3.99.88 \(\int \frac {32 x^4-64 x^4 \log (3)+48 x^4 \log ^2(3)-16 x^4 \log ^3(3)+2 x^4 \log ^4(3)+(-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)) \log ^2(\log (3 e^3))+\log ^4(\log (3 e^3))}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+(-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)) \log ^2(\log (3 e^3))+\log ^4(\log (3 e^3))} \, dx\)

Optimal. Leaf size=31 \[ -8+x+\frac {x}{1-\frac {\log ^2\left (\log \left (3 e^3\right )\right )}{(2 x-x \log (3))^2}} \]

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Rubi [A]  time = 0.18, antiderivative size = 37, normalized size of antiderivative = 1.19, number of steps used = 13, number of rules used = 6, integrand size = 164, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.037, Rules used = {6, 1994, 28, 1814, 21, 8} \begin {gather*} \frac {x \log ^2(3+\log (3))}{x^2 (2-\log (3))^2-\log ^2(3+\log (3))}+2 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(32*x^4 - 64*x^4*Log[3] + 48*x^4*Log[3]^2 - 16*x^4*Log[3]^3 + 2*x^4*Log[3]^4 + (-20*x^2 + 20*x^2*Log[3] -
5*x^2*Log[3]^2)*Log[Log[3*E^3]]^2 + Log[Log[3*E^3]]^4)/(16*x^4 - 32*x^4*Log[3] + 24*x^4*Log[3]^2 - 8*x^4*Log[3
]^3 + x^4*Log[3]^4 + (-8*x^2 + 8*x^2*Log[3] - 2*x^2*Log[3]^2)*Log[Log[3*E^3]]^2 + Log[Log[3*E^3]]^4),x]

[Out]

2*x + (x*Log[3 + Log[3]]^2)/(x^2*(2 - Log[3])^2 - Log[3 + Log[3]]^2)

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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
 n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
 d*x, a + b*x])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1994

Int[(Pq_)*(u_)^(p_.), x_Symbol] :> Int[Pq*ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && PolyQ[Pq, x] && TrinomialQ
[u, x] &&  !TrinomialMatchQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {x^4 (32-64 \log (3))+48 x^4 \log ^2(3)-16 x^4 \log ^3(3)+2 x^4 \log ^4(3)+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {-16 x^4 \log ^3(3)+2 x^4 \log ^4(3)+x^4 \left (32-64 \log (3)+48 \log ^2(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)\right )+x^4 \left (-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{16 x^4-32 x^4 \log (3)+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{x^4 (16-32 \log (3))+24 x^4 \log ^2(3)-8 x^4 \log ^3(3)+x^4 \log ^4(3)+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{-8 x^4 \log ^3(3)+x^4 \log ^4(3)+x^4 \left (16-32 \log (3)+24 \log ^2(3)\right )+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{x^4 \left (16-32 \log (3)+24 \log ^2(3)\right )+x^4 \left (-8 \log ^3(3)+\log ^4(3)\right )+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{x^4 \left (16-32 \log (3)+24 \log ^2(3)-8 \log ^3(3)+\log ^4(3)\right )+\left (-8 x^2+8 x^2 \log (3)-2 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )} \, dx\\ &=\int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{x^4 (2-\log (3))^4-2 x^2 (2-\log (3))^2 \log ^2(3+\log (3))+\log ^4(3+\log (3))} \, dx\\ &=(2-\log (3))^4 \int \frac {x^4 \left (32-64 \log (3)+48 \log ^2(3)-16 \log ^3(3)+2 \log ^4(3)\right )+\left (-20 x^2+20 x^2 \log (3)-5 x^2 \log ^2(3)\right ) \log ^2\left (\log \left (3 e^3\right )\right )+\log ^4\left (\log \left (3 e^3\right )\right )}{\left (x^2 (2-\log (3))^4-(2-\log (3))^2 \log ^2(3+\log (3))\right )^2} \, dx\\ &=\frac {x \log ^2(3+\log (3))}{x^2 (2-\log (3))^2-\log ^2(3+\log (3))}+\frac {(2-\log (3))^2 \int \frac {4 x^2 (2-\log (3))^2 \log ^2(3+\log (3))-4 \log ^4(3+\log (3))}{x^2 (2-\log (3))^4-(2-\log (3))^2 \log ^2(3+\log (3))} \, dx}{2 \log ^2(3+\log (3))}\\ &=\frac {x \log ^2(3+\log (3))}{x^2 (2-\log (3))^2-\log ^2(3+\log (3))}+2 \int 1 \, dx\\ &=2 x+\frac {x \log ^2(3+\log (3))}{x^2 (2-\log (3))^2-\log ^2(3+\log (3))}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 34, normalized size = 1.10 \begin {gather*} x \left (2+\frac {\log ^2(3+\log (3))}{x^2 (-2+\log (3))^2-\log ^2(3+\log (3))}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32*x^4 - 64*x^4*Log[3] + 48*x^4*Log[3]^2 - 16*x^4*Log[3]^3 + 2*x^4*Log[3]^4 + (-20*x^2 + 20*x^2*Log
[3] - 5*x^2*Log[3]^2)*Log[Log[3*E^3]]^2 + Log[Log[3*E^3]]^4)/(16*x^4 - 32*x^4*Log[3] + 24*x^4*Log[3]^2 - 8*x^4
*Log[3]^3 + x^4*Log[3]^4 + (-8*x^2 + 8*x^2*Log[3] - 2*x^2*Log[3]^2)*Log[Log[3*E^3]]^2 + Log[Log[3*E^3]]^4),x]

