Integrand size = 63, antiderivative size = 25 \[ \int \frac {8 x^4-4 x^5+\left (-20 x^4+8 x^5\right ) \log (2 x)+\left (72-72 x+18 x^2\right ) \log ^3(2 x)}{\left (36-36 x+9 x^2\right ) \log ^3(2 x)} \, dx=-4+2 x-\frac {2 x^5}{9 (2-x) \log ^2(2 x)} \] Output:
2*x-4-2/9*x^5/ln(2*x)^2/(2-x)
Time = 2.17 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.88 \[ \int \frac {8 x^4-4 x^5+\left (-20 x^4+8 x^5\right ) \log (2 x)+\left (72-72 x+18 x^2\right ) \log ^3(2 x)}{\left (36-36 x+9 x^2\right ) \log ^3(2 x)} \, dx=2 x+\frac {2 x^5}{9 (-2+x) \log ^2(2 x)} \] Input:
Integrate[(8*x^4 - 4*x^5 + (-20*x^4 + 8*x^5)*Log[2*x] + (72 - 72*x + 18*x^ 2)*Log[2*x]^3)/((36 - 36*x + 9*x^2)*Log[2*x]^3),x]
Output:
2*x + (2*x^5)/(9*(-2 + x)*Log[2*x]^2)
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.
\(\displaystyle \int \frac {-4 x^5+8 x^4+\left (18 x^2-72 x+72\right ) \log ^3(2 x)+\left (8 x^5-20 x^4\right ) \log (2 x)}{\left (9 x^2-36 x+36\right ) \log ^3(2 x)} \, dx\) |
\(\Big \downarrow \) 7277 |
\(\displaystyle 36 \int \frac {-2 x^5+4 x^4+9 \left (x^2-4 x+4\right ) \log ^3(2 x)-2 \left (5 x^4-2 x^5\right ) \log (2 x)}{162 (2-x)^2 \log ^3(2 x)}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2}{9} \int \frac {-2 x^5+4 x^4+9 \left (x^2-4 x+4\right ) \log ^3(2 x)-2 \left (5 x^4-2 x^5\right ) \log (2 x)}{(2-x)^2 \log ^3(2 x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {2}{9} \int \left (\frac {2 (2 x-5) x^4}{(x-2)^2 \log ^2(2 x)}-\frac {2 x^4}{(x-2) \log ^3(2 x)}+9\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{9} \left (-2 \int \frac {x^4}{(x-2) \log ^3(2 x)}dx+2 \int \frac {x^4 (2 x-5)}{(x-2)^2 \log ^2(2 x)}dx+9 x\right )\) |
Input:
Int[(8*x^4 - 4*x^5 + (-20*x^4 + 8*x^5)*Log[2*x] + (72 - 72*x + 18*x^2)*Log [2*x]^3)/((36 - 36*x + 9*x^2)*Log[2*x]^3),x]
Output:
$Aborted
Time = 3.13 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84
method | result | size |
risch | \(2 x +\frac {2 x^{5}}{9 \left (-2+x \right ) \ln \left (2 x \right )^{2}}\) | \(21\) |
norman | \(\frac {-8 \ln \left (2 x \right )^{2}+\frac {2 x^{5}}{9}+2 x^{2} \ln \left (2 x \right )^{2}}{\left (-2+x \right ) \ln \left (2 x \right )^{2}}\) | \(38\) |
parallelrisch | \(\frac {2 x^{5}+18 x^{2} \ln \left (2 x \right )^{2}-72 \ln \left (2 x \right )^{2}}{9 \ln \left (2 x \right )^{2} \left (-2+x \right )}\) | \(39\) |
