Integrand size = 97, antiderivative size = 29 \[ \int \frac {24 x+6 x^2-21 x^3+5 x^4+5 x^5+\left (4+6 x-5 x^2-5 x^3\right ) \log \left (\frac {-100+100 x}{\left (16+80 x+140 x^2+100 x^3+25 x^4\right ) \log (2)}\right )}{-4 x^2-6 x^3+5 x^4+5 x^5} \, dx=5+x+\frac {\log \left (\frac {4 (-1+x)}{\left (\frac {4}{5}+2 x+x^2\right )^2 \log (2)}\right )}{x} \] Output:
ln(4*(-1+x)/ln(2)/(2*x+x^2+4/5)^2)/x+5+x
Time = 0.04 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {24 x+6 x^2-21 x^3+5 x^4+5 x^5+\left (4+6 x-5 x^2-5 x^3\right ) \log \left (\frac {-100+100 x}{\left (16+80 x+140 x^2+100 x^3+25 x^4\right ) \log (2)}\right )}{-4 x^2-6 x^3+5 x^4+5 x^5} \, dx=x+\frac {\log \left (\frac {100 (-1+x)}{\left (4+10 x+5 x^2\right )^2 \log (2)}\right )}{x} \] Input:
Integrate[(24*x + 6*x^2 - 21*x^3 + 5*x^4 + 5*x^5 + (4 + 6*x - 5*x^2 - 5*x^ 3)*Log[(-100 + 100*x)/((16 + 80*x + 140*x^2 + 100*x^3 + 25*x^4)*Log[2])])/ (-4*x^2 - 6*x^3 + 5*x^4 + 5*x^5),x]
Output:
x + Log[(100*(-1 + x))/((4 + 10*x + 5*x^2)^2*Log[2])]/x
Time = 1.97 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, 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.
\(\displaystyle \int \frac {5 x^5+5 x^4-21 x^3+6 x^2+\left (-5 x^3-5 x^2+6 x+4\right ) \log \left (\frac {100 x-100}{\left (25 x^4+100 x^3+140 x^2+80 x+16\right ) \log (2)}\right )+24 x}{5 x^5+5 x^4-6 x^3-4 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {5 x^5+5 x^4-21 x^3+6 x^2+\left (-5 x^3-5 x^2+6 x+4\right ) \log \left (\frac {100 x-100}{\left (25 x^4+100 x^3+140 x^2+80 x+16\right ) \log (2)}\right )+24 x}{x^2 \left (5 x^3+5 x^2-6 x-4\right )}dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {5 x^5+5 x^4-21 x^3+6 x^2+\left (-5 x^3-5 x^2+6 x+4\right ) \log \left (\frac {100 x-100}{\left (25 x^4+100 x^3+140 x^2+80 x+16\right ) \log (2)}\right )+24 x}{19 (x-1) x^2}-\frac {5 (x+3) \left (5 x^5+5 x^4-21 x^3+6 x^2+\left (-5 x^3-5 x^2+6 x+4\right ) \log \left (\frac {100 x-100}{\left (25 x^4+100 x^3+140 x^2+80 x+16\right ) \log (2)}\right )+24 x\right )}{19 x^2 \left (5 x^2+10 x+4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log \left (-\frac {100 (1-x)}{\left (5 x^2+10 x+4\right )^2 \log (2)}\right )}{x}+x\) |
Input:
Int[(24*x + 6*x^2 - 21*x^3 + 5*x^4 + 5*x^5 + (4 + 6*x - 5*x^2 - 5*x^3)*Log [(-100 + 100*x)/((16 + 80*x + 140*x^2 + 100*x^3 + 25*x^4)*Log[2])])/(-4*x^ 2 - 6*x^3 + 5*x^4 + 5*x^5),x]
Output:
x + Log[(-100*(1 - x))/((4 + 10*x + 5*x^2)^2*Log[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 = 1.10 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38
method | result | size |
risch | \(\frac {\ln \left (\frac {100 x -100}{\left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right ) \ln \left (2\right )}\right )}{x}+x\) | \(40\) |
norman | \(\frac {x^{2}+\ln \left (\frac {100 x -100}{\left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right ) \ln \left (2\right )}\right )}{x}\) | \(42\) |
default | \(-\frac {\ln \left (\ln \left (2\right )\right )}{x}+x +\frac {2 \ln \left (10\right )}{x}+\frac {\ln \left (\frac {-1+x}{25 x^{4}+100 x^{3}+140 x^{2}+80 x +16}\right )}{x}\) | \(49\) |
parallelrisch | \(-\frac {-50 x^{2}+100 x -50 \ln \left (\frac {100 x -100}{\left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right ) \ln \left (2\right )}\right )}{50 x}\) | \(49\) |
parts | \(-\frac {\ln \left (\ln \left (2\right )\right )}{x}+x +\frac {2 \ln \left (10\right )}{x}+\frac {\ln \left (\frac {-1+x}{25 x^{4}+100 x^{3}+140 x^{2}+80 x +16}\right )}{x}\) | \(49\) |
