3.86.48 \(\int \frac {-128 x+4 x^2+(-64-2 x^2) \log (x)+(-64+64 x-2 x^2+2 x^3+(64+2 x^2) \log (x)) \log (-1+x+\log (x))}{-32 x+32 x^2+x^3-x^4+(32 x-x^3) \log (x)+(32 x-32 x^2-x^3+x^4+(-32 x+x^3) \log (x)) \log (-1+x+\log (x))} \, dx\)

Optimal. Leaf size=21 \[ \log \left (\left (-\frac {32}{x}+x\right )^2 (-1+\log (-1+x+\log (x)))^2\right ) \]

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Rubi [A]  time = 3.42, antiderivative size = 28, normalized size of antiderivative = 1.33, number of steps used = 8, number of rules used = 6, integrand size = 117, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6688, 12, 6725, 446, 72, 6684} \begin {gather*} 2 \log \left (32-x^2\right )-2 \log (x)+2 \log (1-\log (x+\log (x)-1)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-128*x + 4*x^2 + (-64 - 2*x^2)*Log[x] + (-64 + 64*x - 2*x^2 + 2*x^3 + (64 + 2*x^2)*Log[x])*Log[-1 + x + L
og[x]])/(-32*x + 32*x^2 + x^3 - x^4 + (32*x - x^3)*Log[x] + (32*x - 32*x^2 - x^3 + x^4 + (-32*x + x^3)*Log[x])
*Log[-1 + x + Log[x]]),x]

[Out]

-2*Log[x] + 2*Log[32 - x^2] + 2*Log[1 - Log[-1 + x + Log[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 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6684

Int[(u_)/(y_), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[q*Log[RemoveContent[y, x]], x] /;  !Fa
lseQ[q]]

Rule 6688

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

Rule 6725

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2 \left (-2 (-32+x) x-\left (32+x^2\right ) \log (x) (-1+\log (-1+x+\log (x)))-\left (-32+32 x-x^2+x^3\right ) \log (-1+x+\log (x))\right )}{x \left (32-x^2\right ) (1-x-\log (x)) (1-\log (-1+x+\log (x)))} \, dx\\ &=2 \int \frac {-2 (-32+x) x-\left (32+x^2\right ) \log (x) (-1+\log (-1+x+\log (x)))-\left (-32+32 x-x^2+x^3\right ) \log (-1+x+\log (x))}{x \left (32-x^2\right ) (1-x-\log (x)) (1-\log (-1+x+\log (x)))} \, dx\\ &=2 \int \left (\frac {32+x^2}{x \left (-32+x^2\right )}+\frac {1+x}{x (-1+x+\log (x)) (-1+\log (-1+x+\log (x)))}\right ) \, dx\\ &=2 \int \frac {32+x^2}{x \left (-32+x^2\right )} \, dx+2 \int \frac {1+x}{x (-1+x+\log (x)) (-1+\log (-1+x+\log (x)))} \, dx\\ &=2 \log (1-\log (-1+x+\log (x)))+\operatorname {Subst}\left (\int \frac {32+x}{(-32+x) x} \, dx,x,x^2\right )\\ &=2 \log (1-\log (-1+x+\log (x)))+\operatorname {Subst}\left (\int \left (\frac {2}{-32+x}-\frac {1}{x}\right ) \, dx,x,x^2\right )\\ &=-2 \log (x)+2 \log \left (32-x^2\right )+2 \log (1-\log (-1+x+\log (x)))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 26, normalized size = 1.24 \begin {gather*} 2 \left (-\log (x)+\log \left (32-x^2\right )+\log (1-\log (-1+x+\log (x)))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-128*x + 4*x^2 + (-64 - 2*x^2)*Log[x] + (-64 + 64*x - 2*x^2 + 2*x^3 + (64 + 2*x^2)*Log[x])*Log[-1 +
x + Log[x]])/(-32*x + 32*x^2 + x^3 - x^4 + (32*x - x^3)*Log[x] + (32*x - 32*x^2 - x^3 + x^4 + (-32*x + x^3)*L
og[x])*Log[-1 + x + Log[x]]),x]

[Out]

2*(-Log[x] + Log[32 - x^2] + Log[1 - Log[-1 + x + Log[x]]])

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fricas [A]  time = 1.05, size = 24, normalized size = 1.14 \begin {gather*} 2 \, \log \left (x^{2} - 32\right ) - 2 \, \log \relax (x) + 2 \, \log \left (\log \left (x + \log \relax (x) - 1\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+64)*log(x)+2*x^3-2*x^2+64*x-64)*log(-1+log(x)+x)+(-2*x^2-64)*log(x)+4*x^2-128*x)/(((x^3-32*
x)*log(x)+x^4-x^3-32*x^2+32*x)*log(-1+log(x)+x)+(-x^3+32*x)*log(x)-x^4+x^3+32*x^2-32*x),x, algorithm="fricas")

