3.100.30 \(\int \frac {-100-40 x-20 \log (5)}{625+1000 x+600 x^2+160 x^3+16 x^4+(500+600 x+240 x^2+32 x^3) \log (5)+(150+120 x+24 x^2) \log ^2(5)+(20+8 x) \log ^3(5)+\log ^4(5)+(-50-40 x-8 x^2+(-20-8 x) \log (5)-2 \log ^2(5)) \log ^2(\log (4))+\log ^4(\log (4))} \, dx\)

Optimal. Leaf size=21 \[ \frac {5}{(5+2 x+\log (5))^2-\log ^2(\log (4))} \]

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Rubi [A]  time = 0.07, antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 3, number of rules used = 3, integrand size = 114, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.026, Rules used = {1680, 12, 261} \begin {gather*} \frac {5}{4 \left (x+\frac {1}{64} (160+32 \log (5))\right )^2-\log ^2(\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-100 - 40*x - 20*Log[5])/(625 + 1000*x + 600*x^2 + 160*x^3 + 16*x^4 + (500 + 600*x + 240*x^2 + 32*x^3)*Lo
g[5] + (150 + 120*x + 24*x^2)*Log[5]^2 + (20 + 8*x)*Log[5]^3 + Log[5]^4 + (-50 - 40*x - 8*x^2 + (-20 - 8*x)*Lo
g[5] - 2*Log[5]^2)*Log[Log[4]]^2 + Log[Log[4]]^4),x]

[Out]

5/(4*(x + (160 + 32*Log[5])/64)^2 - Log[Log[4]]^2)

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} &=\operatorname {Subst}\left (\int -\frac {40 x}{\left (4 x^2-\log ^2(\log (4))\right )^2} \, dx,x,x+\frac {1}{64} (160+32 \log (5))\right )\\ &=-\left (40 \operatorname {Subst}\left (\int \frac {x}{\left (4 x^2-\log ^2(\log (4))\right )^2} \, dx,x,x+\frac {1}{64} (160+32 \log (5))\right )\right )\\ &=\frac {5}{(5+2 x+\log (5))^2-\log ^2(\log (4))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 34, normalized size = 1.62 \begin {gather*} \frac {5}{25+20 x+4 x^2+10 \log (5)+4 x \log (5)+\log ^2(5)-\log ^2(\log (4))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-100 - 40*x - 20*Log[5])/(625 + 1000*x + 600*x^2 + 160*x^3 + 16*x^4 + (500 + 600*x + 240*x^2 + 32*x
^3)*Log[5] + (150 + 120*x + 24*x^2)*Log[5]^2 + (20 + 8*x)*Log[5]^3 + Log[5]^4 + (-50 - 40*x - 8*x^2 + (-20 - 8
*x)*Log[5] - 2*Log[5]^2)*Log[Log[4]]^2 + Log[Log[4]]^4),x]

[Out]

5/(25 + 20*x + 4*x^2 + 10*Log[5] + 4*x*Log[5] + Log[5]^2 - Log[Log[4]]^2)

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fricas [A]  time = 3.23, size = 36, normalized size = 1.71 \begin {gather*} \frac {5}{4 \, x^{2} + 2 \, {\left (2 \, x + 5\right )} \log \relax (5) + \log \relax (5)^{2} - \log \left (2 \, \log \relax (2)\right )^{2} + 20 \, x + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(5)-40*x-100)/(log(2*log(2))^4+(-2*log(5)^2+(-8*x-20)*log(5)-8*x^2-40*x-50)*log(2*log(2))^2+
log(5)^4+(8*x+20)*log(5)^3+(24*x^2+120*x+150)*log(5)^2+(32*x^3+240*x^2+600*x+500)*log(5)+16*x^4+160*x^3+600*x^
2+1000*x+625),x, algorithm="fricas")

[Out]

5/(4*x^2 + 2*(2*x + 5)*log(5) + log(5)^2 - log(2*log(2))^2 + 20*x + 25)

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giac [A]  time = 2.68, size = 36, normalized size = 1.71 \begin {gather*} \frac {5}{4 \, x^{2} + 4 \, x \log \relax (5) + \log \relax (5)^{2} - \log \left (2 \, \log \relax (2)\right )^{2} + 20 \, x + 10 \, \log \relax (5) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(5)-40*x-100)/(log(2*log(2))^4+(-2*log(5)^2+(-8*x-20)*log(5)-8*x^2-40*x-50)*log(2*log(2))^2+
log(5)^4+(8*x+20)*log(5)^3+(24*x^2+120*x+150)*log(5)^2+(32*x^3+240*x^2+600*x+500)*log(5)+16*x^4+160*x^3+600*x^
2+1000*x+625),x, algorithm="giac")

[Out]

5/(4*x^2 + 4*x*log(5) + log(5)^2 - log(2*log(2))^2 + 20*x + 10*log(5) + 25)

