Integrand size = 111, antiderivative size = 32 \[ \int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{6 x^3+5 x^4} \, dx=e^{\frac {-\frac {2}{x}+\log (x)+(-x+\log (-x+3 (2+2 x)))^2}{x}} \] Output:
exp(((ln(5*x+6)-x)^2+ln(x)-2/x)/x)
Time = 0.09 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{6 x^3+5 x^4} \, dx=\frac {e^{\frac {-2+x^3+x \log ^2(6+5 x)}{x^2}} x^{\frac {1}{x}}}{(6+5 x)^2} \] Input:
Integrate[(E^((-2 + x^3 + x*Log[x] - 2*x^2*Log[6 + 5*x] + x*Log[6 + 5*x]^2 )/x^2)*(24 + 26*x + 5*x^2 - 4*x^3 + 5*x^4 + (-6*x - 5*x^2)*Log[x] + 10*x^2 *Log[6 + 5*x] + (-6*x - 5*x^2)*Log[6 + 5*x]^2))/(6*x^3 + 5*x^4),x]
Output:
(E^((-2 + x^3 + x*Log[6 + 5*x]^2)/x^2)*x^x^(-1))/(6 + 5*x)^2
Time = 2.44 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.19, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.018, Rules used = {2026, 7257}
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.
\(\displaystyle \int \frac {\left (5 x^4-4 x^3+5 x^2+\left (-5 x^2-6 x\right ) \log ^2(5 x+6)+10 x^2 \log (5 x+6)+\left (-5 x^2-6 x\right ) \log (x)+26 x+24\right ) \exp \left (\frac {x^3-2 x^2 \log (5 x+6)+x \log ^2(5 x+6)+x \log (x)-2}{x^2}\right )}{5 x^4+6 x^3} \, dx\) |
\(\Big \downarrow \) 2026 |
\(\displaystyle \int \frac {\left (5 x^4-4 x^3+5 x^2+\left (-5 x^2-6 x\right ) \log ^2(5 x+6)+10 x^2 \log (5 x+6)+\left (-5 x^2-6 x\right ) \log (x)+26 x+24\right ) \exp \left (\frac {x^3-2 x^2 \log (5 x+6)+x \log ^2(5 x+6)+x \log (x)-2}{x^2}\right )}{x^3 (5 x+6)}dx\) |
\(\Big \downarrow \) 7257 |
\(\displaystyle \frac {x^{\frac {1}{x}} e^{-\frac {-x^3-x \log ^2(5 x+6)+2}{x^2}}}{(5 x+6)^2}\) |
Input:
Int[(E^((-2 + x^3 + x*Log[x] - 2*x^2*Log[6 + 5*x] + x*Log[6 + 5*x]^2)/x^2) *(24 + 26*x + 5*x^2 - 4*x^3 + 5*x^4 + (-6*x - 5*x^2)*Log[x] + 10*x^2*Log[6 + 5*x] + (-6*x - 5*x^2)*Log[6 + 5*x]^2))/(6*x^3 + 5*x^4),x]
Output:
x^x^(-1)/(E^((2 - x^3 - x*Log[6 + 5*x]^2)/x^2)*(6 + 5*x)^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[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Sim p[q*(F^v/Log[F]), x] /; !FalseQ[q]] /; FreeQ[F, x]
Time = 15.79 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06
method | result | size |
risch | \(\frac {x^{\frac {1}{x}} {\mathrm e}^{\frac {x \ln \left (5 x +6\right )^{2}+x^{3}-2}{x^{2}}}}{\left (5 x +6\right )^{2}}\) | \(34\) |
parallelrisch | \({\mathrm e}^{\frac {x \ln \left (5 x +6\right )^{2}-2 x^{2} \ln \left (5 x +6\right )+x \ln \left (x \right )+x^{3}-2}{x^{2}}}\) | \(36\) |
Input:
int(((-5*x^2-6*x)*ln(5*x+6)^2+10*x^2*ln(5*x+6)+(-5*x^2-6*x)*ln(x)+5*x^4-4* x^3+5*x^2+26*x+24)*exp((x*ln(5*x+6)^2-2*x^2*ln(5*x+6)+x*ln(x)+x^3-2)/x^2)/ (5*x^4+6*x^3),x,method=_RETURNVERBOSE)
Output:
1/(5*x+6)^2*x^(1/x)*exp((x*ln(5*x+6)^2+x^3-2)/x^2)
Time = 0.07 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.09 \[ \int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{6 x^3+5 x^4} \, dx=e^{\left (\frac {x^{3} - 2 \, x^{2} \log \left (5 \, x + 6\right ) + x \log \left (5 \, x + 6\right )^{2} + x \log \left (x\right ) - 2}{x^{2}}\right )} \] Input:
integrate(((-5*x^2-6*x)*log(5*x+6)^2+10*x^2*log(5*x+6)+(-5*x^2-6*x)*log(x) +5*x^4-4*x^3+5*x^2+26*x+24)*exp((x*log(5*x+6)^2-2*x^2*log(5*x+6)+x*log(x)+ x^3-2)/x^2)/(5*x^4+6*x^3),x, algorithm="fricas")
Output:
e^((x^3 - 2*x^2*log(5*x + 6) + x*log(5*x + 6)^2 + x*log(x) - 2)/x^2)
Time = 0.51 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{6 x^3+5 x^4} \, dx=e^{\frac {x^{3} - 2 x^{2} \log {\left (5 x + 6 \right )} + x \log {\left (x \right )} + x \log {\left (5 x + 6 \right )}^{2} - 2}{x^{2}}} \] Input:
integrate(((-5*x**2-6*x)*ln(5*x+6)**2+10*x**2*ln(5*x+6)+(-5*x**2-6*x)*ln(x )+5*x**4-4*x**3+5*x**2+26*x+24)*exp((x*ln(5*x+6)**2-2*x**2*ln(5*x+6)+x*ln( x)+x**3-2)/x**2)/(5*x**4+6*x**3),x)
Output:
exp((x**3 - 2*x**2*log(5*x + 6) + x*log(x) + x*log(5*x + 6)**2 - 2)/x**2)
Time = 0.13 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.22 \[ \int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{6 x^3+5 x^4} \, dx=\frac {e^{\left (x + \frac {\log \left (5 \, x + 6\right )^{2}}{x} + \frac {\log \left (x\right )}{x} - \frac {2}{x^{2}}\right )}}{25 \, x^{2} + 60 \, x + 36} \] Input:
integrate(((-5*x^2-6*x)*log(5*x+6)^2+10*x^2*log(5*x+6)+(-5*x^2-6*x)*log(x) +5*x^4-4*x^3+5*x^2+26*x+24)*exp((x*log(5*x+6)^2-2*x^2*log(5*x+6)+x*log(x)+ x^3-2)/x^2)/(5*x^4+6*x^3),x, algorithm="maxima")
Output:
e^(x + log(5*x + 6)^2/x + log(x)/x - 2/x^2)/(25*x^2 + 60*x + 36)
Time = 0.18 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{6 x^3+5 x^4} \, dx=e^{\left (x + \frac {\log \left (5 \, x + 6\right )^{2}}{x} + \frac {\log \left (x\right )}{x} - \frac {2}{x^{2}} - 2 \, \log \left (5 \, x + 6\right )\right )} \] Input:
integrate(((-5*x^2-6*x)*log(5*x+6)^2+10*x^2*log(5*x+6)+(-5*x^2-6*x)*log(x) +5*x^4-4*x^3+5*x^2+26*x+24)*exp((x*log(5*x+6)^2-2*x^2*log(5*x+6)+x*log(x)+ x^3-2)/x^2)/(5*x^4+6*x^3),x, algorithm="giac")
Output:
e^(x + log(5*x + 6)^2/x + log(x)/x - 2/x^2 - 2*log(5*x + 6))
Time = 1.42 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.06 \[ \int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{6 x^3+5 x^4} \, dx=\frac {x^{1/x}\,{\mathrm {e}}^{-\frac {2}{x^2}}\,{\mathrm {e}}^{\frac {{\ln \left (5\,x+6\right )}^2}{x}}\,{\mathrm {e}}^x}{{\left (5\,x+6\right )}^2} \] Input:
int((exp((x*log(x) + x^3 + x*log(5*x + 6)^2 - 2*x^2*log(5*x + 6) - 2)/x^2) *(26*x - log(x)*(6*x + 5*x^2) - log(5*x + 6)^2*(6*x + 5*x^2) + 5*x^2 - 4*x ^3 + 5*x^4 + 10*x^2*log(5*x + 6) + 24))/(6*x^3 + 5*x^4),x)
Output:
(x^(1/x)*exp(-2/x^2)*exp(log(5*x + 6)^2/x)*exp(x))/(5*x + 6)^2
Time = 0.42 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int \frac {e^{\frac {-2+x^3+x \log (x)-2 x^2 \log (6+5 x)+x \log ^2(6+5 x)}{x^2}} \left (24+26 x+5 x^2-4 x^3+5 x^4+\left (-6 x-5 x^2\right ) \log (x)+10 x^2 \log (6+5 x)+\left (-6 x-5 x^2\right ) \log ^2(6+5 x)\right )}{6 x^3+5 x^4} \, dx=\frac {e^{\frac {\mathrm {log}\left (5 x +6\right )^{2}+\mathrm {log}\left (x \right )+x^{2}}{x}}}{e^{\frac {2}{x^{2}}} \left (25 x^{2}+60 x +36\right )} \] Input:
int(((-5*x^2-6*x)*log(5*x+6)^2+10*x^2*log(5*x+6)+(-5*x^2-6*x)*log(x)+5*x^4 -4*x^3+5*x^2+26*x+24)*exp((x*log(5*x+6)^2-2*x^2*log(5*x+6)+x*log(x)+x^3-2) /x^2)/(5*x^4+6*x^3),x)
Output:
e**((log(5*x + 6)**2 + log(x) + x**2)/x)/(e**(2/x**2)*(25*x**2 + 60*x + 36 ))