orering | \(\frac {\left (50 x^{7}-495 x^{5}-10 x^{4}+1232 x^{3}+472 x^{2}-632 x -256\right ) \left (\left (-5 x^{3}-5 x^{2}+6 x +4\right ) \ln \left (\frac {100 x -100}{\left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right ) \ln \left (2\right )}\right )+5 x^{5}+5 x^{4}-21 x^{3}+6 x^{2}+24 x \right )}{\left (50 x^{6}+100 x^{5}+5 x^{4}-300 x^{3}-328 x^{2}-24 x +136\right ) \left (5 x^{5}+5 x^{4}-6 x^{3}-4 x^{2}\right )}+\frac {\left (10 x^{4}-37 x^{2}+14 x +32\right ) x \left (5 x^{2}+10 x +4\right ) \left (-1+x \right ) \left (\frac {\left (-15 x^{2}-10 x +6\right ) \ln \left (\frac {100 x -100}{\left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right ) \ln \left (2\right )}\right )+\frac {\left (-5 x^{3}-5 x^{2}+6 x +4\right ) \left (\frac {100}{\left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right ) \ln \left (2\right )}-\frac {\left (100 x -100\right ) \left (100 x^{3}+300 x^{2}+280 x +80\right )}{\left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right )^{2} \ln \left (2\right )}\right ) \left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right ) \ln \left (2\right )}{100 x -100}+25 x^{4}+20 x^{3}-63 x^{2}+12 x +24}{5 x^{5}+5 x^{4}-6 x^{3}-4 x^{2}}-\frac {\left (\left (-5 x^{3}-5 x^{2}+6 x +4\right ) \ln \left (\frac {100 x -100}{\left (25 x^{4}+100 x^{3}+140 x^{2}+80 x +16\right ) \ln \left (2\right )}\right )+5 x^{5}+5 x^{4}-21 x^{3}+6 x^{2}+24 x \right ) \left (25 x^{4}+20 x^{3}-18 x^{2}-8 x \right )}{\left (5 x^{5}+5 x^{4}-6 x^{3}-4 x^{2}\right )^{2}}\right )}{50 x^{6}+100 x^{5}+5 x^{4}-300 x^{3}-328 x^{2}-24 x +136}\) | \(551\) |
Input:
int(((-5*x^3-5*x^2+6*x+4)*ln((100*x-100)/(25*x^4+100*x^3+140*x^2+80*x+16)/ ln(2))+5*x^5+5*x^4-21*x^3+6*x^2+24*x)/(5*x^5+5*x^4-6*x^3-4*x^2),x,method=_ RETURNVERBOSE)
Output:
1/x*ln((100*x-100)/(25*x^4+100*x^3+140*x^2+80*x+16)/ln(2))+x
Time = 0.09 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.38 \[ \int \frac {24 x+6 x^2-21 x^3+5 x^4+5 x^5+\left (4+6 x-5 x^2-5 x^3\right ) \log \left (\frac {-100+100 x}{\left (16+80 x+140 x^2+100 x^3+25 x^4\right ) \log (2)}\right )}{-4 x^2-6 x^3+5 x^4+5 x^5} \, dx=\frac {x^{2} + \log \left (\frac {100 \, {\left (x - 1\right )}}{{\left (25 \, x^{4} + 100 \, x^{3} + 140 \, x^{2} + 80 \, x + 16\right )} \log \left (2\right )}\right )}{x} \] Input:
integrate(((-5*x^3-5*x^2+6*x+4)*log((100*x-100)/(25*x^4+100*x^3+140*x^2+80 *x+16)/log(2))+5*x^5+5*x^4-21*x^3+6*x^2+24*x)/(5*x^5+5*x^4-6*x^3-4*x^2),x, algorithm="fricas")
Output:
(x^2 + log(100*(x - 1)/((25*x^4 + 100*x^3 + 140*x^2 + 80*x + 16)*log(2)))) /x
Time = 0.12 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {24 x+6 x^2-21 x^3+5 x^4+5 x^5+\left (4+6 x-5 x^2-5 x^3\right ) \log \left (\frac {-100+100 x}{\left (16+80 x+140 x^2+100 x^3+25 x^4\right ) \log (2)}\right )}{-4 x^2-6 x^3+5 x^4+5 x^5} \, dx=x + \frac {\log {\left (\frac {100 x - 100}{\left (25 x^{4} + 100 x^{3} + 140 x^{2} + 80 x + 16\right ) \log {\left (2 \right )}} \right )}}{x} \] Input:
integrate(((-5*x**3-5*x**2+6*x+4)*ln((100*x-100)/(25*x**4+100*x**3+140*x** 2+80*x+16)/ln(2))+5*x**5+5*x**4-21*x**3+6*x**2+24*x)/(5*x**5+5*x**4-6*x**3 -4*x**2),x)
Output:
x + log((100*x - 100)/((25*x**4 + 100*x**3 + 140*x**2 + 80*x + 16)*log(2)) )/x
Leaf count of result is larger than twice the leaf count of optimal. 64 vs. \(2 (29) = 58\).
Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 2.21 \[ \int \frac {24 x+6 x^2-21 x^3+5 x^4+5 x^5+\left (4+6 x-5 x^2-5 x^3\right ) \log \left (\frac {-100+100 x}{\left (16+80 x+140 x^2+100 x^3+25 x^4\right ) \log (2)}\right )}{-4 x^2-6 x^3+5 x^4+5 x^5} \, dx=x - \frac {{\left (5 \, x + 4\right )} \log \left (5 \, x^{2} + 10 \, x + 4\right ) + 2 \, {\left (x - 1\right )} \log \left (x - 1\right ) - 4 \, \log \left (5\right ) - 4 \, \log \left (2\right ) + 2 \, \log \left (\log \left (2\right )\right )}{2 \, x} + \frac {5}{2} \, \log \left (5 \, x^{2} + 10 \, x + 4\right ) + \log \left (x - 1\right ) \] Input:
integrate(((-5*x^3-5*x^2+6*x+4)*log((100*x-100)/(25*x^4+100*x^3+140*x^2+80 *x+16)/log(2))+5*x^5+5*x^4-21*x^3+6*x^2+24*x)/(5*x^5+5*x^4-6*x^3-4*x^2),x, algorithm="maxima")
Output:
x - 1/2*((5*x + 4)*log(5*x^2 + 10*x + 4) + 2*(x - 1)*log(x - 1) - 4*log(5) - 4*log(2) + 2*log(log(2)))/x + 5/2*log(5*x^2 + 10*x + 4) + log(x - 1)
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.93 \[ \int \frac {24 x+6 x^2-21 x^3+5 x^4+5 x^5+\left (4+6 x-5 x^2-5 x^3\right ) \log \left (\frac {-100+100 x}{\left (16+80 x+140 x^2+100 x^3+25 x^4\right ) \log (2)}\right )}{-4 x^2-6 x^3+5 x^4+5 x^5} \, dx=x + \frac {2 \, \log \left (2\right )}{x} - \frac {\log \left (25 \, x^{4} \log \left (2\right ) + 100 \, x^{3} \log \left (2\right ) + 140 \, x^{2} \log \left (2\right ) + 80 \, x \log \left (2\right ) + 16 \, \log \left (2\right )\right )}{x} + \frac {\log \left (25 \, x - 25\right )}{x} \] Input:
integrate(((-5*x^3-5*x^2+6*x+4)*log((100*x-100)/(25*x^4+100*x^3+140*x^2+80 *x+16)/log(2))+5*x^5+5*x^4-21*x^3+6*x^2+24*x)/(5*x^5+5*x^4-6*x^3-4*x^2),x, algorithm="giac")
Output:
x + 2*log(2)/x - log(25*x^4*log(2) + 100*x^3*log(2) + 140*x^2*log(2) + 80* x*log(2) + 16*log(2))/x + log(25*x - 25)/x
Time = 6.87 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.31 \[ \int \frac {24 x+6 x^2-21 x^3+5 x^4+5 x^5+\left (4+6 x-5 x^2-5 x^3\right ) \log \left (\frac {-100+100 x}{\left (16+80 x+140 x^2+100 x^3+25 x^4\right ) \log (2)}\right )}{-4 x^2-6 x^3+5 x^4+5 x^5} \, dx=x+\frac {\ln \left (\frac {100\,\left (x-1\right )}{\ln \left (2\right )\,\left (25\,x^4+100\,x^3+140\,x^2+80\,x+16\right )}\right )}{x} \] Input:
