Integrand size = 74, antiderivative size = 31 \[ \int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{75 x^3+30 x^4+3 x^5} \, dx=4+\frac {\left (x+\frac {x}{3 (5+x)}\right ) \left (2+x-\frac {\log (\log (3))}{x}\right )^2}{x} \]
Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{75 x^3+30 x^4+3 x^5} \, dx=\frac {1}{75} \left (325 x+75 x^2+\frac {(-15+\log (\log (3)))^2}{5+x}+\frac {80 \log ^2(\log (3))}{x^2}-\frac {\log (\log (3)) (320+\log (\log (3)))}{x}\right ) \]
Integrate[(316*x^3 + 280*x^4 + 73*x^5 + 6*x^6 + (320*x + 128*x^2 + 14*x^3) *Log[Log[3]] + (-160 - 63*x - 6*x^2)*Log[Log[3]]^2)/(75*x^3 + 30*x^4 + 3*x ^5),x]
(325*x + 75*x^2 + (-15 + Log[Log[3]])^2/(5 + x) + (80*Log[Log[3]]^2)/x^2 - (Log[Log[3]]*(320 + Log[Log[3]]))/x)/75
Time = 0.38 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {2026, 2007, 2123, 2009}
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.
\(\displaystyle \int \frac {6 x^6+73 x^5+280 x^4+316 x^3+\left (-6 x^2-63 x-160\right ) \log ^2(\log (3))+\left (14 x^3+128 x^2+320 x\right ) \log (\log (3))}{3 x^5+30 x^4+75 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {6 x^6+73 x^5+280 x^4+316 x^3+\left (-6 x^2-63 x-160\right ) \log ^2(\log (3))+\left (14 x^3+128 x^2+320 x\right ) \log (\log (3))}{x^3 \left (3 x^2+30 x+75\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {6 x^6+73 x^5+280 x^4+316 x^3+\left (-6 x^2-63 x-160\right ) \log ^2(\log (3))+\left (14 x^3+128 x^2+320 x\right ) \log (\log (3))}{x^3 \left (\sqrt {3} x+5 \sqrt {3}\right )^2}dx\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \int \left (-\frac {32 \log ^2(\log (3))}{15 x^3}+\frac {\log (\log (3)) (320+\log (\log (3)))}{75 x^2}+2 x-\frac {(\log (\log (3))-15)^2}{75 (x+5)^2}+\frac {13}{3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle x^2+\frac {16 \log ^2(\log (3))}{15 x^2}+\frac {13 x}{3}+\frac {(15-\log (\log (3)))^2}{75 (x+5)}-\frac {\log (\log (3)) (320+\log (\log (3)))}{75 x}\) |
Int[(316*x^3 + 280*x^4 + 73*x^5 + 6*x^6 + (320*x + 128*x^2 + 14*x^3)*Log[L og[3]] + (-160 - 63*x - 6*x^2)*Log[Log[3]]^2)/(75*x^3 + 30*x^4 + 3*x^5),x]
(13*x)/3 + x^2 + (15 - Log[Log[3]])^2/(75*(5 + x)) + (16*Log[Log[3]]^2)/(1 5*x^2) - (Log[Log[3]]*(320 + Log[Log[3]]))/(75*x)
3.6.70.3.1 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])
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.61
method | result | size |
norman | \(\frac {x^{5}+\left (\ln \left (\ln \left (3\right )\right )^{2}-\frac {64 \ln \left (\ln \left (3\right )\right )}{3}\right ) x +\left (-\frac {316}{3}-\frac {14 \ln \left (\ln \left (3\right )\right )}{3}\right ) x^{2}+\frac {28 x^{4}}{3}+\frac {16 \ln \left (\ln \left (3\right )\right )^{2}}{3}}{x^{2} \left (5+x \right )}\) | \(50\) |
default | \(\frac {13 x}{3}+x^{2}+\frac {16 \ln \left (\ln \left (3\right )\right )^{2}}{15 x^{2}}-\frac {\ln \left (\ln \left (3\right )\right ) \left (\ln \left (\ln \left (3\right )\right )+320\right )}{75 x}-\frac {-\frac {\ln \left (\ln \left (3\right )\right )^{2}}{25}+\frac {6 \ln \left (\ln \left (3\right )\right )}{5}-9}{3 \left (5+x \right )}\) | \(52\) |
risch | \(x^{2}+\frac {13 x}{3}+\frac {\frac {\left (9-14 \ln \left (\ln \left (3\right )\right )\right ) x^{2}}{3}+\frac {\left (3 \ln \left (\ln \left (3\right )\right )^{2}-64 \ln \left (\ln \left (3\right )\right )\right ) x}{3}+\frac {16 \ln \left (\ln \left (3\right )\right )^{2}}{3}}{x^{2} \left (5+x \right )}\) | \(53\) |
gosper | \(\frac {3 x^{5}+28 x^{4}+3 \ln \left (\ln \left (3\right )\right )^{2} x -14 x^{2} \ln \left (\ln \left (3\right )\right )+16 \ln \left (\ln \left (3\right )\right )^{2}-64 \ln \left (\ln \left (3\right )\right ) x -316 x^{2}}{3 x^{2} \left (5+x \right )}\) | \(56\) |
parallelrisch | \(\frac {3 x^{5}+28 x^{4}+3 \ln \left (\ln \left (3\right )\right )^{2} x -14 x^{2} \ln \left (\ln \left (3\right )\right )+16 \ln \left (\ln \left (3\right )\right )^{2}-64 \ln \left (\ln \left (3\right )\right ) x -316 x^{2}}{3 x^{2} \left (5+x \right )}\) | \(56\) |
int(((-6*x^2-63*x-160)*ln(ln(3))^2+(14*x^3+128*x^2+320*x)*ln(ln(3))+6*x^6+ 73*x^5+280*x^4+316*x^3)/(3*x^5+30*x^4+75*x^3),x,method=_RETURNVERBOSE)
(x^5+(ln(ln(3))^2-64/3*ln(ln(3)))*x+(-316/3-14/3*ln(ln(3)))*x^2+28/3*x^4+1 6/3*ln(ln(3))^2)/x^2/(5+x)
Time = 0.24 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.90 \[ \int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{75 x^3+30 x^4+3 x^5} \, dx=\frac {3 \, x^{5} + 28 \, x^{4} + 65 \, x^{3} + {\left (3 \, x + 16\right )} \log \left (\log \left (3\right )\right )^{2} + 9 \, x^{2} - 2 \, {\left (7 \, x^{2} + 32 \, x\right )} \log \left (\log \left (3\right )\right )}{3 \, {\left (x^{3} + 5 \, x^{2}\right )}} \]
integrate(((-6*x^2-63*x-160)*log(log(3))^2+(14*x^3+128*x^2+320*x)*log(log( 3))+6*x^6+73*x^5+280*x^4+316*x^3)/(3*x^5+30*x^4+75*x^3),x, algorithm=\
1/3*(3*x^5 + 28*x^4 + 65*x^3 + (3*x + 16)*log(log(3))^2 + 9*x^2 - 2*(7*x^2 + 32*x)*log(log(3)))/(x^3 + 5*x^2)
Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (22) = 44\).
