Integrand size = 86, antiderivative size = 27 \[ \int \frac {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(x) \log (2 x)} \, dx=\frac {16 \log \left (\frac {e^x}{\log (2 x)}\right )}{(4-x)^2 x \log (x)} \]
Time = 0.45 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int \frac {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(x) \log (2 x)} \, dx=\frac {16 \log \left (\frac {e^x}{\log (2 x)}\right )}{(-4+x)^2 x \log (x)} \]
Integrate[((64 - 16*x)*Log[x] + (-64*x + 16*x^2)*Log[x]*Log[2*x] + (64 - 1 6*x + (64 - 48*x)*Log[x])*Log[2*x]*Log[E^x/Log[2*x]])/((-64*x^2 + 48*x^3 - 12*x^4 + x^5)*Log[x]^2*Log[2*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.
\(\displaystyle \int \frac {\left (16 x^2-64 x\right ) \log (2 x) \log (x)+(64-16 x) \log (x)+(-16 x+(64-48 x) \log (x)+64) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (x^5-12 x^4+48 x^3-64 x^2\right ) \log ^2(x) \log (2 x)} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (16 x^2-64 x\right ) \log (2 x) \log (x)+(64-16 x) \log (x)+(-16 x+(64-48 x) \log (x)+64) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \left (x^3-12 x^2+48 x-64\right ) \log ^2(x) \log (2 x)}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {\left (16 x^2-64 x\right ) \log (2 x) \log (x)+(64-16 x) \log (x)+(-16 x+(64-48 x) \log (x)+64) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{(x-4)^3 x^2 \log ^2(x) \log (2 x)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {16 (x \log (2 x)-1)}{(x-4)^2 x^2 \log (x) \log (2 x)}-\frac {16 (x+3 x \log (x)-4 \log (x)-4) \log \left (\frac {e^x}{\log (2 x)}\right )}{(x-4)^3 x^2 \log ^2(x)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log ^2(x)}dx-\int \frac {1}{x^2 \log (x) \log (2 x)}dx-\int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x^2 \log (x)}dx-\int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(x-4)^2 \log ^2(x)}dx+\frac {1}{2} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(x-4) \log ^2(x)}dx-\frac {1}{2} \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{x \log ^2(x)}dx+16 \int \frac {1}{(x-4)^2 x \log (x)}dx-\int \frac {1}{(x-4)^2 \log (x) \log (2 x)}dx+\frac {1}{2} \int \frac {1}{(x-4) \log (x) \log (2 x)}dx-\frac {1}{2} \int \frac {1}{x \log (x) \log (2 x)}dx-8 \int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(x-4)^3 \log (x)}dx+\int \frac {\log \left (\frac {e^x}{\log (2 x)}\right )}{(x-4)^2 \log (x)}dx\) |
Int[((64 - 16*x)*Log[x] + (-64*x + 16*x^2)*Log[x]*Log[2*x] + (64 - 16*x + (64 - 48*x)*Log[x])*Log[2*x]*Log[E^x/Log[2*x]])/((-64*x^2 + 48*x^3 - 12*x^ 4 + x^5)*Log[x]^2*Log[2*x]),x]
3.14.87.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Int[u*(a + b*x)^(Ex pon[Px, x]*p), x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /; IntegerQ[p] && Pol yQ[Px, x] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
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])
Time = 129.77 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11
method | result | size |
parallelrisch | \(\frac {16 \ln \left (\frac {{\mathrm e}^{x}}{\ln \left (2 x \right )}\right )}{x \ln \left (x \right ) \left (x^{2}-8 x +16\right )}\) | \(30\) |
