Integrand size = 69, antiderivative size = 24 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (5)+\left (4+12 x+12 x^2+4 x^3\right ) \log \left (\frac {18+9 x}{x}\right )}{2 x+7 x^2+9 x^3+5 x^4+x^5} \, dx=3+\frac {x \log (5)}{(1+x)^2}-\log ^2\left (\frac {9 (2+x)}{x}\right ) \] Output:
3-ln(27/x*(2/3+1/3*x))^2+ln(5)*x/(1+x)^2
Time = 0.21 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.29 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (5)+\left (4+12 x+12 x^2+4 x^3\right ) \log \left (\frac {18+9 x}{x}\right )}{2 x+7 x^2+9 x^3+5 x^4+x^5} \, dx=-\frac {\log (5)}{(1+x)^2}+\frac {\log (5)}{1+x}-\log ^2\left (\frac {9 (2+x)}{x}\right ) \] Input:
Integrate[((2*x - x^2 - x^3)*Log[5] + (4 + 12*x + 12*x^2 + 4*x^3)*Log[(18 + 9*x)/x])/(2*x + 7*x^2 + 9*x^3 + 5*x^4 + x^5),x]
Output:
-(Log[5]/(1 + x)^2) + Log[5]/(1 + x) - Log[(9*(2 + x))/x]^2
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.55 (sec) , antiderivative size = 262, normalized size of antiderivative = 10.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {2026, 2463, 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.
\(\displaystyle \int \frac {\left (-x^3-x^2+2 x\right ) \log (5)+\left (4 x^3+12 x^2+12 x+4\right ) \log \left (\frac {9 x+18}{x}\right )}{x^5+5 x^4+9 x^3+7 x^2+2 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (-x^3-x^2+2 x\right ) \log (5)+\left (4 x^3+12 x^2+12 x+4\right ) \log \left (\frac {9 x+18}{x}\right )}{x \left (x^4+5 x^3+9 x^2+7 x+2\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {\left (-x^3-x^2+2 x\right ) \log (5)+\left (4 x^3+12 x^2+12 x+4\right ) \log \left (\frac {9 x+18}{x}\right )}{(-x-2) x}+\frac {\left (-x^3-x^2+2 x\right ) \log (5)+\left (4 x^3+12 x^2+12 x+4\right ) \log \left (\frac {9 x+18}{x}\right )}{x (x+1)}-\frac {\left (-x^3-x^2+2 x\right ) \log (5)+\left (4 x^3+12 x^2+12 x+4\right ) \log \left (\frac {9 x+18}{x}\right )}{x (x+1)^2}+\frac {\left (-x^3-x^2+2 x\right ) \log (5)+\left (4 x^3+12 x^2+12 x+4\right ) \log \left (\frac {9 x+18}{x}\right )}{x (x+1)^3}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 4 \operatorname {PolyLog}\left (2,-\frac {2}{x}\right )+4 \operatorname {PolyLog}\left (2,-\frac {x}{2}\right )+2 x^2 \log \left (\frac {18}{x}+9\right )+2 x^2 \log (x)-2 x^2 \left (\log (x)+\log \left (\frac {9 (x+2)}{x}\right )-\log (9 x+18)\right )-2 x^2 \log (9 x+18)-\frac {1}{2} x^2 \log (5)+\log ^2(x)-\log ^2(9 (x+2))+\frac {1}{2} (1-x)^2 \log (5)+4 x \log \left (\frac {18}{x}+9\right )+4 x \log (x)+2 \log \left (\frac {x}{2}+1\right ) \log (x)-2 \log (18) \log (x)+4 \log (9) \log (x)+8 \log (x+2)-4 (x+2) \log (9 (x+2))-4 x \left (\log (x)+\log \left (\frac {9 (x+2)}{x}\right )-\log (9 x+18)\right )-2 \log (x) \left (\log (x)+\log \left (\frac {9 (x+2)}{x}\right )-\log (9 x+18)\right )-2 \log (x+2) \left (\log (x)+\log \left (\frac {9 (x+2)}{x}\right )-\log (9 x+18)\right )+x \log (5)+\frac {\log (5)}{x+1}-\frac {\log (5)}{(x+1)^2}\) |
Input:
Int[((2*x - x^2 - x^3)*Log[5] + (4 + 12*x + 12*x^2 + 4*x^3)*Log[(18 + 9*x) /x])/(2*x + 7*x^2 + 9*x^3 + 5*x^4 + x^5),x]
Output:
