Integrand size = 46, antiderivative size = 25 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=x+\frac {5 \left (-3-x+\log \left (\frac {x}{\log (5)}\right )\right )}{(4+x) \log (19)} \] Output:
5*(ln(x/ln(5))-3-x)/(4+x)/ln(19)+x
Time = 0.05 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.04 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {x \log (19)+\frac {5 \left (1+\log \left (\frac {x}{\log (5)}\right )\right )}{4+x}}{\log (19)} \] Input:
Integrate[(20 + (16*x + 8*x^2 + x^3)*Log[19] - 5*x*Log[x/Log[5]])/((16*x + 8*x^2 + x^3)*Log[19]),x]
Output:
(x*Log[19] + (5*(1 + Log[x/Log[5]]))/(4 + x))/Log[19]
Time = 0.48 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.60, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.109, Rules used = {27, 2026, 2007, 7293, 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.
\(\displaystyle \int \frac {\left (x^3+8 x^2+16 x\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )+20}{\left (x^3+8 x^2+16 x\right ) \log (19)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\log (19) \left (x^3+8 x^2+16 x\right )-5 x \log \left (\frac {x}{\log (5)}\right )+20}{x^3+8 x^2+16 x}dx}{\log (19)}\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \frac {\int \frac {\log (19) \left (x^3+8 x^2+16 x\right )-5 x \log \left (\frac {x}{\log (5)}\right )+20}{x \left (x^2+8 x+16\right )}dx}{\log (19)}\) |
\(\Big \downarrow \) 2007 |
\(\displaystyle \frac {\int \frac {\log (19) \left (x^3+8 x^2+16 x\right )-5 x \log \left (\frac {x}{\log (5)}\right )+20}{x (x+4)^2}dx}{\log (19)}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {\int \left (\frac {\log (19) x^3+8 \log (19) x^2+16 \log (19) x+20}{x (x+4)^2}-\frac {5 \log \left (\frac {x}{\log (5)}\right )}{(x+4)^2}\right )dx}{\log (19)}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {5}{x+4}-\frac {5 x \log \left (\frac {x}{\log (5)}\right )}{4 (x+4)}+x \log (19)+\frac {5 \log (x)}{4}}{\log (19)}\) |
Input:
Int[(20 + (16*x + 8*x^2 + x^3)*Log[19] - 5*x*Log[x/Log[5]])/((16*x + 8*x^2 + x^3)*Log[19]),x]
Output:
(5/(4 + x) + x*Log[19] + (5*Log[x])/4 - (5*x*Log[x/Log[5]])/(4*(4 + x)))/L og[19]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
Time = 0.26 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.28
method | result | size |
parallelrisch | \(\frac {\ln \left (19\right ) x^{2}+5-16 \ln \left (19\right )+5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{\ln \left (19\right ) \left (4+x \right )}\) | \(32\) |
norman | \(\frac {x^{2}+\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{\ln \left (19\right )}-\frac {16 \ln \left (19\right )-5}{\ln \left (19\right )}}{4+x}\) | \(36\) |
risch | \(\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{\ln \left (19\right ) \left (4+x \right )}+\frac {\ln \left (19\right ) x^{2}+4 x \ln \left (19\right )+5}{\ln \left (19\right ) \left (4+x \right )}\) | \(43\) |
parts | \(\frac {x \ln \left (19\right )+\frac {5 \ln \left (x \right )}{4}+\frac {5}{4+x}-\frac {5 \ln \left (4+x \right )}{4}}{\ln \left (19\right )}+\frac {5 \ln \left (4+x \right )}{4 \ln \left (19\right )}-\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right ) x}{4 \ln \left (19\right ) \left (4+x \right )}\) | \(58\) |
derivativedivides | \(\frac {\ln \left (5\right ) \left (\frac {5 \ln \left (4+x \right )}{4 \ln \left (5\right )}-\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right ) x}{4 \ln \left (5\right ) \left (4+x \right )}+\frac {\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{4}+\frac {20}{4 x +16}-\frac {5 \ln \left (4+x \right )}{4}}{\ln \left (5\right )}+\frac {x \ln \left (19\right )}{\ln \left (5\right )}\right )}{\ln \left (19\right )}\) | \(75\) |
