Integrand size = 80, antiderivative size = 29 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=8-2 x-e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \log \left (\frac {5 x}{2}\right ) \] Output:
8-2*x-ln(5/2*x)/exp(4*ln(1/3*x^2-5)^2)
Time = 1.05 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-2 x-e^{-4 \log ^2\left (-5+\frac {x^2}{3}\right )} \log \left (\frac {5 x}{2}\right ) \] Input:
Integrate[(15 - x^2 + E^(4*Log[(-15 + x^2)/3]^2)*(30*x - 2*x^3) + 16*x^2*L og[(5*x)/2]*Log[(-15 + x^2)/3])/(E^(4*Log[(-15 + x^2)/3]^2)*(-15*x + x^3)) ,x]
Output:
-2*x - Log[(5*x)/2]/E^(4*Log[-5 + x^2/3]^2)
Time = 2.65 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.038, 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.
\(\displaystyle \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (x^2-15\right )\right )} \left (-x^2+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (x^2-15\right )\right )+\left (30 x-2 x^3\right ) e^{4 \log ^2\left (\frac {1}{3} \left (x^2-15\right )\right )}+15\right )}{x^3-15 x} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (x^2-15\right )\right )} \left (-x^2+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (x^2-15\right )\right )+\left (30 x-2 x^3\right ) e^{4 \log ^2\left (\frac {1}{3} \left (x^2-15\right )\right )}+15\right )}{x \left (x^2-15\right )}dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {e^{-4 \log ^2\left (\frac {x^2}{3}-5\right )} \left (-x^2+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {x^2}{3}-5\right )+15\right )}{x \left (x^2-15\right )}-2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -e^{-4 \log ^2\left (\frac {x^2}{3}-5\right )} \log \left (\frac {5 x}{2}\right )-2 x\) |
Input:
Int[(15 - x^2 + E^(4*Log[(-15 + x^2)/3]^2)*(30*x - 2*x^3) + 16*x^2*Log[(5* x)/2]*Log[(-15 + x^2)/3])/(E^(4*Log[(-15 + x^2)/3]^2)*(-15*x + x^3)),x]
Output:
-2*x - Log[(5*x)/2]/E^(4*Log[-5 + x^2/3]^2)
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 = 0.61 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.83
method | result | size |
risch | \(-2 x -\ln \left (\frac {5 x}{2}\right ) {\mathrm e}^{-4 \ln \left (\frac {x^{2}}{3}-5\right )^{2}}\) | \(24\) |
parallelrisch | \(-\frac {\left (900 x \,{\mathrm e}^{4 \ln \left (\frac {x^{2}}{3}-5\right )^{2}}+450 \ln \left (\frac {5 x}{2}\right )\right ) {\mathrm e}^{-4 \ln \left (\frac {x^{2}}{3}-5\right )^{2}}}{450}\) | \(41\) |
Input:
int(((-2*x^3+30*x)*exp(4*ln(1/3*x^2-5)^2)+16*x^2*ln(5/2*x)*ln(1/3*x^2-5)-x ^2+15)/(x^3-15*x)/exp(4*ln(1/3*x^2-5)^2),x,method=_RETURNVERBOSE)
Output:
-2*x-ln(5/2*x)*exp(-4*ln(1/3*x^2-5)^2)
Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-{\left (2 \, x e^{\left (4 \, \log \left (\frac {1}{3} \, x^{2} - 5\right )^{2}\right )} + \log \left (\frac {5}{2} \, x\right )\right )} e^{\left (-4 \, \log \left (\frac {1}{3} \, x^{2} - 5\right )^{2}\right )} \] Input:
integrate(((-2*x^3+30*x)*exp(4*log(1/3*x^2-5)^2)+16*x^2*log(5/2*x)*log(1/3 *x^2-5)-x^2+15)/(x^3-15*x)/exp(4*log(1/3*x^2-5)^2),x, algorithm="fricas")
Output:
-(2*x*e^(4*log(1/3*x^2 - 5)^2) + log(5/2*x))*e^(-4*log(1/3*x^2 - 5)^2)
Timed out. \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=\text {Timed out} \] Input:
integrate(((-2*x**3+30*x)*exp(4*ln(1/3*x**2-5)**2)+16*x**2*ln(5/2*x)*ln(1/ 3*x**2-5)-x**2+15)/(x**3-15*x)/exp(4*ln(1/3*x**2-5)**2),x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (24) = 48\).
