Integrand size = 72, antiderivative size = 22 \[ \int \frac {425-275 x-125 x^2-25 x^3+\left (300+600 x+300 x^2\right ) \log (5-x)+\left (-125-225 x-75 x^2+25 x^3\right ) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx=25 (1+x) \left (-1+\frac {2}{1+x}+\log (5-x)\right )^2 \] Output:
25*(2/(1+x)+ln(5-x)-1)^2*(1+x)
Time = 0.04 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.82 \[ \int \frac {425-275 x-125 x^2-25 x^3+\left (300+600 x+300 x^2\right ) \log (5-x)+\left (-125-225 x-75 x^2+25 x^3\right ) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx=\frac {25 \left (4+x+x^2-2 \left (-1+x^2\right ) \log (5-x)+(1+x)^2 \log ^2(5-x)\right )}{1+x} \] Input:
Integrate[(425 - 275*x - 125*x^2 - 25*x^3 + (300 + 600*x + 300*x^2)*Log[5 - x] + (-125 - 225*x - 75*x^2 + 25*x^3)*Log[5 - x]^2)/(-5 - 9*x - 3*x^2 + x^3),x]
Output:
(25*(4 + x + x^2 - 2*(-1 + x^2)*Log[5 - x] + (1 + x)^2*Log[5 - x]^2))/(1 + x)
Leaf count is larger than twice the leaf count of optimal. \(136\) vs. \(2(22)=44\).
Time = 0.85 (sec) , antiderivative size = 136, normalized size of antiderivative = 6.18, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {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.
\(\displaystyle \int \frac {-25 x^3-125 x^2+\left (300 x^2+600 x+300\right ) \log (5-x)+\left (25 x^3-75 x^2-225 x-125\right ) \log ^2(5-x)-275 x+425}{x^3-3 x^2-9 x-5} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {-25 x^3-125 x^2+\left (300 x^2+600 x+300\right ) \log (5-x)+\left (25 x^3-75 x^2-225 x-125\right ) \log ^2(5-x)-275 x+425}{36 (x-5)}-\frac {-25 x^3-125 x^2+\left (300 x^2+600 x+300\right ) \log (5-x)+\left (25 x^3-75 x^2-225 x-125\right ) \log ^2(5-x)-275 x+425}{36 (x+1)}-\frac {-25 x^3-125 x^2+\left (300 x^2+600 x+300\right ) \log (5-x)+\left (25 x^3-75 x^2-225 x-125\right ) \log ^2(5-x)-275 x+425}{6 (x+1)^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {25 x^2}{12}-\frac {25}{12} (5-x)^2+\frac {25 x}{6}+\frac {100}{x+1}+\frac {25}{108} (5-x)^3 \log ^2(5-x)-\frac {25}{6} (5-x)^2 \log ^2(5-x)+\frac {25}{108} (x+1)^3 \log ^2(5-x)+100 \log ^2(5-x)+\frac {25}{6} (5-x)^2 \log (5-x)-\frac {25}{6} (x+1)^2 \log (5-x)-50 \log (5-x)\) |
Input:
Int[(425 - 275*x - 125*x^2 - 25*x^3 + (300 + 600*x + 300*x^2)*Log[5 - x] + (-125 - 225*x - 75*x^2 + 25*x^3)*Log[5 - x]^2)/(-5 - 9*x - 3*x^2 + x^3),x ]
Output:
(-25*(5 - x)^2)/12 + (25*x)/6 + (25*x^2)/12 + 100/(1 + x) - 50*Log[5 - x] + (25*(5 - x)^2*Log[5 - x])/6 - (25*(1 + x)^2*Log[5 - x])/6 + 100*Log[5 - x]^2 - (25*(5 - x)^2*Log[5 - x]^2)/6 + (25*(5 - x)^3*Log[5 - x]^2)/108 + ( 25*(1 + x)^3*Log[5 - x]^2)/108
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]
Leaf count of result is larger than twice the leaf count of optimal. \(50\) vs. \(2(22)=44\).