default | \(\frac {\ln \left (5\right ) \left (\frac {5 \ln \left (4+x \right )}{4 \ln \left (5\right )}-\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right ) x}{4 \ln \left (5\right ) \left (4+x \right )}+\frac {\frac {5 \ln \left (\frac {x}{\ln \left (5\right )}\right )}{4}+\frac {20}{4 x +16}-\frac {5 \ln \left (4+x \right )}{4}}{\ln \left (5\right )}+\frac {x \ln \left (19\right )}{\ln \left (5\right )}\right )}{\ln \left (19\right )}\) | \(75\) |
Input:
int((-5*x*ln(x/ln(5))+(x^3+8*x^2+16*x)*ln(19)+20)/(x^3+8*x^2+16*x)/ln(19), x,method=_RETURNVERBOSE)
Output:
1/ln(19)*(ln(19)*x^2+5-16*ln(19)+5*ln(x/ln(5)))/(4+x)
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {{\left (x^{2} + 4 \, x\right )} \log \left (19\right ) + 5 \, \log \left (\frac {x}{\log \left (5\right )}\right ) + 5}{{\left (x + 4\right )} \log \left (19\right )} \] Input:
integrate((-5*x*log(x/log(5))+(x^3+8*x^2+16*x)*log(19)+20)/(x^3+8*x^2+16*x )/log(19),x, algorithm="fricas")
Output:
((x^2 + 4*x)*log(19) + 5*log(x/log(5)) + 5)/((x + 4)*log(19))
Time = 0.14 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=x + \frac {5 \log {\left (\frac {x}{\log {\left (5 \right )}} \right )}}{x \log {\left (19 \right )} + 4 \log {\left (19 \right )}} + \frac {5}{x \log {\left (19 \right )} + 4 \log {\left (19 \right )}} \] Input:
integrate((-5*x*ln(x/ln(5))+(x**3+8*x**2+16*x)*ln(19)+20)/(x**3+8*x**2+16* x)/ln(19),x)
Output:
x + 5*log(x/log(5))/(x*log(19) + 4*log(19)) + 5/(x*log(19) + 4*log(19))
Leaf count of result is larger than twice the leaf count of optimal. 70 vs. \(2 (25) = 50\).
Time = 0.04 (sec) , antiderivative size = 70, normalized size of antiderivative = 2.80 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {{\left (x - \frac {16}{x + 4} - 8 \, \log \left (x + 4\right )\right )} \log \left (19\right ) + 8 \, {\left (\frac {4}{x + 4} + \log \left (x + 4\right )\right )} \log \left (19\right ) - \frac {16 \, \log \left (19\right )}{x + 4} + \frac {5 \, \log \left (\frac {x}{\log \left (5\right )}\right )}{x + 4} + \frac {5}{x + 4}}{\log \left (19\right )} \] Input:
integrate((-5*x*log(x/log(5))+(x^3+8*x^2+16*x)*log(19)+20)/(x^3+8*x^2+16*x )/log(19),x, algorithm="maxima")
Output:
((x - 16/(x + 4) - 8*log(x + 4))*log(19) + 8*(4/(x + 4) + log(x + 4))*log( 19) - 16*log(19)/(x + 4) + 5*log(x/log(5))/(x + 4) + 5/(x + 4))/log(19)
Time = 0.11 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.24 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {x \log \left (19\right ) - \frac {5 \, {\left (\log \left (\log \left (5\right )\right ) - 1\right )}}{x + 4} + \frac {5 \, \log \left (x\right )}{x + 4}}{\log \left (19\right )} \] Input:
integrate((-5*x*log(x/log(5))+(x^3+8*x^2+16*x)*log(19)+20)/(x^3+8*x^2+16*x )/log(19),x, algorithm="giac")
Output:
(x*log(19) - 5*(log(log(5)) - 1)/(x + 4) + 5*log(x)/(x + 4))/log(19)
Time = 7.04 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=x+\frac {5\,\ln \left (\frac {x}{\ln \left (5\right )}\right )+5}{\ln \left (19\right )\,\left (x+4\right )} \] Input:
int((log(19)*(16*x + 8*x^2 + x^3) - 5*x*log(x/log(5)) + 20)/(log(19)*(16*x + 8*x^2 + x^3)),x)
Output:
x + (5*log(x/log(5)) + 5)/(log(19)*(x + 4))
Time = 0.16 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {20+\left (16 x+8 x^2+x^3\right ) \log (19)-5 x \log \left (\frac {x}{\log (5)}\right )}{\left (16 x+8 x^2+x^3\right ) \log (19)} \, dx=\frac {-5 \,\mathrm {log}\left (\frac {x}{\mathrm {log}\left (5\right )}\right ) x +5 \,\mathrm {log}\left (x \right ) x +20 \,\mathrm {log}\left (x \right )+4 \,\mathrm {log}\left (19\right ) x^{2}+16 \,\mathrm {log}\left (19\right ) x -5 x}{4 \,\mathrm {log}\left (19\right ) \left (x +4\right )} \] Input:
int((-5*x*log(x/log(5))+(x^3+8*x^2+16*x)*log(19)+20)/(x^3+8*x^2+16*x)/log( 19),x)
Output:
( - 5*log(x/log(5))*x + 5*log(x)*x + 20*log(x) + 4*log(19)*x**2 + 16*log(1 9)*x - 5*x)/(4*log(19)*(x + 4))