Integrand size = 133, antiderivative size = 30 \[ \int \frac {\left (\left (-320-16 x-1280 x^2-64 x^3\right ) \log \left (\frac {20+x}{4}\right )+\left (1280 x^2+64 x^3\right ) \log \left (x^2\right ) \log \left (\frac {20+x}{4}\right )\right ) \log \left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )+\left (-4 x-16 x^3\right ) \log \left (x^2\right ) \log ^2\left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )}{\left (20 x+x^2+80 x^3+4 x^4\right ) \log \left (x^2\right ) \log ^2\left (\frac {20+x}{4}\right )} \, dx=\frac {4 \log ^2\left (\frac {4 \left (4+16 x^2\right )}{\log \left (x^2\right )}\right )}{\log \left (5+\frac {x}{4}\right )} \]
\[ \int \frac {\left (\left (-320-16 x-1280 x^2-64 x^3\right ) \log \left (\frac {20+x}{4}\right )+\left (1280 x^2+64 x^3\right ) \log \left (x^2\right ) \log \left (\frac {20+x}{4}\right )\right ) \log \left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )+\left (-4 x-16 x^3\right ) \log \left (x^2\right ) \log ^2\left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )}{\left (20 x+x^2+80 x^3+4 x^4\right ) \log \left (x^2\right ) \log ^2\left (\frac {20+x}{4}\right )} \, dx=\int \frac {\left (\left (-320-16 x-1280 x^2-64 x^3\right ) \log \left (\frac {20+x}{4}\right )+\left (1280 x^2+64 x^3\right ) \log \left (x^2\right ) \log \left (\frac {20+x}{4}\right )\right ) \log \left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )+\left (-4 x-16 x^3\right ) \log \left (x^2\right ) \log ^2\left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )}{\left (20 x+x^2+80 x^3+4 x^4\right ) \log \left (x^2\right ) \log ^2\left (\frac {20+x}{4}\right )} \, dx \]
Integrate[(((-320 - 16*x - 1280*x^2 - 64*x^3)*Log[(20 + x)/4] + (1280*x^2 + 64*x^3)*Log[x^2]*Log[(20 + x)/4])*Log[(16 + 64*x^2)/Log[x^2]] + (-4*x - 16*x^3)*Log[x^2]*Log[(16 + 64*x^2)/Log[x^2]]^2)/((20*x + x^2 + 80*x^3 + 4* x^4)*Log[x^2]*Log[(20 + x)/4]^2),x]
Integrate[(((-320 - 16*x - 1280*x^2 - 64*x^3)*Log[(20 + x)/4] + (1280*x^2 + 64*x^3)*Log[x^2]*Log[(20 + x)/4])*Log[(16 + 64*x^2)/Log[x^2]] + (-4*x - 16*x^3)*Log[x^2]*Log[(16 + 64*x^2)/Log[x^2]]^2)/((20*x + x^2 + 80*x^3 + 4* x^4)*Log[x^2]*Log[(20 + x)/4]^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.
| \(\displaystyle \int \frac {\left (-16 x^3-4 x\right ) \log \left (x^2\right ) \log ^2\left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )+\left (\left (-64 x^3-1280 x^2-16 x-320\right ) \log \left (\frac {x+20}{4}\right )+\left (64 x^3+1280 x^2\right ) \log \left (x^2\right ) \log \left (\frac {x+20}{4}\right )\right ) \log \left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )}{\left (4 x^4+80 x^3+x^2+20 x\right ) \log \left (x^2\right ) \log ^2\left (\frac {x+20}{4}\right )} \, dx\) |
| \(\Big \downarrow \) 2026 |
| \(\displaystyle \int \frac {\left (-16 x^3-4 x\right ) \log \left (x^2\right ) \log ^2\left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )+\left (\left (-64 x^3-1280 x^2-16 x-320\right ) \log \left (\frac {x+20}{4}\right )+\left (64 x^3+1280 x^2\right ) \log \left (x^2\right ) \log \left (\frac {x+20}{4}\right )\right ) \log \left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )}{x \left (4 x^3+80 x^2+x+20\right ) \log \left (x^2\right ) \log ^2\left (\frac {x+20}{4}\right )}dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle \int \left (\frac {\left (-16 x^3-4 x\right ) \log \left (x^2\right ) \log ^2\left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )+\left (\left (-64 x^3-1280 x^2-16 x-320\right ) \log \left (\frac {x+20}{4}\right )+\left (64 x^3+1280 x^2\right ) \log \left (x^2\right ) \log \left (\frac {x+20}{4}\right )\right ) \log \left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )}{1601 x (x+20) \log \left (x^2\right ) \log ^2\left (\frac {x+20}{4}\right )}-\frac {4 (x-20) \left (\left (-16 x^3-4 x\right ) \log \left (x^2\right ) \log ^2\left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )+\left (\left (-64 x^3-1280 x^2-16 x-320\right ) \log \left (\frac {x+20}{4}\right )+\left (64 x^3+1280 x^2\right ) \log \left (x^2\right ) \log \left (\frac {x+20}{4}\right )\right ) \log \left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )\right )}{1601 x \left (4 x^2+1\right ) \log \left (x^2\right ) \log ^2\left (\frac {x+20}{4}\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -4 \int \frac {\log ^2\left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )}{(x+20) \log ^2\left (\frac {x}{4}+5\right )}dx-16 \int \frac {\log \left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )}{(i-2 x) \log \left (\frac {x}{4}+5\right )}dx+16 \int \frac {\log \left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )}{(2 x+i) \log \left (\frac {x}{4}+5\right )}dx-16 \int \frac {\log \left (\frac {64 x^2+16}{\log \left (x^2\right )}\right )}{x \log \left (\frac {x}{4}+5\right ) \log \left (x^2\right )}dx\) |
Int[(((-320 - 16*x - 1280*x^2 - 64*x^3)*Log[(20 + x)/4] + (1280*x^2 + 64*x ^3)*Log[x^2]*Log[(20 + x)/4])*Log[(16 + 64*x^2)/Log[x^2]] + (-4*x - 16*x^3 )*Log[x^2]*Log[(16 + 64*x^2)/Log[x^2]]^2)/((20*x + x^2 + 80*x^3 + 4*x^4)*L og[x^2]*Log[(20 + x)/4]^2),x]
3.14.50.3.1 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_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 79.13 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.97
| method | result | size |
| parallelrisch | \(\frac {4 \ln \left (\frac {64 x^{2}+16}{\ln \left (x^{2}\right )}\right )^{2}}{\ln \left (5+\frac {x}{4}\right )}\) | \(29\) |
| risch | \(\text {Expression too large to display}\) | \(7928\) |
int(((-16*x^3-4*x)*ln(x^2)*ln((64*x^2+16)/ln(x^2))^2+((64*x^3+1280*x^2)*ln (5+1/4*x)*ln(x^2)+(-64*x^3-1280*x^2-16*x-320)*ln(5+1/4*x))*ln((64*x^2+16)/ ln(x^2)))/(4*x^4+80*x^3+x^2+20*x)/ln(5+1/4*x)^2/ln(x^2),x,method=_RETURNVE RBOSE)
Time = 0.26 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\left (\left (-320-16 x-1280 x^2-64 x^3\right ) \log \left (\frac {20+x}{4}\right )+\left (1280 x^2+64 x^3\right ) \log \left (x^2\right ) \log \left (\frac {20+x}{4}\right )\right ) \log \left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )+\left (-4 x-16 x^3\right ) \log \left (x^2\right ) \log ^2\left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )}{\left (20 x+x^2+80 x^3+4 x^4\right ) \log \left (x^2\right ) \log ^2\left (\frac {20+x}{4}\right )} \, dx=\frac {4 \, \log \left (\frac {16 \, {\left (4 \, x^{2} + 1\right )}}{\log \left (x^{2}\right )}\right )^{2}}{\log \left (\frac {1}{4} \, x + 5\right )} \]
integrate(((-16*x^3-4*x)*log(x^2)*log((64*x^2+16)/log(x^2))^2+((64*x^3+128 0*x^2)*log(5+1/4*x)*log(x^2)+(-64*x^3-1280*x^2-16*x-320)*log(5+1/4*x))*log ((64*x^2+16)/log(x^2)))/(4*x^4+80*x^3+x^2+20*x)/log(5+1/4*x)^2/log(x^2),x, algorithm=\
Exception generated. \[ \int \frac {\left (\left (-320-16 x-1280 x^2-64 x^3\right ) \log \left (\frac {20+x}{4}\right )+\left (1280 x^2+64 x^3\right ) \log \left (x^2\right ) \log \left (\frac {20+x}{4}\right )\right ) \log \left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )+\left (-4 x-16 x^3\right ) \log \left (x^2\right ) \log ^2\left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )}{\left (20 x+x^2+80 x^3+4 x^4\right ) \log \left (x^2\right ) \log ^2\left (\frac {20+x}{4}\right )} \, dx=\text {Exception raised: TypeError} \]
integrate(((-16*x**3-4*x)*ln(x**2)*ln((64*x**2+16)/ln(x**2))**2+((64*x**3+ 1280*x**2)*ln(5+1/4*x)*ln(x**2)+(-64*x**3-1280*x**2-16*x-320)*ln(5+1/4*x)) *ln((64*x**2+16)/ln(x**2)))/(4*x**4+80*x**3+x**2+20*x)/ln(5+1/4*x)**2/ln(x **2),x)
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (28) = 56\).