Time = 0.71 (sec) , antiderivative size = 51, normalized size of antiderivative = 2.32
method | result | size |
risch | \(\left (25 x +25\right ) \ln \left (5-x \right )^{2}-50 \ln \left (5-x \right ) x +\frac {50 \ln \left (-5+x \right ) x +25 x^{2}+50 \ln \left (-5+x \right )+25 x +100}{1+x}\) | \(51\) |
parts | \(-25 \ln \left (5-x \right )^{2} \left (5-x \right )+50 \left (5-x \right ) \ln \left (5-x \right )-250+25 x +150 \ln \left (5-x \right )^{2}+\frac {100}{1+x}-200 \ln \left (-5+x \right )\) | \(57\) |
derivativedivides | \(-25 \ln \left (5-x \right )^{2} \left (5-x \right )+50 \left (5-x \right ) \ln \left (5-x \right )-125+25 x +150 \ln \left (5-x \right )^{2}-\frac {100}{-1-x}-200 \ln \left (5-x \right )\) | \(61\) |
default | \(-25 \ln \left (5-x \right )^{2} \left (5-x \right )+50 \left (5-x \right ) \ln \left (5-x \right )-125+25 x +150 \ln \left (5-x \right )^{2}-\frac {100}{-1-x}-200 \ln \left (5-x \right )\) | \(61\) |
norman | \(\frac {50 \ln \left (5-x \right )+25 x^{2}+25 \ln \left (5-x \right )^{2}-50 x^{2} \ln \left (5-x \right )+50 \ln \left (5-x \right )^{2} x +25 \ln \left (5-x \right )^{2} x^{2}+75}{1+x}\) | \(67\) |
parallelrisch | \(\frac {25 \ln \left (5-x \right )^{2} x^{2}+325-50 x^{2} \ln \left (5-x \right )+50 \ln \left (5-x \right )^{2} x +25 x^{2}+25 \ln \left (5-x \right )^{2}+250 x +50 \ln \left (5-x \right )}{1+x}\) | \(70\) |
orering | \(\frac {\left (1+x \right ) \left (\left (25 x^{3}-75 x^{2}-225 x -125\right ) \ln \left (5-x \right )^{2}+\left (300 x^{2}+600 x +300\right ) \ln \left (5-x \right )-25 x^{3}-125 x^{2}-275 x +425\right )}{x^{3}-3 x^{2}-9 x -5}+\frac {12 \left (-5+x \right ) \left (x^{4}+8 x^{3}+88 x -121\right ) \left (\frac {\left (75 x^{2}-150 x -225\right ) \ln \left (5-x \right )^{2}-\frac {2 \left (25 x^{3}-75 x^{2}-225 x -125\right ) \ln \left (5-x \right )}{5-x}+\left (600 x +600\right ) \ln \left (5-x \right )-\frac {300 x^{2}+600 x +300}{5-x}-75 x^{2}-250 x -275}{x^{3}-3 x^{2}-9 x -5}-\frac {\left (\left (25 x^{3}-75 x^{2}-225 x -125\right ) \ln \left (5-x \right )^{2}+\left (300 x^{2}+600 x +300\right ) \ln \left (5-x \right )-25 x^{3}-125 x^{2}-275 x +425\right ) \left (3 x^{2}-6 x -9\right )}{\left (x^{3}-3 x^{2}-9 x -5\right )^{2}}\right )}{x^{4}-28 x^{3}+30 x^{2}+1012 x -1639}+\frac {\left (-5+x \right )^{2} \left (1+x \right ) \left (x^{4}+22 x^{2}+121\right ) \left (\frac {\left (150 x -150\right ) \ln \left (5-x \right )^{2}-\frac {4 \left (75 x^{2}-150 x -225\right ) \ln \left (5-x \right )}{5-x}+\frac {50 x^{3}-150 x^{2}-450 x -250}{\left (5-x \right )^{2}}-\frac {2 \left (25 x^{3}-75 x^{2}-225 x -125\right ) \ln \left (5-x \right )}{\left (5-x \right )^{2}}+600 \ln \left (5-x \right )-\frac {2 \left (600 x +600\right )}{5-x}-\frac {300 x^{2}+600 x +300}{\left (5-x \right )^{2}}-150 x -250}{x^{3}-3 x^{2}-9 x -5}-\frac {2 \left (\left (75 x^{2}-150 x -225\right ) \ln \left (5-x \right )^{2}-\frac {2 \left (25 x^{3}-75 x^{2}-225 x -125\right ) \ln \left (5-x \right )}{5-x}+\left (600 x +600\right ) \ln \left (5-x \right )-\frac {300 x^{2}+600 x +300}{5-x}-75 x^{2}-250 x -275\right ) \left (3 x^{2}-6 x -9\right )}{\left (x^{3}-3 x^{2}-9 x -5\right )^{2}}+\frac {2 \left (\left (25 x^{3}-75 x^{2}-225 x -125\right ) \ln \left (5-x \right )^{2}+\left (300 x^{2}+600 x +300\right ) \ln \left (5-x \right )-25 x^{3}-125 x^{2}-275 x +425\right ) \left (3 x^{2}-6 x -9\right )^{2}}{\left (x^{3}-3 x^{2}-9 x -5\right )^{3}}-\frac {\left (\left (25 x^{3}-75 x^{2}-225 x -125\right ) \ln \left (5-x \right )^{2}+\left (300 x^{2}+600 x +300\right ) \ln \left (5-x \right )-25 x^{3}-125 x^{2}-275 x +425\right ) \left (6 x -6\right )}{\left (x^{3}-3 x^{2}-9 x -5\right )^{2}}\right )}{x^{4}-28 x^{3}+30 x^{2}+1012 x -1639}\) | \(780\) |
Input:
int(((25*x^3-75*x^2-225*x-125)*ln(5-x)^2+(300*x^2+600*x+300)*ln(5-x)-25*x^ 3-125*x^2-275*x+425)/(x^3-3*x^2-9*x-5),x,method=_RETURNVERBOSE)
Output:
(25*x+25)*ln(5-x)^2-50*ln(5-x)*x+25*(2*ln(-5+x)*x+x^2+2*ln(-5+x)+x+4)/(1+x )
Time = 0.09 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.95 \[ \int \frac {425-275 x-125 x^2-25 x^3+\left (300+600 x+300 x^2\right ) \log (5-x)+\left (-125-225 x-75 x^2+25 x^3\right ) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx=\frac {25 \, {\left ({\left (x^{2} + 2 \, x + 1\right )} \log \left (-x + 5\right )^{2} + x^{2} - 2 \, {\left (x^{2} - 1\right )} \log \left (-x + 5\right ) + x + 4\right )}}{x + 1} \] Input:
integrate(((25*x^3-75*x^2-225*x-125)*log(5-x)^2+(300*x^2+600*x+300)*log(5- x)-25*x^3-125*x^2-275*x+425)/(x^3-3*x^2-9*x-5),x, algorithm="fricas")
Output:
25*((x^2 + 2*x + 1)*log(-x + 5)^2 + x^2 - 2*(x^2 - 1)*log(-x + 5) + x + 4) /(x + 1)
Time = 0.11 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.55 \[ \int \frac {425-275 x-125 x^2-25 x^3+\left (300+600 x+300 x^2\right ) \log (5-x)+\left (-125-225 x-75 x^2+25 x^3\right ) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx=- 50 x \log {\left (5 - x \right )} + 25 x + \left (25 x + 25\right ) \log {\left (5 - x \right )}^{2} + 50 \log {\left (x - 5 \right )} + \frac {100}{x + 1} \] Input:
integrate(((25*x**3-75*x**2-225*x-125)*ln(5-x)**2+(300*x**2+600*x+300)*ln( 5-x)-25*x**3-125*x**2-275*x+425)/(x**3-3*x**2-9*x-5),x)
Output:
-50*x*log(5 - x) + 25*x + (25*x + 25)*log(5 - x)**2 + 50*log(x - 5) + 100/ (x + 1)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (22) = 44\).