[Out]

x*(2 + Log[3 + Log[3]]^2/(x^2*(-2 + Log[3])^2 - Log[3 + Log[3]]^2))

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fricas [B]  time = 0.78, size = 65, normalized size = 2.10 \begin {gather*} \frac {2 \, x^{3} \log \relax (3)^{2} - 8 \, x^{3} \log \relax (3) + 8 \, x^{3} - x \log \left (\log \relax (3) + 3\right )^{2}}{x^{2} \log \relax (3)^{2} - 4 \, x^{2} \log \relax (3) + 4 \, x^{2} - \log \left (\log \relax (3) + 3\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(log(3*exp(3)))^4+(-5*x^2*log(3)^2+20*x^2*log(3)-20*x^2)*log(log(3*exp(3)))^2+2*x^4*log(3)^4-16*
x^4*log(3)^3+48*x^4*log(3)^2-64*x^4*log(3)+32*x^4)/(log(log(3*exp(3)))^4+(-2*x^2*log(3)^2+8*x^2*log(3)-8*x^2)*
log(log(3*exp(3)))^2+x^4*log(3)^4-8*x^4*log(3)^3+24*x^4*log(3)^2-32*x^4*log(3)+16*x^4),x, algorithm="fricas")

[Out]

(2*x^3*log(3)^2 - 8*x^3*log(3) + 8*x^3 - x*log(log(3) + 3)^2)/(x^2*log(3)^2 - 4*x^2*log(3) + 4*x^2 - log(log(3
) + 3)^2)

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giac [B]  time = 0.22, size = 99, normalized size = 3.19 \begin {gather*} \frac {x \log \left (\log \left (3 \, e^{3}\right )\right )^{2}}{x^{2} \log \relax (3)^{2} - 4 \, x^{2} \log \relax (3) + 4 \, x^{2} - \log \left (\log \left (3 \, e^{3}\right )\right )^{2}} + \frac {2 \, {\left (x \log \relax (3)^{4} - 8 \, x \log \relax (3)^{3} + 24 \, x \log \relax (3)^{2} - 32 \, x \log \relax (3) + 16 \, x\right )}}{\log \relax (3)^{4} - 8 \, \log \relax (3)^{3} + 24 \, \log \relax (3)^{2} - 32 \, \log \relax (3) + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(log(3*exp(3)))^4+(-5*x^2*log(3)^2+20*x^2*log(3)-20*x^2)*log(log(3*exp(3)))^2+2*x^4*log(3)^4-16*
x^4*log(3)^3+48*x^4*log(3)^2-64*x^4*log(3)+32*x^4)/(log(log(3*exp(3)))^4+(-2*x^2*log(3)^2+8*x^2*log(3)-8*x^2)*
log(log(3*exp(3)))^2+x^4*log(3)^4-8*x^4*log(3)^3+24*x^4*log(3)^2-32*x^4*log(3)+16*x^4),x, algorithm="giac")

[Out]

x*log(log(3*e^3))^2/(x^2*log(3)^2 - 4*x^2*log(3) + 4*x^2 - log(log(3*e^3))^2) + 2*(x*log(3)^4 - 8*x*log(3)^3 +
 24*x*log(3)^2 - 32*x*log(3) + 16*x)/(log(3)^4 - 8*log(3)^3 + 24*log(3)^2 - 32*log(3) + 16)

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maple [A]  time = 0.25, size = 46, normalized size = 1.48




