Integrand size = 94, antiderivative size = 23 \[ \int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^4+8 x^3 \log \left (4+4 \log (5)+\log ^2(5)\right )+16 x^2 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )} \, dx=25-\frac {5}{x}-5 x-\frac {5}{x+8 \log (2+\log (5))} \] Output:
25-5/(x+8*ln(2+ln(5)))-5*x-5/x
Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78 \[ \int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^4+8 x^3 \log \left (4+4 \log (5)+\log ^2(5)\right )+16 x^2 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )} \, dx=-5 \left (\frac {1}{x}+x+\frac {1}{x+8 \log (2+\log (5))}\right ) \] Input:
Integrate[(10*x^2 - 5*x^4 + (40*x - 40*x^3)*Log[4 + 4*Log[5] + Log[5]^2] + (80 - 80*x^2)*Log[4 + 4*Log[5] + Log[5]^2]^2)/(x^4 + 8*x^3*Log[4 + 4*Log[ 5] + Log[5]^2] + 16*x^2*Log[4 + 4*Log[5] + Log[5]^2]^2),x]
Output:
-5*(x^(-1) + x + (x + 8*Log[2 + Log[5]])^(-1))
Time = 0.36 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, 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.
\(\displaystyle \int \frac {-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+\log ^2(5)+4 \log (5)\right )+10 x^2+\left (80-80 x^2\right ) \log ^2\left (4+\log ^2(5)+4 \log (5)\right )}{x^4+8 x^3 \log \left (4+\log ^2(5)+4 \log (5)\right )+16 x^2 \log ^2\left (4+\log ^2(5)+4 \log (5)\right )} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+\log ^2(5)+4 \log (5)\right )+10 x^2+\left (80-80 x^2\right ) \log ^2\left (4+\log ^2(5)+4 \log (5)\right )}{x^2 \left (x^2+16 x \log (2+\log (5))+64 \log ^2(2+\log (5))\right )}dx\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \int \frac {-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+\log ^2(5)+4 \log (5)\right )+10 x^2+\left (80-80 x^2\right ) \log ^2\left (4+\log ^2(5)+4 \log (5)\right )}{x^2 (x+8 \log (2+\log (5)))^2}dx\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \int \left (\frac {5}{x^2}+\frac {5}{(x+8 \log (2+\log (5)))^2}-5\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -5 x-\frac {5}{x}-\frac {5}{x+8 \log (2+\log (5))}\) |
Input:
Int[(10*x^2 - 5*x^4 + (40*x - 40*x^3)*Log[4 + 4*Log[5] + Log[5]^2] + (80 - 80*x^2)*Log[4 + 4*Log[5] + Log[5]^2]^2)/(x^4 + 8*x^3*Log[4 + 4*Log[5] + L og[5]^2] + 16*x^2*Log[4 + 4*Log[5] + Log[5]^2]^2),x]
Output:
-5/x - 5*x - 5/(x + 8*Log[2 + Log[5]])
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 = 0.18 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.26
method | result | size |
default | \(-5 x -\frac {5}{4 \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )+x}-\frac {5}{x}\) | \(29\) |
risch | \(-5 x +\frac {-40 \ln \left (2+\ln \left (5\right )\right )-10 x}{\left (x +8 \ln \left (2+\ln \left (5\right )\right )\right ) x}\) | \(32\) |
norman | \(\frac {\left (80 \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )^{2}-10\right ) x -5 x^{3}-20 \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )}{x \left (4 \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )+x \right )}\) | \(60\) |
parallelrisch | \(\frac {80 x \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )^{2}-5 x^{3}-20 \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )-10 x}{x \left (4 \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )+x \right )}\) | \(60\) |
gosper | \(\frac {80 x \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )^{2}-5 x^{3}-20 \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )-10 x}{x \left (4 \ln \left (\ln \left (5\right )^{2}+4 \ln \left (5\right )+4\right )+x \right )}\) | \(61\) |
Input:
int(((-80*x^2+80)*ln(ln(5)^2+4*ln(5)+4)^2+(-40*x^3+40*x)*ln(ln(5)^2+4*ln(5 )+4)-5*x^4+10*x^2)/(16*x^2*ln(ln(5)^2+4*ln(5)+4)^2+8*x^3*ln(ln(5)^2+4*ln(5 )+4)+x^4),x,method=_RETURNVERBOSE)
Output:
-5*x-5/(4*ln(ln(5)^2+4*ln(5)+4)+x)-5/x
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (23) = 46\).
