3.70.79 \(\int \frac {(1-x) \log (5)+(-4 x+2 x^2+(4+2 x-2 x^2) \log (5)) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (5) \log (-2+x)} \, dx\) [6979]

Optimal. Leaf size=31 \[ 2-2 x-x^2+\frac {1+x^2}{\log (5)}+(1-x) \log (\log (-2+x)) \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 45, normalized size of antiderivative = 1.45, number of steps used = 13, number of rules used = 9, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {12, 6820, 6, 2458, 2395, 2335, 2339, 29, 2600} \begin {gather*} \frac {x^2 (2-\log (25))}{2 \log (5)}-\frac {x \log (25)}{\log (5)}+(2-x) \log (\log (x-2))-\log (\log (x-2)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 - x)*Log[5] + (-4*x + 2*x^2 + (4 + 2*x - 2*x^2)*Log[5])*Log[-2 + x] + (2 - x)*Log[5]*Log[-2 + x]*Log[L
og[-2 + x]])/((-2 + x)*Log[5]*Log[-2 + x]),x]

[Out]

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

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 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2339

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/(b*n), Subst[Int[x^p, x], x, a + b*L
og[c*x^n]], x] /; FreeQ[{a, b, c, n, p}, x]

Rule 2395

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol]
:> With[{u = ExpandIntegrand[(a + b*Log[c*x^n])^p, (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[
{a, b, c, d, e, f, m, n, p, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IGtQ[p, 0] && IntegerQ[m] && IntegerQ[r
]))

Rule 2458

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[(g*(x/e))^q*((e*h - d*i)/e + i*(x/e))^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

Rule 2600

Int[Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)], x_Symbol] :> Simp[x*Log[c*Log[d*x^n]^p], x] - Dist[n*p, Int[1/Log[
d*x^n], x], x] /; FreeQ[{c, d, n, p}, x]

Rule 6820

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {\int \frac {(1-x) \log (5)+\left (-4 x+2 x^2+\left (4+2 x-2 x^2\right ) \log (5)\right ) \log (-2+x)+(2-x) \log (5) \log (-2+x) \log (\log (-2+x))}{(-2+x) \log (-2+x)} \, dx}{\log (5)}\\ &=\frac {\int \left (2 x-\log (25)-x \log (25)-\frac {(-1+x) \log (5)}{(-2+x) \log (-2+x)}-\log (5) \log (\log (-2+x))\right ) \, dx}{\log (5)}\\ &=\frac {\int \left (x (2-\log (25))-\log (25)-\frac {(-1+x) \log (5)}{(-2+x) \log (-2+x)}-\log (5) \log (\log (-2+x))\right ) \, dx}{\log (5)}\\ &=\frac {x^2 (2-\log (25))}{2 \log (5)}-\frac {x \log (25)}{\log (5)}-\int \frac {-1+x}{(-2+x) \log (-2+x)} \, dx-\int \log (\log (-2+x)) \, dx\\ &=\frac {x^2 (2-\log (25))}{2 \log (5)}-\frac {x \log (25)}{\log (5)}-\text {Subst}\left (\int \frac {1+x}{x \log (x)} \, dx,x,-2+x\right )-\text {Subst}(\int \log (\log (x)) \, dx,x,-2+x)\\ &=\frac {x^2 (2-\log (25))}{2 \log (5)}-\frac {x \log (25)}{\log (5)}-(-2+x) \log (\log (-2+x))-\text {Subst}\left (\int \left (\frac {1}{\log (x)}+\frac {1}{x \log (x)}\right ) \, dx,x,-2+x\right )+\text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )\\ &=\frac {x^2 (2-\log (25))}{2 \log (5)}-\frac {x \log (25)}{\log (5)}-(-2+x) \log (\log (-2+x))+\text {li}(-2+x)-\text {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-2+x\right )-\text {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-2+x\right )\\ &=\frac {x^2 (2-\log (25))}{2 \log (5)}-\frac {x \log (25)}{\log (5)}-(-2+x) \log (\log (-2+x))-\text {Subst}\left (\int \frac {1}{x} \, dx,x,\log (-2+x)\right )\\ &=\frac {x^2 (2-\log (25))}{2 \log (5)}-\frac {x \log (25)}{\log (5)}-\log (\log (-2+x))-(-2+x) \log (\log (-2+x))\\ \end {aligned} \end {gather*}