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 2.17 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-{\left (2 \, x e^{\left (4 \, \log \left (3\right )^{2} + 4 \, \log \left (x^{2} - 15\right )^{2}\right )} + {\left (\log \left (5\right ) - \log \left (2\right ) + \log \left (x\right )\right )} e^{\left (8 \, \log \left (3\right ) \log \left (x^{2} - 15\right )\right )}\right )} e^{\left (-4 \, \log \left (3\right )^{2} - 4 \, \log \left (x^{2} - 15\right )^{2}\right )} \] Input:
integrate(((-2*x^3+30*x)*exp(4*log(1/3*x^2-5)^2)+16*x^2*log(5/2*x)*log(1/3 *x^2-5)-x^2+15)/(x^3-15*x)/exp(4*log(1/3*x^2-5)^2),x, algorithm="maxima")
Output:
-(2*x*e^(4*log(3)^2 + 4*log(x^2 - 15)^2) + (log(5) - log(2) + log(x))*e^(8 *log(3)*log(x^2 - 15)))*e^(-4*log(3)^2 - 4*log(x^2 - 15)^2)
\[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=\int { \frac {{\left (16 \, x^{2} \log \left (\frac {1}{3} \, x^{2} - 5\right ) \log \left (\frac {5}{2} \, x\right ) - x^{2} - 2 \, {\left (x^{3} - 15 \, x\right )} e^{\left (4 \, \log \left (\frac {1}{3} \, x^{2} - 5\right )^{2}\right )} + 15\right )} e^{\left (-4 \, \log \left (\frac {1}{3} \, x^{2} - 5\right )^{2}\right )}}{x^{3} - 15 \, x} \,d x } \] Input:
integrate(((-2*x^3+30*x)*exp(4*log(1/3*x^2-5)^2)+16*x^2*log(5/2*x)*log(1/3 *x^2-5)-x^2+15)/(x^3-15*x)/exp(4*log(1/3*x^2-5)^2),x, algorithm="giac")
Output:
integrate((16*x^2*log(1/3*x^2 - 5)*log(5/2*x) - x^2 - 2*(x^3 - 15*x)*e^(4* log(1/3*x^2 - 5)^2) + 15)*e^(-4*log(1/3*x^2 - 5)^2)/(x^3 - 15*x), x)
Time = 3.27 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.79 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=-2\,x-\ln \left (\frac {5\,x}{2}\right )\,{\mathrm {e}}^{-4\,{\ln \left (\frac {x^2}{3}-5\right )}^2} \] Input:
int(-(exp(-4*log(x^2/3 - 5)^2)*(exp(4*log(x^2/3 - 5)^2)*(30*x - 2*x^3) - x ^2 + 16*x^2*log(x^2/3 - 5)*log((5*x)/2) + 15))/(15*x - x^3),x)
Output:
- 2*x - log((5*x)/2)*exp(-4*log(x^2/3 - 5)^2)
Time = 0.16 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.41 \[ \int \frac {e^{-4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (15-x^2+e^{4 \log ^2\left (\frac {1}{3} \left (-15+x^2\right )\right )} \left (30 x-2 x^3\right )+16 x^2 \log \left (\frac {5 x}{2}\right ) \log \left (\frac {1}{3} \left (-15+x^2\right )\right )\right )}{-15 x+x^3} \, dx=\frac {-2 e^{4 \mathrm {log}\left (\frac {x^{2}}{3}-5\right )^{2}} x -\mathrm {log}\left (\frac {5 x}{2}\right )}{e^{4 \mathrm {log}\left (\frac {x^{2}}{3}-5\right )^{2}}} \] Input:
int(((-2*x^3+30*x)*exp(4*log(1/3*x^2-5)^2)+16*x^2*log(5/2*x)*log(1/3*x^2-5 )-x^2+15)/(x^3-15*x)/exp(4*log(1/3*x^2-5)^2),x)
Output:
( - 2*e**(4*log((x**2 - 15)/3)**2)*x - log((5*x)/2))/e**(4*log((x**2 - 15) /3)**2)