Integrand size = 81, antiderivative size = 24 \[ \int \frac {15+5 x+e^{-4+x} \left (-15+9 x+5 x^2\right )+\left (3+x+e^{-4+x} \left (-3+2 x+x^2\right )\right ) \log \left (\frac {1}{15+5 x}\right )+\log (x) \left (-15-6 x+(-3-x) \log \left (\frac {1}{15+5 x}\right )\right )}{3 x^2+x^3} \, dx=\frac {\left (e^{-4+x}+\log (x)\right ) \left (5+\log \left (\frac {1}{5 (3+x)}\right )\right )}{x} \]
Time = 5.15 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.12 \[ \int \frac {15+5 x+e^{-4+x} \left (-15+9 x+5 x^2\right )+\left (3+x+e^{-4+x} \left (-3+2 x+x^2\right )\right ) \log \left (\frac {1}{15+5 x}\right )+\log (x) \left (-15-6 x+(-3-x) \log \left (\frac {1}{15+5 x}\right )\right )}{3 x^2+x^3} \, dx=\frac {\left (e^x+e^4 \log (x)\right ) \left (5+\log \left (\frac {1}{15+5 x}\right )\right )}{e^4 x} \]
Integrate[(15 + 5*x + E^(-4 + x)*(-15 + 9*x + 5*x^2) + (3 + x + E^(-4 + x) *(-3 + 2*x + x^2))*Log[(15 + 5*x)^(-1)] + Log[x]*(-15 - 6*x + (-3 - x)*Log [(15 + 5*x)^(-1)]))/(3*x^2 + x^3),x]
Leaf count is larger than twice the leaf count of optimal. \(136\) vs. \(2(24)=48\).
Time = 2.02 (sec) , antiderivative size = 136, normalized size of antiderivative = 5.67, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {2026, 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.
\(\displaystyle \int \frac {e^{x-4} \left (5 x^2+9 x-15\right )+\left (e^{x-4} \left (x^2+2 x-3\right )+x+3\right ) \log \left (\frac {1}{5 x+15}\right )+5 x+\log (x) \left (-6 x+(-x-3) \log \left (\frac {1}{5 x+15}\right )-15\right )+15}{x^3+3 x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{x-4} \left (5 x^2+9 x-15\right )+\left (e^{x-4} \left (x^2+2 x-3\right )+x+3\right ) \log \left (\frac {1}{5 x+15}\right )+5 x+\log (x) \left (-6 x+(-x-3) \log \left (\frac {1}{5 x+15}\right )-15\right )+15}{x^2 (x+3)}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {e^{x-4} \left (5 x^2+x^2 \log \left (\frac {1}{5 x+15}\right )+9 x+2 x \log \left (\frac {1}{5 x+15}\right )-3 \log \left (\frac {1}{5 x+15}\right )-15\right )}{x^2 (x+3)}+\frac {5 x-6 x \log (x)-x \log (x) \log \left (\frac {1}{5 x+15}\right )+x \log \left (\frac {1}{5 x+15}\right )-15 \log (x)-3 \log (x) \log \left (\frac {1}{5 x+15}\right )+3 \log \left (\frac {1}{5 x+15}\right )+15}{x^2 (x+3)}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 e^{x-4}}{x}-\frac {1}{3} \log ^2(x)+\frac {1}{3} \log \left (\frac {x}{3}+1\right ) \log (x)-\frac {1}{3} (1-\log (x)) \log (x)+\frac {5 \log (x)}{x}+\frac {1}{3} \log (3) \log (x)+\frac {\log (x)}{3}+\frac {1}{3} (1-\log (x)) \log (x+3)-\frac {1}{3} \log (x+3)-\frac {(1-\log (x)) \log \left (\frac {1}{5 x+15}\right )}{x}+\frac {e^{x-4} \log \left (\frac {1}{5 x+15}\right )}{x}+\frac {\log \left (\frac {1}{5 x+15}\right )}{x}\) |
Int[(15 + 5*x + E^(-4 + x)*(-15 + 9*x + 5*x^2) + (3 + x + E^(-4 + x)*(-3 + 2*x + x^2))*Log[(15 + 5*x)^(-1)] + Log[x]*(-15 - 6*x + (-3 - x)*Log[(15 + 5*x)^(-1)]))/(3*x^2 + x^3),x]
(5*E^(-4 + x))/x + Log[x]/3 + (5*Log[x])/x + (Log[3]*Log[x])/3 + (Log[1 + x/3]*Log[x])/3 - ((1 - Log[x])*Log[x])/3 - Log[x]^2/3 - Log[3 + x]/3 + ((1 - Log[x])*Log[3 + x])/3 + Log[(15 + 5*x)^(-1)]/x + (E^(-4 + x)*Log[(15 + 5*x)^(-1)])/x - ((1 - Log[x])*Log[(15 + 5*x)^(-1)])/x
3.9.87.3.1 Defintions of rubi rules used
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 = 4.62 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50
method | result | size |