method result size



risch \(2 x +\frac {\ln \left (3+\ln \relax (3)\right )^{2} x}{x^{2} \ln \relax (3)^{2}-4 x^{2} \ln \relax (3)-\ln \left (3+\ln \relax (3)\right )^{2}+4 x^{2}}\) \(46\)
norman \(\frac {\left (2 \ln \relax (3)^{2}-8 \ln \relax (3)+8\right ) x^{3}-\ln \left (3+\ln \relax (3)\right )^{2} x}{x^{2} \ln \relax (3)^{2}-4 x^{2} \ln \relax (3)-\ln \left (3+\ln \relax (3)\right )^{2}+4 x^{2}}\) \(61\)
gosper \(\frac {x \left (2 x^{2} \ln \relax (3)^{2}-8 x^{2} \ln \relax (3)-\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )^{2}+8 x^{2}\right )}{x^{2} \ln \relax (3)^{2}-4 x^{2} \ln \relax (3)-\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )^{2}+4 x^{2}}\) \(68\)
default \(2 x +\frac {\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )^{2}}{2 \left (2-\ln \relax (3)\right ) \left (-x \ln \relax (3)+\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )+2 x \right )}+\frac {\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )^{2}}{2 \left (2-\ln \relax (3)\right ) \left (-x \ln \relax (3)-\ln \left (\ln \left (3 \,{\mathrm e}^{3}\right )\right )+2 x \right )}\) \(77\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(ln(3*exp(3)))^4+(-5*x^2*ln(3)^2+20*x^2*ln(3)-20*x^2)*ln(ln(3*exp(3)))^2+2*x^4*ln(3)^4-16*x^4*ln(3)^3+4
8*x^4*ln(3)^2-64*x^4*ln(3)+32*x^4)/(ln(ln(3*exp(3)))^4+(-2*x^2*ln(3)^2+8*x^2*ln(3)-8*x^2)*ln(ln(3*exp(3)))^2+x
^4*ln(3)^4-8*x^4*ln(3)^3+24*x^4*ln(3)^2-32*x^4*ln(3)+16*x^4),x,method=_RETURNVERBOSE)

[Out]

2*x+ln(3+ln(3))^2*x/(x^2*ln(3)^2-4*x^2*ln(3)-ln(3+ln(3))^2+4*x^2)

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maxima [A]  time = 0.35, size = 41, normalized size = 1.32 \begin {gather*} \frac {x \log \left (\log \left (3 \, e^{3}\right )\right )^{2}}{{\left (\log \relax (3)^{2} - 4 \, \log \relax (3) + 4\right )} x^{2} - \log \left (\log \left (3 \, e^{3}\right )\right )^{2}} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(log(3*exp(3)))^4+(-5*x^2*log(3)^2+20*x^2*log(3)-20*x^2)*log(log(3*exp(3)))^2+2*x^4*log(3)^4-16*
x^4*log(3)^3+48*x^4*log(3)^2-64*x^4*log(3)+32*x^4)/(log(log(3*exp(3)))^4+(-2*x^2*log(3)^2+8*x^2*log(3)-8*x^2)*
log(log(3*exp(3)))^2+x^4*log(3)^4-8*x^4*log(3)^3+24*x^4*log(3)^2-32*x^4*log(3)+16*x^4),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.80, size = 65, normalized size = 2.10 \begin {gather*} \frac {x\,\left (2\,x^2\,{\ln \relax (3)}^2-{\ln \left (\ln \relax (3)+3\right )}^2-8\,x^2\,\ln \relax (3)+8\,x^2\right )}{x^2\,{\ln \relax (3)}^2-{\ln \left (\ln \relax (3)+3\right )}^2-4\,x^2\,\ln \relax (3)+4\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((48*x^4*log(3)^2 - 16*x^4*log(3)^3 + 2*x^4*log(3)^4 - log(log(3*exp(3)))^2*(5*x^2*log(3)^2 - 20*x^2*log(3)
 + 20*x^2) - 64*x^4*log(3) + 32*x^4 + log(log(3*exp(3)))^4)/(24*x^4*log(3)^2 - 8*x^4*log(3)^3 + x^4*log(3)^4 -
 log(log(3*exp(3)))^2*(2*x^2*log(3)^2 - 8*x^2*log(3) + 8*x^2) - 32*x^4*log(3) + 16*x^4 + log(log(3*exp(3)))^4)
,x)

[Out]

(x*(2*x^2*log(3)^2 - log(log(3) + 3)^2 - 8*x^2*log(3) + 8*x^2))/(x^2*log(3)^2 - log(log(3) + 3)^2 - 4*x^2*log(
3) + 4*x^2)

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sympy [A]  time = 1.71, size = 36, normalized size = 1.16 \begin {gather*} 2 x + \frac {x \log {\left (\log {\relax (3 )} + 3 \right )}^{2}}{x^{2} \left (- 4 \log {\relax (3 )} + \log {\relax (3 )}^{2} + 4\right ) - \log {\left (\log {\relax (3 )} + 3 \right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(ln(3*exp(3)))**4+(-5*x**2*ln(3)**2+20*x**2*ln(3)-20*x**2)*ln(ln(3*exp(3)))**2+2*x**4*ln(3)**4-16
*x**4*ln(3)**3+48*x**4*ln(3)**2-64*x**4*ln(3)+32*x**4)/(ln(ln(3*exp(3)))**4+(-2*x**2*ln(3)**2+8*x**2*ln(3)-8*x
**2)*ln(ln(3*exp(3)))**2+x**4*ln(3)**4-8*x**4*ln(3)**3+24*x**4*ln(3)**2-32*x**4*ln(3)+16*x**4),x)

[Out]

2*x + x*log(log(3) + 3)**2/(x**2*(-4*log(3) + log(3)**2 + 4) - log(log(3) + 3)**2)

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