3.44.48 \(\int \frac {4-4 x-8 x^2+4 x^3+16 x^4+16 x^5+(4+16 x-4 x^2-24 x^3-32 x^4) \log (\log (5))+(-8+8 x^2+16 x^3) \log ^2(\log (5))+(8 x^2+32 x^3+32 x^4+(-8 x-48 x^2-64 x^3) \log (\log (5))+(16 x+32 x^2) \log ^2(\log (5))) \log (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))})+(4 x+16 x^2+16 x^3+(-4-24 x-32 x^2) \log (\log (5))+(8+16 x) \log ^2(\log (5))) \log ^2(\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))})}{x^3+2 x^4+(-x^2-4 x^3) \log (\log (5))+2 x^2 \log ^2(\log (5))+(2 x^2+4 x^3+(-2 x-8 x^2) \log (\log (5))+4 x \log ^2(\log (5))) \log (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))})+(x+2 x^2+(-1-4 x) \log (\log (5))+2 \log ^2(\log (5))) \log ^2(\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))})} \, dx\)

Optimal. Leaf size=23 \[ 4 \left (x+x^2+\frac {1}{x+\log \left (-2+\frac {1}{-x+\log (\log (5))}\right )}\right ) \]

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Rubi [A]  time = 1.62, antiderivative size = 37, normalized size of antiderivative = 1.61, number of steps used = 5, number of rules used = 4, integrand size = 339, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.012, Rules used = {6688, 12, 6728, 6686} \begin {gather*} 4 x^2+4 x+\frac {4}{x+\log \left (-\frac {2 x+1-2 \log (\log (5))}{x-\log (\log (5))}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4 - 4*x - 8*x^2 + 4*x^3 + 16*x^4 + 16*x^5 + (4 + 16*x - 4*x^2 - 24*x^3 - 32*x^4)*Log[Log[5]] + (-8 + 8*x^
2 + 16*x^3)*Log[Log[5]]^2 + (8*x^2 + 32*x^3 + 32*x^4 + (-8*x - 48*x^2 - 64*x^3)*Log[Log[5]] + (16*x + 32*x^2)*
Log[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])] + (4*x + 16*x^2 + 16*x^3 + (-4 - 24*x - 32*x^
2)*Log[Log[5]] + (8 + 16*x)*Log[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])]^2)/(x^3 + 2*x^4 +
 (-x^2 - 4*x^3)*Log[Log[5]] + 2*x^2*Log[Log[5]]^2 + (2*x^2 + 4*x^3 + (-2*x - 8*x^2)*Log[Log[5]] + 4*x*Log[Log[
5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])] + (x + 2*x^2 + (-1 - 4*x)*Log[Log[5]] + 2*Log[Log[5]]
^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])]^2),x]

[Out]

4*x + 4*x^2 + 4/(x + Log[-((1 + 2*x - 2*Log[Log[5]])/(x - Log[Log[5]]))])