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.04 \[ \int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^4+8 x^3 \log \left (4+4 \log (5)+\log ^2(5)\right )+16 x^2 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )} \, dx=-\frac {5 \, {\left (x^{3} + 4 \, {\left (x^{2} + 1\right )} \log \left (\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4\right ) + 2 \, x\right )}}{x^{2} + 4 \, x \log \left (\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4\right )} \] Input:
integrate(((-80*x^2+80)*log(log(5)^2+4*log(5)+4)^2+(-40*x^3+40*x)*log(log( 5)^2+4*log(5)+4)-5*x^4+10*x^2)/(16*x^2*log(log(5)^2+4*log(5)+4)^2+8*x^3*lo g(log(5)^2+4*log(5)+4)+x^4),x, algorithm="fricas")
Output:
-5*(x^3 + 4*(x^2 + 1)*log(log(5)^2 + 4*log(5) + 4) + 2*x)/(x^2 + 4*x*log(l og(5)^2 + 4*log(5) + 4))
Leaf count of result is larger than twice the leaf count of optimal. 42 vs. \(2 (19) = 38\).
Time = 0.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.83 \[ \int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^4+8 x^3 \log \left (4+4 \log (5)+\log ^2(5)\right )+16 x^2 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )} \, dx=- 5 x - \frac {10 x + 20 \log {\left (\log {\left (5 \right )}^{2} + 4 + 4 \log {\left (5 \right )} \right )}}{x^{2} + 4 x \log {\left (\log {\left (5 \right )}^{2} + 4 + 4 \log {\left (5 \right )} \right )}} \] Input:
integrate(((-80*x**2+80)*ln(ln(5)**2+4*ln(5)+4)**2+(-40*x**3+40*x)*ln(ln(5 )**2+4*ln(5)+4)-5*x**4+10*x**2)/(16*x**2*ln(ln(5)**2+4*ln(5)+4)**2+8*x**3* ln(ln(5)**2+4*ln(5)+4)+x**4),x)
Output:
-5*x - (10*x + 20*log(log(5)**2 + 4 + 4*log(5)))/(x**2 + 4*x*log(log(5)**2 + 4 + 4*log(5)))
Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^4+8 x^3 \log \left (4+4 \log (5)+\log ^2(5)\right )+16 x^2 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )} \, dx=-5 \, x - \frac {10 \, {\left (x + 2 \, \log \left (\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4\right )\right )}}{x^{2} + 4 \, x \log \left (\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4\right )} \] Input:
integrate(((-80*x^2+80)*log(log(5)^2+4*log(5)+4)^2+(-40*x^3+40*x)*log(log( 5)^2+4*log(5)+4)-5*x^4+10*x^2)/(16*x^2*log(log(5)^2+4*log(5)+4)^2+8*x^3*lo g(log(5)^2+4*log(5)+4)+x^4),x, algorithm="maxima")
Output:
-5*x - 10*(x + 2*log(log(5)^2 + 4*log(5) + 4))/(x^2 + 4*x*log(log(5)^2 + 4 *log(5) + 4))
Time = 0.11 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.78 \[ \int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^4+8 x^3 \log \left (4+4 \log (5)+\log ^2(5)\right )+16 x^2 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )} \, dx=-5 \, x - \frac {10 \, {\left (x + 2 \, \log \left (\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4\right )\right )}}{x^{2} + 4 \, x \log \left (\log \left (5\right )^{2} + 4 \, \log \left (5\right ) + 4\right )} \] Input:
