Integrand size = 76, antiderivative size = 25 \[ \int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{-4 x+x^3} \, dx=\log \left (\frac {-2+x}{e^2}\right ) \left (5+e^4+x-\log (x)+\log (4+2 x)\right ) \] Output:
ln((-2+x)/exp(2))*(exp(4)+x-ln(x)+5+ln(4+2*x))
Time = 0.08 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.84 \[ \int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{-4 x+x^3} \, dx=-2 x+\log (4)+\left (7+e^4\right ) \log (2-x)+2 \log (x)-2 \log (2+x)+\log (-2+x) (-2+x-\log (x)+\log (2 (2+x))) \] Input:
Integrate[(10*x + 7*x^2 + x^3 + E^4*(2*x + x^2) + (4 - 6*x + x^3)*Log[(-2 + x)/E^2] + (-2*x - x^2)*Log[x] + (2*x + x^2)*Log[4 + 2*x])/(-4*x + x^3),x ]
Output:
-2*x + Log[4] + (7 + E^4)*Log[2 - x] + 2*Log[x] - 2*Log[2 + x] + Log[-2 + x]*(-2 + x - Log[x] + Log[2*(2 + x)])
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 1.20 (sec) , antiderivative size = 123, normalized size of antiderivative = 4.92, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {2026, 7276, 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.
\(\displaystyle \int \frac {x^3+\left (x^3-6 x+4\right ) \log \left (\frac {x-2}{e^2}\right )+7 x^2+e^4 \left (x^2+2 x\right )+\left (-x^2-2 x\right ) \log (x)+\left (x^2+2 x\right ) \log (2 x+4)+10 x}{x^3-4 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {x^3+\left (x^3-6 x+4\right ) \log \left (\frac {x-2}{e^2}\right )+7 x^2+e^4 \left (x^2+2 x\right )+\left (-x^2-2 x\right ) \log (x)+\left (x^2+2 x\right ) \log (2 x+4)+10 x}{x \left (x^2-4\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {x^2}{x^2-4}+\frac {7 x}{x^2-4}+\frac {10}{x^2-4}+\frac {\left (x^2+2 x-2\right ) (\log (x-2)-2)}{(x+2) x}+\frac {e^4}{x-2}-\frac {\log (x)}{x-2}+\frac {\log (2 x+4)}{x-2}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -7 \text {arctanh}\left (\frac {x}{2}\right )+\operatorname {PolyLog}\left (2,\frac {2-x}{4}\right )+\operatorname {PolyLog}\left (2,\frac {x+2}{4}\right )+\frac {7}{2} \log \left (4-x^2\right )-2 x+e^4 \log (2-x)-(2-x) \log (x-2)-\log (2) \log (x-2)+(2-\log (x-2)) \log \left (\frac {x}{2}\right )-(2-\log (x-2)) \log \left (\frac {x+2}{4}\right )+\log \left (\frac {2-x}{4}\right ) \log (2 x+4)\) |
Input:
Int[(10*x + 7*x^2 + x^3 + E^4*(2*x + x^2) + (4 - 6*x + x^3)*Log[(-2 + x)/E ^2] + (-2*x - x^2)*Log[x] + (2*x + x^2)*Log[4 + 2*x])/(-4*x + x^3),x]
Output:
-2*x - 7*ArcTanh[x/2] + E^4*Log[2 - x] - (2 - x)*Log[-2 + x] - Log[2]*Log[ -2 + x] + (2 - Log[-2 + x])*Log[x/2] - (2 - Log[-2 + x])*Log[(2 + x)/4] + Log[(2 - x)/4]*Log[4 + 2*x] + (7*Log[4 - x^2])/2 + PolyLog[2, (2 - x)/4] + PolyLog[2, (2 + x)/4]
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_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 4.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.40
method | result | size |
risch | \(\left (x -\ln \left (x \right )+\ln \left (4+2 x \right )\right ) \ln \left (\left (-2+x \right ) {\mathrm e}^{-2}\right )+\ln \left (-2+x \right ) {\mathrm e}^{4}+5 \ln \left (-2+x \right )\) | \(35\) |
parallelrisch | \(\ln \left (\left (-2+x \right ) {\mathrm e}^{-2}\right ) {\mathrm e}^{4}+\ln \left (\left (-2+x \right ) {\mathrm e}^{-2}\right ) x -\ln \left (\left (-2+x \right ) {\mathrm e}^{-2}\right ) \ln \left (x \right )+\ln \left (\left (-2+x \right ) {\mathrm e}^{-2}\right ) \ln \left (4+2 x \right )+5 \ln \left (\left (-2+x \right ) {\mathrm e}^{-2}\right )\) | \(65\) |