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 \left (1+4 x^5+x^4 (4-8 \log (\log (5)))+(1-2 \log (\log (5))) \log (\log (5))+x (-1+4 \log (\log (5)))+x^2 \left (-2-\log (\log (5))+2 \log ^2(\log (5))\right )+x^3 \left (1-6 \log (\log (5))+4 \log ^2(\log (5))\right )+2 x (1+2 x) \left (x+2 x^2-4 x \log (\log (5))+\log (\log (5)) (-1+2 \log (\log (5)))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+(1+2 x) \left (x+2 x^2-4 x \log (\log (5))+\log (\log (5)) (-1+2 \log (\log (5)))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )\right )}{\left (2 x^2+x (1-4 \log (\log (5)))-(1-2 \log (\log (5))) \log (\log (5))\right ) \left (x+\log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )\right )^2} \, dx\\ &=4 \int \frac {1+4 x^5+x^4 (4-8 \log (\log (5)))+(1-2 \log (\log (5))) \log (\log (5))+x (-1+4 \log (\log (5)))+x^2 \left (-2-\log (\log (5))+2 \log ^2(\log (5))\right )+x^3 \left (1-6 \log (\log (5))+4 \log ^2(\log (5))\right )+2 x (1+2 x) \left (x+2 x^2-4 x \log (\log (5))+\log (\log (5)) (-1+2 \log (\log (5)))\right ) \log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )+(1+2 x) \left (x+2 x^2-4 x \log (\log (5))+\log (\log (5)) (-1+2 \log (\log (5)))\right ) \log ^2\left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}{\left (2 x^2+x (1-4 \log (\log (5)))-(1-2 \log (\log (5))) \log (\log (5))\right ) \left (x+\log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )\right )^2} \, dx\\ &=4 \int \left (1+2 x+\frac {1-2 x^2-x (1-4 \log (\log (5)))+\log (\log (5))-2 \log ^2(\log (5))}{(1+2 x-2 \log (\log (5))) (x-\log (\log (5))) \left (x+\log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )\right )^2}\right ) \, dx\\ &=4 x+4 x^2+4 \int \frac {1-2 x^2-x (1-4 \log (\log (5)))+\log (\log (5))-2 \log ^2(\log (5))}{(1+2 x-2 \log (\log (5))) (x-\log (\log (5))) \left (x+\log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )\right )^2} \, dx\\ &=4 x+4 x^2+\frac {4}{x+\log \left (-\frac {1+2 x-2 \log (\log (5))}{x-\log (\log (5))}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.14, size = 32, normalized size = 1.39 \begin {gather*} 4 \left (x+x^2+\frac {1}{x+\log \left (\frac {1+2 x-2 \log (\log (5))}{-x+\log (\log (5))}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4 - 4*x - 8*x^2 + 4*x^3 + 16*x^4 + 16*x^5 + (4 + 16*x - 4*x^2 - 24*x^3 - 32*x^4)*Log[Log[5]] + (-8
+ 8*x^2 + 16*x^3)*Log[Log[5]]^2 + (8*x^2 + 32*x^3 + 32*x^4 + (-8*x - 48*x^2 - 64*x^3)*Log[Log[5]] + (16*x + 32
*x^2)*Log[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])] + (4*x + 16*x^2 + 16*x^3 + (-4 - 24*x -
 32*x^2)*Log[Log[5]] + (8 + 16*x)*Log[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])]^2)/(x^3 + 2
*x^4 + (-x^2 - 4*x^3)*Log[Log[5]] + 2*x^2*Log[Log[5]]^2 + (2*x^2 + 4*x^3 + (-2*x - 8*x^2)*Log[Log[5]] + 4*x*Lo
g[Log[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])] + (x + 2*x^2 + (-1 - 4*x)*Log[Log[5]] + 2*Log[L
og[5]]^2)*Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])]^2),x]

[Out]

4*(x + x^2 + (x + Log[(1 + 2*x - 2*Log[Log[5]])/(-x + Log[Log[5]])])^(-1))

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fricas [B]  time = 0.59, size = 64, normalized size = 2.78 \begin {gather*} \frac {4 \, {\left (x^{3} + x^{2} + {\left (x^{2} + x\right )} \log \left (-\frac {2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1}{x - \log \left (\log \relax (5)\right )}\right ) + 1\right )}}{x + \log \left (-\frac {2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1}{x - \log \left (\log \relax (5)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+8)*log(log(5))^2+(-32*x^2-24*x-4)*log(log(5))+16*x^3+16*x^2+4*x)*log((-2*log(log(5))+2*x+1)/
(log(log(5))-x))^2+((32*x^2+16*x)*log(log(5))^2+(-64*x^3-48*x^2-8*x)*log(log(5))+32*x^4+32*x^3+8*x^2)*log((-2*
log(log(5))+2*x+1)/(log(log(5))-x))+(16*x^3+8*x^2-8)*log(log(5))^2+(-32*x^4-24*x^3-4*x^2+16*x+4)*log(log(5))+1
6*x^5+16*x^4+4*x^3-8*x^2-4*x+4)/((2*log(log(5))^2+(-4*x-1)*log(log(5))+2*x^2+x)*log((-2*log(log(5))+2*x+1)/(lo
g(log(5))-x))^2+(4*x*log(log(5))^2+(-8*x^2-2*x)*log(log(5))+4*x^3+2*x^2)*log((-2*log(log(5))+2*x+1)/(log(log(5
))-x))+2*x^2*log(log(5))^2+(-4*x^3-x^2)*log(log(5))+2*x^4+x^3),x, algorithm="fricas")