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maple [A]  time = 0.35, size = 37, normalized size = 1.76

method	result	size
gosper	\(-\frac {5}{\ln \left (2 \ln \relax (2)\right )^{2}-\ln \relax (5)^{2}-4 x \ln \relax (5)-4 x^{2}-10 \ln \relax (5)-20 x -25}\)	\(37\)
norman	\(-\frac {5}{\ln \left (2 \ln \relax (2)\right )^{2}-\ln \relax (5)^{2}-4 x \ln \relax (5)-4 x^{2}-10 \ln \relax (5)-20 x -25}\)	\(37\)
risch	\(-\frac {5}{\ln \relax (2)^{2}+2 \ln \relax (2) \ln \left (\ln \relax (2)\right )-\ln \relax (5)^{2}-4 x \ln \relax (5)+\ln \left (\ln \relax (2)\right )^{2}-4 x^{2}-10 \ln \relax (5)-20 x -25}\)	\(46\)
default	\(-\frac {5}{2 \ln \left (2 \ln \relax (2)\right ) \left (2 x +5+\ln \left (2 \ln \relax (2)\right )+\ln \relax (5)\right )}+\frac {5}{2 \ln \left (2 \ln \relax (2)\right ) \left (2 x +5-\ln \left (2 \ln \relax (2)\right )+\ln \relax (5)\right )}\)	\(50\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-20*ln(5)-40*x-100)/(ln(2*ln(2))^4+(-2*ln(5)^2+(-8*x-20)*ln(5)-8*x^2-40*x-50)*ln(2*ln(2))^2+ln(5)^4+(8*x+
20)*ln(5)^3+(24*x^2+120*x+150)*ln(5)^2+(32*x^3+240*x^2+600*x+500)*ln(5)+16*x^4+160*x^3+600*x^2+1000*x+625),x,m
ethod=_RETURNVERBOSE)

[Out]

-5/(ln(2*ln(2))^2-ln(5)^2-4*x*ln(5)-4*x^2-10*ln(5)-20*x-25)

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maxima [A]  time = 0.34, size = 35, normalized size = 1.67 \begin {gather*} \frac {5}{4 \, x^{2} + 4 \, x {\left (\log \relax (5) + 5\right )} + \log \relax (5)^{2} - \log \left (2 \, \log \relax (2)\right )^{2} + 10 \, \log \relax (5) + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*log(5)-40*x-100)/(log(2*log(2))^4+(-2*log(5)^2+(-8*x-20)*log(5)-8*x^2-40*x-50)*log(2*log(2))^2+
log(5)^4+(8*x+20)*log(5)^3+(24*x^2+120*x+150)*log(5)^2+(32*x^3+240*x^2+600*x+500)*log(5)+16*x^4+160*x^3+600*x^
2+1000*x+625),x, algorithm="maxima")

[Out]

5/(4*x^2 + 4*x*(log(5) + 5) + log(5)^2 - log(2*log(2))^2 + 10*log(5) + 25)

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mupad [B]  time = 0.32, size = 34, normalized size = 1.62 \begin {gather*} \frac {5}{4\,x^2+\left (4\,\ln \relax (5)+20\right )\,x+10\,\ln \relax (5)-{\ln \left (\ln \relax (4)\right )}^2+{\ln \relax (5)}^2+25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(40*x + 20*log(5) + 100)/(1000*x + log(5)^3*(8*x + 20) + log(5)*(600*x + 240*x^2 + 32*x^3 + 500) - log(2*
log(2))^2*(40*x + log(5)*(8*x + 20) + 2*log(5)^2 + 8*x^2 + 50) + log(2*log(2))^4 + log(5)^2*(120*x + 24*x^2 +
150) + log(5)^4 + 600*x^2 + 160*x^3 + 16*x^4 + 625),x)

[Out]

5/(10*log(5) + x*(4*log(5) + 20) - log(log(4))^2 + log(5)^2 + 4*x^2 + 25)

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sympy [B]  time = 1.39, size = 48, normalized size = 2.29 \begin {gather*} \frac {5}{4 x^{2} + x \left (4 \log {\relax (5 )} + 20\right ) - \log {\relax (2 )}^{2} - \log {\left (\log {\relax (2 )} \right )}^{2} - 2 \log {\relax (2 )} \log {\left (\log {\relax (2 )} \right )} + \log {\relax (5 )}^{2} + 10 \log {\relax (5 )} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-20*ln(5)-40*x-100)/(ln(2*ln(2))**4+(-2*ln(5)**2+(-8*x-20)*ln(5)-8*x**2-40*x-50)*ln(2*ln(2))**2+ln(
5)**4+(8*x+20)*ln(5)**3+(24*x**2+120*x+150)*ln(5)**2+(32*x**3+240*x**2+600*x+500)*ln(5)+16*x**4+160*x**3+600*x
**2+1000*x+625),x)

[Out]

5/(4*x**2 + x*(4*log(5) + 20) - log(2)**2 - log(log(2))**2 - 2*log(2)*log(log(2)) + log(5)**2 + 10*log(5) + 25
)

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