Integrand size = 111, antiderivative size = 29 \[ \int \frac {45 e^5+x^2+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \log ^2(2)+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx=\frac {1}{9} (x-\log (3))+\frac {5 e^5}{x (1+\log (2)-2 \log (x))} \]
Time = 0.48 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {45 e^5+x^2+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \log ^2(2)+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx=\frac {90 e^5+x^2 (2+\log (4))-4 x^2 \log (x)}{18 x (1+\log (2)-2 \log (x))} \]
Integrate[(45*E^5 + x^2 + (-45*E^5 + 2*x^2)*Log[2] + x^2*Log[2]^2 + (90*E^ 5 - 4*x^2 - 4*x^2*Log[2])*Log[x] + 4*x^2*Log[x]^2)/(9*x^2 + 18*x^2*Log[2] + 9*x^2*Log[2]^2 + (-36*x^2 - 36*x^2*Log[2])*Log[x] + 36*x^2*Log[x]^2),x]
Time = 0.78 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {6, 6, 6, 7292, 27, 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.
\(\displaystyle \int \frac {x^2+4 x^2 \log ^2(x)+x^2 \log ^2(2)+\left (-4 x^2-4 x^2 \log (2)+90 e^5\right ) \log (x)+\left (2 x^2-45 e^5\right ) \log (2)+45 e^5}{9 x^2+36 x^2 \log ^2(x)+9 x^2 \log ^2(2)+18 x^2 \log (2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {4 x^2 \log ^2(x)+x^2 \left (1+\log ^2(2)\right )+\left (-4 x^2-4 x^2 \log (2)+90 e^5\right ) \log (x)+\left (2 x^2-45 e^5\right ) \log (2)+45 e^5}{9 x^2+36 x^2 \log ^2(x)+9 x^2 \log ^2(2)+18 x^2 \log (2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {4 x^2 \log ^2(x)+x^2 \left (1+\log ^2(2)\right )+\left (-4 x^2-4 x^2 \log (2)+90 e^5\right ) \log (x)+\left (2 x^2-45 e^5\right ) \log (2)+45 e^5}{36 x^2 \log ^2(x)+9 x^2 \log ^2(2)+x^2 (9+18 \log (2))+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)}dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {4 x^2 \log ^2(x)+x^2 \left (1+\log ^2(2)\right )+\left (-4 x^2-4 x^2 \log (2)+90 e^5\right ) \log (x)+\left (2 x^2-45 e^5\right ) \log (2)+45 e^5}{36 x^2 \log ^2(x)+x^2 \left (9+9 \log ^2(2)+18 \log (2)\right )+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7292 |
\(\displaystyle \int \frac {4 x^2 \log ^2(x)+x^2 \left (1+\log ^2(2)\right )+\left (-4 x^2-4 x^2 \log (2)+90 e^5\right ) \log (x)+\left (2 x^2-45 e^5\right ) \log (2)+45 e^5}{9 x^2 (-2 \log (x)+1+\log (2))^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{9} \int \frac {4 \log ^2(x) x^2+\left (1+\log ^2(2)\right ) x^2+2 \left (-2 \log (2) x^2-2 x^2+45 e^5\right ) \log (x)-\left (45 e^5-2 x^2\right ) \log (2)+45 e^5}{x^2 (-2 \log (x)+\log (2)+1)^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {1}{9} \int \left (1-\frac {45 e^5}{x^2 (-2 \log (x)+\log (2)+1)}+\frac {90 e^5}{x^2 (-2 \log (x)+\log (2)+1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{9} \left (x+\frac {45 e^5}{x (-2 \log (x)+1+\log (2))}\right )\) |
Int[(45*E^5 + x^2 + (-45*E^5 + 2*x^2)*Log[2] + x^2*Log[2]^2 + (90*E^5 - 4* x^2 - 4*x^2*Log[2])*Log[x] + 4*x^2*Log[x]^2)/(9*x^2 + 18*x^2*Log[2] + 9*x^ 2*Log[2]^2 + (-36*x^2 - 36*x^2*Log[2])*Log[x] + 36*x^2*Log[x]^2),x]
3.22.44.3.1 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]]
Time = 0.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76
method | result | size |
default | \(\frac {5 \,{\mathrm e}^{5}}{\left (1+\ln \left (2\right )-2 \ln \left (x \right )\right ) x}+\frac {x}{9}\) | \(22\) |
risch | \(\frac {5 \,{\mathrm e}^{5}}{\left (1+\ln \left (2\right )-2 \ln \left (x \right )\right ) x}+\frac {x}{9}\) | \(22\) |
norman | \(\frac {\left (\frac {\ln \left (2\right )}{9}+\frac {1}{9}\right ) x^{2}-\frac {2 x^{2} \ln \left (x \right )}{9}+5 \,{\mathrm e}^{5}}{x \left (1+\ln \left (2\right )-2 \ln \left (x \right )\right )}\) | \(37\) |