[Out]

4*(x^3 + x^2 + (x^2 + x)*log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5)))) + 1)/(x + log(-(2*x - 2*log(log(5))
 + 1)/(x - log(log(5)))))

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giac [B]  time = 1.09, size = 223, normalized size = 9.70 \begin {gather*} \frac {4 \, {\left (\frac {2 \, {\left (2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1\right )} \log \left (\log \relax (5)\right )}{x - \log \left (\log \relax (5)\right )} + \frac {2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1}{x - \log \left (\log \relax (5)\right )} - 4 \, \log \left (\log \relax (5)\right ) - 1\right )}}{\frac {{\left (2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1\right )}^{2}}{{\left (x - \log \left (\log \relax (5)\right )\right )}^{2}} - \frac {4 \, {\left (2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1\right )}}{x - \log \left (\log \relax (5)\right )} + 4} + \frac {4 \, {\left (\frac {2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1}{x - \log \left (\log \relax (5)\right )} - 2\right )}}{\frac {{\left (2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1\right )} \log \left (-\frac {2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1}{x - \log \left (\log \relax (5)\right )}\right )}{x - \log \left (\log \relax (5)\right )} + \frac {{\left (2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1\right )} \log \left (\log \relax (5)\right )}{x - \log \left (\log \relax (5)\right )} - 2 \, \log \left (-\frac {2 \, x - 2 \, \log \left (\log \relax (5)\right ) + 1}{x - \log \left (\log \relax (5)\right )}\right ) - 2 \, \log \left (\log \relax (5)\right ) + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+8)*log(log(5))^2+(-32*x^2-24*x-4)*log(log(5))+16*x^3+16*x^2+4*x)*log((-2*log(log(5))+2*x+1)/
(log(log(5))-x))^2+((32*x^2+16*x)*log(log(5))^2+(-64*x^3-48*x^2-8*x)*log(log(5))+32*x^4+32*x^3+8*x^2)*log((-2*
log(log(5))+2*x+1)/(log(log(5))-x))+(16*x^3+8*x^2-8)*log(log(5))^2+(-32*x^4-24*x^3-4*x^2+16*x+4)*log(log(5))+1
6*x^5+16*x^4+4*x^3-8*x^2-4*x+4)/((2*log(log(5))^2+(-4*x-1)*log(log(5))+2*x^2+x)*log((-2*log(log(5))+2*x+1)/(lo
g(log(5))-x))^2+(4*x*log(log(5))^2+(-8*x^2-2*x)*log(log(5))+4*x^3+2*x^2)*log((-2*log(log(5))+2*x+1)/(log(log(5
))-x))+2*x^2*log(log(5))^2+(-4*x^3-x^2)*log(log(5))+2*x^4+x^3),x, algorithm="giac")

[Out]

4*(2*(2*x - 2*log(log(5)) + 1)*log(log(5))/(x - log(log(5))) + (2*x - 2*log(log(5)) + 1)/(x - log(log(5))) - 4
*log(log(5)) - 1)/((2*x - 2*log(log(5)) + 1)^2/(x - log(log(5)))^2 - 4*(2*x - 2*log(log(5)) + 1)/(x - log(log(
5))) + 4) + 4*((2*x - 2*log(log(5)) + 1)/(x - log(log(5))) - 2)/((2*x - 2*log(log(5)) + 1)*log(-(2*x - 2*log(l
og(5)) + 1)/(x - log(log(5))))/(x - log(log(5))) + (2*x - 2*log(log(5)) + 1)*log(log(5))/(x - log(log(5))) - 2
*log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5)))) - 2*log(log(5)) + 1)

