Integrand size = 87, antiderivative size = 23 \[ \int \frac {6 x+3 x^2+2 x \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx=\frac {x^2}{3+3 x+\log \left (\frac {1}{4} \left (1+\frac {1}{e^3}\right )^2\right )} \]
Leaf count is larger than twice the leaf count of optimal. \(54\) vs. \(2(23)=46\).
Time = 0.03 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.35 \[ \int \frac {6 x+3 x^2+2 x \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx=\frac {1}{9} \left (-3+3 x-\log (4)-\frac {\left (3+\log (4)-2 \log \left (1+e^3\right )\right )^2}{3-3 x+\log (4)-2 \log \left (1+e^3\right )}+2 \log \left (1+e^3\right )\right ) \]
Integrate[(6*x + 3*x^2 + 2*x*Log[(1 + 2*E^3 + E^6)/(4*E^6)])/(9 + 18*x + 9 *x^2 + (6 + 6*x)*Log[(1 + 2*E^3 + E^6)/(4*E^6)] + Log[(1 + 2*E^3 + E^6)/(4 *E^6)]^2),x]
(-3 + 3*x - Log[4] - (3 + Log[4] - 2*Log[1 + E^3])^2/(3 - 3*x + Log[4] - 2 *Log[1 + E^3]) + 2*Log[1 + E^3])/9
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6, 2027, 2082, 1184, 27, 83}
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.
| \(\displaystyle \int \frac {3 x^2+6 x+2 x \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )}{9 x^2+18 x+(6 x+6) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+9+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx\) |
| \(\Big \downarrow \) 6 |
| \(\displaystyle \int \frac {3 x^2+x \left (6+2 \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )\right )}{9 x^2+18 x+(6 x+6) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+9+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )}dx\) |
| \(\Big \downarrow \) 2027 |
| \(\displaystyle \int \frac {x \left (3 x+6+2 \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )\right )}{9 x^2+18 x+(6 x+6) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+9+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )}dx\) |
| \(\Big \downarrow \) 2082 |
| \(\displaystyle \int \frac {x \left (3 x-2 \left (3+\log (4)-2 \log \left (1+e^3\right )\right )\right )}{9 x^2-6 x \left (3+\log (4)-2 \log \left (1+e^3\right )\right )+\left (3+\log (4)-2 \log \left (1+e^3\right )\right )^2}dx\) |
| \(\Big \downarrow \) 1184 |
| \(\displaystyle 9 \int \frac {x \left (3 x-2 \left (3+\log (4)-2 \log \left (1+e^3\right )\right )\right )}{9 \left (-3 x-2 \log \left (1+e^3\right )+\log (4)+3\right )^2}dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {x \left (3 x-2 \left (3+\log (4)-2 \log \left (1+e^3\right )\right )\right )}{\left (-3 x+3-2 \log \left (1+e^3\right )+\log (4)\right )^2}dx\) |
| \(\Big \downarrow \) 83 |
| \(\displaystyle -\frac {\left (3 x-2 \left (3+\log (4)-2 \log \left (1+e^3\right )\right )\right )^2}{9 \left (-3 x+3-2 \log \left (1+e^3\right )+\log (4)\right )}\) |
Int[(6*x + 3*x^2 + 2*x*Log[(1 + 2*E^3 + E^6)/(4*E^6)])/(9 + 18*x + 9*x^2 + (6 + 6*x)*Log[(1 + 2*E^3 + E^6)/(4*E^6)] + Log[(1 + 2*E^3 + E^6)/(4*E^6)] ^2),x]
3.14.27.3.1 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[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] && EqQ[a*d*f *(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_ ) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/c^p Int[(d + e*x)^m*(f + g*x )^n*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && E qQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Int[(Fx_.)*((a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.))^(p_.), x_Symbol] :> Int[x^ (p*r)*(a + b*x^(s - r))^p*Fx, x] /; FreeQ[{a, b, r, s}, x] && IntegerQ[p] & & PosQ[s - r] && !(EqQ[p, 1] && EqQ[u, 1])
Int[(u_)^(m_.)*(v_)^(n_.)*(w_)^(p_.), x_Symbol] :> Int[ExpandToSum[u, x]^m* ExpandToSum[v, x]^n*ExpandToSum[w, x]^p, x] /; FreeQ[{m, n, p}, x] && Linea rQ[{u, v}, x] && QuadraticQ[w, x] && !(LinearMatchQ[{u, v}, x] && Quadrati cMatchQ[w, x])
Time = 0.09 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.09
| method | result | size |
| norman | \(\frac {x^{2}}{\ln \left (\frac {{\mathrm e}^{6}}{4}+\frac {{\mathrm e}^{3}}{2}+\frac {1}{4}\right )+3 x -3}\) | \(25\) |
| gosper | \(\frac {x^{2}}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3+3 x}\) | \(29\) |
