Integrand size = 109, antiderivative size = 25 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=\frac {3+x}{x-x \left (4+x^2 \log ^2(2)\right )^2+\log (6)} \]
Time = 0.03 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=\frac {-3-x}{15 x+8 x^3 \log ^2(2)+x^5 \log ^4(2)-\log (6)} \]
Integrate[(45 + (72*x^2 + 16*x^3)*Log[2]^2 + (15*x^4 + 4*x^5)*Log[2]^4 + L og[6])/(225*x^2 + 240*x^4*Log[2]^2 + 94*x^6*Log[2]^4 + 16*x^8*Log[2]^6 + x ^10*Log[2]^8 + (-30*x - 16*x^3*Log[2]^2 - 2*x^5*Log[2]^4)*Log[6] + Log[6]^ 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.
\(\displaystyle \int \frac {\left (4 x^5+15 x^4\right ) \log ^4(2)+\left (16 x^3+72 x^2\right ) \log ^2(2)+45+\log (6)}{x^{10} \log ^8(2)+16 x^8 \log ^6(2)+94 x^6 \log ^4(2)+240 x^4 \log ^2(2)+225 x^2+\log (6) \left (-2 x^5 \log ^4(2)-16 x^3 \log ^2(2)-30 x\right )+\log ^2(6)} \, dx\) |
\(\Big \downarrow \) 2462 |
\(\displaystyle \int \left (\frac {4}{x^5 \log ^4(2)+8 x^3 \log ^2(2)+15 x-\log (6)}+\frac {15 x^4 \log ^4(2)-16 x^3 \log ^2(2)+72 x^2 \log ^2(2)-60 x+5 (9+\log (6))}{\left (x^5 \log ^4(2)+8 x^3 \log ^2(2)+15 x-\log (6)\right )^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 5 \log (6) \int \frac {1}{\left (\log ^4(2) x^5+8 \log ^2(2) x^3+15 x-\log (6)\right )^2}dx-60 \int \frac {x}{\left (\log ^4(2) x^5+8 \log ^2(2) x^3+15 x-\log (6)\right )^2}dx-16 \log ^2(2) \int \frac {x^3}{\left (\log ^4(2) x^5+8 \log ^2(2) x^3+15 x-\log (6)\right )^2}dx+4 \int \frac {1}{\log ^4(2) x^5+8 \log ^2(2) x^3+15 x-\log (6)}dx-\frac {3}{x^5 \log ^4(2)+8 x^3 \log ^2(2)+15 x-\log (6)}\) |
Int[(45 + (72*x^2 + 16*x^3)*Log[2]^2 + (15*x^4 + 4*x^5)*Log[2]^4 + Log[6]) /(225*x^2 + 240*x^4*Log[2]^2 + 94*x^6*Log[2]^4 + 16*x^8*Log[2]^6 + x^10*Lo g[2]^8 + (-30*x - 16*x^3*Log[2]^2 - 2*x^5*Log[2]^4)*Log[6] + Log[6]^2),x]
3.27.38.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u*Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && GtQ [Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0 ] && RationalFunctionQ[u, x]
Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.32
method | result | size |
gosper | \(-\frac {3+x}{x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (6\right )+15 x}\) | \(33\) |
default | \(\frac {-3-x}{x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (6\right )+15 x}\) | \(34\) |
risch | \(\frac {-3-x}{x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (2\right )-\ln \left (3\right )+15 x}\) | \(38\) |
parallelrisch | \(\frac {-45+x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (6\right )}{15 x^{5} \ln \left (2\right )^{4}+120 x^{3} \ln \left (2\right )^{2}-15 \ln \left (6\right )+225 x}\) | \(53\) |