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maple [A]  time = 0.43, size = 37, normalized size = 1.61




method result size



risch \(4 x^{2}+4 x +\frac {4}{x +\ln \left (\frac {-2 \ln \left (\ln \relax (5)\right )+2 x +1}{\ln \left (\ln \relax (5)\right )-x}\right )}\) \(37\)
norman \(\frac {4+4 x \ln \left (\frac {-2 \ln \left (\ln \relax (5)\right )+2 x +1}{\ln \left (\ln \relax (5)\right )-x}\right )+4 x^{2}+4 x^{3}+4 \ln \left (\frac {-2 \ln \left (\ln \relax (5)\right )+2 x +1}{\ln \left (\ln \relax (5)\right )-x}\right ) x^{2}}{x +\ln \left (\frac {-2 \ln \left (\ln \relax (5)\right )+2 x +1}{\ln \left (\ln \relax (5)\right )-x}\right )}\) \(89\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((16*x+8)*ln(ln(5))^2+(-32*x^2-24*x-4)*ln(ln(5))+16*x^3+16*x^2+4*x)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x)
)^2+((32*x^2+16*x)*ln(ln(5))^2+(-64*x^3-48*x^2-8*x)*ln(ln(5))+32*x^4+32*x^3+8*x^2)*ln((-2*ln(ln(5))+2*x+1)/(ln
(ln(5))-x))+(16*x^3+8*x^2-8)*ln(ln(5))^2+(-32*x^4-24*x^3-4*x^2+16*x+4)*ln(ln(5))+16*x^5+16*x^4+4*x^3-8*x^2-4*x
+4)/((2*ln(ln(5))^2+(-4*x-1)*ln(ln(5))+2*x^2+x)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x))^2+(4*x*ln(ln(5))^2+(-8*
x^2-2*x)*ln(ln(5))+4*x^3+2*x^2)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x))+2*x^2*ln(ln(5))^2+(-4*x^3-x^2)*ln(ln(5)
)+2*x^4+x^3),x,method=_RETURNVERBOSE)

[Out]

4*x^2+4*x+4/(x+ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x)))

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maxima [B]  time = 0.54, size = 67, normalized size = 2.91 \begin {gather*} \frac {4 \, {\left (x^{3} + x^{2} - {\left (x^{2} + x\right )} \log \left (x - \log \left (\log \relax (5)\right )\right ) + {\left (x^{2} + x\right )} \log \left (-2 \, x + 2 \, \log \left (\log \relax (5)\right ) - 1\right ) + 1\right )}}{x - \log \left (x - \log \left (\log \relax (5)\right )\right ) + \log \left (-2 \, x + 2 \, \log \left (\log \relax (5)\right ) - 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+8)*log(log(5))^2+(-32*x^2-24*x-4)*log(log(5))+16*x^3+16*x^2+4*x)*log((-2*log(log(5))+2*x+1)/
(log(log(5))-x))^2+((32*x^2+16*x)*log(log(5))^2+(-64*x^3-48*x^2-8*x)*log(log(5))+32*x^4+32*x^3+8*x^2)*log((-2*
log(log(5))+2*x+1)/(log(log(5))-x))+(16*x^3+8*x^2-8)*log(log(5))^2+(-32*x^4-24*x^3-4*x^2+16*x+4)*log(log(5))+1
6*x^5+16*x^4+4*x^3-8*x^2-4*x+4)/((2*log(log(5))^2+(-4*x-1)*log(log(5))+2*x^2+x)*log((-2*log(log(5))+2*x+1)/(lo
g(log(5))-x))^2+(4*x*log(log(5))^2+(-8*x^2-2*x)*log(log(5))+4*x^3+2*x^2)*log((-2*log(log(5))+2*x+1)/(log(log(5
))-x))+2*x^2*log(log(5))^2+(-4*x^3-x^2)*log(log(5))+2*x^4+x^3),x, algorithm="maxima")

