Integrand size = 79, antiderivative size = 33 \[ \int \frac {e^{-\frac {5}{-3+e^2}} \left (-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx=-x+\frac {e^{\frac {5}{3-e^2}}}{x \left (x^2+\frac {\log (3)}{4}\right )} \]
Time = 0.07 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.24 \[ \int \frac {e^{-\frac {5}{-3+e^2}} \left (-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx=e^{-\frac {5}{-3+e^2}} \left (-e^{\frac {5}{-3+e^2}} x+\frac {4}{4 x^3+x \log (3)}\right ) \]
Integrate[(-48*x^2 - 4*Log[3] + E^(5/(-3 + E^2))*(-16*x^6 - 8*x^4*Log[3] - x^2*Log[3]^2))/(E^(5/(-3 + E^2))*(16*x^6 + 8*x^4*Log[3] + x^2*Log[3]^2)), x]
Time = 0.56 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.94, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {27, 25, 2026, 1380, 2336, 27, 1387, 244, 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.
\(\displaystyle \int \frac {e^{-\frac {5}{e^2-3}} \left (-48 x^2+e^{\frac {5}{e^2-3}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )-4 \log (3)\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle e^{\frac {5}{3-e^2}} \int -\frac {48 x^2+e^{-\frac {5}{3-e^2}} \left (16 x^6+8 \log (3) x^4+\log ^2(3) x^2\right )+4 \log (3)}{16 x^6+8 \log (3) x^4+\log ^2(3) x^2}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -e^{\frac {5}{3-e^2}} \int \frac {48 x^2+e^{-\frac {5}{3-e^2}} \left (16 x^6+8 \log (3) x^4+\log ^2(3) x^2\right )+\log (81)}{16 x^6+8 \log (3) x^4+\log ^2(3) x^2}dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle -e^{\frac {5}{3-e^2}} \int \frac {48 x^2+e^{-\frac {5}{3-e^2}} \left (16 x^6+8 \log (3) x^4+\log ^2(3) x^2\right )+\log (81)}{x^2 \left (16 x^4+8 \log (3) x^2+\log ^2(3)\right )}dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle -16 e^{\frac {5}{3-e^2}} \int \frac {48 x^2+e^{-\frac {5}{3-e^2}} \left (16 x^6+8 \log (3) x^4+\log ^2(3) x^2\right )+\log (81)}{x^2 \left (16 x^2+\log (81)\right )^2}dx\) |
\(\Big \downarrow \) 2336 |
\(\displaystyle -16 e^{\frac {5}{3-e^2}} \left (\frac {16 x}{\log (81) \left (16 x^2+\log (81)\right )}-\frac {\int -\frac {32 e^{-\frac {5}{3-e^2}} \log (81) x^4+\left (512+e^{-\frac {5}{3-e^2}} \left (16 \log ^2(3)+\log ^2(81)\right )\right ) x^2+32 \log (81)}{16 x^2 \left (16 x^2+\log (81)\right )}dx}{2 \log (81)}\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -16 e^{\frac {5}{3-e^2}} \left (\frac {\int \frac {32 e^{-\frac {5}{3-e^2}} \log (81) x^4+\left (512+e^{-\frac {5}{3-e^2}} \left (16 \log ^2(3)+\log ^2(81)\right )\right ) x^2+32 \log (81)}{x^2 \left (16 x^2+\log (81)\right )}dx}{32 \log (81)}+\frac {16 x}{\log (81) \left (16 x^2+\log (81)\right )}\right )\) |
\(\Big \downarrow \) 1387 |
\(\displaystyle -16 e^{\frac {5}{3-e^2}} \left (\frac {\int \frac {2 e^{-\frac {5}{3-e^2}} \log (81) x^2+32}{x^2}dx}{32 \log (81)}+\frac {16 x}{\log (81) \left (16 x^2+\log (81)\right )}\right )\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -16 e^{\frac {5}{3-e^2}} \left (\frac {\int \left (e^{\frac {5}{-3+e^2}} \log (6561)+\frac {32}{x^2}\right )dx}{32 \log (81)}+\frac {16 x}{\log (81) \left (16 x^2+\log (81)\right )}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -16 e^{\frac {5}{3-e^2}} \left (\frac {16 x}{\log (81) \left (16 x^2+\log (81)\right )}+\frac {e^{-\frac {5}{3-e^2}} x \log (6561)-\frac {32}{x}}{32 \log (81)}\right )\) |
Int[(-48*x^2 - 4*Log[3] + E^(5/(-3 + E^2))*(-16*x^6 - 8*x^4*Log[3] - x^2*L og[3]^2))/(E^(5/(-3 + E^2))*(16*x^6 + 8*x^4*Log[3] + x^2*Log[3]^2)),x]
-16*E^(5/(3 - E^2))*((16*x)/(Log[81]*(16*x^2 + Log[81])) + (-32/x + (x*Log [6561])/E^(5/(3 - E^2)))/(32*Log[81]))
3.28.44.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
