Integrand size = 98, antiderivative size = 26 \[ \int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{-8 x^2-8 x^3-2 x^4+e^4 \left (2 x^2+x^3\right ) \log (4)} \, dx=\frac {4-2 x+\log \left (-5+\frac {5 e^4 \log (4)}{4+2 x}\right )}{x} \] Output:
(ln(10*exp(4)/(4+2*x)*ln(2)-5)+4-2*x)/x
Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{-8 x^2-8 x^3-2 x^4+e^4 \left (2 x^2+x^3\right ) \log (4)} \, dx=\frac {4}{x}+\frac {\log \left (\frac {5 \left (-4-2 x+e^4 \log (4)\right )}{2 (2+x)}\right )}{x} \] Input:
Integrate[(32 + 32*x + 8*x^2 + E^4*(-8 - 5*x)*Log[4] + (8 + 8*x + 2*x^2 + E^4*(-2 - x)*Log[4])*Log[(-20 - 10*x + 5*E^4*Log[4])/(4 + 2*x)])/(-8*x^2 - 8*x^3 - 2*x^4 + E^4*(2*x^2 + x^3)*Log[4]),x]
Output:
4/x + Log[(5*(-4 - 2*x + E^4*Log[4]))/(2*(2 + x))]/x
Leaf count is larger than twice the leaf count of optimal. \(295\) vs. \(2(26)=52\).
Time = 1.37 (sec) , antiderivative size = 295, normalized size of antiderivative = 11.35, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, 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.
| \(\displaystyle \int \frac {8 x^2+\left (2 x^2+8 x+e^4 (-x-2) \log (4)+8\right ) \log \left (\frac {-10 x-20+5 e^4 \log (4)}{2 x+4}\right )+32 x+e^4 (-5 x-8) \log (4)+32}{-2 x^4-8 x^3-8 x^2+e^4 \left (x^3+2 x^2\right ) \log (4)} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {8 x^2+\left (2 x^2+8 x+e^4 (-x-2) \log (4)+8\right ) \log \left (\frac {-10 x-20+5 e^4 \log (4)}{2 x+4}\right )+32 x+e^4 (-5 x-8) \log (4)+32}{x^2 \left (-2 x^2-x \left (8-e^4 \log (4)\right )-8+e^4 \log (16)\right )}dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {-8 x^2-x \left (32-5 e^4 \log (4)\right )-8 \left (4-e^4 \log (4)\right )}{x^2 \left (2 x^2+x \left (8-e^4 \log (4)\right )+8-e^4 \log (16)\right )}+\frac {(x+2) \left (-2 x-4+e^4 \log (4)\right ) \log \left (\frac {5 \left (-2 x-4+e^4 \log (4)\right )}{2 (x+2)}\right )}{x^2 \left (2 x^2+x \left (8-e^4 \log (4)\right )+8-e^4 \log (16)\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {e^4 \left (e^4 \left (8 \log ^2(4)-\log (16) \log (1024)\right )+\log (65536)\right ) \log (x)}{\left (8-e^4 \log (16)\right )^2}+\frac {\left (e^8 \log ^2(4) \log (16)+4 e^4 \left (22 \log ^2(4)-22 \log (4) \log (16)+\log (16) \log (262144)\right )+64 \log (4)-4 \log (65536)\right ) \log (x+2)}{\log (4) \left (8-e^4 \log (16)\right )^2}-\frac {\left (e^8 \log (4) \left (8 \log ^2(4)+\log (4) \log (16)-\log (16) \log (1024)\right )+e^4 \left (88 \log ^2(4)-\log (4) (88 \log (16)-\log (65536))+4 \log (16) \log (262144)\right )+64 \log (4)-4 \log (65536)\right ) \log \left (2 x+4-e^4 \log (4)\right )}{\log (4) \left (8-e^4 \log (16)\right )^2}-\frac {e^4 \log (4) \log \left (\frac {x}{x+2}\right )}{2 \left (4-e^4 \log (4)\right )}+\frac {\left (2 x+4-e^4 \log (4)\right ) \log \left (-\frac {5 \left (2 x+4-e^4 \log (4)\right )}{2 (x+2)}\right )}{x \left (4-e^4 \log (4)\right )}+\frac {8 \left (4-e^4 \log (4)\right )}{x \left (8-e^4 \log (16)\right )}\) |
Input:
Int[(32 + 32*x + 8*x^2 + E^4*(-8 - 5*x)*Log[4] + (8 + 8*x + 2*x^2 + E^4*(- 2 - x)*Log[4])*Log[(-20 - 10*x + 5*E^4*Log[4])/(4 + 2*x)])/(-8*x^2 - 8*x^3 - 2*x^4 + E^4*(2*x^2 + x^3)*Log[4]),x]
Output:
(8*(4 - E^4*Log[4]))/(x*(8 - E^4*Log[16])) + (E^4*(E^4*(8*Log[4]^2 - Log[1 6]*Log[1024]) + Log[65536])*Log[x])/(8 - E^4*Log[16])^2 - (E^4*Log[4]*Log[ x/(2 + x)])/(2*(4 - E^4*Log[4])) + ((64*Log[4] + E^8*Log[4]^2*Log[16] - 4* Log[65536] + 4*E^4*(22*Log[4]^2 - 22*Log[4]*Log[16] + Log[16]*Log[262144]) )*Log[2 + x])/(Log[4]*(8 - E^4*Log[16])^2) - ((64*Log[4] + E^8*Log[4]*(8*L og[4]^2 + Log[4]*Log[16] - Log[16]*Log[1024]) - 4*Log[65536] + E^4*(88*Log [4]^2 - Log[4]*(88*Log[16] - Log[65536]) + 4*Log[16]*Log[262144]))*Log[4 + 2*x - E^4*Log[4]])/(Log[4]*(8 - E^4*Log[16])^2) + ((4 + 2*x - E^4*Log[4]) *Log[(-5*(4 + 2*x - E^4*Log[4]))/(2*(2 + x))])/(x*(4 - E^4*Log[4]))
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 = 0.81 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04
| method | result | size |
| norman | \(\frac {4+\ln \left (\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )-10 x -20}{4+2 x}\right )}{x}\) | \(27\) |
| parallelrisch | \(-\frac {-4-\ln \left (\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )-5 x -10}{2+x}\right )}{x}\) | \(28\) |
| risch | \(\frac {\ln \left (\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )-10 x -20}{4+2 x}\right )}{x}+\frac {4}{x}\) | \(31\) |
| parts | \(5 \,{\mathrm e}^{4} \ln \left (2\right ) \left (\frac {\ln \left (5 \,{\mathrm e}^{4} \ln \left (2\right )-\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{10 \,{\mathrm e}^{4} \ln \left (2\right )-20}+\frac {\ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right ) \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \left (2\right )-2\right ) \left (5 \,{\mathrm e}^{4} \ln \left (2\right )-\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}\right )+\frac {\ln \left (-{\mathrm e}^{4} \ln \left (2\right )+x +2\right )}{{\mathrm e}^{4} \ln \left (2\right )-2}-\frac {{\mathrm e}^{4} \ln \left (2\right ) \ln \left (x \right )}{2 \,{\mathrm e}^{4} \ln \left (2\right )-4}+\frac {4}{x}+\frac {\ln \left (2+x \right )}{2}\) | \(146\) |
| derivativedivides | \(-\frac {5 \,{\mathrm e}^{4} \ln \left (2\right ) \left (-\frac {\ln \left (5 \,{\mathrm e}^{4} \ln \left (2\right )-\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \left (2\right )-2\right )}-\frac {2 \ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right ) \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \left (2\right )-2\right ) \left (5 \,{\mathrm e}^{4} \ln \left (2\right )-\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}-\frac {2 \,{\mathrm e}^{-4} \ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \ln \left (2\right ) \left ({\mathrm e}^{4} \ln \left (2\right )-2\right )}+\frac {4}{-5 \,{\mathrm e}^{4} \ln \left (2\right )+\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}}+\frac {\ln \left (-5 \,{\mathrm e}^{4} \ln \left (2\right )+\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \,{\mathrm e}^{4} \ln \left (2\right )-10}\right )}{2}\) | \(181\) |