Time = 0.09 (sec) , antiderivative size = 67, normalized size of antiderivative = 3.05 \[ \int \frac {425-275 x-125 x^2-25 x^3+\left (300+600 x+300 x^2\right ) \log (5-x)+\left (-125-225 x-75 x^2+25 x^3\right ) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx=\frac {25 \, {\left (36 \, {\left (x^{2} + 2 \, x + 1\right )} \log \left (-x + 5\right )^{2} + 36 \, x^{2} - {\left (72 \, x^{2} + 17 \, x - 55\right )} \log \left (-x + 5\right ) + 36 \, x + 42\right )}}{36 \, {\left (x + 1\right )}} + \frac {425}{6 \, {\left (x + 1\right )}} + \frac {425}{36} \, \log \left (x - 5\right ) \] Input:
integrate(((25*x^3-75*x^2-225*x-125)*log(5-x)^2+(300*x^2+600*x+300)*log(5- x)-25*x^3-125*x^2-275*x+425)/(x^3-3*x^2-9*x-5),x, algorithm="maxima")
Output:
25/36*(36*(x^2 + 2*x + 1)*log(-x + 5)^2 + 36*x^2 - (72*x^2 + 17*x - 55)*lo g(-x + 5) + 36*x + 42)/(x + 1) + 425/6/(x + 1) + 425/36*log(x - 5)
Time = 0.13 (sec) , antiderivative size = 44, normalized size of antiderivative = 2.00 \[ \int \frac {425-275 x-125 x^2-25 x^3+\left (300+600 x+300 x^2\right ) \log (5-x)+\left (-125-225 x-75 x^2+25 x^3\right ) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx=25 \, {\left (x + 1\right )} \log \left (-x + 5\right )^{2} - 50 \, {\left (x - 5\right )} \log \left (-x + 5\right ) + 25 \, x + \frac {100}{x + 1} - 200 \, \log \left (-x + 5\right ) - 125 \] Input:
integrate(((25*x^3-75*x^2-225*x-125)*log(5-x)^2+(300*x^2+600*x+300)*log(5- x)-25*x^3-125*x^2-275*x+425)/(x^3-3*x^2-9*x-5),x, algorithm="giac")
Output:
25*(x + 1)*log(-x + 5)^2 - 50*(x - 5)*log(-x + 5) + 25*x + 100/(x + 1) - 2 00*log(-x + 5) - 125
Time = 0.18 (sec) , antiderivative size = 46, normalized size of antiderivative = 2.09 \[ \int \frac {425-275 x-125 x^2-25 x^3+\left (300+600 x+300 x^2\right ) \log (5-x)+\left (-125-225 x-75 x^2+25 x^3\right ) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx=50\,\ln \left (x-5\right )+25\,{\ln \left (5-x\right )}^2+\frac {100}{x+1}+x\,\left (25\,{\ln \left (5-x\right )}^2-50\,\ln \left (5-x\right )+25\right ) \] Input:
int((275*x + log(5 - x)^2*(225*x + 75*x^2 - 25*x^3 + 125) - log(5 - x)*(60 0*x + 300*x^2 + 300) + 125*x^2 + 25*x^3 - 425)/(9*x + 3*x^2 - x^3 + 5),x)
Output:
50*log(x - 5) + 25*log(5 - x)^2 + 100/(x + 1) + x*(25*log(5 - x)^2 - 50*lo g(5 - x) + 25)
Time = 0.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 3.91 \[ \int \frac {425-275 x-125 x^2-25 x^3+\left (300+600 x+300 x^2\right ) \log (5-x)+\left (-125-225 x-75 x^2+25 x^3\right ) \log ^2(5-x)}{-5-9 x-3 x^2+x^3} \, dx=\frac {25 \mathrm {log}\left (-x +5\right )^{2} x^{2}+50 \mathrm {log}\left (-x +5\right )^{2} x +25 \mathrm {log}\left (-x +5\right )^{2}-50 \,\mathrm {log}\left (-x +5\right ) x^{2}+200 \,\mathrm {log}\left (-x +5\right ) x +250 \,\mathrm {log}\left (-x +5\right )-200 \,\mathrm {log}\left (-5+x \right ) x -200 \,\mathrm {log}\left (-5+x \right )+25 x^{2}-75 x}{x +1} \] Input:
int(((25*x^3-75*x^2-225*x-125)*log(5-x)^2+(300*x^2+600*x+300)*log(5-x)-25* x^3-125*x^2-275*x+425)/(x^3-3*x^2-9*x-5),x)
Output:
(25*(log( - x + 5)**2*x**2 + 2*log( - x + 5)**2*x + log( - x + 5)**2 - 2*l og( - x + 5)*x**2 + 8*log( - x + 5)*x + 10*log( - x + 5) - 8*log(x - 5)*x - 8*log(x - 5) + x**2 - 3*x))/(x + 1)