parallelrisch | \(\frac {2 x^{2} \ln \left (2\right )-4 x^{2} \ln \left (x \right )+2 x^{2}+90 \,{\mathrm e}^{5}}{18 x \left (1+\ln \left (2\right )-2 \ln \left (x \right )\right )}\) | \(40\) |
int((4*x^2*ln(x)^2+(-4*x^2*ln(2)+90*exp(5)-4*x^2)*ln(x)+x^2*ln(2)^2+(-45*e xp(5)+2*x^2)*ln(2)+45*exp(5)+x^2)/(36*x^2*ln(x)^2+(-36*x^2*ln(2)-36*x^2)*l n(x)+9*x^2*ln(2)^2+18*x^2*ln(2)+9*x^2),x,method=_RETURNVERBOSE)
Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {45 e^5+x^2+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \log ^2(2)+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx=\frac {x^{2} \log \left (2\right ) - 2 \, x^{2} \log \left (x\right ) + x^{2} + 45 \, e^{5}}{9 \, {\left (x \log \left (2\right ) - 2 \, x \log \left (x\right ) + x\right )}} \]
integrate((4*x^2*log(x)^2+(-4*x^2*log(2)+90*exp(5)-4*x^2)*log(x)+x^2*log(2 )^2+(-45*exp(5)+2*x^2)*log(2)+45*exp(5)+x^2)/(36*x^2*log(x)^2+(-36*x^2*log (2)-36*x^2)*log(x)+9*x^2*log(2)^2+18*x^2*log(2)+9*x^2),x, algorithm=\
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.69 \[ \int \frac {45 e^5+x^2+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \log ^2(2)+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx=\frac {x}{9} - \frac {5 e^{5}}{2 x \log {\left (x \right )} - x - x \log {\left (2 \right )}} \]
integrate((4*x**2*ln(x)**2+(-4*x**2*ln(2)+90*exp(5)-4*x**2)*ln(x)+x**2*ln( 2)**2+(-45*exp(5)+2*x**2)*ln(2)+45*exp(5)+x**2)/(36*x**2*ln(x)**2+(-36*x** 2*ln(2)-36*x**2)*ln(x)+9*x**2*ln(2)**2+18*x**2*ln(2)+9*x**2),x)
Time = 0.31 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {45 e^5+x^2+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \log ^2(2)+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx=\frac {x^{2} {\left (\log \left (2\right ) + 1\right )} - 2 \, x^{2} \log \left (x\right ) + 45 \, e^{5}}{9 \, {\left (x {\left (\log \left (2\right ) + 1\right )} - 2 \, x \log \left (x\right )\right )}} \]
integrate((4*x^2*log(x)^2+(-4*x^2*log(2)+90*exp(5)-4*x^2)*log(x)+x^2*log(2 )^2+(-45*exp(5)+2*x^2)*log(2)+45*exp(5)+x^2)/(36*x^2*log(x)^2+(-36*x^2*log (2)-36*x^2)*log(x)+9*x^2*log(2)^2+18*x^2*log(2)+9*x^2),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {45 e^5+x^2+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \log ^2(2)+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx=\frac {x^{2} \log \left (2\right ) - 2 \, x^{2} \log \left (x\right ) + x^{2} + 45 \, e^{5}}{9 \, {\left (x \log \left (2\right ) - 2 \, x \log \left (x\right ) + x\right )}} \]
integrate((4*x^2*log(x)^2+(-4*x^2*log(2)+90*exp(5)-4*x^2)*log(x)+x^2*log(2 )^2+(-45*exp(5)+2*x^2)*log(2)+45*exp(5)+x^2)/(36*x^2*log(x)^2+(-36*x^2*log (2)-36*x^2)*log(x)+9*x^2*log(2)^2+18*x^2*log(2)+9*x^2),x, algorithm=\
Timed out. \[ \int \frac {45 e^5+x^2+\left (-45 e^5+2 x^2\right ) \log (2)+x^2 \log ^2(2)+\left (90 e^5-4 x^2-4 x^2 \log (2)\right ) \log (x)+4 x^2 \log ^2(x)}{9 x^2+18 x^2 \log (2)+9 x^2 \log ^2(2)+\left (-36 x^2-36 x^2 \log (2)\right ) \log (x)+36 x^2 \log ^2(x)} \, dx=\int \frac {45\,{\mathrm {e}}^5+x^2\,{\ln \left (2\right )}^2-\ln \left (2\right )\,\left (45\,{\mathrm {e}}^5-2\,x^2\right )-\ln \left (x\right )\,\left (4\,x^2\,\ln \left (2\right )-90\,{\mathrm {e}}^5+4\,x^2\right )+4\,x^2\,{\ln \left (x\right )}^2+x^2}{9\,x^2\,{\ln \left (2\right )}^2+36\,x^2\,{\ln \left (x\right )}^2-\ln \left (x\right )\,\left (36\,x^2\,\ln \left (2\right )+36\,x^2\right )+18\,x^2\,\ln \left (2\right )+9\,x^2} \,d x \]
int((45*exp(5) + x^2*log(2)^2 - log(2)*(45*exp(5) - 2*x^2) - log(x)*(4*x^2 *log(2) - 90*exp(5) + 4*x^2) + 4*x^2*log(x)^2 + x^2)/(9*x^2*log(2)^2 + 36* x^2*log(x)^2 - log(x)*(36*x^2*log(2) + 36*x^2) + 18*x^2*log(2) + 9*x^2),x)