Integrand size = 86, antiderivative size = 25 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {e^x+x-\log (x)}{2 x (-4+x-\log (3))} \] Output:
1/2*(x+exp(x)-ln(x))/x/(x-4-ln(3))
Time = 0.47 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {e^x+x-\log (x)}{2 x (-4+x-\log (3))} \] Input:
Integrate[(4 - x - x^2 + Log[3] + E^x*(4 - 6*x + x^2 + (1 - x)*Log[3]) + ( -4 + 2*x - Log[3])*Log[x])/(32*x^2 - 16*x^3 + 2*x^4 + (16*x^2 - 4*x^3)*Log [3] + 2*x^2*Log[3]^2),x]
Output:
(E^x + x - Log[x])/(2*x*(-4 + x - Log[3]))
Leaf count is larger than twice the leaf count of optimal. \(88\) vs. \(2(25)=50\).
Time = 2.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.52, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {6, 2026, 7277, 27, 7292, 7293, 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.
| \(\displaystyle \int \frac {-x^2+e^x \left (x^2-6 x+(1-x) \log (3)+4\right )-x+(2 x-4-\log (3)) \log (x)+4+\log (3)}{2 x^4-16 x^3+32 x^2+2 x^2 \log ^2(3)+\left (16 x^2-4 x^3\right ) \log (3)} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {-x^2+e^x \left (x^2-6 x+(1-x) \log (3)+4\right )-x+(2 x-4-\log (3)) \log (x)+4+\log (3)}{2 x^4-16 x^3+x^2 \left (32+2 \log ^2(3)\right )+\left (16 x^2-4 x^3\right ) \log (3)}dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {-x^2+e^x \left (x^2-6 x+(1-x) \log (3)+4\right )-x+(2 x-4-\log (3)) \log (x)+4+\log (3)}{x^2 \left (2 x^2-4 x (4+\log (3))+2 (4+\log (3))^2\right )}dx\) |
| \(\Big \downarrow \) 7277 |
| \(\displaystyle 8 \int \frac {-x^2-x+e^x \left (x^2-6 x+(1-x) \log (3)+4\right )-(-2 x+\log (3)+4) \log (x)+\log (3)+4}{16 x^2 (-x+\log (3)+4)^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} \int \frac {-x^2-x+e^x \left (x^2-6 x+(1-x) \log (3)+4\right )-(-2 x+\log (3)+4) \log (x)+\log (3)+4}{x^2 (-x+\log (3)+4)^2}dx\) |
| \(\Big \downarrow \) 7292 |
| \(\displaystyle \frac {1}{2} \int \frac {-x^2-x+e^x \left (x^2-6 x+(1-x) \log (3)+4\right )-(-2 x+\log (3)+4) \log (x)+4 \left (1+\frac {\log (3)}{4}\right )}{x^2 (-x+\log (3)+4)^2}dx\) |
| \(\Big \downarrow \) 7293 |
| \(\displaystyle \frac {1}{2} \int \left (\frac {e^x \left (x^2-(6+\log (3)) x+\log (3)+4\right )}{x^2 (-x+\log (3)+4)^2}+\frac {(2 x-\log (3)-4) \log (x)}{x^2 (x-\log (3)-4)^2}-\frac {1}{x (x-\log (3)-4)^2}-\frac {1}{(x-\log (3)-4)^2}+\frac {4+\log (3)}{x^2 (x-\log (3)-4)^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {x \log (x)}{(4+\log (3))^2 (-x+4+\log (3))}+\frac {\log (x)}{x (4+\log (3))}+\frac {\log (x)}{(4+\log (3))^2}+\frac {1}{x-4-\log (3)}-\frac {e^x}{(4+\log (3)) (-x+4+\log (3))}-\frac {e^x}{x (4+\log (3))}\right )\) |
Input:
Int[(4 - x - x^2 + Log[3] + E^x*(4 - 6*x + x^2 + (1 - x)*Log[3]) + (-4 + 2 *x - Log[3])*Log[x])/(32*x^2 - 16*x^3 + 2*x^4 + (16*x^2 - 4*x^3)*Log[3] + 2*x^2*Log[3]^2),x]
Output:
((-4 + x - Log[3])^(-1) - E^x/(x*(4 + Log[3])) - E^x/((4 + Log[3])*(4 - x + Log[3])) + Log[x]/(4 + Log[3])^2 + Log[x]/(x*(4 + Log[3])) + (x*Log[x])/ ((4 + Log[3])^2*(4 - x + Log[3])))/2
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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_.))^(p_.), x_Symbol] :> Simp[1/(4^p*c^p) Int[u*(b + 2*c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n} , x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p] && !AlgebraicFu nctionQ[u, x]
Time = 1.13 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00
| method | result | size |
| parallelrisch | \(\frac {-{\mathrm e}^{x}-x +\ln \left (x \right )}{2 x \left (4+\ln \left (3\right )-x \right )}\) | \(25\) |
| norman | \(\frac {-\frac {x}{2}-\frac {{\mathrm e}^{x}}{2}+\frac {\ln \left (x \right )}{2}}{x \left (4+\ln \left (3\right )-x \right )}\) | \(26\) |
