Integrand size = 68, antiderivative size = 34 \[ \int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2-2 x^4+x^6} \, dx=\frac {5 e^{1-e^{2/x}-\frac {4 \log (15)}{x}}}{x \left (-\frac {1}{x}+x\right )} \] Output:
5*exp(1-4*ln(15)/x-exp(2/x))/x/(x-1/x)
Time = 0.15 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2-2 x^4+x^6} \, dx=\frac {5^{\frac {-4+x}{x}} 81^{-1/x} e^{1-e^{2/x}}}{-1+x^2} \] Input:
Integrate[(E^((x - E^(2/x)*x - 4*Log[15])/x)*(-10*x^3 + E^(2/x)*(-10 + 10* x^2) + (-20 + 20*x^2)*Log[15]))/(x^2 - 2*x^4 + x^6),x]
Output:
(5^((-4 + x)/x)*E^(1 - E^(2/x)))/(81^x^(-1)*(-1 + x^2))
Leaf count is larger than twice the leaf count of optimal. \(129\) vs. \(2(34)=68\).
Time = 1.39 (sec) , antiderivative size = 129, normalized size of antiderivative = 3.79, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {2026, 1380, 27, 2725, 27, 2726}
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.
\(\displaystyle \int \frac {e^{\frac {-e^{2/x} x+x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (10 x^2-10\right )+\left (20 x^2-20\right ) \log (15)\right )}{x^6-2 x^4+x^2} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {e^{\frac {-e^{2/x} x+x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (10 x^2-10\right )+\left (20 x^2-20\right ) \log (15)\right )}{x^2 \left (x^4-2 x^2+1\right )}dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle \int -\frac {2\ 3^{-4/x} 5^{1-\frac {4}{x}} e^{\frac {x-e^{2/x} x}{x}} \left (x^3+e^{2/x} \left (1-x^2\right )+2 \left (1-x^2\right ) \log (15)\right )}{x^2 \left (1-x^2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -2 \int \frac {3^{-4/x} 5^{1-\frac {4}{x}} e^{\frac {x-e^{2/x} x}{x}} \left (x^3+e^{2/x} \left (1-x^2\right )+2 \left (1-x^2\right ) \log (15)\right )}{x^2 \left (1-x^2\right )^2}dx\) |
\(\Big \downarrow \) 2725 |
\(\displaystyle -2 \int \frac {5 e^{\frac {x-e^{2/x} x}{x}+\frac {-4 \log (3)-4 \log (5)}{x}} \left (x^3+e^{2/x} \left (1-x^2\right )+2 \left (1-x^2\right ) \log (15)\right )}{x^2 \left (1-x^2\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -10 \int \frac {e^{\frac {x-e^{2/x} x}{x}-\frac {4 \log (15)}{x}} \left (x^3+e^{2/x} \left (1-x^2\right )+2 \left (1-x^2\right ) \log (15)\right )}{x^2 \left (1-x^2\right )^2}dx\) |
\(\Big \downarrow \) 2726 |
\(\displaystyle -\frac {2\ 3^{-4/x} 5^{1-\frac {4}{x}} e^{\frac {x-e^{2/x} x}{x}} \left (e^{2/x} \left (1-x^2\right )+2 \left (1-x^2\right ) \log (15)\right )}{x^2 \left (1-x^2\right )^2 \left (-\frac {x-e^{2/x} x}{x^2}+\frac {4 \log (15)}{x^2}+\frac {-e^{2/x}+\frac {2 e^{2/x}}{x}+1}{x}\right )}\) |
Input:
Int[(E^((x - E^(2/x)*x - 4*Log[15])/x)*(-10*x^3 + E^(2/x)*(-10 + 10*x^2) + (-20 + 20*x^2)*Log[15]))/(x^2 - 2*x^4 + x^6),x]
Output:
(-2*5^(1 - 4/x)*E^((x - E^(2/x)*x)/x)*(E^(2/x)*(1 - x^2) + 2*(1 - x^2)*Log [15]))/(3^(4/x)*x^2*(1 - x^2)^2*((1 - E^(2/x) + (2*E^(2/x))/x)/x - (x - E^ (2/x)*x)/x^2 + (4*Log[15])/x^2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> S imp[1/c^p Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
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_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x], x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]
Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = v*(y/(Log[F]*D[u, x]))}, Simp[F^u*z, x] /; EqQ[D[z, x], w*y]] /; FreeQ[F, x]
Time = 1.43 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88
method | result | size |