risch | \(-\frac {\left (\ln \left (x \right )+{\mathrm e}^{x -4}\right ) \ln \left (3+x \right )}{x}-\frac {\left (-10+2 \ln \left (5\right )\right ) \left (\ln \left (x \right )+{\mathrm e}^{x -4}\right )}{2 x}\) | \(36\) |
parallelrisch | \(-\frac {-6 \ln \left (x \right ) \ln \left (\frac {1}{5 x +15}\right )-6 \ln \left (\frac {1}{5 x +15}\right ) {\mathrm e}^{x -4}-30 \ln \left (x \right )-30 \,{\mathrm e}^{x -4}}{6 x}\) | \(43\) |
default | \(-\frac {\ln \left (-1+\frac {3}{3+x}\right )}{3}+\frac {\ln \left (\frac {1}{5 x +15}\right )}{\left (3+x \right ) \left (-1+\frac {3}{3+x}\right )}-\frac {\ln \left (x \right ) \ln \left (3+x \right )+\left (-5-\ln \left (\frac {1}{3+x}\right )-\ln \left (3+x \right )\right ) \ln \left (x \right )+\frac {x \ln \left (3+x \right )}{3}-\frac {x \ln \left (x \right )}{3}-5-\ln \left (\frac {1}{3+x}\right )}{x}+\ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+\frac {\left (\ln \left (\frac {1}{5 x +15}\right )+\ln \left (5 x +15\right )+5\right ) {\mathrm e}^{x -4}-\ln \left (5 x +15\right ) {\mathrm e}^{x -4}}{x}-\frac {5}{x}\) | \(151\) |
parts | \(-\frac {\ln \left (-1+\frac {3}{3+x}\right )}{3}+\frac {\ln \left (\frac {1}{5 x +15}\right )}{\left (3+x \right ) \left (-1+\frac {3}{3+x}\right )}-\frac {\ln \left (x \right ) \ln \left (3+x \right )+\left (-5-\ln \left (\frac {1}{3+x}\right )-\ln \left (3+x \right )\right ) \ln \left (x \right )+\frac {x \ln \left (3+x \right )}{3}-\frac {x \ln \left (x \right )}{3}-5-\ln \left (\frac {1}{3+x}\right )}{x}+\ln \left (5\right ) \left (-\frac {\ln \left (x \right )}{x}-\frac {1}{x}\right )+\frac {\left (\ln \left (\frac {1}{5 x +15}\right )+\ln \left (5 x +15\right )+5\right ) {\mathrm e}^{x -4}-\ln \left (5 x +15\right ) {\mathrm e}^{x -4}}{x}-\frac {5}{x}\) | \(151\) |
int((((-3-x)*ln(1/(5*x+15))-6*x-15)*ln(x)+((x^2+2*x-3)*exp(x-4)+3+x)*ln(1/ (5*x+15))+(5*x^2+9*x-15)*exp(x-4)+5*x+15)/(x^3+3*x^2),x,method=_RETURNVERB OSE)
Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.54 \[ \int \frac {15+5 x+e^{-4+x} \left (-15+9 x+5 x^2\right )+\left (3+x+e^{-4+x} \left (-3+2 x+x^2\right )\right ) \log \left (\frac {1}{15+5 x}\right )+\log (x) \left (-15-6 x+(-3-x) \log \left (\frac {1}{15+5 x}\right )\right )}{3 x^2+x^3} \, dx=\frac {{\left (\log \left (\frac {1}{5 \, {\left (x + 3\right )}}\right ) + 5\right )} \log \left (x\right ) + e^{\left (x - 4\right )} \log \left (\frac {1}{5 \, {\left (x + 3\right )}}\right ) + 5 \, e^{\left (x - 4\right )}}{x} \]
integrate((((-3-x)*log(1/(5*x+15))-6*x-15)*log(x)+((x^2+2*x-3)*exp(x-4)+3+ x)*log(1/(5*x+15))+(5*x^2+9*x-15)*exp(x-4)+5*x+15)/(x^3+3*x^2),x, algorith m=\
Time = 0.44 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.50 \[ \int \frac {15+5 x+e^{-4+x} \left (-15+9 x+5 x^2\right )+\left (3+x+e^{-4+x} \left (-3+2 x+x^2\right )\right ) \log \left (\frac {1}{15+5 x}\right )+\log (x) \left (-15-6 x+(-3-x) \log \left (\frac {1}{15+5 x}\right )\right )}{3 x^2+x^3} \, dx=\frac {\left (\log {\left (\frac {1}{5 x + 15} \right )} + 5\right ) e^{x - 4}}{x} + \frac {\log {\left (x \right )} \log {\left (\frac {1}{5 x + 15} \right )}}{x} + \frac {5 \log {\left (x \right )}}{x} \]
integrate((((-3-x)*ln(1/(5*x+15))-6*x-15)*ln(x)+((x**2+2*x-3)*exp(x-4)+3+x )*ln(1/(5*x+15))+(5*x**2+9*x-15)*exp(x-4)+5*x+15)/(x**3+3*x**2),x)
Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (21) = 42\).