[Out]

2*log(x^2 - 32) - 2*log(x) + 2*log(log(x + log(x) - 1) - 1)

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giac [A]  time = 0.18, size = 24, normalized size = 1.14 \begin {gather*} 2 \, \log \left (x^{2} - 32\right ) - 2 \, \log \relax (x) + 2 \, \log \left (\log \left (x + \log \relax (x) - 1\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+64)*log(x)+2*x^3-2*x^2+64*x-64)*log(-1+log(x)+x)+(-2*x^2-64)*log(x)+4*x^2-128*x)/(((x^3-32*
x)*log(x)+x^4-x^3-32*x^2+32*x)*log(-1+log(x)+x)+(-x^3+32*x)*log(x)-x^4+x^3+32*x^2-32*x),x, algorithm="giac")

[Out]

2*log(x^2 - 32) - 2*log(x) + 2*log(log(x + log(x) - 1) - 1)

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maple [A]  time = 0.04, size = 25, normalized size = 1.19

method	result	size
risch	\(-2 \ln \relax (x )+2 \ln \left (x^{2}-32\right )+2 \ln \left (\ln \left (-1+\ln \relax (x )+x \right )-1\right )\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^2+64)*ln(x)+2*x^3-2*x^2+64*x-64)*ln(-1+ln(x)+x)+(-2*x^2-64)*ln(x)+4*x^2-128*x)/(((x^3-32*x)*ln(x)+x
^4-x^3-32*x^2+32*x)*ln(-1+ln(x)+x)+(-x^3+32*x)*ln(x)-x^4+x^3+32*x^2-32*x),x,method=_RETURNVERBOSE)

[Out]

-2*ln(x)+2*ln(x^2-32)+2*ln(ln(-1+ln(x)+x)-1)

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maxima [A]  time = 0.43, size = 24, normalized size = 1.14 \begin {gather*} 2 \, \log \left (x^{2} - 32\right ) - 2 \, \log \relax (x) + 2 \, \log \left (\log \left (x + \log \relax (x) - 1\right ) - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^2+64)*log(x)+2*x^3-2*x^2+64*x-64)*log(-1+log(x)+x)+(-2*x^2-64)*log(x)+4*x^2-128*x)/(((x^3-32*
x)*log(x)+x^4-x^3-32*x^2+32*x)*log(-1+log(x)+x)+(-x^3+32*x)*log(x)-x^4+x^3+32*x^2-32*x),x, algorithm="maxima")

[Out]

2*log(x^2 - 32) - 2*log(x) + 2*log(log(x + log(x) - 1) - 1)

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mupad [B]  time = 6.42, size = 24, normalized size = 1.14 \begin {gather*} 2\,\ln \left (x^2-32\right )-2\,\ln \relax (x)+2\,\ln \left (\ln \left (x+\ln \relax (x)-1\right )-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((128*x - log(x + log(x) - 1)*(64*x - 2*x^2 + 2*x^3 + log(x)*(2*x^2 + 64) - 64) - 4*x^2 + log(x)*(2*x^2 + 6
4))/(32*x - log(x)*(32*x - x^3) + log(x + log(x) - 1)*(log(x)*(32*x - x^3) - 32*x + 32*x^2 + x^3 - x^4) - 32*x
^2 - x^3 + x^4),x)

[Out]

2*log(x^2 - 32) - 2*log(x) + 2*log(log(x + log(x) - 1) - 1)

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sympy [A]  time = 0.49, size = 26, normalized size = 1.24 \begin {gather*} - 2 \log {\relax (x )} + 2 \log {\left (x^{2} - 32 \right )} + 2 \log {\left (\log {\left (x + \log {\relax (x )} - 1 \right )} - 1 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**2+64)*ln(x)+2*x**3-2*x**2+64*x-64)*ln(-1+ln(x)+x)+(-2*x**2-64)*ln(x)+4*x**2-128*x)/(((x**3-3
2*x)*ln(x)+x**4-x**3-32*x**2+32*x)*ln(-1+ln(x)+x)+(-x**3+32*x)*ln(x)-x**4+x**3+32*x**2-32*x),x)

[Out]

-2*log(x) + 2*log(x**2 - 32) + 2*log(log(x + log(x) - 1) - 1)

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