norman | \(\frac {5 \,{\mathrm e}^{\frac {-x \,{\mathrm e}^{\frac {2}{x}}-4 \ln \left (15\right )+x}{x}}}{x^{2}-1}\) | \(30\) |
risch | \(\frac {5 \left (\frac {1}{81}\right )^{\frac {1}{x}} \left (\frac {1}{625}\right )^{\frac {1}{x}} {\mathrm e}^{-{\mathrm e}^{\frac {2}{x}}+1}}{x^{2}-1}\) | \(31\) |
parallelrisch | \(\frac {5 \,{\mathrm e}^{-\frac {x \,{\mathrm e}^{\frac {2}{x}}+4 \ln \left (15\right )-x}{x}}}{x^{2}-1}\) | \(32\) |
Input:
int(((10*x^2-10)*exp(2/x)+(20*x^2-20)*ln(15)-10*x^3)*exp((-x*exp(2/x)-4*ln (15)+x)/x)/(x^6-2*x^4+x^2),x,method=_RETURNVERBOSE)
Output:
5*exp((-x*exp(2/x)-4*ln(15)+x)/x)/(x^2-1)
Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.91 \[ \int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2-2 x^4+x^6} \, dx=\frac {5 \, e^{\left (-\frac {x e^{\frac {2}{x}} - x + 4 \, \log \left (15\right )}{x}\right )}}{x^{2} - 1} \] Input:
integrate(((10*x^2-10)*exp(2/x)+(20*x^2-20)*log(15)-10*x^3)*exp((-x*exp(2/ x)-4*log(15)+x)/x)/(x^6-2*x^4+x^2),x, algorithm="fricas")
Output:
5*e^(-(x*e^(2/x) - x + 4*log(15))/x)/(x^2 - 1)
Time = 0.11 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65 \[ \int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2-2 x^4+x^6} \, dx=\frac {5 e^{\frac {- x e^{\frac {2}{x}} + x - 4 \log {\left (15 \right )}}{x}}}{x^{2} - 1} \] Input:
integrate(((10*x**2-10)*exp(2/x)+(20*x**2-20)*ln(15)-10*x**3)*exp((-x*exp( 2/x)-4*ln(15)+x)/x)/(x**6-2*x**4+x**2),x)
Output:
5*exp((-x*exp(2/x) + x - 4*log(15))/x)/(x**2 - 1)
Time = 0.19 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2-2 x^4+x^6} \, dx=\frac {5 \, e^{\left (-\frac {4 \, \log \left (5\right )}{x} - \frac {4 \, \log \left (3\right )}{x} - e^{\frac {2}{x}} + 1\right )}}{x^{2} - 1} \] Input:
integrate(((10*x^2-10)*exp(2/x)+(20*x^2-20)*log(15)-10*x^3)*exp((-x*exp(2/ x)-4*log(15)+x)/x)/(x^6-2*x^4+x^2),x, algorithm="maxima")
Output:
5*e^(-4*log(5)/x - 4*log(3)/x - e^(2/x) + 1)/(x^2 - 1)
\[ \int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2-2 x^4+x^6} \, dx=\int { -\frac {10 \, {\left (x^{3} - {\left (x^{2} - 1\right )} e^{\frac {2}{x}} - 2 \, {\left (x^{2} - 1\right )} \log \left (15\right )\right )} e^{\left (-\frac {x e^{\frac {2}{x}} - x + 4 \, \log \left (15\right )}{x}\right )}}{x^{6} - 2 \, x^{4} + x^{2}} \,d x } \] Input:
integrate(((10*x^2-10)*exp(2/x)+(20*x^2-20)*log(15)-10*x^3)*exp((-x*exp(2/ x)-4*log(15)+x)/x)/(x^6-2*x^4+x^2),x, algorithm="giac")
Output:
integrate(-10*(x^3 - (x^2 - 1)*e^(2/x) - 2*(x^2 - 1)*log(15))*e^(-(x*e^(2/ x) - x + 4*log(15))/x)/(x^6 - 2*x^4 + x^2), x)
Time = 3.85 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2-2 x^4+x^6} \, dx=\frac {5\,{\mathrm {e}}^{-{\mathrm {e}}^{2/x}}\,\mathrm {e}}{{15}^{4/x}\,\left (x^2-1\right )} \] Input:
int((exp(-(4*log(15) - x + x*exp(2/x))/x)*(log(15)*(20*x^2 - 20) + exp(2/x )*(10*x^2 - 10) - 10*x^3))/(x^2 - 2*x^4 + x^6),x)
Output:
(5*exp(-exp(2/x))*exp(1))/(15^(4/x)*(x^2 - 1))
Time = 0.17 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.94 \[ \int \frac {e^{\frac {x-e^{2/x} x-4 \log (15)}{x}} \left (-10 x^3+e^{2/x} \left (-10+10 x^2\right )+\left (-20+20 x^2\right ) \log (15)\right )}{x^2-2 x^4+x^6} \, dx=\frac {5 e}{e^{\frac {e^{\frac {2}{x}} x +4 \,\mathrm {log}\left (15\right )}{x}} \left (x^{2}-1\right )} \] Input:
int(((10*x^2-10)*exp(2/x)+(20*x^2-20)*log(15)-10*x^3)*exp((-x*exp(2/x)-4*l og(15)+x)/x)/(x^6-2*x^4+x^2),x)
Output:
(5*e)/(e**((e**(2/x)*x + 4*log(15))/x)*(x**2 - 1))