| parallelrisch | \(\frac {x^{2}}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3+3 x}\) | \(29\) |
| default | \(\frac {x}{3}-\frac {-\frac {{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )}^{2}}{9}-\frac {2 \ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )}{3}-1}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3+3 x}\) | \(61\) |
| risch | \(\frac {x}{3}+\frac {2 \ln \left ({\mathrm e}^{3}+1\right )^{2}}{9 \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}-\frac {4 \ln \left (2\right ) \ln \left ({\mathrm e}^{3}+1\right )}{9 \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}+\frac {2 \ln \left (2\right )^{2}}{9 \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}-\frac {2 \ln \left ({\mathrm e}^{3}+1\right )}{3 \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}+\frac {2 \ln \left (2\right )}{3 \left (-\ln \left (2\right )+\ln \left ({\mathrm e}^{3}+1\right )-\frac {3}{2}+\frac {3 x}{2}\right )}+\frac {1}{-2 \ln \left (2\right )+2 \ln \left ({\mathrm e}^{3}+1\right )-3+3 x}\) | \(138\) |
| meijerg | \(\left (\frac {2 \ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )}{9}+\frac {2}{3}\right ) \left (-\frac {3 x}{\left (\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3\right ) \left (1+\frac {3 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}\right )}+\ln \left (1+\frac {3 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}\right )\right )+\frac {{\left (\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3\right )}^{2} \left (\frac {x \left (\frac {9 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}+6\right )}{\left (\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3\right ) \left (1+\frac {3 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}\right )}-2 \ln \left (1+\frac {3 x}{\ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+3}\right )\right )}{9 \ln \left (\frac {\left ({\mathrm e}^{6}+2 \,{\mathrm e}^{3}+1\right ) {\mathrm e}^{-6}}{4}\right )+27}\) | \(217\) |
int((2*x*ln(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+3*x^2+6*x)/(ln(1/4*(exp(3) ^2+2*exp(3)+1)/exp(3)^2)^2+(6+6*x)*ln(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+ 9*x^2+18*x+9),x,method=_RETURNVERBOSE)
Leaf count of result is larger than twice the leaf count of optimal. 65 vs. \(2 (20) = 40\).
Time = 0.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 2.83 \[ \int \frac {6 x+3 x^2+2 x \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx=\frac {9 \, x^{2} + 3 \, {\left (x + 2\right )} \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right )^{2} + 9 \, x + 9}{9 \, {\left (3 \, x + \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 3\right )}} \]
integrate((2*x*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+3*x^2+6*x)/(log(1/4 *(exp(3)^2+2*exp(3)+1)/exp(3)^2)^2+(6+6*x)*log(1/4*(exp(3)^2+2*exp(3)+1)/e xp(3)^2)+9*x^2+18*x+9),x, algorithm=\
1/9*(9*x^2 + 3*(x + 2)*log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + log(1/4*(e^6 + 2*e^3 + 1)*e^(-6))^2 + 9*x + 9)/(3*x + log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + 3)
Leaf count of result is larger than twice the leaf count of optimal. 83 vs. \(2 (19) = 38\).
Time = 0.39 (sec) , antiderivative size = 83, normalized size of antiderivative = 3.61 \[ \int \frac {6 x+3 x^2+2 x \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx=\frac {x}{3} + \frac {- 6 \log {\left (1 + 2 e^{3} + e^{6} \right )} - 4 \log {\left (2 \right )} \log {\left (1 + 2 e^{3} + e^{6} \right )} + 4 \log {\left (2 \right )}^{2} + 12 \log {\left (2 \right )} + 9 + \log {\left (1 + 2 e^{3} + e^{6} \right )}^{2}}{27 x - 27 - 18 \log {\left (2 \right )} + 9 \log {\left (1 + 2 e^{3} + e^{6} \right )}} \]
integrate((2*x*ln(1/4*(exp(3)**2+2*exp(3)+1)/exp(3)**2)+3*x**2+6*x)/(ln(1/ 4*(exp(3)**2+2*exp(3)+1)/exp(3)**2)**2+(6+6*x)*ln(1/4*(exp(3)**2+2*exp(3)+ 1)/exp(3)**2)+9*x**2+18*x+9),x)
x/3 + (-6*log(1 + 2*exp(3) + exp(6)) - 4*log(2)*log(1 + 2*exp(3) + exp(6)) + 4*log(2)**2 + 12*log(2) + 9 + log(1 + 2*exp(3) + exp(6))**2)/(27*x - 27 - 18*log(2) + 9*log(1 + 2*exp(3) + exp(6)))
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (20) = 40\).