int(-(24*x + log((100*x - 100)/(log(2)*(80*x + 140*x^2 + 100*x^3 + 25*x^4 + 16)))*(6*x - 5*x^2 - 5*x^3 + 4) + 6*x^2 - 21*x^3 + 5*x^4 + 5*x^5)/(4*x^2 + 6*x^3 - 5*x^4 - 5*x^5),x)
Output:
x + log((100*(x - 1))/(log(2)*(80*x + 140*x^2 + 100*x^3 + 25*x^4 + 16)))/x
Time = 0.16 (sec) , antiderivative size = 208, normalized size of antiderivative = 7.17 \[ \int \frac {24 x+6 x^2-21 x^3+5 x^4+5 x^5+\left (4+6 x-5 x^2-5 x^3\right ) \log \left (\frac {-100+100 x}{\left (16+80 x+140 x^2+100 x^3+25 x^4\right ) \log (2)}\right )}{-4 x^2-6 x^3+5 x^4+5 x^5} \, dx=\frac {-2 \sqrt {5}\, \mathrm {log}\left (-\sqrt {5}+5 x +5\right ) x -2 \sqrt {5}\, \mathrm {log}\left (\sqrt {5}+5 x +5\right ) x +\sqrt {5}\, \mathrm {log}\left (x -1\right ) x -\sqrt {5}\, \mathrm {log}\left (\frac {100 x -100}{25 \,\mathrm {log}\left (2\right ) x^{4}+100 \,\mathrm {log}\left (2\right ) x^{3}+140 \,\mathrm {log}\left (2\right ) x^{2}+80 \,\mathrm {log}\left (2\right ) x +16 \,\mathrm {log}\left (2\right )}\right ) x +10 \,\mathrm {log}\left (-\sqrt {5}+5 x +5\right ) x +10 \,\mathrm {log}\left (\sqrt {5}+5 x +5\right ) x -5 \,\mathrm {log}\left (x -1\right ) x +5 \,\mathrm {log}\left (\frac {100 x -100}{25 \,\mathrm {log}\left (2\right ) x^{4}+100 \,\mathrm {log}\left (2\right ) x^{3}+140 \,\mathrm {log}\left (2\right ) x^{2}+80 \,\mathrm {log}\left (2\right ) x +16 \,\mathrm {log}\left (2\right )}\right ) x +4 \,\mathrm {log}\left (\frac {100 x -100}{25 \,\mathrm {log}\left (2\right ) x^{4}+100 \,\mathrm {log}\left (2\right ) x^{3}+140 \,\mathrm {log}\left (2\right ) x^{2}+80 \,\mathrm {log}\left (2\right ) x +16 \,\mathrm {log}\left (2\right )}\right )+4 x^{2}}{4 x} \] Input:
int(((-5*x^3-5*x^2+6*x+4)*log((100*x-100)/(25*x^4+100*x^3+140*x^2+80*x+16) /log(2))+5*x^5+5*x^4-21*x^3+6*x^2+24*x)/(5*x^5+5*x^4-6*x^3-4*x^2),x)
Output:
( - 2*sqrt(5)*log( - sqrt(5) + 5*x + 5)*x - 2*sqrt(5)*log(sqrt(5) + 5*x + 5)*x + sqrt(5)*log(x - 1)*x - sqrt(5)*log((100*x - 100)/(25*log(2)*x**4 + 100*log(2)*x**3 + 140*log(2)*x**2 + 80*log(2)*x + 16*log(2)))*x + 10*log( - sqrt(5) + 5*x + 5)*x + 10*log(sqrt(5) + 5*x + 5)*x - 5*log(x - 1)*x + 5* log((100*x - 100)/(25*log(2)*x**4 + 100*log(2)*x**3 + 140*log(2)*x**2 + 80 *log(2)*x + 16*log(2)))*x + 4*log((100*x - 100)/(25*log(2)*x**4 + 100*log( 2)*x**3 + 140*log(2)*x**2 + 80*log(2)*x + 16*log(2))) + 4*x**2)/(4*x)