Time = 0.31 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.96 \[ \int \frac {15+5 x+e^{-4+x} \left (-15+9 x+5 x^2\right )+\left (3+x+e^{-4+x} \left (-3+2 x+x^2\right )\right ) \log \left (\frac {1}{15+5 x}\right )+\log (x) \left (-15-6 x+(-3-x) \log \left (\frac {1}{15+5 x}\right )\right )}{3 x^2+x^3} \, dx=-\frac {{\left ({\left (\log \left (5\right ) - 5\right )} e^{4} \log \left (x\right ) + {\left (\log \left (5\right ) - 5\right )} e^{x} + {\left (e^{4} \log \left (x\right ) + e^{x}\right )} \log \left (x + 3\right ) - 5 \, e^{4}\right )} e^{\left (-4\right )}}{x} - \frac {5}{x} \]
integrate((((-3-x)*log(1/(5*x+15))-6*x-15)*log(x)+((x^2+2*x-3)*exp(x-4)+3+ x)*log(1/(5*x+15))+(5*x^2+9*x-15)*exp(x-4)+5*x+15)/(x^3+3*x^2),x, algorith m=\
-((log(5) - 5)*e^4*log(x) + (log(5) - 5)*e^x + (e^4*log(x) + e^x)*log(x + 3) - 5*e^4)*e^(-4)/x - 5/x
Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.58 \[ \int \frac {15+5 x+e^{-4+x} \left (-15+9 x+5 x^2\right )+\left (3+x+e^{-4+x} \left (-3+2 x+x^2\right )\right ) \log \left (\frac {1}{15+5 x}\right )+\log (x) \left (-15-6 x+(-3-x) \log \left (\frac {1}{15+5 x}\right )\right )}{3 x^2+x^3} \, dx=-\frac {{\left (e^{4} \log \left (5 \, x + 15\right ) \log \left (x\right ) + e^{x} \log \left (5 \, x + 15\right ) - 5 \, e^{4} \log \left (x\right ) - 5 \, e^{x}\right )} e^{\left (-4\right )}}{x} \]
integrate((((-3-x)*log(1/(5*x+15))-6*x-15)*log(x)+((x^2+2*x-3)*exp(x-4)+3+ x)*log(1/(5*x+15))+(5*x^2+9*x-15)*exp(x-4)+5*x+15)/(x^3+3*x^2),x, algorith m=\
Timed out. \[ \int \frac {15+5 x+e^{-4+x} \left (-15+9 x+5 x^2\right )+\left (3+x+e^{-4+x} \left (-3+2 x+x^2\right )\right ) \log \left (\frac {1}{15+5 x}\right )+\log (x) \left (-15-6 x+(-3-x) \log \left (\frac {1}{15+5 x}\right )\right )}{3 x^2+x^3} \, dx=\int \frac {5\,x+{\mathrm {e}}^{x-4}\,\left (5\,x^2+9\,x-15\right )-\ln \left (x\right )\,\left (6\,x+\ln \left (\frac {1}{5\,x+15}\right )\,\left (x+3\right )+15\right )+\ln \left (\frac {1}{5\,x+15}\right )\,\left (x+{\mathrm {e}}^{x-4}\,\left (x^2+2\,x-3\right )+3\right )+15}{x^3+3\,x^2} \,d x \]
int((5*x + exp(x - 4)*(9*x + 5*x^2 - 15) - log(x)*(6*x + log(1/(5*x + 15)) *(x + 3) + 15) + log(1/(5*x + 15))*(x + exp(x - 4)*(2*x + x^2 - 3) + 3) + 15)/(3*x^2 + x^3),x)