Integrand size = 79, antiderivative size = 26 \[ \int \frac {6-2 x-6 x^2+2 e^3 x^2+4 x^3+\left (-6-x+e^3 x+2 x^2\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )}{4 x^2+2 x \log (x)} \, dx=(-3+x) \left (\frac {1}{2} \left (e^3+x\right )+\log \left (\frac {2 x}{2 x+\log (x)}\right )\right ) \]
Time = 0.24 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65 \[ \int \frac {6-2 x-6 x^2+2 e^3 x^2+4 x^3+\left (-6-x+e^3 x+2 x^2\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )}{4 x^2+2 x \log (x)} \, dx=\frac {1}{2} \left (\left (-3+e^3\right ) x+x^2-6 \log (x)+2 x \log \left (\frac {2 x}{2 x+\log (x)}\right )+6 \log (2 x+\log (x))\right ) \]
Integrate[(6 - 2*x - 6*x^2 + 2*E^3*x^2 + 4*x^3 + (-6 - x + E^3*x + 2*x^2)* Log[x] + (4*x^2 + 2*x*Log[x])*Log[(2*x)/(2*x + Log[x])])/(4*x^2 + 2*x*Log[ x]),x]
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.
\(\displaystyle \int \frac {4 x^3+2 e^3 x^2-6 x^2+\left (2 x^2+e^3 x-x-6\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )-2 x+6}{4 x^2+2 x \log (x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {4 x^3+\left (2 e^3-6\right ) x^2+\left (2 x^2+e^3 x-x-6\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )-2 x+6}{4 x^2+2 x \log (x)}dx\) |
\(\Big \downarrow \) 3041 |
\(\displaystyle \int \frac {4 x^3+\left (2 e^3-6\right ) x^2+\left (2 x^2+e^3 x-x-6\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )-2 x+6}{x (4 x+2 \log (x))}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 x^3+\left (2 e^3-6\right ) x^2+\left (2 x^2+e^3 x-x-6\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )-2 x+6}{2 x (2 x+\log (x))}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int \frac {4 x^3-2 \left (3-e^3\right ) x^2-2 x-\left (-2 x^2-e^3 x+x+6\right ) \log (x)+2 \left (2 x^2+\log (x) x\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )+6}{x (2 x+\log (x))}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{2} \int \left (\frac {4 x^3+2 \log (x) x^2-6 \left (1-\frac {e^3}{3}\right ) x^2-\left (1-e^3\right ) \log (x) x-2 x-6 \log (x)+6}{x (2 x+\log (x))}+2 \log \left (\frac {2 x}{2 x+\log (x)}\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (-2 \int \frac {1}{-2 x-\log (x)}dx+10 \int \frac {1}{2 x+\log (x)}dx+6 \int \frac {1}{x (2 x+\log (x))}dx+x^2-\left (1-e^3\right ) x-2 x+2 x \log \left (\frac {2 x}{2 x+\log (x)}\right )-6 \log (x)\right )\) |
Int[(6 - 2*x - 6*x^2 + 2*E^3*x^2 + 4*x^3 + (-6 - x + E^3*x + 2*x^2)*Log[x] + (4*x^2 + 2*x*Log[x])*Log[(2*x)/(2*x + Log[x])])/(4*x^2 + 2*x*Log[x]),x]
3.5.71.3.1 Defintions of rubi rules used
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[(u_.)*((a_.)*(x_)^(m_.) + Log[(c_.)*(x_)^(n_.)]^(q_.)*(b_.)*(x_)^(r_.)) ^(p_.), x_Symbol] :> Int[u*x^(p*r)*(a*x^(m - r) + b*Log[c*x^n]^q)^p, x] /; FreeQ[{a, b, c, m, n, p, q, r}, x] && IntegerQ[p]
Time = 3.23 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.65
method | result | size |
parallelrisch | \(\frac {x \,{\mathrm e}^{3}}{2}+\frac {x^{2}}{2}+x \ln \left (\frac {2 x}{2 x +\ln \left (x \right )}\right )-\frac {3 x}{2}-3 \ln \left (\frac {2 x}{2 x +\ln \left (x \right )}\right )\) | \(43\) |