((1 - x)^2*Log[5])/2 + x*Log[5] - (x^2*Log[5])/2 - Log[5]/(1 + x)^2 + Log[ 5]/(1 + x) + 4*x*Log[9 + 18/x] + 2*x^2*Log[9 + 18/x] + 4*x*Log[x] + 2*x^2* Log[x] + 4*Log[9]*Log[x] - 2*Log[18]*Log[x] + 2*Log[1 + x/2]*Log[x] + Log[ x]^2 + 8*Log[2 + x] - 4*(2 + x)*Log[9*(2 + x)] - Log[9*(2 + x)]^2 - 4*x*(L og[x] + Log[(9*(2 + x))/x] - Log[18 + 9*x]) - 2*x^2*(Log[x] + Log[(9*(2 + x))/x] - Log[18 + 9*x]) - 2*Log[x]*(Log[x] + Log[(9*(2 + x))/x] - Log[18 + 9*x]) - 2*Log[2 + x]*(Log[x] + Log[(9*(2 + x))/x] - Log[18 + 9*x]) - 2*x^ 2*Log[18 + 9*x] + 4*PolyLog[2, -2/x] + 4*PolyLog[2, -1/2*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])
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 = 0.26 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.38
method | result | size |
risch | \(\frac {\ln \left (5\right )}{1+x}-\frac {\ln \left (5\right )}{\left (1+x \right )^{2}}-\ln \left (\frac {9 x +18}{x}\right )^{2}\) | \(33\) |
parts | \(-\ln \left (5\right ) \left (-\frac {1}{1+x}+\frac {1}{\left (1+x \right )^{2}}\right )-\ln \left (\frac {9 x +18}{x}\right )^{2}\) | \(33\) |
derivativedivides | \(-18 \ln \left (5\right ) \left (\frac {18}{\left (\frac {18}{x}+18\right )^{2}}-\frac {1}{\frac {18}{x}+18}\right )-\ln \left (9+\frac {18}{x}\right )^{2}\) | \(41\) |
default | \(-18 \ln \left (5\right ) \left (\frac {18}{\left (\frac {18}{x}+18\right )^{2}}-\frac {1}{\frac {18}{x}+18}\right )-\ln \left (9+\frac {18}{x}\right )^{2}\) | \(41\) |
norman | \(\frac {x \ln \left (5\right )-\ln \left (\frac {9 x +18}{x}\right )^{2}-2 x \ln \left (\frac {9 x +18}{x}\right )^{2}-x^{2} \ln \left (\frac {9 x +18}{x}\right )^{2}}{\left (1+x \right )^{2}}\) | \(58\) |
parallelrisch | \(-\frac {x^{2} \ln \left (\frac {9 x +18}{x}\right )^{2}+2 x \ln \left (\frac {9 x +18}{x}\right )^{2}-x \ln \left (5\right )+\ln \left (\frac {9 x +18}{x}\right )^{2}}{x^{2}+2 x +1}\) | \(59\) |
Input:
int(((4*x^3+12*x^2+12*x+4)*ln((9*x+18)/x)+(-x^3-x^2+2*x)*ln(5))/(x^5+5*x^4 +9*x^3+7*x^2+2*x),x,method=_RETURNVERBOSE)
Output:
ln(5)/(1+x)-ln(5)/(1+x)^2-ln((9*x+18)/x)^2
Time = 0.09 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (5)+\left (4+12 x+12 x^2+4 x^3\right ) \log \left (\frac {18+9 x}{x}\right )}{2 x+7 x^2+9 x^3+5 x^4+x^5} \, dx=-\frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (\frac {9 \, {\left (x + 2\right )}}{x}\right )^{2} - x \log \left (5\right )}{x^{2} + 2 \, x + 1} \] Input:
integrate(((4*x^3+12*x^2+12*x+4)*log((9*x+18)/x)+(-x^3-x^2+2*x)*log(5))/(x ^5+5*x^4+9*x^3+7*x^2+2*x),x, algorithm="fricas")
Output:
-((x^2 + 2*x + 1)*log(9*(x + 2)/x)^2 - x*log(5))/(x^2 + 2*x + 1)
Time = 0.13 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.92 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (5)+\left (4+12 x+12 x^2+4 x^3\right ) \log \left (\frac {18+9 x}{x}\right )}{2 x+7 x^2+9 x^3+5 x^4+x^5} \, dx=\frac {x \log {\left (5 \right )}}{x^{2} + 2 x + 1} - \log {\left (\frac {9 x + 18}{x} \right )}^{2} \] Input:
integrate(((4*x**3+12*x**2+12*x+4)*ln((9*x+18)/x)+(-x**3-x**2+2*x)*ln(5))/ (x**5+5*x**4+9*x**3+7*x**2+2*x),x)
Output:
x*log(5)/(x**2 + 2*x + 1) - log((9*x + 18)/x)**2
Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (24) = 48\).