derivativedivides | \(\frac {32}{9 \ln \left (2 x \right )^{2}}+\frac {2 x^{4}}{9 \ln \left (2 x \right )^{2}}+\frac {16 x}{9 \ln \left (2 x \right )^{2}}+\frac {4 x^{3}}{9 \ln \left (2 x \right )^{2}}+\frac {8 x^{2}}{9 \ln \left (2 x \right )^{2}}+2 x +\frac {128}{9 \left (2 x -4\right ) \ln \left (2 x \right )^{2}}\) | \(70\) |
default | \(\frac {32}{9 \ln \left (2 x \right )^{2}}+\frac {2 x^{4}}{9 \ln \left (2 x \right )^{2}}+\frac {16 x}{9 \ln \left (2 x \right )^{2}}+\frac {4 x^{3}}{9 \ln \left (2 x \right )^{2}}+\frac {8 x^{2}}{9 \ln \left (2 x \right )^{2}}+2 x +\frac {128}{9 \left (2 x -4\right ) \ln \left (2 x \right )^{2}}\) | \(70\) |
parts | \(2 x +\frac {64}{9 \left (-2+x \right ) \left (\ln \left (2\right )+\ln \left (x \right )\right )^{2}}+\frac {2 x^{4}}{9 \left (\ln \left (2\right )+\ln \left (x \right )\right )^{2}}+\frac {16 x}{9 \left (\ln \left (2\right )+\ln \left (x \right )\right )^{2}}+\frac {8 x^{2}}{9 \left (\ln \left (2\right )+\ln \left (x \right )\right )^{2}}+\frac {4 x^{3}}{9 \left (\ln \left (2\right )+\ln \left (x \right )\right )^{2}}+\frac {32}{9 \left (\ln \left (2\right )+\ln \left (x \right )\right )^{2}}\) | \(74\) |
Input:
int(((18*x^2-72*x+72)*ln(2*x)^3+(8*x^5-20*x^4)*ln(2*x)-4*x^5+8*x^4)/(9*x^2 -36*x+36)/ln(2*x)^3,x,method=_RETURNVERBOSE)
Output:
2*x+2/9*x^5/(-2+x)/ln(2*x)^2
Time = 0.10 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28 \[ \int \frac {8 x^4-4 x^5+\left (-20 x^4+8 x^5\right ) \log (2 x)+\left (72-72 x+18 x^2\right ) \log ^3(2 x)}{\left (36-36 x+9 x^2\right ) \log ^3(2 x)} \, dx=\frac {2 \, {\left (x^{5} + 9 \, {\left (x^{2} - 2 \, x\right )} \log \left (2 \, x\right )^{2}\right )}}{9 \, {\left (x - 2\right )} \log \left (2 \, x\right )^{2}} \] Input:
integrate(((18*x^2-72*x+72)*log(2*x)^3+(8*x^5-20*x^4)*log(2*x)-4*x^5+8*x^4 )/(9*x^2-36*x+36)/log(2*x)^3,x, algorithm="fricas")
Output:
2/9*(x^5 + 9*(x^2 - 2*x)*log(2*x)^2)/((x - 2)*log(2*x)^2)
Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.76 \[ \int \frac {8 x^4-4 x^5+\left (-20 x^4+8 x^5\right ) \log (2 x)+\left (72-72 x+18 x^2\right ) \log ^3(2 x)}{\left (36-36 x+9 x^2\right ) \log ^3(2 x)} \, dx=\frac {2 x^{5}}{\left (9 x - 18\right ) \log {\left (2 x \right )}^{2}} + 2 x \] Input:
integrate(((18*x**2-72*x+72)*ln(2*x)**3+(8*x**5-20*x**4)*ln(2*x)-4*x**5+8* x**4)/(9*x**2-36*x+36)/ln(2*x)**3,x)
Output:
2*x**5/((9*x - 18)*log(2*x)**2) + 2*x
Leaf count of result is larger than twice the leaf count of optimal. 87 vs. \(2 (21) = 42\).