Time = 0.38 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.13 \[ \int \frac {\left (\left (-320-16 x-1280 x^2-64 x^3\right ) \log \left (\frac {20+x}{4}\right )+\left (1280 x^2+64 x^3\right ) \log \left (x^2\right ) \log \left (\frac {20+x}{4}\right )\right ) \log \left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )+\left (-4 x-16 x^3\right ) \log \left (x^2\right ) \log ^2\left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )}{\left (20 x+x^2+80 x^3+4 x^4\right ) \log \left (x^2\right ) \log ^2\left (\frac {20+x}{4}\right )} \, dx=-\frac {4 \, {\left (9 \, \log \left (2\right )^{2} + 2 \, {\left (3 \, \log \left (2\right ) - \log \left (\log \left (x\right )\right )\right )} \log \left (4 \, x^{2} + 1\right ) + \log \left (4 \, x^{2} + 1\right )^{2} - 6 \, \log \left (2\right ) \log \left (\log \left (x\right )\right ) + \log \left (\log \left (x\right )\right )^{2}\right )}}{2 \, \log \left (2\right ) - \log \left (x + 20\right )} \]
integrate(((-16*x^3-4*x)*log(x^2)*log((64*x^2+16)/log(x^2))^2+((64*x^3+128 0*x^2)*log(5+1/4*x)*log(x^2)+(-64*x^3-1280*x^2-16*x-320)*log(5+1/4*x))*log ((64*x^2+16)/log(x^2)))/(4*x^4+80*x^3+x^2+20*x)/log(5+1/4*x)^2/log(x^2),x, algorithm=\
-4*(9*log(2)^2 + 2*(3*log(2) - log(log(x)))*log(4*x^2 + 1) + log(4*x^2 + 1 )^2 - 6*log(2)*log(log(x)) + log(log(x))^2)/(2*log(2) - log(x + 20))
Leaf count of result is larger than twice the leaf count of optimal. 137 vs. \(2 (28) = 56\).
Time = 0.60 (sec) , antiderivative size = 137, normalized size of antiderivative = 4.57 \[ \int \frac {\left (\left (-320-16 x-1280 x^2-64 x^3\right ) \log \left (\frac {20+x}{4}\right )+\left (1280 x^2+64 x^3\right ) \log \left (x^2\right ) \log \left (\frac {20+x}{4}\right )\right ) \log \left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )+\left (-4 x-16 x^3\right ) \log \left (x^2\right ) \log ^2\left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )}{\left (20 x+x^2+80 x^3+4 x^4\right ) \log \left (x^2\right ) \log ^2\left (\frac {20+x}{4}\right )} \, dx=-8 \, {\left (\frac {4 \, \log \left (2\right )}{2 \, \log \left (2\right ) - \log \left (x + 20\right )} - \frac {\log \left (\log \left (x^{2}\right )\right )}{2 \, \log \left (2\right ) - \log \left (x + 20\right )}\right )} \log \left (4 \, x^{2} + 1\right ) - \frac {64 \, \log \left (2\right )^{2}}{2 \, \log \left (2\right ) - \log \left (x + 20\right )} - \frac {4 \, \log \left (4 \, x^{2} + 1\right )^{2}}{2 \, \log \left (2\right ) - \log \left (x + 20\right )} + \frac {32 \, \log \left (2\right ) \log \left (\log \left (x^{2}\right )\right )}{2 \, \log \left (2\right ) - \log \left (x + 20\right )} - \frac {4 \, \log \left (\log \left (x^{2}\right )\right )^{2}}{2 \, \log \left (2\right ) - \log \left (x + 20\right )} \]
integrate(((-16*x^3-4*x)*log(x^2)*log((64*x^2+16)/log(x^2))^2+((64*x^3+128 0*x^2)*log(5+1/4*x)*log(x^2)+(-64*x^3-1280*x^2-16*x-320)*log(5+1/4*x))*log ((64*x^2+16)/log(x^2)))/(4*x^4+80*x^3+x^2+20*x)/log(5+1/4*x)^2/log(x^2),x, algorithm=\
-8*(4*log(2)/(2*log(2) - log(x + 20)) - log(log(x^2))/(2*log(2) - log(x + 20)))*log(4*x^2 + 1) - 64*log(2)^2/(2*log(2) - log(x + 20)) - 4*log(4*x^2 + 1)^2/(2*log(2) - log(x + 20)) + 32*log(2)*log(log(x^2))/(2*log(2) - log( x + 20)) - 4*log(log(x^2))^2/(2*log(2) - log(x + 20))
Time = 11.95 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {\left (\left (-320-16 x-1280 x^2-64 x^3\right ) \log \left (\frac {20+x}{4}\right )+\left (1280 x^2+64 x^3\right ) \log \left (x^2\right ) \log \left (\frac {20+x}{4}\right )\right ) \log \left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )+\left (-4 x-16 x^3\right ) \log \left (x^2\right ) \log ^2\left (\frac {16+64 x^2}{\log \left (x^2\right )}\right )}{\left (20 x+x^2+80 x^3+4 x^4\right ) \log \left (x^2\right ) \log ^2\left (\frac {20+x}{4}\right )} \, dx=\frac {4\,{\ln \left (\frac {16\,\left (4\,x^2+1\right )}{\ln \left (x^2\right )}\right )}^2}{\ln \left (\frac {x}{4}+5\right )} \]