5}{12}+\frac {x}{12}-\frac {\left (8 \,{\mathrm e}+2\right ) {\mathrm e}^{-1} \ln \left (4 \,{\mathrm e}+1+\frac {5}{-5+x}\right )}{60}\) | \(156\) |
| parts | \(-\frac {-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{20}-\frac {{\mathrm e}^{-1} \ln \left (16 \,{\mathrm e}^{2}+\frac {40 \,{\mathrm e}}{-5+x}+\frac {25}{\left (-5+x \right )^{2}}+8 \,{\mathrm e}+\frac {10}{-5+x}+1\right )}{4 \left (-5+x \right )}}{3 \left (1+\frac {5}{-5+x}\right )}-\frac {{\mathrm e}^{-1} \ln \left (1+\frac {5}{-5+x}\right )}{30}-\frac {4 \ln \left (2\right )}{15 \left (1+\frac {5}{-5+x}\right )}+\frac {x}{12}+\frac {2 \ln \left (-5+x \right )}{15}+\frac {{\mathrm e}^{-1} \ln \left (x \right )}{30}-\frac {\left (4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1} \ln \left (4 x \,{\mathrm e}-20 \,{\mathrm e}+x \right )}{30}\) | \(183\) |
Input:
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^2-160*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,meth od=_RETURNVERBOSE)
Output:
-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
Time = 0.09 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.38 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {x^{2} - 4 \, \log \left (\frac {x^{2} + 16 \, {\left (x^{2} - 10 \, x + 25\right )} e^{2} + 8 \, {\left (x^{2} - 5 \, x\right )} e}{16 \, {\left (x^{2} - 10 \, x + 25\right )}}\right )}{12 \, x} \] Input:
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)/(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, algorithm="fricas")
Output:
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
Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.24 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {x}{12} - \frac {\log {\left (\frac {x^{2} + e \left (8 x^{2} - 40 x\right ) + \left (16 x^{2} - 160 x + 400\right ) e^{2}}{16 x^{2} - 160 x + 400} \right )}}{3 x} \] Input:
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)
Output:
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)
Leaf count of result is larger than twice the leaf count of optimal. 269 vs. \(2 (31) = 62\).
Time = 0.20 (sec) , antiderivative size = 269, normalized size of antiderivative = 7.27 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=-\frac {1}{30} \, {\left (4 \, e + 1\right )} e^{\left (-1\right )} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - \frac {1}{3} \, {\left (\frac {80 \, e^{2} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{16 \, e^{2} + 8 \, e + 1} - \frac {x}{4 \, e + 1} - 5 \, \log \left (x - 5\right )\right )} e + \frac {10}{3} \, {\left (\frac {4 \, e \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{4 \, e + 1} - \log \left (x - 5\right )\right )} e - \frac {5}{3} \, {\left (\log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - \log \left (x - 5\right )\right )} e + \frac {{\left (40 \, e \log \left (2\right ) + {\left (x {\left (4 \, e + 1\right )} - 20 \, e\right )} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right ) - 4 \, {\left (x e - 5 \, e\right )} \log \left (x - 5\right )\right )} e^{\left (-1\right )}}{30 \, x} - \frac {20 \, e^{2} \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{3 \, {\left (16 \, e^{2} + 8 \, e + 1\right )}} + \frac {5 \, e \log \left (x {\left (4 \, e + 1\right )} - 20 \, e\right )}{3 \, {\left (4 \, e + 1\right )}} + \frac {x}{12 \, {\left (4 \, e + 1\right )}} + \frac {2}{15} \, \log \left (x - 5\right ) \] Input:
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)/(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, algorithm="maxima")
Output:
-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)*log(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)
Time = 0.29 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.51 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {x^{2} - 4 \, \log \left (\frac {16 \, x^{2} e^{2} + 8 \, x^{2} e + x^{2} - 160 \, x e^{2} - 40 \, x e + 400 \, e^{2}}{16 \, {\left (x^{2} - 10 \, x + 25\right )}}\right )}{12 \, x} \] Input:
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)/(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, algorithm="giac")
Output:
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
Time = 2.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.43 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {x}{12}-\frac {\ln \left (\frac {{\mathrm {e}}^2\,\left (16\,x^2-160\,x+400\right )-\mathrm {e}\,\left (40\,x-8\,x^2\right )+x^2}{16\,x^2-160\,x+400}\right )}{3\,x} \] Input:
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 - 48 0*x^3 + 48*x^4) - 60*x^3 + 12*x^4),x)
Output:
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)
Time = 0.17 (sec) , antiderivative size = 207, normalized size of antiderivative = 5.59 \[ \int \frac {40 x-5 x^3+x^4+e \left (100 x^2-40 x^3+4 x^4\right )+\left (-20 x+4 x^2+e \left (400-160 x+16 x^2\right )\right ) \log \left (\frac {x^2+e \left (-40 x+8 x^2\right )+e^2 \left (400-160 x+16 x^2\right )}{400-160 x+16 x^2}\right )}{-60 x^3+12 x^4+e \left (1200 x^2-480 x^3+48 x^4\right )} \, dx=\frac {-8 \,\mathrm {log}\left (4 e x -20 e +x \right ) e x -2 \,\mathrm {log}\left (4 e x -20 e +x \right ) x +8 \,\mathrm {log}\left (-5+x \right ) e x +2 \,\mathrm {log}\left (-5+x \right ) x +4 \,\mathrm {log}\left (\frac {16 e^{2} x^{2}-160 e^{2} x +8 e \,x^{2}+400 e^{2}-40 e x +x^{2}}{16 x^{2}-160 x +400}\right ) e x -20 \,\mathrm {log}\left (\frac {16 e^{2} x^{2}-160 e^{2} x +8 e \,x^{2}+400 e^{2}-40 e x +x^{2}}{16 x^{2}-160 x +400}\right ) e +\mathrm {log}\left (\frac {16 e^{2} x^{2}-160 e^{2} x +8 e \,x^{2}+400 e^{2}-40 e x +x^{2}}{16 x^{2}-160 x +400}\right ) x +5 e \,x^{2}}{60 e x} \] Input:
int((((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)/(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)
Output:
( - 8*log(4*e*x - 20*e + x)*e*x - 2*log(4*e*x - 20*e + x)*x + 8*log(x - 5) *e*x + 2*log(x - 5)*x + 4*log((16*e**2*x**2 - 160*e**2*x + 400*e**2 + 8*e* x**2 - 40*e*x + x**2)/(16*x**2 - 160*x + 400))*e*x - 20*log((16*e**2*x**2 - 160*e**2*x + 400*e**2 + 8*e*x**2 - 40*e*x + x**2)/(16*x**2 - 160*x + 400 ))*e + log((16*e**2*x**2 - 160*e**2*x + 400*e**2 + 8*e*x**2 - 40*e*x + x** 2)/(16*x**2 - 160*x + 400))*x + 5*e*x**2)/(60*e*x)