Integrand size = 112, antiderivative size = 33 \begin {dmath*} \int \frac {-36 x+42 x^2-12 x^3+e^2 \left (-18+24 x-6 x^2\right )+\left (-72 x+90 x^2-24 x^3\right ) \log (x)}{e^4 \left (9 x^2-12 x^3+4 x^4\right )+e^2 \left (36 x^3-48 x^4+16 x^5\right ) \log (x)+\left (36 x^4-48 x^5+16 x^6\right ) \log ^2(x)} \, dx=\frac {\frac {1}{x}+\frac {3-x}{3 x-2 x^2}}{e^2+2 x \log (x)} \end {dmath*}
Time = 0.37 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \begin {dmath*} \int \frac {-36 x+42 x^2-12 x^3+e^2 \left (-18+24 x-6 x^2\right )+\left (-72 x+90 x^2-24 x^3\right ) \log (x)}{e^4 \left (9 x^2-12 x^3+4 x^4\right )+e^2 \left (36 x^3-48 x^4+16 x^5\right ) \log (x)+\left (36 x^4-48 x^5+16 x^6\right ) \log ^2(x)} \, dx=-\frac {3 (2-x)}{x (-3+2 x) \left (e^2+2 x \log (x)\right )} \end {dmath*}
Integrate[(-36*x + 42*x^2 - 12*x^3 + E^2*(-18 + 24*x - 6*x^2) + (-72*x + 9 0*x^2 - 24*x^3)*Log[x])/(E^4*(9*x^2 - 12*x^3 + 4*x^4) + E^2*(36*x^3 - 48*x ^4 + 16*x^5)*Log[x] + (36*x^4 - 48*x^5 + 16*x^6)*Log[x]^2),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.
\(\displaystyle \int \frac {-12 x^3+42 x^2+e^2 \left (-6 x^2+24 x-18\right )+\left (-24 x^3+90 x^2-72 x\right ) \log (x)-36 x}{\left (16 x^6-48 x^5+36 x^4\right ) \log ^2(x)+e^2 \left (16 x^5-48 x^4+36 x^3\right ) \log (x)+e^4 \left (4 x^4-12 x^3+9 x^2\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {6 \left (-e^2 \left (x^2-4 x+3\right )-x \left (2 x^2-7 x+6\right )-x \left (4 x^2-15 x+12\right ) \log (x)\right )}{(3-2 x)^2 x^2 \left (2 x \log (x)+e^2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 6 \int -\frac {e^2 \left (x^2-4 x+3\right )+x \left (2 x^2-7 x+6\right )+x \left (4 x^2-15 x+12\right ) \log (x)}{(3-2 x)^2 x^2 \left (2 x \log (x)+e^2\right )^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -6 \int \frac {e^2 \left (x^2-4 x+3\right )+x \left (2 x^2-7 x+6\right )+x \left (4 x^2-15 x+12\right ) \log (x)}{(3-2 x)^2 x^2 \left (2 x \log (x)+e^2\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -6 \int \left (\frac {4 x^2-15 x+12}{2 x^2 (2 x-3)^2 \left (2 x \log (x)+e^2\right )}-\frac {\left (e^2-2 x\right ) (x-2)}{2 x^2 (2 x-3) \left (2 x \log (x)+e^2\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -6 \left (-\frac {1}{3} e^2 \int \frac {1}{x^2 \left (2 x \log (x)+e^2\right )^2}dx+\frac {2}{3} \int \frac {1}{x^2 \left (2 x \log (x)+e^2\right )}dx+\frac {1}{18} \left (12-e^2\right ) \int \frac {1}{x \left (2 x \log (x)+e^2\right )^2}dx-\frac {1}{9} \left (3-e^2\right ) \int \frac {1}{(2 x-3) \left (2 x \log (x)+e^2\right )^2}dx+\frac {1}{18} \int \frac {1}{x \left (2 x \log (x)+e^2\right )}dx-\frac {1}{3} \int \frac {1}{(2 x-3)^2 \left (2 x \log (x)+e^2\right )}dx-\frac {1}{9} \int \frac {1}{(2 x-3) \left (2 x \log (x)+e^2\right )}dx\right )\) |
Int[(-36*x + 42*x^2 - 12*x^3 + E^2*(-18 + 24*x - 6*x^2) + (-72*x + 90*x^2 - 24*x^3)*Log[x])/(E^4*(9*x^2 - 12*x^3 + 4*x^4) + E^2*(36*x^3 - 48*x^4 + 1 6*x^5)*Log[x] + (36*x^4 - 48*x^5 + 16*x^6)*Log[x]^2),x]
3.9.96.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 3.68 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {-6+3 x}{x \left (-3+2 x \right ) \left (2 x \ln \left (x \right )+{\mathrm e}^{2}\right )}\) | \(26\) |
norman | \(\frac {-6+3 x}{x \left (-3+2 x \right ) \left (2 x \ln \left (x \right )+{\mathrm e}^{2}\right )}\) | \(27\) |
default | \(-\frac {6 \left (1-\frac {x}{2}\right )}{x \left (-3+2 x \right ) \left (2 x \ln \left (x \right )+{\mathrm e}^{2}\right )}\) | \(28\) |
