Integrand size = 57, antiderivative size = 24 \[ \int \frac {5+x+\left (5 x^2+x^3\right ) \log (4)+\left (20 x+8 x^2\right ) \log (4) \log ^3\left (\frac {1}{2} \left (5 x+x^2\right )\right )}{\left (5 x^2+x^3\right ) \log (4)} \, dx=x+\frac {-1+x}{x \log (4)}+\log ^4\left (\frac {1}{2} x (5+x)\right ) \]
Time = 0.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {5+x+\left (5 x^2+x^3\right ) \log (4)+\left (20 x+8 x^2\right ) \log (4) \log ^3\left (\frac {1}{2} \left (5 x+x^2\right )\right )}{\left (5 x^2+x^3\right ) \log (4)} \, dx=x-\frac {1}{x \log (4)}+\log ^4\left (\frac {1}{2} x (5+x)\right ) \]
Integrate[(5 + x + (5*x^2 + x^3)*Log[4] + (20*x + 8*x^2)*Log[4]*Log[(5*x + x^2)/2]^3)/((5*x^2 + x^3)*Log[4]),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.
\(\displaystyle \int \frac {\left (8 x^2+20 x\right ) \log (4) \log ^3\left (\frac {1}{2} \left (x^2+5 x\right )\right )+\left (x^3+5 x^2\right ) \log (4)+x+5}{\left (x^3+5 x^2\right ) \log (4)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {4 \left (2 x^2+5 x\right ) \log (4) \log ^3\left (\frac {1}{2} \left (x^2+5 x\right )\right )+x+\left (x^3+5 x^2\right ) \log (4)+5}{x^3+5 x^2}dx}{\log (4)}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \frac {\int \frac {4 \left (2 x^2+5 x\right ) \log (4) \log ^3\left (\frac {1}{2} \left (x^2+5 x\right )\right )+x+\left (x^3+5 x^2\right ) \log (4)+5}{x^2 (x+5)}dx}{\log (4)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {4 (2 x+5) \log (4) \log ^3\left (\frac {1}{2} x (x+5)\right )}{x (x+5)}+\frac {\log (4) x^2+1}{x^2}\right )dx}{\log (4)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \log (4) \int \frac {\log ^3\left (\frac {1}{2} x (x+5)\right )}{x}dx+4 \log (4) \int \frac {\log ^3\left (\frac {1}{2} x (x+5)\right )}{x+5}dx-\frac {1}{x}+x \log (4)}{\log (4)}\) |
Int[(5 + x + (5*x^2 + x^3)*Log[4] + (20*x + 8*x^2)*Log[4]*Log[(5*x + x^2)/ 2]^3)/((5*x^2 + x^3)*Log[4]),x]
3.9.54.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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 = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00
method | result | size |
parts | \(x -\frac {1}{2 x \ln \left (2\right )}+\ln \left (\frac {1}{2} x^{2}+\frac {5}{2} x \right )^{4}\) | \(24\) |
norman | \(\frac {x^{2}+x \ln \left (\frac {1}{2} x^{2}+\frac {5}{2} x \right )^{4}-\frac {1}{2 \ln \left (2\right )}}{x}\) | \(29\) |
risch | \(\ln \left (\frac {1}{2} x^{2}+\frac {5}{2} x \right )^{4}+\frac {2 x^{2} \ln \left (2\right )-1}{2 \ln \left (2\right ) x}\) | \(32\) |
default | \(\frac {-\frac {1}{x}+2 \ln \left (2\right ) \left (x +\ln \left (x^{2}+5 x \right )^{4}\right )-8 \ln \left (2\right )^{4} \ln \left (\left (5+x \right ) x \right )+12 \ln \left (2\right )^{3} \ln \left (x^{2}+5 x \right )^{2}-8 \ln \left (2\right )^{2} \ln \left (x^{2}+5 x \right )^{3}}{2 \ln \left (2\right )}\) | \(73\) |
int(1/2*(2*(8*x^2+20*x)*ln(2)*ln(1/2*x^2+5/2*x)^3+2*(x^3+5*x^2)*ln(2)+5+x) /(x^3+5*x^2)/ln(2),x,method=_RETURNVERBOSE)
Time = 0.26 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.46 \[ \int \frac {5+x+\left (5 x^2+x^3\right ) \log (4)+\left (20 x+8 x^2\right ) \log (4) \log ^3\left (\frac {1}{2} \left (5 x+x^2\right )\right )}{\left (5 x^2+x^3\right ) \log (4)} \, dx=\frac {2 \, x \log \left (2\right ) \log \left (\frac {1}{2} \, x^{2} + \frac {5}{2} \, x\right )^{4} + 2 \, x^{2} \log \left (2\right ) - 1}{2 \, x \log \left (2\right )} \]
integrate(1/2*(2*(8*x^2+20*x)*log(2)*log(1/2*x^2+5/2*x)^3+2*(x^3+5*x^2)*lo g(2)+5+x)/(x^3+5*x^2)/log(2),x, algorithm=\
Time = 0.09 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {5+x+\left (5 x^2+x^3\right ) \log (4)+\left (20 x+8 x^2\right ) \log (4) \log ^3\left (\frac {1}{2} \left (5 x+x^2\right )\right )}{\left (5 x^2+x^3\right ) \log (4)} \, dx=\frac {2 x \log {\left (2 \right )} - \frac {1}{x}}{2 \log {\left (2 \right )}} + \log {\left (\frac {x^{2}}{2} + \frac {5 x}{2} \right )}^{4} \]
integrate(1/2*(2*(8*x**2+20*x)*ln(2)*ln(1/2*x**2+5/2*x)**3+2*(x**3+5*x**2) *ln(2)+5+x)/(x**3+5*x**2)/ln(2),x)
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (23) = 46\).