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[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.)*((d_) + (e_.)* (x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^ p, x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 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])
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) ^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
Time = 0.14 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.88
method | result | size |
risch | \(-x +\frac {4 \,{\mathrm e}^{-\frac {5}{{\mathrm e}^{2}-3}}}{x \left (4 x^{2}+\ln \left (3\right )\right )}\) | \(29\) |
default | \({\mathrm e}^{-\frac {5}{{\mathrm e}^{2}-3}} \left (-x \,{\mathrm e}^{\frac {5}{{\mathrm e}^{2}-3}}-\frac {16 x}{\ln \left (3\right ) \left (4 x^{2}+\ln \left (3\right )\right )}+\frac {4}{x \ln \left (3\right )}\right )\) | \(52\) |
gosper | \(-\frac {\left (4 \,{\mathrm e}^{\frac {5}{{\mathrm e}^{2}-3}} x^{4}-4+\ln \left (3\right ) {\mathrm e}^{\frac {5}{{\mathrm e}^{2}-3}} x^{2}\right ) {\mathrm e}^{-\frac {5}{{\mathrm e}^{2}-3}}}{x \left (4 x^{2}+\ln \left (3\right )\right )}\) | \(58\) |
parallelrisch | \(-\frac {{\mathrm e}^{-\frac {5}{{\mathrm e}^{2}-3}} \left (16 \,{\mathrm e}^{\frac {5}{{\mathrm e}^{2}-3}} x^{4}+4 \ln \left (3\right ) {\mathrm e}^{\frac {5}{{\mathrm e}^{2}-3}} x^{2}-16\right )}{4 x \left (4 x^{2}+\ln \left (3\right )\right )}\) | \(59\) |
int(((-x^2*ln(3)^2-8*x^4*ln(3)-16*x^6)*exp(5/(exp(2)-3))-4*ln(3)-48*x^2)/( x^2*ln(3)^2+8*x^4*ln(3)+16*x^6)/exp(5/(exp(2)-3)),x,method=_RETURNVERBOSE)
Time = 0.24 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.42 \[ \int \frac {e^{-\frac {5}{-3+e^2}} \left (-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx=-\frac {{\left ({\left (4 \, x^{4} + x^{2} \log \left (3\right )\right )} e^{\left (\frac {5}{e^{2} - 3}\right )} - 4\right )} e^{\left (-\frac {5}{e^{2} - 3}\right )}}{4 \, x^{3} + x \log \left (3\right )} \]
integrate(((-x^2*log(3)^2-8*x^4*log(3)-16*x^6)*exp(5/(exp(2)-3))-4*log(3)- 48*x^2)/(x^2*log(3)^2+8*x^4*log(3)+16*x^6)/exp(5/(exp(2)-3)),x, algorithm= \
Time = 0.24 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.97 \[ \int \frac {e^{-\frac {5}{-3+e^2}} \left (-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx=- x + \frac {4}{4 x^{3} e^{- \frac {5}{3 - e^{2}}} + x e^{- \frac {5}{3 - e^{2}}} \log {\left (3 \right )}} \]
integrate(((-x**2*ln(3)**2-8*x**4*ln(3)-16*x**6)*exp(5/(exp(2)-3))-4*ln(3) -48*x**2)/(x**2*ln(3)**2+8*x**4*ln(3)+16*x**6)/exp(5/(exp(2)-3)),x)
Time = 0.19 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {5}{-3+e^2}} \left (-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx=-{\left (x e^{\left (\frac {5}{e^{2} - 3}\right )} - \frac {4}{4 \, x^{3} + x \log \left (3\right )}\right )} e^{\left (-\frac {5}{e^{2} - 3}\right )} \]
integrate(((-x^2*log(3)^2-8*x^4*log(3)-16*x^6)*exp(5/(exp(2)-3))-4*log(3)- 48*x^2)/(x^2*log(3)^2+8*x^4*log(3)+16*x^6)/exp(5/(exp(2)-3)),x, algorithm= \
Time = 0.26 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.12 \[ \int \frac {e^{-\frac {5}{-3+e^2}} \left (-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx=-{\left (x e^{\left (\frac {5}{e^{2} - 3}\right )} - \frac {4}{4 \, x^{3} + x \log \left (3\right )}\right )} e^{\left (-\frac {5}{e^{2} - 3}\right )} \]
integrate(((-x^2*log(3)^2-8*x^4*log(3)-16*x^6)*exp(5/(exp(2)-3))-4*log(3)- 48*x^2)/(x^2*log(3)^2+8*x^4*log(3)+16*x^6)/exp(5/(exp(2)-3)),x, algorithm= \
Time = 9.94 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.85 \[ \int \frac {e^{-\frac {5}{-3+e^2}} \left (-48 x^2-4 \log (3)+e^{\frac {5}{-3+e^2}} \left (-16 x^6-8 x^4 \log (3)-x^2 \log ^2(3)\right )\right )}{16 x^6+8 x^4 \log (3)+x^2 \log ^2(3)} \, dx=\frac {4\,{\mathrm {e}}^{-\frac {5}{{\mathrm {e}}^2-3}}}{x\,\left (4\,x^2+\ln \left (3\right )\right )}-x \]