[Out]

4*(x^3 + x^2 - (x^2 + x)*log(x - log(log(5))) + (x^2 + x)*log(-2*x + 2*log(log(5)) - 1) + 1)/(x - log(x - log(
log(5))) + log(-2*x + 2*log(log(5)) - 1))

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mupad [B]  time = 62.18, size = 37, normalized size = 1.61 \begin {gather*} 4\,x+4\,x^2+\frac {4}{x+\ln \left (-\frac {2\,x-2\,\ln \left (\ln \relax (5)\right )+1}{x-\ln \left (\ln \relax (5)\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(log(5))^2*(8*x^2 + 16*x^3 - 8) - log(log(5))*(4*x^2 - 16*x + 24*x^3 + 32*x^4 - 4) - 4*x + log(-(2*x -
 2*log(log(5)) + 1)/(x - log(log(5))))^2*(4*x - log(log(5))*(24*x + 32*x^2 + 4) + log(log(5))^2*(16*x + 8) + 1
6*x^2 + 16*x^3) - 8*x^2 + 4*x^3 + 16*x^4 + 16*x^5 + log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5))))*(log(log
(5))^2*(16*x + 32*x^2) + 8*x^2 + 32*x^3 + 32*x^4 - log(log(5))*(8*x + 48*x^2 + 64*x^3)) + 4)/(2*x^2*log(log(5)
)^2 + log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5))))^2*(x + 2*log(log(5))^2 - log(log(5))*(4*x + 1) + 2*x^2
) + log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5))))*(4*x*log(log(5))^2 - log(log(5))*(2*x + 8*x^2) + 2*x^2 +
 4*x^3) - log(log(5))*(x^2 + 4*x^3) + x^3 + 2*x^4),x)

[Out]

4*x + 4*x^2 + 4/(x + log(-(2*x - 2*log(log(5)) + 1)/(x - log(log(5)))))

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sympy [A]  time = 0.42, size = 31, normalized size = 1.35 \begin {gather*} 4 x^{2} + 4 x + \frac {4}{x + \log {\left (\frac {2 x - 2 \log {\left (\log {\relax (5 )} \right )} + 1}{- x + \log {\left (\log {\relax (5 )} \right )}} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((16*x+8)*ln(ln(5))**2+(-32*x**2-24*x-4)*ln(ln(5))+16*x**3+16*x**2+4*x)*ln((-2*ln(ln(5))+2*x+1)/(ln
(ln(5))-x))**2+((32*x**2+16*x)*ln(ln(5))**2+(-64*x**3-48*x**2-8*x)*ln(ln(5))+32*x**4+32*x**3+8*x**2)*ln((-2*ln
(ln(5))+2*x+1)/(ln(ln(5))-x))+(16*x**3+8*x**2-8)*ln(ln(5))**2+(-32*x**4-24*x**3-4*x**2+16*x+4)*ln(ln(5))+16*x*
*5+16*x**4+4*x**3-8*x**2-4*x+4)/((2*ln(ln(5))**2+(-4*x-1)*ln(ln(5))+2*x**2+x)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5
))-x))**2+(4*x*ln(ln(5))**2+(-8*x**2-2*x)*ln(ln(5))+4*x**3+2*x**2)*ln((-2*ln(ln(5))+2*x+1)/(ln(ln(5))-x))+2*x*
*2*ln(ln(5))**2+(-4*x**3-x**2)*ln(ln(5))+2*x**4+x**3),x)

[Out]

4*x**2 + 4*x + 4/(x + log((2*x - 2*log(log(5)) + 1)/(-x + log(log(5)))))

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