Integrand size = 99, antiderivative size = 28 \[ \int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx=5-\frac {3}{-x^2+3 \left (e^{\frac {1}{e^5 x}}-\log (5)\right )} \] Output:
5-3/(3*exp(1/x/exp(5))-3*ln(5)-x^2)
Time = 0.04 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx=\frac {3}{-3 e^{\frac {1}{e^5 x}}+x^2+\log (125)} \] Input:
Integrate[(-9*E^(1/(E^5*x)) - 6*E^5*x^3)/(9*E^(5 + 2/(E^5*x))*x^2 + E^5*x^ 6 + 6*E^5*x^4*Log[5] + 9*E^5*x^2*Log[5]^2 + E^(1/(E^5*x))*(-6*E^5*x^4 - 18 *E^5*x^2*Log[5])),x]
Output:
3/(-3*E^(1/(E^5*x)) + x^2 + Log[125])
Time = 0.59 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {7239, 27, 25, 7237}
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.
\(\displaystyle \int \frac {-6 e^5 x^3-9 e^{\frac {1}{e^5 x}}}{e^5 x^6+6 e^5 x^4 \log (5)+9 e^{\frac {2}{e^5 x}+5} x^2+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx\) |
\(\Big \downarrow \) 7239 |
\(\displaystyle \int \frac {3 \left (-2 e^5 x^3-3 e^{\frac {1}{e^5 x}}\right )}{e^5 x^2 \left (-x^2+3 e^{\frac {1}{e^5 x}}-\log (125)\right )^2}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \int -\frac {2 e^5 x^3+3 e^{\frac {1}{e^5 x}}}{x^2 \left (-x^2+3 e^{\frac {1}{e^5 x}}-\log (125)\right )^2}dx}{e^5}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {3 \int \frac {2 e^5 x^3+3 e^{\frac {1}{e^5 x}}}{x^2 \left (-x^2+3 e^{\frac {1}{e^5 x}}-\log (125)\right )^2}dx}{e^5}\) |
\(\Big \downarrow \) 7237 |
\(\displaystyle -\frac {3}{-x^2+3 e^{\frac {1}{e^5 x}}-\log (125)}\) |
Input:
Int[(-9*E^(1/(E^5*x)) - 6*E^5*x^3)/(9*E^(5 + 2/(E^5*x))*x^2 + E^5*x^6 + 6* E^5*x^4*Log[5] + 9*E^5*x^2*Log[5]^2 + E^(1/(E^5*x))*(-6*E^5*x^4 - 18*E^5*x ^2*Log[5])),x]
Output:
-3/(3*E^(1/(E^5*x)) - x^2 - Log[125])
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Si mp[q*(y^(m + 1)/(m + 1)), x] /; !FalseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]
Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; Simpl erIntegrandQ[v, u, x]]
Time = 2.38 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {3}{x^{2}+3 \ln \left (5\right )-3 \,{\mathrm e}^{\frac {{\mathrm e}^{-5}}{x}}}\) | \(22\) |
norman | \(\frac {3}{x^{2}+3 \ln \left (5\right )-3 \,{\mathrm e}^{\frac {{\mathrm e}^{-5}}{x}}}\) | \(24\) |
parallelrisch | \(\frac {3}{x^{2}+3 \ln \left (5\right )-3 \,{\mathrm e}^{\frac {{\mathrm e}^{-5}}{x}}}\) | \(24\) |
Input:
int((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^2+(-18 *x^2*exp(5)*ln(5)-6*x^4*exp(5))*exp(1/x/exp(5))+9*x^2*exp(5)*ln(5)^2+6*x^4 *exp(5)*ln(5)+x^6*exp(5)),x,method=_RETURNVERBOSE)
Output:
3/(x^2+3*ln(5)-3*exp(1/x*exp(-5)))
Time = 0.09 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx=\frac {3}{x^{2} - 3 \, e^{\left (\frac {e^{\left (-5\right )}}{x}\right )} + 3 \, \log \left (5\right )} \] Input:
integrate((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^ 2+(-18*x^2*exp(5)*log(5)-6*x^4*exp(5))*exp(1/x/exp(5))+9*x^2*exp(5)*log(5) ^2+6*x^4*exp(5)*log(5)+x^6*exp(5)),x, algorithm="fricas")
Output:
3/(x^2 - 3*e^(e^(-5)/x) + 3*log(5))
Time = 0.07 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.71 \[ \int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx=- \frac {3}{- x^{2} + 3 e^{\frac {1}{x e^{5}}} - 3 \log {\left (5 \right )}} \] Input:
integrate((-9*exp(1/x/exp(5))-6*x**3*exp(5))/(9*x**2*exp(5)*exp(1/x/exp(5) )**2+(-18*x**2*exp(5)*ln(5)-6*x**4*exp(5))*exp(1/x/exp(5))+9*x**2*exp(5)*l n(5)**2+6*x**4*exp(5)*ln(5)+x**6*exp(5)),x)
Output:
-3/(-x**2 + 3*exp(exp(-5)/x) - 3*log(5))
Time = 0.16 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.75 \[ \int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx=\frac {3}{x^{2} - 3 \, e^{\left (\frac {e^{\left (-5\right )}}{x}\right )} + 3 \, \log \left (5\right )} \] Input:
integrate((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^ 2+(-18*x^2*exp(5)*log(5)-6*x^4*exp(5))*exp(1/x/exp(5))+9*x^2*exp(5)*log(5) ^2+6*x^4*exp(5)*log(5)+x^6*exp(5)),x, algorithm="maxima")
Output:
3/(x^2 - 3*e^(e^(-5)/x) + 3*log(5))
Time = 0.15 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx=-\frac {3}{x^{2} {\left (\frac {3 \, e^{\left (\frac {e^{\left (-5\right )}}{x}\right )}}{x^{2}} - \frac {3 \, \log \left (5\right )}{x^{2}} - 1\right )}} \] Input:
integrate((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^ 2+(-18*x^2*exp(5)*log(5)-6*x^4*exp(5))*exp(1/x/exp(5))+9*x^2*exp(5)*log(5) ^2+6*x^4*exp(5)*log(5)+x^6*exp(5)),x, algorithm="giac")
Output:
-3/(x^2*(3*e^(e^(-5)/x)/x^2 - 3*log(5)/x^2 - 1))
Timed out. \[ \int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx=\int -\frac {9\,{\mathrm {e}}^{\frac {{\mathrm {e}}^{-5}}{x}}+6\,x^3\,{\mathrm {e}}^5}{x^6\,{\mathrm {e}}^5-{\mathrm {e}}^{\frac {{\mathrm {e}}^{-5}}{x}}\,\left (6\,{\mathrm {e}}^5\,x^4+18\,{\mathrm {e}}^5\,\ln \left (5\right )\,x^2\right )+6\,x^4\,{\mathrm {e}}^5\,\ln \left (5\right )+9\,x^2\,{\mathrm {e}}^5\,{\ln \left (5\right )}^2+9\,x^2\,{\mathrm {e}}^{\frac {2\,{\mathrm {e}}^{-5}}{x}}\,{\mathrm {e}}^5} \,d x \] Input:
int(-(9*exp(exp(-5)/x) + 6*x^3*exp(5))/(x^6*exp(5) - exp(exp(-5)/x)*(6*x^4 *exp(5) + 18*x^2*exp(5)*log(5)) + 6*x^4*exp(5)*log(5) + 9*x^2*exp(5)*log(5 )^2 + 9*x^2*exp((2*exp(-5))/x)*exp(5)),x)
Output:
int(-(9*exp(exp(-5)/x) + 6*x^3*exp(5))/(x^6*exp(5) - exp(exp(-5)/x)*(6*x^4 *exp(5) + 18*x^2*exp(5)*log(5)) + 6*x^4*exp(5)*log(5) + 9*x^2*exp(5)*log(5 )^2 + 9*x^2*exp((2*exp(-5))/x)*exp(5)), x)
Time = 0.19 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.89 \[ \int \frac {-9 e^{\frac {1}{e^5 x}}-6 e^5 x^3}{9 e^{5+\frac {2}{e^5 x}} x^2+e^5 x^6+6 e^5 x^4 \log (5)+9 e^5 x^2 \log ^2(5)+e^{\frac {1}{e^5 x}} \left (-6 e^5 x^4-18 e^5 x^2 \log (5)\right )} \, dx=-\frac {3}{3 e^{\frac {1}{e^{5} x}}-3 \,\mathrm {log}\left (5\right )-x^{2}} \] Input:
int((-9*exp(1/x/exp(5))-6*x^3*exp(5))/(9*x^2*exp(5)*exp(1/x/exp(5))^2+(-18 *x^2*exp(5)*log(5)-6*x^4*exp(5))*exp(1/x/exp(5))+9*x^2*exp(5)*log(5)^2+6*x ^4*exp(5)*log(5)+x^6*exp(5)),x)
Output:
( - 3)/(3*e**(1/(e**5*x)) - 3*log(5) - x**2)