Time = 0.61 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.74 \[ \int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{75 x^3+30 x^4+3 x^5} \, dx=x^{2} + \frac {13 x}{3} + \frac {x^{2} \cdot \left (9 - 14 \log {\left (\log {\left (3 \right )} \right )}\right ) + x \left (- 64 \log {\left (\log {\left (3 \right )} \right )} + 3 \log {\left (\log {\left (3 \right )} \right )}^{2}\right ) + 16 \log {\left (\log {\left (3 \right )} \right )}^{2}}{3 x^{3} + 15 x^{2}} \]
integrate(((-6*x**2-63*x-160)*ln(ln(3))**2+(14*x**3+128*x**2+320*x)*ln(ln( 3))+6*x**6+73*x**5+280*x**4+316*x**3)/(3*x**5+30*x**4+75*x**3),x)
x**2 + 13*x/3 + (x**2*(9 - 14*log(log(3))) + x*(-64*log(log(3)) + 3*log(lo g(3))**2) + 16*log(log(3))**2)/(3*x**3 + 15*x**2)
Time = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.77 \[ \int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{75 x^3+30 x^4+3 x^5} \, dx=x^{2} + \frac {13}{3} \, x - \frac {x^{2} {\left (14 \, \log \left (\log \left (3\right )\right ) - 9\right )} - {\left (3 \, \log \left (\log \left (3\right )\right )^{2} - 64 \, \log \left (\log \left (3\right )\right )\right )} x - 16 \, \log \left (\log \left (3\right )\right )^{2}}{3 \, {\left (x^{3} + 5 \, x^{2}\right )}} \]
integrate(((-6*x^2-63*x-160)*log(log(3))^2+(14*x^3+128*x^2+320*x)*log(log( 3))+6*x^6+73*x^5+280*x^4+316*x^3)/(3*x^5+30*x^4+75*x^3),x, algorithm=\
x^2 + 13/3*x - 1/3*(x^2*(14*log(log(3)) - 9) - (3*log(log(3))^2 - 64*log(l og(3)))*x - 16*log(log(3))^2)/(x^3 + 5*x^2)
Time = 0.28 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{75 x^3+30 x^4+3 x^5} \, dx=x^{2} + \frac {13}{3} \, x + \frac {\log \left (\log \left (3\right )\right )^{2} - 30 \, \log \left (\log \left (3\right )\right ) + 225}{75 \, {\left (x + 5\right )}} - \frac {x \log \left (\log \left (3\right )\right )^{2} + 320 \, x \log \left (\log \left (3\right )\right ) - 80 \, \log \left (\log \left (3\right )\right )^{2}}{75 \, x^{2}} \]
integrate(((-6*x^2-63*x-160)*log(log(3))^2+(14*x^3+128*x^2+320*x)*log(log( 3))+6*x^6+73*x^5+280*x^4+316*x^3)/(3*x^5+30*x^4+75*x^3),x, algorithm=\
x^2 + 13/3*x + 1/75*(log(log(3))^2 - 30*log(log(3)) + 225)/(x + 5) - 1/75* (x*log(log(3))^2 + 320*x*log(log(3)) - 80*log(log(3))^2)/x^2
Time = 0.18 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.81 \[ \int \frac {316 x^3+280 x^4+73 x^5+6 x^6+\left (320 x+128 x^2+14 x^3\right ) \log (\log (3))+\left (-160-63 x-6 x^2\right ) \log ^2(\log (3))}{75 x^3+30 x^4+3 x^5} \, dx=\frac {13\,x}{3}-\frac {\left (14\,\ln \left (\ln \left (3\right )\right )-9\right )\,x^2+\left (64\,\ln \left (\ln \left (3\right )\right )-3\,{\ln \left (\ln \left (3\right )\right )}^2\right )\,x-16\,{\ln \left (\ln \left (3\right )\right )}^2}{3\,x^3+15\,x^2}+x^2 \]
int((316*x^3 - log(log(3))^2*(63*x + 6*x^2 + 160) + 280*x^4 + 73*x^5 + 6*x ^6 + log(log(3))*(320*x + 128*x^2 + 14*x^3))/(75*x^3 + 30*x^4 + 3*x^5),x)