integrate(((-80*x^2+80)*log(log(5)^2+4*log(5)+4)^2+(-40*x^3+40*x)*log(log( 5)^2+4*log(5)+4)-5*x^4+10*x^2)/(16*x^2*log(log(5)^2+4*log(5)+4)^2+8*x^3*lo g(log(5)^2+4*log(5)+4)+x^4),x, algorithm="giac")
Output:
-5*x - 10*(x + 2*log(log(5)^2 + 4*log(5) + 4))/(x^2 + 4*x*log(log(5)^2 + 4 *log(5) + 4))
Time = 0.22 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.70 \[ \int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^4+8 x^3 \log \left (4+4 \log (5)+\log ^2(5)\right )+16 x^2 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )} \, dx=-5\,x-\frac {10\,x+20\,\ln \left (\ln \left (625\right )+{\ln \left (5\right )}^2+4\right )}{x\,\left (x+4\,\ln \left (\ln \left (625\right )+{\ln \left (5\right )}^2+4\right )\right )} \] Input:
int((log(4*log(5) + log(5)^2 + 4)*(40*x - 40*x^3) - log(4*log(5) + log(5)^ 2 + 4)^2*(80*x^2 - 80) + 10*x^2 - 5*x^4)/(8*x^3*log(4*log(5) + log(5)^2 + 4) + x^4 + 16*x^2*log(4*log(5) + log(5)^2 + 4)^2),x)
Output:
- 5*x - (10*x + 20*log(log(625) + log(5)^2 + 4))/(x*(x + 4*log(log(625) + log(5)^2 + 4)))
Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 3.83 \[ \int \frac {10 x^2-5 x^4+\left (40 x-40 x^3\right ) \log \left (4+4 \log (5)+\log ^2(5)\right )+\left (80-80 x^2\right ) \log ^2\left (4+4 \log (5)+\log ^2(5)\right )}{x^4+8 x^3 \log \left (4+4 \log (5)+\log ^2(5)\right )+16 x^2 \log ^2\left (4+4 \log (5)+\log ^2(5)\right )} \, dx=\frac {-20 \mathrm {log}\left (\mathrm {log}\left (5\right )^{2}+4 \,\mathrm {log}\left (5\right )+4\right )^{2} x^{2}-20 \mathrm {log}\left (\mathrm {log}\left (5\right )^{2}+4 \,\mathrm {log}\left (5\right )+4\right )^{2}-5 \,\mathrm {log}\left (\mathrm {log}\left (5\right )^{2}+4 \,\mathrm {log}\left (5\right )+4\right ) x^{3}+\frac {5 x^{2}}{2}}{\mathrm {log}\left (\mathrm {log}\left (5\right )^{2}+4 \,\mathrm {log}\left (5\right )+4\right ) x \left (4 \,\mathrm {log}\left (\mathrm {log}\left (5\right )^{2}+4 \,\mathrm {log}\left (5\right )+4\right )+x \right )} \] Input:
int(((-80*x^2+80)*log(log(5)^2+4*log(5)+4)^2+(-40*x^3+40*x)*log(log(5)^2+4 *log(5)+4)-5*x^4+10*x^2)/(16*x^2*log(log(5)^2+4*log(5)+4)^2+8*x^3*log(log( 5)^2+4*log(5)+4)+x^4),x)
Output:
(5*( - 8*log(log(5)**2 + 4*log(5) + 4)**2*x**2 - 8*log(log(5)**2 + 4*log(5 ) + 4)**2 - 2*log(log(5)**2 + 4*log(5) + 4)*x**3 + x**2))/(2*log(log(5)**2 + 4*log(5) + 4)*x*(4*log(log(5)**2 + 4*log(5) + 4) + x))