Integrand size = 92, antiderivative size = 26 \[ \int \frac {e^{\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}} \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx=e^{x \left (-x+\frac {4}{x^4+\frac {1}{2} e^3 x \log (5)}\right )} \]
Time = 0.07 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.92 \[ \int \frac {e^{\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}} \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx=e^{-x^2+\frac {8}{2 x^3+e^3 \log (5)}} \]
Integrate[(E^((8 - 2*x^5 - E^3*x^2*Log[5])/(2*x^3 + E^3*Log[5]))*(-48*x^2 - 8*x^7 - 8*E^3*x^4*Log[5] - 2*E^6*x*Log[5]^2))/(4*x^6 + 4*E^3*x^3*Log[5] + E^6*Log[5]^2),x]
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.
\(\displaystyle \int \frac {\left (-8 x^7-8 e^3 x^4 \log (5)-48 x^2-2 e^6 x \log ^2(5)\right ) \exp \left (\frac {-2 x^5-e^3 x^2 \log (5)+8}{2 x^3+e^3 \log (5)}\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle 4 \int -\frac {2\ 5^{-\frac {e^3 x^2}{2 x^3+e^3 \log (5)}} e^{\frac {2 \left (4-x^5\right )}{2 x^3+e^3 \log (5)}} \left (4 x^7+4 e^3 \log (5) x^4+24 x^2+e^6 \log ^2(5) x\right )}{\left (4 x^3+e^3 \log (25)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -8 \int \frac {5^{-\frac {e^3 x^2}{2 x^3+e^3 \log (5)}} e^{\frac {2 \left (4-x^5\right )}{2 x^3+e^3 \log (5)}} \left (4 x^7+4 e^3 \log (5) x^4+24 x^2+e^6 \log ^2(5) x\right )}{\left (4 x^3+e^3 \log (25)\right )^2}dx\) |
\(\Big \downarrow \) 2029 |
\(\displaystyle -8 \int \frac {5^{-\frac {e^3 x^2}{2 x^3+e^3 \log (5)}} e^{\frac {2 \left (4-x^5\right )}{2 x^3+e^3 \log (5)}} x \left (4 x^6+4 e^3 \log (5) x^3+24 x+e^6 \log ^2(5)\right )}{\left (4 x^3+e^3 \log (25)\right )^2}dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -8 \int \left (\frac {24\ 5^{-\frac {e^3 x^2}{2 x^3+e^3 \log (5)}} e^{\frac {2 \left (4-x^5\right )}{2 x^3+e^3 \log (5)}} x^2}{\left (4 x^3+e^3 \log (25)\right )^2}+\frac {1}{4} 5^{-\frac {e^3 x^2}{2 x^3+e^3 \log (5)}} e^{\frac {2 \left (4-x^5\right )}{2 x^3+e^3 \log (5)}} x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -8 \left (\frac {1}{4} \int 5^{-\frac {e^3 x^2}{2 x^3+e^3 \log (5)}} e^{\frac {2 \left (4-x^5\right )}{2 x^3+e^3 \log (5)}} xdx+24 \int \frac {5^{-\frac {e^3 x^2}{2 x^3+e^3 \log (5)}} e^{\frac {2 \left (4-x^5\right )}{2 x^3+e^3 \log (5)}} x^2}{\left (4 x^3+e^3 \log (25)\right )^2}dx\right )\) |
Int[(E^((8 - 2*x^5 - E^3*x^2*Log[5])/(2*x^3 + E^3*Log[5]))*(-48*x^2 - 8*x^ 7 - 8*E^3*x^4*Log[5] - 2*E^6*x*Log[5]^2))/(4*x^6 + 4*E^3*x^3*Log[5] + E^6* Log[5]^2),x]
3.25.20.3.1 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[(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_.)*((d_.)*(x_)^(q_.) + (a_.)*(x_)^(r_.) + (b_.)*(x_)^(s_.) + (c_.)* (x_)^(t_.))^(p_.), x_Symbol] :> Int[x^(p*r)*(a + b*x^(s - r) + c*x^(t - r) + d*x^(q - r))^p*Fx, x] /; FreeQ[{a, b, c, d, r, s, t, q}, x] && IntegerQ[p ] && PosQ[s - r] && PosQ[t - r] && PosQ[q - r] && !(EqQ[p, 1] && EqQ[u, 1] )
Time = 0.59 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.23
method | result | size |
gosper | \({\mathrm e}^{-\frac {2 x^{5}+x^{2} {\mathrm e}^{3} \ln \left (5\right )-8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}\) | \(32\) |
risch | \({\mathrm e}^{-\frac {2 x^{5}+x^{2} {\mathrm e}^{3} \ln \left (5\right )-8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}\) | \(32\) |
parallelrisch | \({\mathrm e}^{-\frac {2 x^{5}+x^{2} {\mathrm e}^{3} \ln \left (5\right )-8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}\) | \(32\) |