| default | \(\frac {\ln \left (x \right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {1}{-2 \ln \left (3\right )+2 x -8}-\frac {\ln \left (x -4-\ln \left (3\right )\right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (4+\ln \left (3\right )-x \right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (x \right ) x}{2 \left (\ln \left (3\right )+4\right )^{2} \left (4+\ln \left (3\right )-x \right )}+\frac {\ln \left (x \right )}{2 \left (\ln \left (3\right )+4\right ) x}+\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right ) \left (x -4-\ln \left (3\right )\right )}+\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{2 \ln \left (3\right )+8}-\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}-\frac {{\mathrm e}^{x}}{2 \left (x -4-\ln \left (3\right )\right )}-\frac {{\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{2}-\frac {2 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right )^{2} \left (x -4-\ln \left (3\right )\right )}-\frac {5 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}-\frac {2 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right )^{2} x}+\frac {\operatorname {expIntegral}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}-\frac {4 \,\operatorname {expIntegral}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}+\frac {4 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}+\frac {3 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right ) \left (x -4-\ln \left (3\right )\right )}+\frac {3 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\ln \left (3\right )+4}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right )^{2} \left (x -4-\ln \left (3\right )\right )}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right )^{2} x}-\frac {\ln \left (3\right ) \operatorname {expIntegral}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}+\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}\) | \(402\) |
| parts | \(\frac {\ln \left (x \right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {1}{-2 \ln \left (3\right )+2 x -8}-\frac {\ln \left (x -4-\ln \left (3\right )\right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (4+\ln \left (3\right )-x \right )}{2 \left (\ln \left (3\right )+4\right )^{2}}+\frac {\ln \left (x \right ) x}{2 \left (\ln \left (3\right )+4\right )^{2} \left (4+\ln \left (3\right )-x \right )}+\frac {\ln \left (x \right )}{2 \left (\ln \left (3\right )+4\right ) x}+\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right ) \left (x -4-\ln \left (3\right )\right )}+\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{2 \ln \left (3\right )+8}-\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}-\frac {{\mathrm e}^{x}}{2 \left (x -4-\ln \left (3\right )\right )}-\frac {{\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{2}-\frac {2 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right )^{2} \left (x -4-\ln \left (3\right )\right )}-\frac {5 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}-\frac {2 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right )^{2} x}+\frac {\operatorname {expIntegral}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{2}}-\frac {4 \,\operatorname {expIntegral}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}+\frac {4 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}+\frac {3 \,{\mathrm e}^{x}}{\left (\ln \left (3\right )+4\right ) \left (x -4-\ln \left (3\right )\right )}+\frac {3 \,{\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\ln \left (3\right )+4}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right )^{2} \left (x -4-\ln \left (3\right )\right )}-\frac {\ln \left (3\right ) {\mathrm e}^{x}}{2 \left (\ln \left (3\right )+4\right )^{2} x}-\frac {\ln \left (3\right ) \operatorname {expIntegral}_{1}\left (-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}+\frac {\ln \left (3\right ) {\mathrm e}^{\ln \left (3\right )+4} \operatorname {expIntegral}_{1}\left (4+\ln \left (3\right )-x \right )}{\left (\ln \left (3\right )+4\right )^{3}}\) | \(402\) |
Input:
int(((-ln(3)+2*x-4)*ln(x)+((1-x)*ln(3)+x^2-6*x+4)*exp(x)+ln(3)-x^2-x+4)/(2 *x^2*ln(3)^2+(-4*x^3+16*x^2)*ln(3)+2*x^4-16*x^3+32*x^2),x,method=_RETURNVE RBOSE)
Output:
1/2/x*(-exp(x)-x+ln(x))/(4+ln(3)-x)
Time = 0.09 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {x + e^{x} - \log \left (x\right )}{2 \, {\left (x^{2} - x \log \left (3\right ) - 4 \, x\right )}} \] Input:
integrate(((-log(3)+2*x-4)*log(x)+((1-x)*log(3)+x^2-6*x+4)*exp(x)+log(3)-x ^2-x+4)/(2*x^2*log(3)^2+(-4*x^3+16*x^2)*log(3)+2*x^4-16*x^3+32*x^2),x, alg orithm="fricas")
Output:
1/2*(x + e^x - log(x))/(x^2 - x*log(3) - 4*x)
Leaf count of result is larger than twice the leaf count of optimal. 48 vs. \(2 (20) = 40\).