risch | \(-x \ln \left (x +\frac {\ln \left (x \right )}{2}\right )+x \ln \left (x \right )-\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i}{x +\frac {\ln \left (x \right )}{2}}\right ) \operatorname {csgn}\left (\frac {i x}{x +\frac {\ln \left (x \right )}{2}}\right )}{2}+\frac {i \pi x \,\operatorname {csgn}\left (i x \right ) \operatorname {csgn}\left (\frac {i x}{x +\frac {\ln \left (x \right )}{2}}\right )^{2}}{2}+\frac {i \pi x \,\operatorname {csgn}\left (\frac {i}{x +\frac {\ln \left (x \right )}{2}}\right ) \operatorname {csgn}\left (\frac {i x}{x +\frac {\ln \left (x \right )}{2}}\right )^{2}}{2}-\frac {i \pi x \operatorname {csgn}\left (\frac {i x}{x +\frac {\ln \left (x \right )}{2}}\right )^{3}}{2}+\frac {x \,{\mathrm e}^{3}}{2}+\frac {x^{2}}{2}-\frac {3 x}{2}-3 \ln \left (x \right )+3 \ln \left (2 x +\ln \left (x \right )\right )\) | \(154\) |
int(((2*x*ln(x)+4*x^2)*ln(2*x/(2*x+ln(x)))+(x*exp(3)+2*x^2-x-6)*ln(x)+2*x^ 2*exp(3)+4*x^3-6*x^2-2*x+6)/(2*x*ln(x)+4*x^2),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.15 \[ \int \frac {6-2 x-6 x^2+2 e^3 x^2+4 x^3+\left (-6-x+e^3 x+2 x^2\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )}{4 x^2+2 x \log (x)} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{2} \, x e^{3} + {\left (x - 3\right )} \log \left (\frac {2 \, x}{2 \, x + \log \left (x\right )}\right ) - \frac {3}{2} \, x \]
integrate(((2*x*log(x)+4*x^2)*log(2*x/(2*x+log(x)))+(x*exp(3)+2*x^2-x-6)*l og(x)+2*x^2*exp(3)+4*x^3-6*x^2-2*x+6)/(2*x*log(x)+4*x^2),x, algorithm=\
Time = 0.21 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {6-2 x-6 x^2+2 e^3 x^2+4 x^3+\left (-6-x+e^3 x+2 x^2\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )}{4 x^2+2 x \log (x)} \, dx=\frac {x^{2}}{2} + x \log {\left (\frac {2 x}{2 x + \log {\left (x \right )}} \right )} + \frac {x \left (-3 + e^{3}\right )}{2} - 3 \log {\left (x \right )} + 3 \log {\left (2 x + \log {\left (x \right )} \right )} \]
integrate(((2*x*ln(x)+4*x**2)*ln(2*x/(2*x+ln(x)))+(x*exp(3)+2*x**2-x-6)*ln (x)+2*x**2*exp(3)+4*x**3-6*x**2-2*x+6)/(2*x*ln(x)+4*x**2),x)
Time = 0.31 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.35 \[ \int \frac {6-2 x-6 x^2+2 e^3 x^2+4 x^3+\left (-6-x+e^3 x+2 x^2\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )}{4 x^2+2 x \log (x)} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{2} \, x {\left (e^{3} + 2 \, \log \left (2\right ) - 3\right )} - {\left (x - 3\right )} \log \left (2 \, x + \log \left (x\right )\right ) + {\left (x - 3\right )} \log \left (x\right ) \]
integrate(((2*x*log(x)+4*x^2)*log(2*x/(2*x+log(x)))+(x*exp(3)+2*x^2-x-6)*l og(x)+2*x^2*exp(3)+4*x^3-6*x^2-2*x+6)/(2*x*log(x)+4*x^2),x, algorithm=\
Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.73 \[ \int \frac {6-2 x-6 x^2+2 e^3 x^2+4 x^3+\left (-6-x+e^3 x+2 x^2\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )}{4 x^2+2 x \log (x)} \, dx=\frac {1}{2} \, x^{2} + \frac {1}{2} \, x e^{3} + x \log \left (2\right ) - x \log \left (2 \, x + \log \left (x\right )\right ) + x \log \left (x\right ) - \frac {3}{2} \, x + 3 \, \log \left (2 \, x + \log \left (x\right )\right ) - 3 \, \log \left (x\right ) \]
integrate(((2*x*log(x)+4*x^2)*log(2*x/(2*x+log(x)))+(x*exp(3)+2*x^2-x-6)*l og(x)+2*x^2*exp(3)+4*x^3-6*x^2-2*x+6)/(2*x*log(x)+4*x^2),x, algorithm=\
1/2*x^2 + 1/2*x*e^3 + x*log(2) - x*log(2*x + log(x)) + x*log(x) - 3/2*x + 3*log(2*x + log(x)) - 3*log(x)
Time = 9.53 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.58 \[ \int \frac {6-2 x-6 x^2+2 e^3 x^2+4 x^3+\left (-6-x+e^3 x+2 x^2\right ) \log (x)+\left (4 x^2+2 x \log (x)\right ) \log \left (\frac {2 x}{2 x+\log (x)}\right )}{4 x^2+2 x \log (x)} \, dx=3\,\ln \left (2\,x+\ln \left (x\right )\right )-3\,\ln \left (x\right )+x\,\ln \left (\frac {2\,x}{2\,x+\ln \left (x\right )}\right )+\frac {x^2}{2}+x\,\left (\frac {{\mathrm {e}}^3}{2}-\frac {3}{2}\right ) \]