Time = 0.19 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.52 \[ \int \frac {6 x+3 x^2+2 x \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx=\frac {1}{3} \, x + \frac {\log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right )^{2} + 6 \, \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 9}{9 \, {\left (3 \, x + \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 3\right )}} \]
integrate((2*x*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+3*x^2+6*x)/(log(1/4 *(exp(3)^2+2*exp(3)+1)/exp(3)^2)^2+(6+6*x)*log(1/4*(exp(3)^2+2*exp(3)+1)/e xp(3)^2)+9*x^2+18*x+9),x, algorithm=\
1/3*x + 1/9*(log(1/4*(e^6 + 2*e^3 + 1)*e^(-6))^2 + 6*log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + 9)/(3*x + log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + 3)
Leaf count of result is larger than twice the leaf count of optimal. 58 vs. \(2 (20) = 40\).
Time = 0.25 (sec) , antiderivative size = 58, normalized size of antiderivative = 2.52 \[ \int \frac {6 x+3 x^2+2 x \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx=\frac {1}{3} \, x + \frac {\log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right )^{2} + 6 \, \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 9}{9 \, {\left (3 \, x + \log \left (\frac {1}{4} \, {\left (e^{6} + 2 \, e^{3} + 1\right )} e^{\left (-6\right )}\right ) + 3\right )}} \]
integrate((2*x*log(1/4*(exp(3)^2+2*exp(3)+1)/exp(3)^2)+3*x^2+6*x)/(log(1/4 *(exp(3)^2+2*exp(3)+1)/exp(3)^2)^2+(6+6*x)*log(1/4*(exp(3)^2+2*exp(3)+1)/e xp(3)^2)+9*x^2+18*x+9),x, algorithm=\
1/3*x + 1/9*(log(1/4*(e^6 + 2*e^3 + 1)*e^(-6))^2 + 6*log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + 9)/(3*x + log(1/4*(e^6 + 2*e^3 + 1)*e^(-6)) + 3)
Time = 11.08 (sec) , antiderivative size = 196, normalized size of antiderivative = 8.52 \[ \int \frac {6 x+3 x^2+2 x \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )}{9+18 x+9 x^2+(6+6 x) \log \left (\frac {1+2 e^3+e^6}{4 e^6}\right )+\log ^2\left (\frac {1+2 e^3+e^6}{4 e^6}\right )} \, dx=\frac {x}{3}-\frac {2\,\mathrm {atan}\left (\frac {\frac {\left (\ln \left (\frac {{\left ({\mathrm {e}}^3+1\right )}^{12}}{4096}\right )-18\right )\,\left (\ln \left (\frac {4096}{{\left ({\mathrm {e}}^3+1\right )}^{12}}\right )+4\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2+9\right )}{3\,\sqrt {144\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2-{\ln \left (\frac {{\left ({\mathrm {e}}^3+1\right )}^{12}}{4096}\right )}^2}}+\frac {x\,\left (\ln \left (\frac {105312291668557186697918027683670432318895095400549111254310977536}{{\left ({\mathrm {e}}^3+1\right )}^{216}}\right )+72\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2+162\right )}{3\,\sqrt {144\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2-{\ln \left (\frac {{\left ({\mathrm {e}}^3+1\right )}^{12}}{4096}\right )}^2}}}{\ln \left (\frac {16}{{\left ({\mathrm {e}}^3+1\right )}^4}\right )+\frac {4\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2}{3}+3}\right )\,\left (\ln \left (\frac {4096}{{\left ({\mathrm {e}}^3+1\right )}^{12}}\right )+4\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2+9\right )}{3\,\sqrt {144\,{\ln \left (\frac {{\mathrm {e}}^3}{2}+\frac {1}{2}\right )}^2-{\ln \left (\frac {{\left ({\mathrm {e}}^3+1\right )}^{12}}{4096}\right )}^2}} \]
int((6*x + 3*x^2 + 2*x*log(exp(-6)*(exp(3)/2 + exp(6)/4 + 1/4)))/(18*x + l og(exp(-6)*(exp(3)/2 + exp(6)/4 + 1/4))^2 + log(exp(-6)*(exp(3)/2 + exp(6) /4 + 1/4))*(6*x + 6) + 9*x^2 + 9),x)
x/3 - (2*atan((((log((exp(3) + 1)^12/4096) - 18)*(log(4096/(exp(3) + 1)^12 ) + 4*log(exp(3)/2 + 1/2)^2 + 9))/(3*(144*log(exp(3)/2 + 1/2)^2 - log((exp (3) + 1)^12/4096)^2)^(1/2)) + (x*(log(105312291668557186697918027683670432 318895095400549111254310977536/(exp(3) + 1)^216) + 72*log(exp(3)/2 + 1/2)^ 2 + 162))/(3*(144*log(exp(3)/2 + 1/2)^2 - log((exp(3) + 1)^12/4096)^2)^(1/ 2)))/(log(16/(exp(3) + 1)^4) + (4*log(exp(3)/2 + 1/2)^2)/3 + 3))*(log(4096 /(exp(3) + 1)^12) + 4*log(exp(3)/2 + 1/2)^2 + 9))/(3*(144*log(exp(3)/2 + 1 /2)^2 - log((exp(3) + 1)^12/4096)^2)^(1/2))