Time = 0.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {e^{x}}{2 x^{2} - 8 x - 2 x \log {\left (3 \right )}} - \frac {\log {\left (x \right )}}{2 x^{2} - 8 x - 2 x \log {\left (3 \right )}} + \frac {1}{2 x - 8 - 2 \log {\left (3 \right )}} \] Input:
integrate(((-ln(3)+2*x-4)*ln(x)+((1-x)*ln(3)+x**2-6*x+4)*exp(x)+ln(3)-x**2 -x+4)/(2*x**2*ln(3)**2+(-4*x**3+16*x**2)*ln(3)+2*x**4-16*x**3+32*x**2),x)
Output:
exp(x)/(2*x**2 - 8*x - 2*x*log(3)) - log(x)/(2*x**2 - 8*x - 2*x*log(3)) + 1/(2*x - 8 - 2*log(3))
Leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (22) = 44\).
Time = 0.16 (sec) , antiderivative size = 362, normalized size of antiderivative = 14.48 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=-\frac {1}{2} \, {\left (\frac {2 \, x - \log \left (3\right ) - 4}{{\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} x^{2} - {\left (\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64\right )} x} + \frac {2 \, \log \left (x - \log \left (3\right ) - 4\right )}{\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64} - \frac {2 \, \log \left (x\right )}{\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64}\right )} \log \left (3\right ) + \frac {x {\left (\log \left (3\right ) + 4\right )} + {\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} e^{x} - \log \left (3\right )^{2} - {\left (x^{2} - x {\left (\log \left (3\right ) + 4\right )} + \log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} \log \left (x\right ) - 8 \, \log \left (3\right ) - 16}{2 \, {\left ({\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} x^{2} - {\left (\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64\right )} x\right )}} - \frac {2 \, {\left (2 \, x - \log \left (3\right ) - 4\right )}}{{\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )} x^{2} - {\left (\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64\right )} x} - \frac {4 \, \log \left (x - \log \left (3\right ) - 4\right )}{\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64} + \frac {\log \left (x - \log \left (3\right ) - 4\right )}{\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16} + \frac {4 \, \log \left (x\right )}{\log \left (3\right )^{3} + 12 \, \log \left (3\right )^{2} + 48 \, \log \left (3\right ) + 64} - \frac {\log \left (x\right )}{2 \, {\left (\log \left (3\right )^{2} + 8 \, \log \left (3\right ) + 16\right )}} + \frac {1}{2 \, {\left (x {\left (\log \left (3\right ) + 4\right )} - \log \left (3\right )^{2} - 8 \, \log \left (3\right ) - 16\right )}} + \frac {1}{2 \, {\left (x - \log \left (3\right ) - 4\right )}} \] Input:
integrate(((-log(3)+2*x-4)*log(x)+((1-x)*log(3)+x^2-6*x+4)*exp(x)+log(3)-x ^2-x+4)/(2*x^2*log(3)^2+(-4*x^3+16*x^2)*log(3)+2*x^4-16*x^3+32*x^2),x, alg orithm="maxima")
Output:
-1/2*((2*x - log(3) - 4)/((log(3)^2 + 8*log(3) + 16)*x^2 - (log(3)^3 + 12* log(3)^2 + 48*log(3) + 64)*x) + 2*log(x - log(3) - 4)/(log(3)^3 + 12*log(3 )^2 + 48*log(3) + 64) - 2*log(x)/(log(3)^3 + 12*log(3)^2 + 48*log(3) + 64) )*log(3) + 1/2*(x*(log(3) + 4) + (log(3)^2 + 8*log(3) + 16)*e^x - log(3)^2 - (x^2 - x*(log(3) + 4) + log(3)^2 + 8*log(3) + 16)*log(x) - 8*log(3) - 1 6)/((log(3)^2 + 8*log(3) + 16)*x^2 - (log(3)^3 + 12*log(3)^2 + 48*log(3) + 64)*x) - 2*(2*x - log(3) - 4)/((log(3)^2 + 8*log(3) + 16)*x^2 - (log(3)^3 + 12*log(3)^2 + 48*log(3) + 64)*x) - 4*log(x - log(3) - 4)/(log(3)^3 + 12 *log(3)^2 + 48*log(3) + 64) + log(x - log(3) - 4)/(log(3)^2 + 8*log(3) + 1 6) + 4*log(x)/(log(3)^3 + 12*log(3)^2 + 48*log(3) + 64) - 1/2*log(x)/(log( 3)^2 + 8*log(3) + 16) + 1/2/(x*(log(3) + 4) - log(3)^2 - 8*log(3) - 16) + 1/2/(x - log(3) - 4)