Time = 0.34 (sec) , antiderivative size = 162, normalized size of antiderivative = 6.75 \[ \int \frac {5+x+\left (5 x^2+x^3\right ) \log (4)+\left (20 x+8 x^2\right ) \log (4) \log ^3\left (\frac {1}{2} \left (5 x+x^2\right )\right )}{\left (5 x^2+x^3\right ) \log (4)} \, dx=\frac {2 \, \log \left (2\right ) \log \left (x + 5\right )^{4} - 8 \, \log \left (2\right )^{4} \log \left (x\right ) + 12 \, \log \left (2\right )^{3} \log \left (x\right )^{2} - 8 \, \log \left (2\right )^{2} \log \left (x\right )^{3} + 2 \, \log \left (2\right ) \log \left (x\right )^{4} - 8 \, {\left (\log \left (2\right )^{2} - \log \left (2\right ) \log \left (x\right )\right )} \log \left (x + 5\right )^{3} + 12 \, {\left (\log \left (2\right )^{3} - 2 \, \log \left (2\right )^{2} \log \left (x\right ) + \log \left (2\right ) \log \left (x\right )^{2}\right )} \log \left (x + 5\right )^{2} + 2 \, {\left (x - 5 \, \log \left (x + 5\right )\right )} \log \left (2\right ) - 8 \, {\left (\log \left (2\right )^{4} - 3 \, \log \left (2\right )^{3} \log \left (x\right ) + 3 \, \log \left (2\right )^{2} \log \left (x\right )^{2} - \log \left (2\right ) \log \left (x\right )^{3}\right )} \log \left (x + 5\right ) + 10 \, \log \left (2\right ) \log \left (x + 5\right ) - \frac {1}{x}}{2 \, \log \left (2\right )} \]
integrate(1/2*(2*(8*x^2+20*x)*log(2)*log(1/2*x^2+5/2*x)^3+2*(x^3+5*x^2)*lo g(2)+5+x)/(x^3+5*x^2)/log(2),x, algorithm=\
1/2*(2*log(2)*log(x + 5)^4 - 8*log(2)^4*log(x) + 12*log(2)^3*log(x)^2 - 8* log(2)^2*log(x)^3 + 2*log(2)*log(x)^4 - 8*(log(2)^2 - log(2)*log(x))*log(x + 5)^3 + 12*(log(2)^3 - 2*log(2)^2*log(x) + log(2)*log(x)^2)*log(x + 5)^2 + 2*(x - 5*log(x + 5))*log(2) - 8*(log(2)^4 - 3*log(2)^3*log(x) + 3*log(2 )^2*log(x)^2 - log(2)*log(x)^3)*log(x + 5) + 10*log(2)*log(x + 5) - 1/x)/l og(2)
\[ \int \frac {5+x+\left (5 x^2+x^3\right ) \log (4)+\left (20 x+8 x^2\right ) \log (4) \log ^3\left (\frac {1}{2} \left (5 x+x^2\right )\right )}{\left (5 x^2+x^3\right ) \log (4)} \, dx=\int { \frac {8 \, {\left (2 \, x^{2} + 5 \, x\right )} \log \left (2\right ) \log \left (\frac {1}{2} \, x^{2} + \frac {5}{2} \, x\right )^{3} + 2 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \left (2\right ) + x + 5}{2 \, {\left (x^{3} + 5 \, x^{2}\right )} \log \left (2\right )} \,d x } \]
integrate(1/2*(2*(8*x^2+20*x)*log(2)*log(1/2*x^2+5/2*x)^3+2*(x^3+5*x^2)*lo g(2)+5+x)/(x^3+5*x^2)/log(2),x, algorithm=\
integrate(1/2*(8*(2*x^2 + 5*x)*log(2)*log(1/2*x^2 + 5/2*x)^3 + 2*(x^3 + 5* x^2)*log(2) + x + 5)/((x^3 + 5*x^2)*log(2)), x)
Time = 10.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.96 \[ \int \frac {5+x+\left (5 x^2+x^3\right ) \log (4)+\left (20 x+8 x^2\right ) \log (4) \log ^3\left (\frac {1}{2} \left (5 x+x^2\right )\right )}{\left (5 x^2+x^3\right ) \log (4)} \, dx=x-\frac {1}{2\,x\,\ln \left (2\right )}+{\ln \left (\frac {x^2}{2}+\frac {5\,x}{2}\right )}^4 \]