norman | \(\frac {{\mathrm e}^{3} \ln \left (5\right ) {\mathrm e}^{\frac {-x^{2} {\mathrm e}^{3} \ln \left (5\right )-2 x^{5}+8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}+2 x^{3} {\mathrm e}^{\frac {-x^{2} {\mathrm e}^{3} \ln \left (5\right )-2 x^{5}+8}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}}}{{\mathrm e}^{3} \ln \left (5\right )+2 x^{3}}\) | \(88\) |
int((-2*x*exp(3)^2*ln(5)^2-8*x^4*exp(3)*ln(5)-8*x^7-48*x^2)*exp((-x^2*exp( 3)*ln(5)-2*x^5+8)/(exp(3)*ln(5)+2*x^3))/(exp(3)^2*ln(5)^2+4*x^3*exp(3)*ln( 5)+4*x^6),x,method=_RETURNVERBOSE)
Time = 0.27 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.19 \[ \int \frac {e^{\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}} \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx=e^{\left (-\frac {2 \, x^{5} + x^{2} e^{3} \log \left (5\right ) - 8}{2 \, x^{3} + e^{3} \log \left (5\right )}\right )} \]
integrate((-2*x*exp(3)^2*log(5)^2-8*x^4*exp(3)*log(5)-8*x^7-48*x^2)*exp((- x^2*exp(3)*log(5)-2*x^5+8)/(exp(3)*log(5)+2*x^3))/(exp(3)^2*log(5)^2+4*x^3 *exp(3)*log(5)+4*x^6),x, algorithm=\
Time = 0.16 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}} \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx=e^{\frac {- 2 x^{5} - x^{2} e^{3} \log {\left (5 \right )} + 8}{2 x^{3} + e^{3} \log {\left (5 \right )}}} \]
integrate((-2*x*exp(3)**2*ln(5)**2-8*x**4*exp(3)*ln(5)-8*x**7-48*x**2)*exp ((-x**2*exp(3)*ln(5)-2*x**5+8)/(exp(3)*ln(5)+2*x**3))/(exp(3)**2*ln(5)**2+ 4*x**3*exp(3)*ln(5)+4*x**6),x)
Time = 0.44 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.85 \[ \int \frac {e^{\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}} \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx=e^{\left (-x^{2} + \frac {8}{2 \, x^{3} + e^{3} \log \left (5\right )}\right )} \]
integrate((-2*x*exp(3)^2*log(5)^2-8*x^4*exp(3)*log(5)-8*x^7-48*x^2)*exp((- x^2*exp(3)*log(5)-2*x^5+8)/(exp(3)*log(5)+2*x^3))/(exp(3)^2*log(5)^2+4*x^3 *exp(3)*log(5)+4*x^6),x, algorithm=\
Leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (22) = 44\).
Time = 0.40 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.15 \[ \int \frac {e^{\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}} \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx=e^{\left (-\frac {2 \, x^{5} e^{3} \log \left (5\right ) + x^{2} e^{6} \log \left (5\right )^{2} + 16 \, x^{3}}{2 \, x^{3} e^{3} \log \left (5\right ) + e^{6} \log \left (5\right )^{2}} + \frac {8 \, e^{\left (-3\right )}}{\log \left (5\right )}\right )} \]
integrate((-2*x*exp(3)^2*log(5)^2-8*x^4*exp(3)*log(5)-8*x^7-48*x^2)*exp((- x^2*exp(3)*log(5)-2*x^5+8)/(exp(3)*log(5)+2*x^3))/(exp(3)^2*log(5)^2+4*x^3 *exp(3)*log(5)+4*x^6),x, algorithm=\
e^(-(2*x^5*e^3*log(5) + x^2*e^6*log(5)^2 + 16*x^3)/(2*x^3*e^3*log(5) + e^6 *log(5)^2) + 8*e^(-3)/log(5))
Time = 9.75 (sec) , antiderivative size = 59, normalized size of antiderivative = 2.27 \[ \int \frac {e^{\frac {8-2 x^5-e^3 x^2 \log (5)}{2 x^3+e^3 \log (5)}} \left (-48 x^2-8 x^7-8 e^3 x^4 \log (5)-2 e^6 x \log ^2(5)\right )}{4 x^6+4 e^3 x^3 \log (5)+e^6 \log ^2(5)} \, dx=\frac {{\mathrm {e}}^{-\frac {2\,x^5}{2\,x^3+{\mathrm {e}}^3\,\ln \left (5\right )}}\,{\mathrm {e}}^{\frac {8}{2\,x^3+{\mathrm {e}}^3\,\ln \left (5\right )}}}{5^{\frac {x^2\,{\mathrm {e}}^3}{2\,x^3+{\mathrm {e}}^3\,\ln \left (5\right )}}} \]
int(-(exp(-(2*x^5 + x^2*exp(3)*log(5) - 8)/(exp(3)*log(5) + 2*x^3))*(48*x^ 2 + 8*x^7 + 8*x^4*exp(3)*log(5) + 2*x*exp(6)*log(5)^2))/(exp(6)*log(5)^2 + 4*x^6 + 4*x^3*exp(3)*log(5)),x)