| default | \(-\frac {5 \,{\mathrm e}^{4} \ln \left (2\right ) \left (-\frac {\ln \left (5 \,{\mathrm e}^{4} \ln \left (2\right )-\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \left (2\right )-2\right )}-\frac {2 \ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right ) \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \left ({\mathrm e}^{4} \ln \left (2\right )-2\right ) \left (5 \,{\mathrm e}^{4} \ln \left (2\right )-\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}-\frac {2 \,{\mathrm e}^{-4} \ln \left (-5+\frac {5 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \ln \left (2\right ) \left ({\mathrm e}^{4} \ln \left (2\right )-2\right )}+\frac {4}{-5 \,{\mathrm e}^{4} \ln \left (2\right )+\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}}+\frac {\ln \left (-5 \,{\mathrm e}^{4} \ln \left (2\right )+\frac {10 \,{\mathrm e}^{4} \ln \left (2\right )}{2+x}\right )}{5 \,{\mathrm e}^{4} \ln \left (2\right )-10}\right )}{2}\) | \(181\) |
Input:
int(((2*(-2-x)*exp(4)*ln(2)+2*x^2+8*x+8)*ln((10*exp(4)*ln(2)-10*x-20)/(4+2 *x))+2*(-5*x-8)*exp(4)*ln(2)+8*x^2+32*x+32)/(2*(x^3+2*x^2)*exp(4)*ln(2)-2* x^4-8*x^3-8*x^2),x,method=_RETURNVERBOSE)
Output:
(4+ln((10*exp(4)*ln(2)-10*x-20)/(4+2*x)))/x
Time = 0.10 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{-8 x^2-8 x^3-2 x^4+e^4 \left (2 x^2+x^3\right ) \log (4)} \, dx=\frac {\log \left (\frac {5 \, {\left (e^{4} \log \left (2\right ) - x - 2\right )}}{x + 2}\right ) + 4}{x} \] Input:
integrate(((2*(-2-x)*exp(4)*log(2)+2*x^2+8*x+8)*log((10*exp(4)*log(2)-10*x -20)/(4+2*x))+2*(-5*x-8)*exp(4)*log(2)+8*x^2+32*x+32)/(2*(x^3+2*x^2)*exp(4 )*log(2)-2*x^4-8*x^3-8*x^2),x, algorithm="fricas")
Output:
(log(5*(e^4*log(2) - x - 2)/(x + 2)) + 4)/x
Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{-8 x^2-8 x^3-2 x^4+e^4 \left (2 x^2+x^3\right ) \log (4)} \, dx=\frac {\log {\left (\frac {- 10 x - 20 + 10 e^{4} \log {\left (2 \right )}}{2 x + 4} \right )}}{x} + \frac {4}{x} \] Input:
integrate(((2*(-2-x)*exp(4)*ln(2)+2*x**2+8*x+8)*ln((10*exp(4)*ln(2)-10*x-2 0)/(4+2*x))+2*(-5*x-8)*exp(4)*ln(2)+8*x**2+32*x+32)/(2*(x**3+2*x**2)*exp(4 )*ln(2)-2*x**4-8*x**3-8*x**2),x)
Output:
log((-10*x - 20 + 10*exp(4)*log(2))/(2*x + 4))/x + 4/x
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 391, normalized size of antiderivative = 15.04 \[ \int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{-8 x^2-8 x^3-2 x^4+e^4 \left (2 x^2+x^3\right ) \log (4)} \, dx=-2 \, {\left (\frac {e^{\left (-4\right )} \log \left (x + 2\right )}{\log \left (2\right )} - \frac {{\left (e^{4} \log \left (2\right ) - 4\right )} \log \left (x\right )}{e^{8} \log \left (2\right )^{2} - 4 \, e^{4} \log \left (2\right ) + 4} - \frac {4 \, \log \left (-e^{4} \log \left (2\right ) + x + 2\right )}{e^{12} \log \left (2\right )^{3} - 4 \, e^{8} \log \left (2\right )^{2} + 