Integrand size = 98, antiderivative size = 23 \[ \int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} \left (10 x+2 x^2+4 x^3\right )+\left (10+2 x+2 e^{71/25} x+4 x^2\right ) \log (\log (5))+\log ^2(\log (5))} \, dx=\frac {3}{x+\frac {5+x+2 x^2+\log (\log (5))}{e^{71/25}}} \] Output:
3/((2*x^2+ln(ln(5))+5+x)*exp(-71/25)+x)
Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.17 \[ \int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} \left (10 x+2 x^2+4 x^3\right )+\left (10+2 x+2 e^{71/25} x+4 x^2\right ) \log (\log (5))+\log ^2(\log (5))} \, dx=\frac {3 e^{71/25}}{5+x+e^{71/25} x+2 x^2+\log (\log (5))} \] Input:
Integrate[(-3*E^(142/25) + E^(71/25)*(-3 - 12*x))/(25 + 10*x + 21*x^2 + E^ (142/25)*x^2 + 4*x^3 + 4*x^4 + E^(71/25)*(10*x + 2*x^2 + 4*x^3) + (10 + 2* x + 2*E^(71/25)*x + 4*x^2)*Log[Log[5]] + Log[Log[5]]^2),x]
Output:
(3*E^(71/25))/(5 + x + E^(71/25)*x + 2*x^2 + Log[Log[5]])
Leaf count is larger than twice the leaf count of optimal. \(49\) vs. \(2(23)=46\).
Time = 0.37 (sec) , antiderivative size = 49, normalized size of antiderivative = 2.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {6, 2459, 27, 27, 1380, 27, 241}
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.
\(\displaystyle \int \frac {e^{71/25} (-12 x-3)-3 e^{142/25}}{4 x^4+4 x^3+e^{142/25} x^2+21 x^2+\left (4 x^2+2 e^{71/25} x+2 x+10\right ) \log (\log (5))+e^{71/25} \left (4 x^3+2 x^2+10 x\right )+10 x+25+\log ^2(\log (5))} \, dx\) |
\(\Big \downarrow \) 6 |
\(\displaystyle \int \frac {e^{71/25} (-12 x-3)-3 e^{142/25}}{4 x^4+4 x^3+\left (21+e^{142/25}\right ) x^2+\left (4 x^2+2 e^{71/25} x+2 x+10\right ) \log (\log (5))+e^{71/25} \left (4 x^3+2 x^2+10 x\right )+10 x+25+\log ^2(\log (5))}dx\) |
\(\Big \downarrow \) 2459 |
\(\displaystyle \int -\frac {12 e^{71/25} \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )}{4 \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^4+\frac {1}{2} \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^2 \left (39-2 e^{71/25}-e^{142/25}+8 \log (\log (5))\right )+\frac {1}{64} \left (39-2 e^{71/25}-e^{142/25}+8 \log (\log (5))\right )^2}d\left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -12 e^{71/25} \int \frac {64 \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )}{256 \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^4+32 \left (39-2 e^{71/25}-e^{142/25}+8 \log (\log (5))\right ) \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^2+\left (-39+2 e^{71/25}+e^{142/25}-8 \log (\log (5))\right )^2}d\left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -768 e^{71/25} \int \frac {x+\frac {1}{16} \left (4+4 e^{71/25}\right )}{256 \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^4+32 \left (39-2 e^{71/25}-e^{142/25}+8 \log (\log (5))\right ) \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^2+\left (-39+2 e^{71/25}+e^{142/25}-8 \log (\log (5))\right )^2}d\left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )\) |
\(\Big \downarrow \) 1380 |
\(\displaystyle -196608 e^{71/25} \int \frac {x+\frac {1}{16} \left (4+4 e^{71/25}\right )}{256 \left (16 \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^2+8 \log (\log (5))-e^{142/25}-2 e^{71/25}+39\right )^2}d\left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -768 e^{71/25} \int \frac {x+\frac {1}{16} \left (4+4 e^{71/25}\right )}{\left (16 \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^2+8 \log (\log (5))-e^{142/25}-2 e^{71/25}+39\right )^2}d\left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {24 e^{71/25}}{16 \left (x+\frac {1}{16} \left (4+4 e^{71/25}\right )\right )^2-e^{142/25}-2 e^{71/25}+39+8 \log (\log (5))}\) |