parallelrisch | \(\frac {12 x -24}{4 x \left (4 x^{2} \ln \left (x \right )+2 \,{\mathrm e}^{2} x -6 x \ln \left (x \right )-3 \,{\mathrm e}^{2}\right )}\) | \(35\) |
int(((-24*x^3+90*x^2-72*x)*ln(x)+(-6*x^2+24*x-18)*exp(2)-12*x^3+42*x^2-36* x)/((16*x^6-48*x^5+36*x^4)*ln(x)^2+(16*x^5-48*x^4+36*x^3)*exp(2)*ln(x)+(4* x^4-12*x^3+9*x^2)*exp(2)^2),x,method=_RETURNVERBOSE)
Time = 0.33 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \begin {dmath*} \int \frac {-36 x+42 x^2-12 x^3+e^2 \left (-18+24 x-6 x^2\right )+\left (-72 x+90 x^2-24 x^3\right ) \log (x)}{e^4 \left (9 x^2-12 x^3+4 x^4\right )+e^2 \left (36 x^3-48 x^4+16 x^5\right ) \log (x)+\left (36 x^4-48 x^5+16 x^6\right ) \log ^2(x)} \, dx=\frac {3 \, {\left (x - 2\right )}}{{\left (2 \, x^{2} - 3 \, x\right )} e^{2} + 2 \, {\left (2 \, x^{3} - 3 \, x^{2}\right )} \log \left (x\right )} \end {dmath*}
integrate(((-24*x^3+90*x^2-72*x)*log(x)+(-6*x^2+24*x-18)*exp(2)-12*x^3+42* x^2-36*x)/((16*x^6-48*x^5+36*x^4)*log(x)^2+(16*x^5-48*x^4+36*x^3)*exp(2)*l og(x)+(4*x^4-12*x^3+9*x^2)*exp(2)^2),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \begin {dmath*} \int \frac {-36 x+42 x^2-12 x^3+e^2 \left (-18+24 x-6 x^2\right )+\left (-72 x+90 x^2-24 x^3\right ) \log (x)}{e^4 \left (9 x^2-12 x^3+4 x^4\right )+e^2 \left (36 x^3-48 x^4+16 x^5\right ) \log (x)+\left (36 x^4-48 x^5+16 x^6\right ) \log ^2(x)} \, dx=\frac {3 x - 6}{2 x^{2} e^{2} - 3 x e^{2} + \left (4 x^{3} - 6 x^{2}\right ) \log {\left (x \right )}} \end {dmath*}
integrate(((-24*x**3+90*x**2-72*x)*ln(x)+(-6*x**2+24*x-18)*exp(2)-12*x**3+ 42*x**2-36*x)/((16*x**6-48*x**5+36*x**4)*ln(x)**2+(16*x**5-48*x**4+36*x**3 )*exp(2)*ln(x)+(4*x**4-12*x**3+9*x**2)*exp(2)**2),x)
Time = 0.25 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.06 \begin {dmath*} \int \frac {-36 x+42 x^2-12 x^3+e^2 \left (-18+24 x-6 x^2\right )+\left (-72 x+90 x^2-24 x^3\right ) \log (x)}{e^4 \left (9 x^2-12 x^3+4 x^4\right )+e^2 \left (36 x^3-48 x^4+16 x^5\right ) \log (x)+\left (36 x^4-48 x^5+16 x^6\right ) \log ^2(x)} \, dx=\frac {3 \, {\left (x - 2\right )}}{2 \, x^{2} e^{2} - 3 \, x e^{2} + 2 \, {\left (2 \, x^{3} - 3 \, x^{2}\right )} \log \left (x\right )} \end {dmath*}
integrate(((-24*x^3+90*x^2-72*x)*log(x)+(-6*x^2+24*x-18)*exp(2)-12*x^3+42* x^2-36*x)/((16*x^6-48*x^5+36*x^4)*log(x)^2+(16*x^5-48*x^4+36*x^3)*exp(2)*l og(x)+(4*x^4-12*x^3+9*x^2)*exp(2)^2),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.03 \begin {dmath*} \int \frac {-36 x+42 x^2-12 x^3+e^2 \left (-18+24 x-6 x^2\right )+\left (-72 x+90 x^2-24 x^3\right ) \log (x)}{e^4 \left (9 x^2-12 x^3+4 x^4\right )+e^2 \left (36 x^3-48 x^4+16 x^5\right ) \log (x)+\left (36 x^4-48 x^5+16 x^6\right ) \log ^2(x)} \, dx=\frac {3 \, {\left (x - 2\right )}}{4 \, x^{3} \log \left (x\right ) + 2 \, x^{2} e^{2} - 6 \, x^{2} \log \left (x\right ) - 3 \, x e^{2}} \end {dmath*}
integrate(((-24*x^3+90*x^2-72*x)*log(x)+(-6*x^2+24*x-18)*exp(2)-12*x^3+42* x^2-36*x)/((16*x^6-48*x^5+36*x^4)*log(x)^2+(16*x^5-48*x^4+36*x^3)*exp(2)*l og(x)+(4*x^4-12*x^3+9*x^2)*exp(2)^2),x, algorithm=\
Time = 17.96 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.76 \begin {dmath*} \int \frac {-36 x+42 x^2-12 x^3+e^2 \left (-18+24 x-6 x^2\right )+\left (-72 x+90 x^2-24 x^3\right ) \log (x)}{e^4 \left (9 x^2-12 x^3+4 x^4\right )+e^2 \left (36 x^3-48 x^4+16 x^5\right ) \log (x)+\left (36 x^4-48 x^5+16 x^6\right ) \log ^2(x)} \, dx=\frac {3\,\left (x-2\right )}{x\,\left (2\,x-3\right )\,\left ({\mathrm {e}}^2+2\,x\,\ln \left (x\right )\right )} \end {dmath*}