Integrand size = 73, antiderivative size = 24 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\left (x+\frac {\log (16)}{3}-\log \left (3-x^3\right )\right ) \log \left (\frac {1}{\log (x)}\right ) \]
Time = 0.20 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\left (x-\log \left (3-x^3\right )\right ) \log \left (\frac {1}{\log (x)}\right )-\frac {1}{3} \log (16) \log (\log (x)) \]
Integrate[(9*x - 3*x^4 + (3 - x^3)*Log[16] + (-9 + 3*x^3)*Log[3 - x^3] + ( -9*x - 9*x^3 + 3*x^4)*Log[x]*Log[Log[x]^(-1)])/((-9*x + 3*x^4)*Log[x]),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.
\(\displaystyle \int \frac {-3 x^4+\left (3 x^3-9\right ) \log \left (3-x^3\right )+\left (3-x^3\right ) \log (16)+\left (3 x^4-9 x^3-9 x\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )+9 x}{\left (3 x^4-9 x\right ) \log (x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-3 x^4+\left (3 x^3-9\right ) \log \left (3-x^3\right )+\left (3-x^3\right ) \log (16)+\left (3 x^4-9 x^3-9 x\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )+9 x}{x \left (3 x^3-9\right ) \log (x)}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {3 \log \left (3-x^3\right )-3 x-\log (16)}{3 x \log (x)}+\frac {\left (x^3-3 x^2-3\right ) \log \left (\frac {1}{\log (x)}\right )}{x^3-3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \int \frac {\log \left (3-x^3\right )}{x \log (x)}dx+\int \frac {\log \left (\frac {1}{\log (x)}\right )}{-x-\sqrt [3]{-3}}dx+\int \frac {\log \left (\frac {1}{\log (x)}\right )}{\sqrt [3]{3}-x}dx+\int \frac {\log \left (\frac {1}{\log (x)}\right )}{(-1)^{2/3} \sqrt [3]{3}-x}dx+x \log \left (\frac {1}{\log (x)}\right )-\frac {1}{3} \log (16) \log (\log (x))\) |
Int[(9*x - 3*x^4 + (3 - x^3)*Log[16] + (-9 + 3*x^3)*Log[3 - x^3] + (-9*x - 9*x^3 + 3*x^4)*Log[x]*Log[Log[x]^(-1)])/((-9*x + 3*x^4)*Log[x]),x]
3.22.51.3.1 Defintions of rubi rules used
Int[(Fx_.)*(Px_)^(p_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Int[x^(p *r)*ExpandToSum[Px/x^r, x]^p*Fx, x] /; IGtQ[r, 0]] /; PolyQ[Px, x] && Integ erQ[p] && !MonomialQ[Px, x] && (ILtQ[p, 0] || !PolyQ[u, x])
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 70.38 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04
method | result | size |
risch | \(\left (-x +\ln \left (-x^{3}+3\right )\right ) \ln \left (\ln \left (x \right )\right )-\frac {4 \ln \left (2\right ) \ln \left (\ln \left (x \right )\right )}{3}\) | \(25\) |
parallelrisch | \(\frac {4 \ln \left (\frac {1}{\ln \left (x \right )}\right ) \ln \left (2\right )}{3}+\ln \left (\frac {1}{\ln \left (x \right )}\right ) x -\ln \left (-x^{3}+3\right ) \ln \left (\frac {1}{\ln \left (x \right )}\right )\) | \(33\) |
int(((3*x^4-9*x^3-9*x)*ln(x)*ln(1/ln(x))+(3*x^3-9)*ln(-x^3+3)+4*(-x^3+3)*l n(2)-3*x^4+9*x)/(3*x^4-9*x)/ln(x),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\frac {1}{3} \, {\left (3 \, x + 4 \, \log \left (2\right ) - 3 \, \log \left (-x^{3} + 3\right )\right )} \log \left (\frac {1}{\log \left (x\right )}\right ) \]
integrate(((3*x^4-9*x^3-9*x)*log(x)*log(1/log(x))+(3*x^3-9)*log(-x^3+3)+4* (-x^3+3)*log(2)-3*x^4+9*x)/(3*x^4-9*x)/log(x),x, algorithm=\
Time = 0.28 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=\left (x - \log {\left (3 - x^{3} \right )}\right ) \log {\left (\frac {1}{\log {\left (x \right )}} \right )} - \frac {4 \log {\left (2 \right )} \log {\left (\log {\left (x \right )} \right )}}{3} \]
integrate(((3*x**4-9*x**3-9*x)*ln(x)*ln(1/ln(x))+(3*x**3-9)*ln(-x**3+3)+4* (-x**3+3)*ln(2)-3*x**4+9*x)/(3*x**4-9*x)/ln(x),x)
Time = 0.32 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=-\frac {1}{3} \, {\left (3 \, x + 4 \, \log \left (2\right )\right )} \log \left (\log \left (x\right )\right ) + \log \left (-x^{3} + 3\right ) \log \left (\log \left (x\right )\right ) \]
integrate(((3*x^4-9*x^3-9*x)*log(x)*log(1/log(x))+(3*x^3-9)*log(-x^3+3)+4* (-x^3+3)*log(2)-3*x^4+9*x)/(3*x^4-9*x)/log(x),x, algorithm=\
Time = 0.34 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.08 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=-x \log \left (\log \left (x\right )\right ) - \frac {4}{3} \, \log \left (2\right ) \log \left (\log \left (x\right )\right ) + \log \left (-x^{3} + 3\right ) \log \left (\log \left (x\right )\right ) \]
integrate(((3*x^4-9*x^3-9*x)*log(x)*log(1/log(x))+(3*x^3-9)*log(-x^3+3)+4* (-x^3+3)*log(2)-3*x^4+9*x)/(3*x^4-9*x)/log(x),x, algorithm=\
Time = 12.90 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.92 \[ \int \frac {9 x-3 x^4+\left (3-x^3\right ) \log (16)+\left (-9+3 x^3\right ) \log \left (3-x^3\right )+\left (-9 x-9 x^3+3 x^4\right ) \log (x) \log \left (\frac {1}{\log (x)}\right )}{\left (-9 x+3 x^4\right ) \log (x)} \, dx=-\ln \left (\frac {1}{\ln \left (x\right )}\right )\,\left (\ln \left (3-x^3\right )+\frac {3\,x^2-x^5}{x\,\left (x^3-3\right )}\right )-\frac {4\,\ln \left (\ln \left (x\right )\right )\,\ln \left (2\right )}{3} \]