Time = 0.16 (sec) , antiderivative size = 99, normalized size of antiderivative = 4.12 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (5)+\left (4+12 x+12 x^2+4 x^3\right ) \log \left (\frac {18+9 x}{x}\right )}{2 x+7 x^2+9 x^3+5 x^4+x^5} \, dx=-\frac {{\left (x^{2} + 2 \, x + 1\right )} \log \left (x + 2\right )^{2} + {\left (x^{2} + 2 \, x + 1\right )} \log \left (x\right )^{2} - x \log \left (5\right ) + 2 \, {\left (2 \, x^{2} \log \left (3\right ) + 4 \, x \log \left (3\right ) - {\left (x^{2} + 2 \, x + 1\right )} \log \left (x\right ) + 2 \, \log \left (3\right )\right )} \log \left (x + 2\right ) - 4 \, {\left (x^{2} \log \left (3\right ) + 2 \, x \log \left (3\right ) + \log \left (3\right )\right )} \log \left (x\right )}{x^{2} + 2 \, x + 1} \] Input:
integrate(((4*x^3+12*x^2+12*x+4)*log((9*x+18)/x)+(-x^3-x^2+2*x)*log(5))/(x ^5+5*x^4+9*x^3+7*x^2+2*x),x, algorithm="maxima")
Output:
-((x^2 + 2*x + 1)*log(x + 2)^2 + (x^2 + 2*x + 1)*log(x)^2 - x*log(5) + 2*( 2*x^2*log(3) + 4*x*log(3) - (x^2 + 2*x + 1)*log(x) + 2*log(3))*log(x + 2) - 4*(x^2*log(3) + 2*x*log(3) + log(3))*log(x))/(x^2 + 2*x + 1)
Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (24) = 48\).
Time = 0.13 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.12 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (5)+\left (4+12 x+12 x^2+4 x^3\right ) \log \left (\frac {18+9 x}{x}\right )}{2 x+7 x^2+9 x^3+5 x^4+x^5} \, dx=-\log \left (\frac {9 \, {\left (x + 2\right )}}{x}\right )^{2} + \frac {2 \, {\left (\frac {{\left (x + 2\right )} \log \left (5\right )}{x} - \log \left (5\right )\right )}}{\frac {{\left (x + 2\right )}^{2}}{x^{2}} + \frac {2 \, {\left (x + 2\right )}}{x} + 1} \] Input:
integrate(((4*x^3+12*x^2+12*x+4)*log((9*x+18)/x)+(-x^3-x^2+2*x)*log(5))/(x ^5+5*x^4+9*x^3+7*x^2+2*x),x, algorithm="giac")
Output:
-log(9*(x + 2)/x)^2 + 2*((x + 2)*log(5)/x - log(5))/((x + 2)^2/x^2 + 2*(x + 2)/x + 1)
Time = 2.82 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (5)+\left (4+12 x+12 x^2+4 x^3\right ) \log \left (\frac {18+9 x}{x}\right )}{2 x+7 x^2+9 x^3+5 x^4+x^5} \, dx=\frac {x\,\ln \left (5\right )}{{\left (x+1\right )}^2}-{\ln \left (\frac {9\,x+18}{x}\right )}^2 \] Input:
int(-(log(5)*(x^2 - 2*x + x^3) - log((9*x + 18)/x)*(12*x + 12*x^2 + 4*x^3 + 4))/(2*x + 7*x^2 + 9*x^3 + 5*x^4 + x^5),x)
Output:
(x*log(5))/(x + 1)^2 - log((9*x + 18)/x)^2
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 2.96 \[ \int \frac {\left (2 x-x^2-x^3\right ) \log (5)+\left (4+12 x+12 x^2+4 x^3\right ) \log \left (\frac {18+9 x}{x}\right )}{2 x+7 x^2+9 x^3+5 x^4+x^5} \, dx=\frac {-2 \mathrm {log}\left (\frac {9 x +18}{x}\right )^{2} x^{2}-4 \mathrm {log}\left (\frac {9 x +18}{x}\right )^{2} x -2 \mathrm {log}\left (\frac {9 x +18}{x}\right )^{2}-\mathrm {log}\left (5\right ) x^{2}-\mathrm {log}\left (5\right )}{2 x^{2}+4 x +2} \] Input:
int(((4*x^3+12*x^2+12*x+4)*log((9*x+18)/x)+(-x^3-x^2+2*x)*log(5))/(x^5+5*x ^4+9*x^3+7*x^2+2*x),x)
Output:
( - 2*log((9*x + 18)/x)**2*x**2 - 4*log((9*x + 18)/x)**2*x - 2*log((9*x + 18)/x)**2 - log(5)*x**2 - log(5))/(2*(x**2 + 2*x + 1))