3.5.66 \(\int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+21 x^2+e^{142/25} x^2+4 x^3+4 x^4+e^{71/25} (10 x+2 x^2+4 x^3)+(10+2 x+2 e^{71/25} x+4 x^2) \log (\log (5))+\log ^2(\log (5))} \, dx\)

Optimal. Leaf size=23 \[ \frac {3}{x+\frac {5+x+2 x^2+\log (\log (5))}{e^{71/25}}} \]

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Rubi [A]  time = 0.10, antiderivative size = 28, normalized size of antiderivative = 1.22, number of steps used = 4, number of rules used = 4, integrand size = 98, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.041, Rules used = {6, 1680, 12, 261} \begin {gather*} \frac {3 e^{71/25}}{2 x^2+\left (1+e^{71/25}\right ) x+5+\log (\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-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]

[Out]

(3*E^(71/25))/(5 + (1 + E^(71/25))*x + 2*x^2 + Log[Log[5]])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 261

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a + b*x^n)^(p + 1)/(b*n*(p + 1)), x] /; FreeQ
[{a, b, m, n, p}, x] && EqQ[m, n - 1] && NeQ[p, -1]

Rule 1680

Int[(Pq_)*(Q4_)^(p_), x_Symbol] :> With[{a = Coeff[Q4, x, 0], b = Coeff[Q4, x, 1], c = Coeff[Q4, x, 2], d = Co
eff[Q4, x, 3], e = Coeff[Q4, x, 4]}, Subst[Int[SimplifyIntegrand[(Pq /. x -> -(d/(4*e)) + x)*(a + d^4/(256*e^3
) - (b*d)/(8*e) + (c - (3*d^2)/(8*e))*x^2 + e*x^4)^p, x], x], x, d/(4*e) + x] /; EqQ[d^3 - 4*c*d*e + 8*b*e^2,
0] && NeQ[d, 0]] /; FreeQ[p, x] && PolyQ[Pq, x] && PolyQ[Q4, x, 4] &&  !IGtQ[p, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3 e^{142/25}+e^{71/25} (-3-12 x)}{25+10 x+\left (21+e^{142/25}\right ) 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\\ &=\operatorname {Subst}\left (\int -\frac {768 e^{71/25} x}{\left (39-2 e^{71/25}-e^{142/25}+16 x^2+8 \log (\log (5))\right )^2} \, dx,x,\frac {1}{16} \left (4+4 e^{71/25}\right )+x\right )\\ &=-\left (\left (768 e^{71/25}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (39-2 e^{71/25}-e^{142/25}+16 x^2+8 \log (\log (5))\right )^2} \, dx,x,\frac {1}{16} \left (4+4 e^{71/25}\right )+x\right )\right )\\ &=\frac {3 e^{71/25}}{5+\left (1+e^{71/25}\right ) x+2 x^2+\log (\log (5))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 27, normalized size = 1.17 \begin {gather*} \frac {3 e^{71/25}}{5+x+e^{71/25} x+2 x^2+\log (\log (5))} \end {gather*}

Antiderivative was successfully verified.

[In]

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]

[Out]

(3*E^(71/25))/(5 + x + E^(71/25)*x + 2*x^2 + Log[Log[5]])

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fricas [A]  time = 0.75, size = 21, normalized size = 0.91 \begin {gather*} \frac {3 \, e^{\frac {71}{25}}}{2 \, x^{2} + x e^{\frac {71}{25}} + x + \log \left (\log \relax (5)\right ) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-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, algorithm="fricas")

[Out]

3*e^(71/25)/(2*x^2 + x*e^(71/25) + x + log(log(5)) + 5)

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giac [A]  time = 0.37, size = 19, normalized size = 0.83 \begin {gather*} -2.32272982294142083457 \times 10^{6} \, \log \left (x + 8.74478923391886229064\right ) + 2.32272935514670354111 \times 10^{6} \, \log \left (x + 8.74478792300523420934\right ) + 1.32459894481542 \times 10^{15} \, \log \left (x + 0.313094190110780\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-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, algorithm="giac")

[Out]

-2.32272982294142083457e6*log(x + 8.74478923391886229064) + 2.32272935514670354111e6*log(x + 8.744787923005234
20934) + 1.32459894481542e15*log(x + 0.313094190110780)