Input:
Int[(-3*E^(142/25) + E^(71/25)*(-3 - 12*x))/(25 + 10*x + 21*x^2 + E^(142/2 5)*x^2 + 4*x^3 + 4*x^4 + E^(71/25)*(10*x + 2*x^2 + 4*x^3) + (10 + 2*x + 2* E^(71/25)*x + 4*x^2)*Log[Log[5]] + Log[Log[5]]^2),x]
Output:
(24*E^(71/25))/(39 - 2*E^(71/25) - E^(142/25) + 16*((4 + 4*E^(71/25))/16 + x)^2 + 8*Log[Log[5]])
Int[(u_.)*((v_.) + (a_.)*(Fx_) + (b_.)*(Fx_))^(p_.), x_Symbol] :> Int[u*(v + (a + b)*Fx)^p, x] /; FreeQ[{a, b}, x] && !FreeQ[Fx, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
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[(Pn_)^(p_.)*(Qx_), x_Symbol] :> With[{S = Coeff[Pn, x, Expon[Pn, x] - 1 ]/(Expon[Pn, x]*Coeff[Pn, x, Expon[Pn, x]])}, Subst[Int[ExpandToSum[Pn /. x -> x - S, x]^p*ExpandToSum[Qx /. x -> x - S, x], x], x, x + S] /; Binomial Q[Pn /. x -> x - S, x] || (IntegerQ[Expon[Pn, x]/2] && TrinomialQ[Pn /. x - > x - S, x])] /; FreeQ[p, x] && PolyQ[Pn, x] && GtQ[Expon[Pn, x], 2] && NeQ [Coeff[Pn, x, Expon[Pn, x] - 1], 0] && PolyQ[Qx, x] && !(MonomialQ[Qx, x] && IGtQ[p, 0])
Time = 0.39 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96
method | result | size |
gosper | \(\frac {3 \,{\mathrm e}^{\frac {71}{25}}}{x \,{\mathrm e}^{\frac {71}{25}}+2 x^{2}+\ln \left (\ln \left (5\right )\right )+x +5}\) | \(22\) |
norman | \(\frac {3 \,{\mathrm e}^{\frac {71}{25}}}{x \,{\mathrm e}^{\frac {71}{25}}+2 x^{2}+\ln \left (\ln \left (5\right )\right )+x +5}\) | \(22\) |
risch | \(\frac {3 \,{\mathrm e}^{\frac {71}{25}}}{x \,{\mathrm e}^{\frac {71}{25}}+2 x^{2}+\ln \left (\ln \left (5\right )\right )+x +5}\) | \(22\) |
parallelrisch | \(\frac {3 \,{\mathrm e}^{\frac {71}{25}}}{x \,{\mathrm e}^{\frac {71}{25}}+2 x^{2}+\ln \left (\ln \left (5\right )\right )+x +5}\) | \(22\) |
default | \(\frac {3 \,{\mathrm e}^{\frac {71}{25}} \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (4 \textit {\_Z}^{4}+\left (4 \,{\mathrm e}^{\frac {71}{25}}+4\right ) \textit {\_Z}^{3}+\left (2 \,{\mathrm e}^{\frac {71}{25}}+4 \ln \left (\ln \left (5\right )\right )+{\mathrm e}^{\frac {142}{25}}+21\right ) \textit {\_Z}^{2}+\left (2 \ln \left (\ln \left (5\right )\right ) {\mathrm e}^{\frac {71}{25}}+10 \,{\mathrm e}^{\frac {71}{25}}+2 \ln \left (\ln \left (5\right )\right )+10\right ) \textit {\_Z} +25+\ln \left (\ln \left (5\right )\right )^{2}+10 \ln \left (\ln \left (5\right )\right )\right )}{\sum }\frac {\left (-{\mathrm e}^{\frac {71}{25}}-4 \textit {\_R} -1\right ) \ln \left (x -\textit {\_R} \right )}{5+\textit {\_R} \,{\mathrm e}^{\frac {142}{25}}+6 \,{\mathrm e}^{\frac {71}{25}} \textit {\_R}^{2}+8 \textit {\_R}^{3}+\ln \left (\ln \left (5\right )\right ) {\mathrm e}^{\frac {71}{25}}+2 \textit {\_R} \,{\mathrm e}^{\frac {71}{25}}+4 \textit {\_R} \ln \left (\ln \left (5\right )\right )+6 \textit {\_R}^{2}+5 \,{\mathrm e}^{\frac {71}{25}}+\ln \left (\ln \left (5\right )\right )+21 \textit {\_R}}\right )}{2}\) | \(141\) |
Input:
int((-3*exp(71/25)^2+(-12*x-3)*exp(71/25))/(ln(ln(5))^2+(2*x*exp(71/25)+4* x^2+2*x+10)*ln(ln(5))+x^2*exp(71/25)^2+(4*x^3+2*x^2+10*x)*exp(71/25)+4*x^4 +4*x^3+21*x^2+10*x+25),x,method=_RETURNVERBOSE)
Output:
3*exp(71/25)/(x*exp(71/25)+2*x^2+ln(ln(5))+x+5)
Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} \left (10 x+2 x^2+4 x^3\right )+\left (10+2 x+2 e^{71/25} x+4 x^2\right ) \log (\log (5))+\log ^2(\log (5))} \, dx=\frac {3 \, e^{\frac {71}{25}}}{2 \, x^{2} + x e^{\frac {71}{25}} + x + \log \left (\log \left (5\right )\right ) + 5} \] Input:
integrate((-3*exp(71/25)^2+(-12*x-3)*exp(71/25))/(log(log(5))^2+(2*x*exp(7 1/25)+4*x^2+2*x+10)*log(log(5))+x^2*exp(71/25)^2+(4*x^3+2*x^2+10*x)*exp(71 /25)+4*x^4+4*x^3+21*x^2+10*x+25),x, algorithm="fricas")
Output:
3*e^(71/25)/(2*x^2 + x*e^(71/25) + x + log(log(5)) + 5)
Time = 0.42 (sec) , antiderivative size = 26, normalized size of antiderivative = 1.13 \[ \int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} \left (10 x+2 x^2+4 x^3\right )+\left (10+2 x+2 e^{71/25} x+4 x^2\right ) \log (\log (5))+\log ^2(\log (5))} \, dx=\frac {3 e^{\frac {71}{25}}}{2 x^{2} + x \left (1 + e^{\frac {71}{25}}\right ) + \log {\left (\log {\left (5 \right )} \right )} + 5} \] Input:
integrate((-3*exp(71/25)**2+(-12*x-3)*exp(71/25))/(ln(ln(5))**2+(2*x*exp(7 1/25)+4*x**2+2*x+10)*ln(ln(5))+x**2*exp(71/25)**2+(4*x**3+2*x**2+10*x)*exp (71/25)+4*x**4+4*x**3+21*x**2+10*x+25),x)
Output:
3*exp(71/25)/(2*x**2 + x*(1 + exp(71/25)) + log(log(5)) + 5)
Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} \left (10 x+2 x^2+4 x^3\right )+\left (10+2 x+2 e^{71/25} x+4 x^2\right ) \log (\log (5))+\log ^2(\log (5))} \, dx=\frac {3 \, e^{\frac {71}{25}}}{2 \, x^{2} + x {\left (e^{\frac {71}{25}} + 1\right )} + \log \left (\log \left (5\right )\right ) + 5} \] Input:
integrate((-3*exp(71/25)^2+(-12*x-3)*exp(71/25))/(log(log(5))^2+(2*x*exp(7 1/25)+4*x^2+2*x+10)*log(log(5))+x^2*exp(71/25)^2+(4*x^3+2*x^2+10*x)*exp(71 /25)+4*x^4+4*x^3+21*x^2+10*x+25),x, algorithm="maxima")
Output:
3*e^(71/25)/(2*x^2 + x*(e^(71/25) + 1) + log(log(5)) + 5)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.35 \[ \int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} \left (10 x+2 x^2+4 x^3\right )+\left (10+2 x+2 e^{71/25} x+4 x^2\right ) \log (\log (5))+\log ^2(\log (5))} \, dx=\frac {3 \, e^{\frac {142}{25}}}{x e^{\frac {142}{25}} + {\left (2 \, x^{2} + x\right )} e^{\frac {71}{25}} + e^{\frac {71}{25}} \log \left (\log \left (5\right )\right ) + 5 \, e^{\frac {71}{25}}} \] Input:
integrate((-3*exp(71/25)^2+(-12*x-3)*exp(71/25))/(log(log(5))^2+(2*x*exp(7 1/25)+4*x^2+2*x+10)*log(log(5))+x^2*exp(71/25)^2+(4*x^3+2*x^2+10*x)*exp(71 /25)+4*x^4+4*x^3+21*x^2+10*x+25),x, algorithm="giac")
Output:
3*e^(142/25)/(x*e^(142/25) + (2*x^2 + x)*e^(71/25) + e^(71/25)*log(log(5)) + 5*e^(71/25))
Time = 0.12 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.96 \[ \int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} \left (10 x+2 x^2+4 x^3\right )+\left (10+2 x+2 e^{71/25} x+4 x^2\right ) \log (\log (5))+\log ^2(\log (5))} \, dx=\frac {3\,{\mathrm {e}}^{71/25}}{2\,x^2+\left ({\mathrm {e}}^{71/25}+1\right )\,x+\ln \left (\ln \left (5\right )\right )+5} \] Input:
int(-(3*exp(142/25) + exp(71/25)*(12*x + 3))/(10*x + log(log(5))^2 + exp(7 1/25)*(10*x + 2*x^2 + 4*x^3) + x^2*exp(142/25) + 21*x^2 + 4*x^3 + 4*x^4 + log(log(5))*(2*x + 2*x*exp(71/25) + 4*x^2 + 10) + 25),x)
Output:
(3*exp(71/25))/(log(log(5)) + x*(exp(71/25) + 1) + 2*x^2 + 5)
Time = 0.26 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} \left (10 x+2 x^2+4 x^3\right )+\left (10+2 x+2 e^{71/25} x+4 x^2\right ) \log (\log (5))+\log ^2(\log (5))} \, dx=\frac {3 e^{\frac {71}{25}}}{e^{\frac {71}{25}} x +\mathrm {log}\left (\mathrm {log}\left (5\right )\right )+2 x^{2}+x +5} \] Input:
int((-3*exp(71/25)^2+(-12*x-3)*exp(71/25))/(log(log(5))^2+(2*x*exp(71/25)+ 4*x^2+2*x+10)*log(log(5))+x^2*exp(71/25)^2+(4*x^3+2*x^2+10*x)*exp(71/25)+4 *x^4+4*x^3+21*x^2+10*x+25),x)
Output:
(3*e**(21/25)*e**2)/(e**(21/25)*e**2*x + log(log(5)) + 2*x**2 + x + 5)