4 \, e^{4} \log \left (2\right )} - \frac {2}{{\left (e^{4} \log \left (2\right ) - 2\right )} x}\right )} e^{4} \log \left (2\right ) + \frac {5}{2} \, {\left (\frac {e^{\left (-4\right )} \log \left (x + 2\right )}{\log \left (2\right )} + \frac {2 \, \log \left (-e^{4} \log \left (2\right ) + x + 2\right )}{e^{8} \log \left (2\right )^{2} - 2 \, e^{4} \log \left (2\right )} - \frac {\log \left (x\right )}{e^{4} \log \left (2\right ) - 2}\right )} e^{4} \log \left (2\right ) + \frac {e^{4} \log \left (2\right ) \log \left (x\right )}{2 \, {\left (e^{4} \log \left (2\right ) - 2\right )}} - \frac {4 \, e^{\left (-4\right )} \log \left (-e^{4} \log \left (2\right ) + x + 2\right )}{\log \left (2\right )} - \frac {4 \, {\left (e^{4} \log \left (2\right ) - 4\right )} \log \left (x\right )}{e^{8} \log \left (2\right )^{2} - 4 \, e^{4} \log \left (2\right ) + 4} - \frac {16 \, \log \left (-e^{4} \log \left (2\right ) + x + 2\right )}{e^{12} \log \left (2\right )^{3} - 4 \, e^{8} \log \left (2\right )^{2} + 4 \, e^{4} \log \left (2\right )} - \frac {16 \, \log \left (-e^{4} \log \left (2\right ) + x + 2\right )}{e^{8} \log \left (2\right )^{2} - 2 \, e^{4} \log \left (2\right )} + \frac {8 \, \log \left (x\right )}{e^{4} \log \left (2\right ) - 2} - \frac {4 i \, \pi - 2 \, {\left (i \, \pi \log \left (2\right ) + \log \left (5\right ) \log \left (2\right )\right )} e^{4} - 2 \, {\left (e^{4} \log \left (2\right ) - x - 2\right )} \log \left (-e^{4} \log \left (2\right ) + x + 2\right ) + {\left ({\left (e^{4} \log \left (2\right ) - 2\right )} x + 2 \, e^{4} \log \left (2\right ) - 4\right )} \log \left (x + 2\right ) + 4 \, \log \left (5\right )}{2 \, {\left (e^{4} \log \left (2\right ) - 2\right )} x} - \frac {8}{{\left (e^{4} \log \left (2\right ) - 2\right )} x} \] Input:
integrate(((2*(-2-x)*exp(4)*log(2)+2*x^2+8*x+8)*log((10*exp(4)*log(2)-10*x -20)/(4+2*x))+2*(-5*x-8)*exp(4)*log(2)+8*x^2+32*x+32)/(2*(x^3+2*x^2)*exp(4 )*log(2)-2*x^4-8*x^3-8*x^2),x, algorithm="maxima")
Output:
-2*(e^(-4)*log(x + 2)/log(2) - (e^4*log(2) - 4)*log(x)/(e^8*log(2)^2 - 4*e ^4*log(2) + 4) - 4*log(-e^4*log(2) + x + 2)/(e^12*log(2)^3 - 4*e^8*log(2)^ 2 + 4*e^4*log(2)) - 2/((e^4*log(2) - 2)*x))*e^4*log(2) + 5/2*(e^(-4)*log(x + 2)/log(2) + 2*log(-e^4*log(2) + x + 2)/(e^8*log(2)^2 - 2*e^4*log(2)) - log(x)/(e^4*log(2) - 2))*e^4*log(2) + 1/2*e^4*log(2)*log(x)/(e^4*log(2) - 2) - 4*e^(-4)*log(-e^4*log(2) + x + 2)/log(2) - 4*(e^4*log(2) - 4)*log(x)/ (e^8*log(2)^2 - 4*e^4*log(2) + 4) - 16*log(-e^4*log(2) + x + 2)/(e^12*log( 2)^3 - 4*e^8*log(2)^2 + 4*e^4*log(2)) - 16*log(-e^4*log(2) + x + 2)/(e^8*l og(2)^2 - 2*e^4*log(2)) + 8*log(x)/(e^4*log(2) - 2) - 1/2*(4*I*pi - 2*(I*p i*log(2) + log(5)*log(2))*e^4 - 2*(e^4*log(2) - x - 2)*log(-e^4*log(2) + x + 2) + ((e^4*log(2) - 2)*x + 2*e^4*log(2) - 4)*log(x + 2) + 4*log(5))/((e ^4*log(2) - 2)*x) - 8/((e^4*log(2) - 2)*x)
Leaf count of result is larger than twice the leaf count of optimal. 120 vs. \(2 (26) = 52\).
Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 4.62 \[ \int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{-8 x^2-8 x^3-2 x^4+e^4 \left (2 x^2+x^3\right ) \log (4)} \, dx=\frac {{\left (2 \, e^{8} \log \left (2\right )^{2} + \frac {{\left (e^{4} \log \left (2\right ) - x - 2\right )} e^{4} \log \left (2\right ) \log \left (\frac {5 \, {\left (e^{4} \log \left (2\right ) - x - 2\right )}}{x + 2}\right )}{x + 2} + e^{4} \log \left (2\right ) \log \left (\frac {5 \, {\left (e^{4} \log \left (2\right ) - x - 2\right )}}{x + 2}\right )\right )} {\left (\frac {{\left (e^{4} \log \left (2\right ) - 2\right )} e^{\left (-8\right )}}{\log \left (2\right )^{2}} + \frac {2 \, e^{\left (-8\right )}}{\log \left (2\right )^{2}}\right )}}{e^{4} \log \left (2\right ) - \frac {2 \, {\left (e^{4} \log \left (2\right ) - x - 2\right )}}{x + 2} - 2} \] Input:
integrate(((2*(-2-x)*exp(4)*log(2)+2*x^2+8*x+8)*log((10*exp(4)*log(2)-10*x -20)/(4+2*x))+2*(-5*x-8)*exp(4)*log(2)+8*x^2+32*x+32)/(2*(x^3+2*x^2)*exp(4 )*log(2)-2*x^4-8*x^3-8*x^2),x, algorithm="giac")
Output:
(2*e^8*log(2)^2 + (e^4*log(2) - x - 2)*e^4*log(2)*log(5*(e^4*log(2) - x - 2)/(x + 2))/(x + 2) + e^4*log(2)*log(5*(e^4*log(2) - x - 2)/(x + 2)))*((e^ 4*log(2) - 2)*e^(-8)/log(2)^2 + 2*e^(-8)/log(2)^2)/(e^4*log(2) - 2*(e^4*lo g(2) - x - 2)/(x + 2) - 2)
Time = 4.71 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{-8 x^2-8 x^3-2 x^4+e^4 \left (2 x^2+x^3\right ) \log (4)} \, dx=\frac {\ln \left (-\frac {5\,\left (x-{\mathrm {e}}^4\,\ln \left (2\right )+2\right )}{x+2}\right )+4}{x} \] Input:
int(-(32*x + log(-(10*x - 10*exp(4)*log(2) + 20)/(2*x + 4))*(8*x + 2*x^2 - 2*exp(4)*log(2)*(x + 2) + 8) + 8*x^2 - 2*exp(4)*log(2)*(5*x + 8) + 32)/(8 *x^2 + 8*x^3 + 2*x^4 - 2*exp(4)*log(2)*(2*x^2 + x^3)),x)
Output:
(log(-(5*(x - exp(4)*log(2) + 2))/(x + 2)) + 4)/x
Time = 0.16 (sec) , antiderivative size = 112, normalized size of antiderivative = 4.31 \[ \int \frac {32+32 x+8 x^2+e^4 (-8-5 x) \log (4)+\left (8+8 x+2 x^2+e^4 (-2-x) \log (4)\right ) \log \left (\frac {-20-10 x+5 e^4 \log (4)}{4+2 x}\right )}{-8 x^2-8 x^3-2 x^4+e^4 \left (2 x^2+x^3\right ) \log (4)} \, dx=\frac {\mathrm {log}\left (\mathrm {log}\left (2\right ) e^{4}-x -2\right ) x -\mathrm {log}\left (x +2\right ) x +\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (2\right ) e^{4}-5 x -10}{x +2}\right ) \mathrm {log}\left (2\right ) e^{4}-\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (2\right ) e^{4}-5 x -10}{x +2}\right ) x -2 \,\mathrm {log}\left (\frac {5 \,\mathrm {log}\left (2\right ) e^{4}-5 x -10}{x +2}\right )+4 \,\mathrm {log}\left (2\right ) e^{4}-8}{x \left (\mathrm {log}\left (2\right ) e^{4}-2\right )} \] Input:
int(((2*(-2-x)*exp(4)*log(2)+2*x^2+8*x+8)*log((10*exp(4)*log(2)-10*x-20)/( 4+2*x))+2*(-5*x-8)*exp(4)*log(2)+8*x^2+32*x+32)/(2*(x^3+2*x^2)*exp(4)*log( 2)-2*x^4-8*x^3-8*x^2),x)
Output:
(log(log(2)*e**4 - x - 2)*x - log(x + 2)*x + log((5*log(2)*e**4 - 5*x - 10 )/(x + 2))*log(2)*e**4 - log((5*log(2)*e**4 - 5*x - 10)/(x + 2))*x - 2*log ((5*log(2)*e**4 - 5*x - 10)/(x + 2)) + 4*log(2)*e**4 - 8)/(x*(log(2)*e**4 - 2))