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maple [A]  time = 0.24, size = 22, normalized size = 0.96

method	result	size
gosper	\(\frac {3 \,{\mathrm e}^{\frac {71}{25}}}{x \,{\mathrm e}^{\frac {71}{25}}+2 x^{2}+\ln \left (\ln \relax (5)\right )+x +5}\)	\(22\)
norman	\(\frac {3 \,{\mathrm e}^{\frac {71}{25}}}{x \,{\mathrm e}^{\frac {71}{25}}+2 x^{2}+\ln \left (\ln \relax (5)\right )+x +5}\)	\(22\)
risch	\(\frac {3 \,{\mathrm e}^{\frac {71}{25}}}{x \,{\mathrm e}^{\frac {71}{25}}+2 x^{2}+\ln \left (\ln \relax (5)\right )+x +5}\)	\(22\)
default	\(\frac {3 \,{\mathrm e}^{\frac {71}{25}} \left (\munderset {\textit {\_R} =\RootOf \left (4 \textit {\_Z}^{4}+\left (4 \,{\mathrm e}^{\frac {71}{25}}+4\right ) \textit {\_Z}^{3}+\left ({\mathrm e}^{\frac {142}{25}}+2 \,{\mathrm e}^{\frac {71}{25}}+4 \ln \left (\ln \relax (5)\right )+21\right ) \textit {\_Z}^{2}+\left (2 \ln \left (\ln \relax (5)\right ) {\mathrm e}^{\frac {71}{25}}+10 \,{\mathrm e}^{\frac {71}{25}}+2 \ln \left (\ln \relax (5)\right )+10\right ) \textit {\_Z} +25+\ln \left (\ln \relax (5)\right )^{2}+10 \ln \left (\ln \relax (5)\right )\right )}{\sum }\frac {\left (-{\mathrm e}^{\frac {71}{25}}-4 \textit {\_R} -1\right ) \ln \left (x -\textit {\_R} \right )}{5+\textit {\_R} \,{\mathrm e}^{\frac {142}{25}}+6 \,{\mathrm e}^{\frac {71}{25}} \textit {\_R}^{2}+8 \textit {\_R}^{3}+\ln \left (\ln \relax (5)\right ) {\mathrm e}^{\frac {71}{25}}+2 \textit {\_R} \,{\mathrm e}^{\frac {71}{25}}+4 \textit {\_R} \ln \left (\ln \relax (5)\right )+6 \textit {\_R}^{2}+5 \,{\mathrm e}^{\frac {71}{25}}+\ln \left (\ln \relax (5)\right )+21 \textit {\_R}}\right )}{2}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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)

[Out]

3*exp(71/25)/(x*exp(71/25)+2*x^2+ln(ln(5))+x+5)

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maxima [A]  time = 0.36, size = 22, normalized size = 0.96 \begin {gather*} \frac {3 \, e^{\frac {71}{25}}}{2 \, x^{2} + x {\left (e^{\frac {71}{25}} + 1\right )} + \log \left (\log \relax (5)\right ) + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-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, algorithm="maxima")

[Out]

3*e^(71/25)/(2*x^2 + x*(e^(71/25) + 1) + log(log(5)) + 5)

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mupad [B]  time = 0.19, size = 22, normalized size = 0.96 \begin {gather*} \frac {3\,{\mathrm {e}}^{71/25}}{2\,x^2+\left ({\mathrm {e}}^{71/25}+1\right )\,x+\ln \left (\ln \relax (5)\right )+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*exp(142/25) + exp(71/25)*(12*x + 3))/(10*x + log(log(5))^2 + exp(71/25)*(10*x + 2*x^2 + 4*x^3) + x^2*e
xp(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)

[Out]

(3*exp(71/25))/(log(log(5)) + x*(exp(71/25) + 1) + 2*x^2 + 5)

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sympy [A]  time = 0.71, size = 26, normalized size = 1.13 \begin {gather*} \frac {3 e^{\frac {71}{25}}}{2 x^{2} + x \left (1 + e^{\frac {71}{25}}\right ) + \log {\left (\log {\relax (5 )} \right )} + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-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)

[Out]

3*exp(71/25)/(2*x**2 + x*(1 + exp(71/25)) + log(log(5)) + 5)

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