norman | \(\frac {-\frac {\left (\ln \left (6\right )+45\right ) x}{\ln \left (6\right )}-\frac {24 \ln \left (2\right )^{2} x^{3}}{\ln \left (6\right )}-\frac {3 \ln \left (2\right )^{4} x^{5}}{\ln \left (6\right )}}{x^{5} \ln \left (2\right )^{4}+8 x^{3} \ln \left (2\right )^{2}-\ln \left (6\right )+15 x}\) | \(67\) |
int((ln(6)+(4*x^5+15*x^4)*ln(2)^4+(16*x^3+72*x^2)*ln(2)^2+45)/(ln(6)^2+(-2 *x^5*ln(2)^4-16*x^3*ln(2)^2-30*x)*ln(6)+x^10*ln(2)^8+16*x^8*ln(2)^6+94*x^6 *ln(2)^4+240*x^4*ln(2)^2+225*x^2),x,method=_RETURNVERBOSE)
Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=-\frac {x + 3}{x^{5} \log \left (2\right )^{4} + 8 \, x^{3} \log \left (2\right )^{2} + 15 \, x - \log \left (6\right )} \]
integrate((log(6)+(4*x^5+15*x^4)*log(2)^4+(16*x^3+72*x^2)*log(2)^2+45)/(lo g(6)^2+(-2*x^5*log(2)^4-16*x^3*log(2)^2-30*x)*log(6)+x^10*log(2)^8+16*x^8* log(2)^6+94*x^6*log(2)^4+240*x^4*log(2)^2+225*x^2),x, algorithm=\
Time = 8.36 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=\frac {- x - 3}{x^{5} \log {\left (2 \right )}^{4} + 8 x^{3} \log {\left (2 \right )}^{2} + 15 x - \log {\left (6 \right )}} \]
integrate((ln(6)+(4*x**5+15*x**4)*ln(2)**4+(16*x**3+72*x**2)*ln(2)**2+45)/ (ln(6)**2+(-2*x**5*ln(2)**4-16*x**3*ln(2)**2-30*x)*ln(6)+x**10*ln(2)**8+16 *x**8*ln(2)**6+94*x**6*ln(2)**4+240*x**4*ln(2)**2+225*x**2),x)
Time = 0.20 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=-\frac {x + 3}{x^{5} \log \left (2\right )^{4} + 8 \, x^{3} \log \left (2\right )^{2} + 15 \, x - \log \left (6\right )} \]
integrate((log(6)+(4*x^5+15*x^4)*log(2)^4+(16*x^3+72*x^2)*log(2)^2+45)/(lo g(6)^2+(-2*x^5*log(2)^4-16*x^3*log(2)^2-30*x)*log(6)+x^10*log(2)^8+16*x^8* log(2)^6+94*x^6*log(2)^4+240*x^4*log(2)^2+225*x^2),x, algorithm=\
Time = 0.29 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=-\frac {x + 3}{x^{5} \log \left (2\right )^{4} + 8 \, x^{3} \log \left (2\right )^{2} + 15 \, x - \log \left (6\right )} \]
integrate((log(6)+(4*x^5+15*x^4)*log(2)^4+(16*x^3+72*x^2)*log(2)^2+45)/(lo g(6)^2+(-2*x^5*log(2)^4-16*x^3*log(2)^2-30*x)*log(6)+x^10*log(2)^8+16*x^8* log(2)^6+94*x^6*log(2)^4+240*x^4*log(2)^2+225*x^2),x, algorithm=\
Time = 0.25 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {45+\left (72 x^2+16 x^3\right ) \log ^2(2)+\left (15 x^4+4 x^5\right ) \log ^4(2)+\log (6)}{225 x^2+240 x^4 \log ^2(2)+94 x^6 \log ^4(2)+16 x^8 \log ^6(2)+x^{10} \log ^8(2)+\left (-30 x-16 x^3 \log ^2(2)-2 x^5 \log ^4(2)\right ) \log (6)+\log ^2(6)} \, dx=-\frac {x+3}{{\ln \left (2\right )}^4\,x^5+8\,{\ln \left (2\right )}^2\,x^3+15\,x-\ln \left (6\right )} \]