Integrand size = 63, antiderivative size = 34 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=e^4-\frac {4+x}{5 x}+\frac {\log \left (\left (-\frac {3}{x^2}+\frac {20+x}{x}\right )^2\right )}{x} \] Output:
exp(4)-1/5*(4+x)/x+ln(((20+x)/x-3/x^2)^2)/x
Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {1}{5} \left (-\frac {4}{x}+\frac {5 \log \left (\frac {\left (-3+20 x+x^2\right )^2}{x^4}\right )}{x}\right ) \] Input:
Integrate[(48 - 120*x + 4*x^2 + (15 - 100*x - 5*x^2)*Log[(9 - 120*x + 394* x^2 + 40*x^3 + x^4)/x^4])/(-15*x^2 + 100*x^3 + 5*x^4),x]
Output:
(-4/x + (5*Log[(-3 + 20*x + x^2)^2/x^4])/x)/5
Time = 0.53 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2026, 7279, 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.
\(\displaystyle \int \frac {4 x^2+\left (-5 x^2-100 x+15\right ) \log \left (\frac {x^4+40 x^3+394 x^2-120 x+9}{x^4}\right )-120 x+48}{5 x^4+100 x^3-15 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {4 x^2+\left (-5 x^2-100 x+15\right ) \log \left (\frac {x^4+40 x^3+394 x^2-120 x+9}{x^4}\right )-120 x+48}{x^2 \left (5 x^2+100 x-15\right )}dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {4 \left (x^2-30 x+12\right )}{5 x^2 \left (x^2+20 x-3\right )}-\frac {\log \left (\frac {\left (x^2+20 x-3\right )^2}{x^4}\right )}{x^2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\log \left (\frac {\left (-x^2-20 x+3\right )^2}{x^4}\right )}{x}-\frac {4}{5 x}\) |
Input:
Int[(48 - 120*x + 4*x^2 + (15 - 100*x - 5*x^2)*Log[(9 - 120*x + 394*x^2 + 40*x^3 + x^4)/x^4])/(-15*x^2 + 100*x^3 + 5*x^4),x]
Output:
-4/(5*x) + Log[(3 - 20*x - x^2)^2/x^4]/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_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.29 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {-\frac {4}{5}+\ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{x}\) | \(30\) |
parallelrisch | \(-\frac {8-10 \ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{10 x}\) | \(33\) |
derivativedivides | \(-\frac {4}{5 x}+\frac {\ln \left (1+\frac {394}{x^{2}}+\frac {40}{x}+\frac {9}{x^{4}}-\frac {120}{x^{3}}\right )}{x}\) | \(34\) |
default | \(-\frac {4}{5 x}+\frac {\ln \left (1+\frac {394}{x^{2}}+\frac {40}{x}+\frac {9}{x^{4}}-\frac {120}{x^{3}}\right )}{x}\) | \(34\) |
risch | \(\frac {\ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )}{x}-\frac {4}{5 x}\) | \(34\) |
parts | \(\frac {\ln \left (1+\frac {394}{x^{2}}+\frac {40}{x}+\frac {9}{x^{4}}-\frac {120}{x^{3}}\right )}{x}-\frac {4}{5 x}-\frac {20 \ln \left (\frac {3}{x^{2}}-\frac {20}{x}-1\right )}{3}+\frac {4 \sqrt {103}\, \operatorname {arctanh}\left (\frac {\left (\frac {6}{x}-20\right ) \sqrt {103}}{206}\right )}{3}-\frac {40 \ln \left (x \right )}{3}+\frac {20 \ln \left (x^{2}+20 x -3\right )}{3}-\frac {4 \sqrt {103}\, \operatorname {arctanh}\left (\frac {\left (2 x +20\right ) \sqrt {103}}{206}\right )}{3}\) | \(98\) |
orering | \(-\frac {x \left (40 x^{3}+585 x^{2}-300 x +27\right ) \left (\left (-5 x^{2}-100 x +15\right ) \ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )+4 x^{2}-120 x +48\right )}{\left (20 x^{3}+191 x^{2}-120 x +9\right ) \left (5 x^{4}+100 x^{3}-15 x^{2}\right )}-\frac {\left (10 x -3\right ) x^{2} \left (x^{2}+20 x -3\right ) \left (\frac {\left (-10 x -100\right ) \ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )+\frac {\left (-5 x^{2}-100 x +15\right ) \left (\frac {4 x^{3}+120 x^{2}+788 x -120}{x^{4}}-\frac {4 \left (x^{4}+40 x^{3}+394 x^{2}-120 x +9\right )}{x^{5}}\right ) x^{4}}{x^{4}+40 x^{3}+394 x^{2}-120 x +9}+8 x -120}{5 x^{4}+100 x^{3}-15 x^{2}}-\frac {\left (\left (-5 x^{2}-100 x +15\right ) \ln \left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )+4 x^{2}-120 x +48\right ) \left (20 x^{3}+300 x^{2}-30 x \right )}{\left (5 x^{4}+100 x^{3}-15 x^{2}\right )^{2}}\right )}{20 x^{3}+191 x^{2}-120 x +9}\) | \(343\) |