default | \(\ln \left (2\right ) \ln \left (-2+x \right )+\left (\ln \left (2+x \right )-\ln \left (\frac {x}{4}+\frac {1}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {x}{4}\right )-2 x +\left ({\mathrm e}^{4}+7\right ) \ln \left (-2+x \right )-2 \ln \left (2+x \right )+2 \ln \left (x \right )+\left (-2+x \right ) \ln \left (-2+x \right )+2-\ln \left (-2+x \right ) \ln \left (\frac {x}{2}\right )+\ln \left (-2+x \right ) \ln \left (\frac {x}{4}+\frac {1}{2}\right )-\left (\ln \left (x \right )-\ln \left (\frac {x}{2}\right )\right ) \ln \left (1-\frac {x}{2}\right )\) | \(98\) |
parts | \(\left (\ln \left (4+2 x \right )-\ln \left (\frac {x}{4}+\frac {1}{2}\right )\right ) \ln \left (\frac {1}{2}-\frac {x}{4}\right )-\operatorname {dilog}\left (\frac {x}{4}+\frac {1}{2}\right )+x +\left ({\mathrm e}^{4}+7\right ) \ln \left (-2+x \right )+{\mathrm e}^{-4} {\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2}\right ) x -2 \,{\mathrm e}^{-4} {\mathrm e}^{4} \ln \left (x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2}\right )-{\mathrm e}^{-4} {\mathrm e}^{4} x +2 \,{\mathrm e}^{-4} {\mathrm e}^{4}+\ln \left (x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2}\right ) \ln \left (\frac {{\mathrm e}^{2} \left (x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2}\right )}{4}+1\right )+\operatorname {dilog}\left (\frac {{\mathrm e}^{2} \left (x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2}\right )}{4}+1\right )-\ln \left (x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2}\right ) \ln \left (\frac {{\mathrm e}^{2} \left (x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2}\right )}{2}+1\right )-\operatorname {dilog}\left (\frac {{\mathrm e}^{2} \left (x \,{\mathrm e}^{-2}-2 \,{\mathrm e}^{-2}\right )}{2}+1\right )-\left (\ln \left (x \right )-\ln \left (\frac {x}{2}\right )\right ) \ln \left (1-\frac {x}{2}\right )+\operatorname {dilog}\left (\frac {x}{2}\right )\) | \(237\) |
Input:
int(((x^2+2*x)*ln(4+2*x)+(-x^2-2*x)*ln(x)+(x^3-6*x+4)*ln((-2+x)/exp(2))+(x ^2+2*x)*exp(4)+x^3+7*x^2+10*x)/(x^3-4*x),x,method=_RETURNVERBOSE)
Output:
(x-ln(x)+ln(4+2*x))*ln((-2+x)*exp(-2))+ln(-2+x)*exp(4)+5*ln(-2+x)
Time = 0.11 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.56 \[ \int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{-4 x+x^3} \, dx={\left (x + e^{4} + 5\right )} \log \left ({\left (x - 2\right )} e^{\left (-2\right )}\right ) + \log \left ({\left (x - 2\right )} e^{\left (-2\right )}\right ) \log \left (2 \, x + 4\right ) - \log \left ({\left (x - 2\right )} e^{\left (-2\right )}\right ) \log \left (x\right ) \] Input:
integrate(((x^2+2*x)*log(4+2*x)+(-x^2-2*x)*log(x)+(x^3-6*x+4)*log((-2+x)/e xp(2))+(x^2+2*x)*exp(4)+x^3+7*x^2+10*x)/(x^3-4*x),x, algorithm="fricas")
Output:
(x + e^4 + 5)*log((x - 2)*e^(-2)) + log((x - 2)*e^(-2))*log(2*x + 4) - log ((x - 2)*e^(-2))*log(x)
Time = 0.25 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{-4 x+x^3} \, dx=\left (x - \log {\left (x \right )} + \log {\left (2 x + 4 \right )}\right ) \log {\left (\frac {x - 2}{e^{2}} \right )} + \left (5 + e^{4}\right ) \log {\left (x - 2 \right )} \] Input:
integrate(((x**2+2*x)*ln(4+2*x)+(-x**2-2*x)*ln(x)+(x**3-6*x+4)*ln((-2+x)/e xp(2))+(x**2+2*x)*exp(4)+x**3+7*x**2+10*x)/(x**3-4*x),x)
Output:
(x - log(x) + log(2*x + 4))*log((x - 2)*exp(-2)) + (5 + exp(4))*log(x - 2)
Leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (23) = 46\).
Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 2.68 \[ \int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{-4 x+x^3} \, dx=\frac {1}{2} \, {\left (\log \left (x + 2\right ) + \log \left (x - 2\right )\right )} e^{4} - \frac {1}{2} \, {\left (\log \left (x + 2\right ) - \log \left (x - 2\right )\right )} e^{4} + {\left (\log \left (x - 2\right ) - 2\right )} \log \left (x + 2\right ) + {\left (x + \log \left (2\right ) - \log \left (x\right ) - 2\right )} \log \left (x - 2\right ) - 2 \, x + 7 \, \log \left (x - 2\right ) + 2 \, \log \left (x\right ) \] Input:
integrate(((x^2+2*x)*log(4+2*x)+(-x^2-2*x)*log(x)+(x^3-6*x+4)*log((-2+x)/e xp(2))+(x^2+2*x)*exp(4)+x^3+7*x^2+10*x)/(x^3-4*x),x, algorithm="maxima")
Output:
1/2*(log(x + 2) + log(x - 2))*e^4 - 1/2*(log(x + 2) - log(x - 2))*e^4 + (l og(x - 2) - 2)*log(x + 2) + (x + log(2) - log(x) - 2)*log(x - 2) - 2*x + 7 *log(x - 2) + 2*log(x)
Leaf count of result is larger than twice the leaf count of optimal. 52 vs. \(2 (23) = 46\).
Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 2.08 \[ \int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{-4 x+x^3} \, dx=x \log \left (x - 2\right ) + e^{4} \log \left (x - 2\right ) + \log \left (2 \, x + 4\right ) \log \left (x - 2\right ) - \log \left (x - 2\right ) \log \left (x\right ) - 2 \, x - 2 \, \log \left (x + 2\right ) + 5 \, \log \left (x - 2\right ) + 2 \, \log \left (x\right ) \] Input:
integrate(((x^2+2*x)*log(4+2*x)+(-x^2-2*x)*log(x)+(x^3-6*x+4)*log((-2+x)/e xp(2))+(x^2+2*x)*exp(4)+x^3+7*x^2+10*x)/(x^3-4*x),x, algorithm="giac")
Output:
x*log(x - 2) + e^4*log(x - 2) + log(2*x + 4)*log(x - 2) - log(x - 2)*log(x ) - 2*x - 2*log(x + 2) + 5*log(x - 2) + 2*log(x)
Time = 3.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.92 \[ \int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{-4 x+x^3} \, dx=5\,\ln \left (x-2\right )+\ln \left (x-2\right )\,{\mathrm {e}}^4+\ln \left ({\mathrm {e}}^{-2}\,\left (x-2\right )\right )\,\ln \left (2\,x+4\right )-\ln \left ({\mathrm {e}}^{-2}\,\left (x-2\right )\right )\,\ln \left (x\right )+x\,\ln \left ({\mathrm {e}}^{-2}\,\left (x-2\right )\right ) \] Input:
int(-(10*x + log(2*x + 4)*(2*x + x^2) - log(x)*(2*x + x^2) + log(exp(-2)*( x - 2))*(x^3 - 6*x + 4) + exp(4)*(2*x + x^2) + 7*x^2 + x^3)/(4*x - x^3),x)
Output:
5*log(x - 2) + log(x - 2)*exp(4) + log(exp(-2)*(x - 2))*log(2*x + 4) - log (exp(-2)*(x - 2))*log(x) + x*log(exp(-2)*(x - 2))
Time = 0.18 (sec) , antiderivative size = 62, normalized size of antiderivative = 2.48 \[ \int \frac {10 x+7 x^2+x^3+e^4 \left (2 x+x^2\right )+\left (4-6 x+x^3\right ) \log \left (\frac {-2+x}{e^2}\right )+\left (-2 x-x^2\right ) \log (x)+\left (2 x+x^2\right ) \log (4+2 x)}{-4 x+x^3} \, dx=\mathrm {log}\left (2 x +4\right ) \mathrm {log}\left (\frac {x -2}{e^{2}}\right )+\mathrm {log}\left (x -2\right ) e^{4}+7 \,\mathrm {log}\left (x -2\right )-\mathrm {log}\left (\frac {x -2}{e^{2}}\right ) \mathrm {log}\left (x \right )+\mathrm {log}\left (\frac {x -2}{e^{2}}\right ) x -2 \,\mathrm {log}\left (\frac {x -2}{e^{2}}\right ) \] Input:
int(((x^2+2*x)*log(4+2*x)+(-x^2-2*x)*log(x)+(x^3-6*x+4)*log((-2+x)/exp(2)) +(x^2+2*x)*exp(4)+x^3+7*x^2+10*x)/(x^3-4*x),x)
Output:
log(2*x + 4)*log((x - 2)/e**2) + log(x - 2)*e**4 + 7*log(x - 2) - log((x - 2)/e**2)*log(x) + log((x - 2)/e**2)*x - 2*log((x - 2)/e**2)