Time = 0.16 (sec) , antiderivative size = 87, normalized size of antiderivative = 3.48 \[ \int \frac {8 x^4-4 x^5+\left (-20 x^4+8 x^5\right ) \log (2 x)+\left (72-72 x+18 x^2\right ) \log ^3(2 x)}{\left (36-36 x+9 x^2\right ) \log ^3(2 x)} \, dx=\frac {2 \, {\left (x^{5} + 9 \, x^{2} \log \left (2\right )^{2} - 18 \, x \log \left (2\right )^{2} + 9 \, {\left (x^{2} - 2 \, x\right )} \log \left (x\right )^{2} + 18 \, {\left (x^{2} \log \left (2\right ) - 2 \, x \log \left (2\right )\right )} \log \left (x\right )\right )}}{9 \, {\left (x \log \left (2\right )^{2} + {\left (x - 2\right )} \log \left (x\right )^{2} - 2 \, \log \left (2\right )^{2} + 2 \, {\left (x \log \left (2\right ) - 2 \, \log \left (2\right )\right )} \log \left (x\right )\right )}} \] Input:
integrate(((18*x^2-72*x+72)*log(2*x)^3+(8*x^5-20*x^4)*log(2*x)-4*x^5+8*x^4 )/(9*x^2-36*x+36)/log(2*x)^3,x, algorithm="maxima")
Output:
2/9*(x^5 + 9*x^2*log(2)^2 - 18*x*log(2)^2 + 9*(x^2 - 2*x)*log(x)^2 + 18*(x ^2*log(2) - 2*x*log(2))*log(x))/(x*log(2)^2 + (x - 2)*log(x)^2 - 2*log(2)^ 2 + 2*(x*log(2) - 2*log(2))*log(x))
Time = 0.13 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.12 \[ \int \frac {8 x^4-4 x^5+\left (-20 x^4+8 x^5\right ) \log (2 x)+\left (72-72 x+18 x^2\right ) \log ^3(2 x)}{\left (36-36 x+9 x^2\right ) \log ^3(2 x)} \, dx=\frac {2 \, x^{5}}{9 \, {\left (x \log \left (2 \, x\right )^{2} - 2 \, \log \left (2 \, x\right )^{2}\right )}} + 2 \, x \] Input:
integrate(((18*x^2-72*x+72)*log(2*x)^3+(8*x^5-20*x^4)*log(2*x)-4*x^5+8*x^4 )/(9*x^2-36*x+36)/log(2*x)^3,x, algorithm="giac")
Output:
2/9*x^5/(x*log(2*x)^2 - 2*log(2*x)^2) + 2*x
Time = 3.39 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {8 x^4-4 x^5+\left (-20 x^4+8 x^5\right ) \log (2 x)+\left (72-72 x+18 x^2\right ) \log ^3(2 x)}{\left (36-36 x+9 x^2\right ) \log ^3(2 x)} \, dx=\frac {2\,x^5}{9\,{\ln \left (2\,x\right )}^2\,\left (x-2\right )}+\frac {2\,x\,\left (9\,x-18\right )}{9\,\left (x-2\right )} \] Input:
int(-(log(2*x)*(20*x^4 - 8*x^5) - log(2*x)^3*(18*x^2 - 72*x + 72) - 8*x^4 + 4*x^5)/(log(2*x)^3*(9*x^2 - 36*x + 36)),x)
Output:
(2*x^5)/(9*log(2*x)^2*(x - 2)) + (2*x*(9*x - 18))/(9*(x - 2))
Time = 0.19 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40 \[ \int \frac {8 x^4-4 x^5+\left (-20 x^4+8 x^5\right ) \log (2 x)+\left (72-72 x+18 x^2\right ) \log ^3(2 x)}{\left (36-36 x+9 x^2\right ) \log ^3(2 x)} \, dx=\frac {2 x \left (9 \mathrm {log}\left (2 x \right )^{2} x -18 \mathrm {log}\left (2 x \right )^{2}+x^{4}\right )}{9 \mathrm {log}\left (2 x \right )^{2} \left (x -2\right )} \] Input:
int(((18*x^2-72*x+72)*log(2*x)^3+(8*x^5-20*x^4)*log(2*x)-4*x^5+8*x^4)/(9*x ^2-36*x+36)/log(2*x)^3,x)
Output:
(2*x*(9*log(2*x)**2*x - 18*log(2*x)**2 + x**4))/(9*log(2*x)**2*(x - 2))