Time = 0.11 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {x + e^{x} - \log \left (x\right )}{2 \, {\left (x^{2} - x \log \left (3\right ) - 4 \, x\right )}} \] Input:
integrate(((-log(3)+2*x-4)*log(x)+((1-x)*log(3)+x^2-6*x+4)*exp(x)+log(3)-x ^2-x+4)/(2*x^2*log(3)^2+(-4*x^3+16*x^2)*log(3)+2*x^4-16*x^3+32*x^2),x, alg orithm="giac")
Output:
1/2*(x + e^x - log(x))/(x^2 - x*log(3) - 4*x)
Time = 7.45 (sec) , antiderivative size = 7592, normalized size of antiderivative = 303.68 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\text {Too large to display} \] Input:
int(-(x - log(3) + exp(x)*(6*x + log(3)*(x - 1) - x^2 - 4) + x^2 + log(x)* (log(3) - 2*x + 4) - 4)/(2*x^2*log(3)^2 + log(3)*(16*x^2 - 4*x^3) + 32*x^2 - 16*x^3 + 2*x^4),x)
Output:
exp(x)/(log(9) - 2*x + 8) - log((12*x - 2*log(81) + (512*log(3) + 64*log(8 1) - x*(128*log(81) - 384*log(3) - 48*log(3)^2 + 4*log(81)^2 + 256) + 32*l og(3)*log(81) + 4*log(3)^2*log(81) + 64*log(3)^2 + 1024)/(2*(log(3) + 4)^2 ) - 32)/(2*(log(3) + 4)^2))/(2*(log(3) + 4)^2) - ((2*exp(x))/(log(3) + 4) - (4*x*exp(x))/(8*log(3) + log(3)^2 + 16))/(4*x + x*log(3) - x^2) - (3*ei( x))/(8*log(3) + log(3)^2 + 16) - 2/(x*(8*log(3) + log(3)^2 + 16)) + log(x - log(3) - 4)/(2*(log(3) + 4)^2) - ((exp(x)*log(3))/(log(3) + 4) + (x*exp( x)*(2*log(3) + log(3)^2))/(8*log(3) + log(3)^2 + 16))/(8*x + 2*x*log(3) - 2*x^2) + (3*exp(4)*ei(x - log(3) - 4))/2 + (x - x^3/(log(3) + 4)^2 + x*log (x) - (x^2*log(x))/(log(3) + 4) + (x^3*log(x))/(log(3) + 4)^2)/(2*x^2*log( 3) + 8*x^2 - 2*x^3) - (3*exp(x))/((log(3) + 4)*(log(3) - x + 4)) - (2*atan h((log(81) - 4*x + 16)/((log(81) - 4*log(3))^(1/2)*(4*log(3) + log(81) + 3 2)^(1/2))))/((log(81) - 4*log(3))^(1/2)*(4*log(3) + log(81) + 32)^(1/2)) + (log((2*x*log(3)^2)/(256*log(3) + 96*log(3)^2 + 16*log(3)^3 + log(3)^4 + 256) - (log(3)^2*log(81) + 16*log(3)^2)/(256*log(3) + 96*log(3)^2 + 16*log (3)^3 + log(3)^4 + 256) + ((log(3)^2*(16*log(81) + 8*(-(4*log(3) - log(81) )*(4*log(3) + log(81) + 32))^(1/2) + 256) - log(3)*(64*log(81) + 4*log(81) *(-(4*log(3) - log(81))*(4*log(3) + log(81) + 32))^(1/2) + 16*(-(4*log(3) - log(81))*(4*log(3) + log(81) + 32))^(1/2) + (log(81)^2*(-(4*log(3) - log (81))*(4*log(3) + log(81) + 32))^(1/2))/8 + 6*log(81)^2 + log(81)^3/8) ...
Time = 0.24 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {4-x-x^2+\log (3)+e^x \left (4-6 x+x^2+(1-x) \log (3)\right )+(-4+2 x-\log (3)) \log (x)}{32 x^2-16 x^3+2 x^4+\left (16 x^2-4 x^3\right ) \log (3)+2 x^2 \log ^2(3)} \, dx=\frac {-e^{x} \mathrm {log}\left (3\right )-4 e^{x}+\mathrm {log}\left (x \right ) \mathrm {log}\left (3\right )+4 \,\mathrm {log}\left (x \right )-x^{2}}{2 x \left (\mathrm {log}\left (3\right )^{2}-\mathrm {log}\left (3\right ) x +8 \,\mathrm {log}\left (3\right )-4 x +16\right )} \] Input:
int(((-log(3)+2*x-4)*log(x)+((1-x)*log(3)+x^2-6*x+4)*exp(x)+log(3)-x^2-x+4 )/(2*x^2*log(3)^2+(-4*x^3+16*x^2)*log(3)+2*x^4-16*x^3+32*x^2),x)
Output:
( - e**x*log(3) - 4*e**x + log(x)*log(3) + 4*log(x) - x**2)/(2*x*(log(3)** 2 - log(3)*x + 8*log(3) - 4*x + 16))