Input:
int(((-5*x^2-100*x+15)*ln((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-120*x+48 )/(5*x^4+100*x^3-15*x^2),x,method=_RETURNVERBOSE)
Output:
(-4/5+ln((x^4+40*x^3+394*x^2-120*x+9)/x^4))/x
Time = 0.08 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {5 \, \log \left (\frac {x^{4} + 40 \, x^{3} + 394 \, x^{2} - 120 \, x + 9}{x^{4}}\right ) - 4}{5 \, x} \] Input:
integrate(((-5*x^2-100*x+15)*log((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-1 20*x+48)/(5*x^4+100*x^3-15*x^2),x, algorithm="fricas")
Output:
1/5*(5*log((x^4 + 40*x^3 + 394*x^2 - 120*x + 9)/x^4) - 4)/x
Time = 0.09 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\log {\left (\frac {x^{4} + 40 x^{3} + 394 x^{2} - 120 x + 9}{x^{4}} \right )}}{x} - \frac {4}{5 x} \] Input:
integrate(((-5*x**2-100*x+15)*ln((x**4+40*x**3+394*x**2-120*x+9)/x**4)+4*x **2-120*x+48)/(5*x**4+100*x**3-15*x**2),x)
Output:
log((x**4 + 40*x**3 + 394*x**2 - 120*x + 9)/x**4)/x - 4/(5*x)
Time = 0.16 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.53 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=-\frac {2 \, {\left ({\left (16 \, x - 3\right )} \log \left (x^{2} + 20 \, x - 3\right ) - 2 \, {\left (16 \, x - 3\right )} \log \left (x\right ) + 6\right )}}{3 \, x} + \frac {16}{5 \, x} + \frac {32}{3} \, \log \left (x^{2} + 20 \, x - 3\right ) - \frac {64}{3} \, \log \left (x\right ) \] Input:
integrate(((-5*x^2-100*x+15)*log((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-1 20*x+48)/(5*x^4+100*x^3-15*x^2),x, algorithm="maxima")
Output:
-2/3*((16*x - 3)*log(x^2 + 20*x - 3) - 2*(16*x - 3)*log(x) + 6)/x + 16/5/x + 32/3*log(x^2 + 20*x - 3) - 64/3*log(x)
Time = 0.16 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.97 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\log \left (\frac {x^{4} + 40 \, x^{3} + 394 \, x^{2} - 120 \, x + 9}{x^{4}}\right )}{x} - \frac {4}{5 \, x} \] Input:
integrate(((-5*x^2-100*x+15)*log((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-1 20*x+48)/(5*x^4+100*x^3-15*x^2),x, algorithm="giac")
Output:
log((x^4 + 40*x^3 + 394*x^2 - 120*x + 9)/x^4)/x - 4/5/x
Time = 7.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {\ln \left (\frac {x^4+40\,x^3+394\,x^2-120\,x+9}{x^4}\right )-\frac {4}{5}}{x} \] Input:
int(-(120*x - 4*x^2 + log((394*x^2 - 120*x + 40*x^3 + x^4 + 9)/x^4)*(100*x + 5*x^2 - 15) - 48)/(100*x^3 - 15*x^2 + 5*x^4),x)
Output:
(log((394*x^2 - 120*x + 40*x^3 + x^4 + 9)/x^4) - 4/5)/x
Time = 0.20 (sec) , antiderivative size = 142, normalized size of antiderivative = 4.18 \[ \int \frac {48-120 x+4 x^2+\left (15-100 x-5 x^2\right ) \log \left (\frac {9-120 x+394 x^2+40 x^3+x^4}{x^4}\right )}{-15 x^2+100 x^3+5 x^4} \, dx=\frac {-10 \sqrt {103}\, \mathrm {log}\left (-\sqrt {103}+x +10\right ) x -10 \sqrt {103}\, \mathrm {log}\left (\sqrt {103}+x +10\right ) x +5 \sqrt {103}\, \mathrm {log}\left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right ) x +20 \sqrt {103}\, \mathrm {log}\left (x \right ) x +100 \,\mathrm {log}\left (-\sqrt {103}+x +10\right ) x +100 \,\mathrm {log}\left (\sqrt {103}+x +10\right ) x -50 \,\mathrm {log}\left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right ) x +15 \,\mathrm {log}\left (\frac {x^{4}+40 x^{3}+394 x^{2}-120 x +9}{x^{4}}\right )-200 \,\mathrm {log}\left (x \right ) x -12}{15 x} \] Input:
int(((-5*x^2-100*x+15)*log((x^4+40*x^3+394*x^2-120*x+9)/x^4)+4*x^2-120*x+4 8)/(5*x^4+100*x^3-15*x^2),x)
Output:
( - 10*sqrt(103)*log( - sqrt(103) + x + 10)*x - 10*sqrt(103)*log(sqrt(103) + x + 10)*x + 5*sqrt(103)*log((x**4 + 40*x**3 + 394*x**2 - 120*x + 9)/x** 4)*x + 20*sqrt(103)*log(x)*x + 100*log( - sqrt(103) + x + 10)*x + 100*log( sqrt(103) + x + 10)*x - 50*log((x**4 + 40*x**3 + 394*x**2 - 120*x + 9)/x** 4)*x + 15*log((x**4 + 40*x**3 + 394*x**2 - 120*x + 9)/x**4) - 200*log(x)*x - 12)/(15*x)