risch | \(\frac {16 \ln \left ({\mathrm e}^{x}\right )}{x \left (x^{2}-8 x +16\right ) \ln \left (x \right )}+\frac {-8 i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right )-8 i \pi \operatorname {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right )^{2}+8 i \pi \,\operatorname {csgn}\left (\frac {i}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right )^{2}-8 i \pi \,\operatorname {csgn}\left (\frac {i}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right ) \operatorname {csgn}\left (i {\mathrm e}^{x}\right )-8 i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right )^{3}+8 i \pi \operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right )^{2} \operatorname {csgn}\left (i {\mathrm e}^{x}\right )+8 i \pi \,\operatorname {csgn}\left (\frac {i {\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right ) \operatorname {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right )^{2}+8 i \pi \operatorname {csgn}\left (\frac {{\mathrm e}^{x}}{2 i \ln \left (2\right )+2 i \ln \left (x \right )}\right )^{3}+8 i \pi +16 \ln \left (2\right )-16 \ln \left (2 i \ln \left (2\right )+2 i \ln \left (x \right )\right )}{x \left (x -4\right )^{2} \ln \left (x \right )}\) | \(333\) |
int((((-48*x+64)*ln(x)-16*x+64)*ln(2*x)*ln(exp(x)/ln(2*x))+(16*x^2-64*x)*l n(x)*ln(2*x)+(-16*x+64)*ln(x))/(x^5-12*x^4+48*x^3-64*x^2)/ln(x)^2/ln(2*x), x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(x) \log (2 x)} \, dx=\frac {16 \, \log \left (\frac {e^{x}}{\log \left (2\right ) + \log \left (x\right )}\right )}{{\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \left (x\right )} \]
integrate((((-48*x+64)*log(x)-16*x+64)*log(2*x)*log(exp(x)/log(2*x))+(16*x ^2-64*x)*log(x)*log(2*x)+(-16*x+64)*log(x))/(x^5-12*x^4+48*x^3-64*x^2)/log (x)^2/log(2*x),x, algorithm=\
Exception generated. \[ \int \frac {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(x) \log (2 x)} \, dx=\text {Exception raised: TypeError} \]
integrate((((-48*x+64)*ln(x)-16*x+64)*ln(2*x)*ln(exp(x)/ln(2*x))+(16*x**2- 64*x)*ln(x)*ln(2*x)+(-16*x+64)*ln(x))/(x**5-12*x**4+48*x**3-64*x**2)/ln(x) **2/ln(2*x),x)
Time = 0.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.11 \[ \int \frac {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(x) \log (2 x)} \, dx=\frac {16 \, {\left (x - \log \left (\log \left (2\right ) + \log \left (x\right )\right )\right )}}{{\left (x^{3} - 8 \, x^{2} + 16 \, x\right )} \log \left (x\right )} \]
integrate((((-48*x+64)*log(x)-16*x+64)*log(2*x)*log(exp(x)/log(2*x))+(16*x ^2-64*x)*log(x)*log(2*x)+(-16*x+64)*log(x))/(x^5-12*x^4+48*x^3-64*x^2)/log (x)^2/log(2*x),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 50 vs. \(2 (24) = 48\).
Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.85 \[ \int \frac {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(x) \log (2 x)} \, dx=-\frac {16 \, \log \left (\log \left (2\right ) + \log \left (x\right )\right )}{x^{3} \log \left (x\right ) - 8 \, x^{2} \log \left (x\right ) + 16 \, x \log \left (x\right )} + \frac {16}{x^{2} \log \left (x\right ) - 8 \, x \log \left (x\right ) + 16 \, \log \left (x\right )} \]
integrate((((-48*x+64)*log(x)-16*x+64)*log(2*x)*log(exp(x)/log(2*x))+(16*x ^2-64*x)*log(x)*log(2*x)+(-16*x+64)*log(x))/(x^5-12*x^4+48*x^3-64*x^2)/log (x)^2/log(2*x),x, algorithm=\
-16*log(log(2) + log(x))/(x^3*log(x) - 8*x^2*log(x) + 16*x*log(x)) + 16/(x ^2*log(x) - 8*x*log(x) + 16*log(x))
Time = 8.76 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.52 \[ \int \frac {(64-16 x) \log (x)+\left (-64 x+16 x^2\right ) \log (x) \log (2 x)+(64-16 x+(64-48 x) \log (x)) \log (2 x) \log \left (\frac {e^x}{\log (2 x)}\right )}{\left (-64 x^2+48 x^3-12 x^4+x^5\right ) \log ^2(x) \log (2 x)} \, dx=-\frac {\left (\ln \left (x\right )\,\left (\frac {48\,x^3-256\,x^2+256\,x}{x^2\,{\left (x-4\right )}^4}-\frac {48\,x-64}{x\,{\left (x-4\right )}^3}\right )-\frac {16}{x\,{\left (x-4\right )}^2}\right )\,\left (x+\ln \left (\frac {1}{\ln \left (2\,x\right )}\right )\right )}{\ln \left (x\right )} \]
int((log(x)*(16*x - 64) + log(2*x)*log(exp(x)/log(2*x))*(16*x + log(x)*(48 *x - 64) - 64) + log(2*x)*log(x)*(64*x - 16*x^2))/(log(2*x)*log(x)^2*(64*x ^2 - 48*x^3 + 12*x^4 - x^5)),x)