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Mathematica [A]
time = 0.07, size = 50, normalized size = 1.61 \begin {gather*} -\frac {(-2+x)^2 (-2+\log (25))}{2 \log (5)}-\frac {(-2+x) (-4+3 \log (25))}{\log (5)}-\log (\log (-2+x))-(-2+x) \log (\log (-2+x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((-2 + x)^2*(-2 + Log[25]))/Log[5] - ((-2 + x)*(-4 + 3*Log[25]))/Log[5] - Log[Log[-2 + x]] - (-2 + x)*Log
[Log[-2 + x]]

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Maple [A]
time = 0.77, size = 40, normalized size = 1.29

method	result	size
norman	\(\ln \left (\ln \left (x -2\right )\right )-2 x -x \ln \left (\ln \left (x -2\right )\right )-\frac {\left (\ln \left (5\right )-1\right ) x^{2}}{\ln \left (5\right )}\)	\(31\)
risch	\(-x \ln \left (\ln \left (x -2\right )\right )-x^{2}+\ln \left (\ln \left (x -2\right )\right )-2 x +\frac {x^{2}}{\ln \left (5\right )}\)	\(31\)
default	\(\frac {-\ln \left (5\right ) \ln \left (\ln \left (x -2\right )\right ) x -x^{2} \ln \left (5\right )+\ln \left (5\right ) \ln \left (\ln \left (x -2\right )\right )-2 x \ln \left (5\right )+x^{2}}{\ln \left (5\right )}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2-x)*ln(5)*ln(x-2)*ln(ln(x-2))+((-2*x^2+2*x+4)*ln(5)+2*x^2-4*x)*ln(x-2)+(1-x)*ln(5))/(x-2)/ln(5)/ln(x-2)
,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (30) = 60\).
time = 0.53, size = 90, normalized size = 2.90 \begin {gather*} -\frac {x^{2} {\left (\log \left (5\right ) - 1\right )} - 6 \, \log \left (5\right ) \log \left (x - 2\right ) \log \left (\log \left (x - 2\right )\right ) + 4 \, {\left (\log \left (x - 2\right ) \log \left (\log \left (x - 2\right )\right ) - \log \left (x - 2\right )\right )} \log \left (5\right ) + 2 \, x \log \left (5\right ) + 4 \, \log \left (5\right ) \log \left (x - 2\right ) + {\left (x \log \left (5\right ) + 2 \, \log \left (5\right ) \log \left (x - 2\right )\right )} \log \left (\log \left (x - 2\right )\right ) - \log \left (5\right ) \log \left (\log \left (x - 2\right )\right )}{\log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2-x)*log(5)*log(-2+x)*log(log(-2+x))+((-2*x^2+2*x+4)*log(5)+2*x^2-4*x)*log(-2+x)+(1-x)*log(5))/(-2
+x)/log(5)/log(-2+x),x, algorithm="maxima")

[Out]

-(x^2*(log(5) - 1) - 6*log(5)*log(x - 2)*log(log(x - 2)) + 4*(log(x - 2)*log(log(x - 2)) - log(x - 2))*log(5)
+ 2*x*log(5) + 4*log(5)*log(x - 2) + (x*log(5) + 2*log(5)*log(x - 2))*log(log(x - 2)) - log(5)*log(log(x - 2))
)/log(5)

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Fricas [A]
time = 0.50, size = 33, normalized size = 1.06 \begin {gather*} -\frac {{\left (x - 1\right )} \log \left (5\right ) \log \left (\log \left (x - 2\right )\right ) - x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (5\right )}{\log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2-x)*log(5)*log(-2+x)*log(log(-2+x))+((-2*x^2+2*x+4)*log(5)+2*x^2-4*x)*log(-2+x)+(1-x)*log(5))/(-2
+x)/log(5)/log(-2+x),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.19, size = 24, normalized size = 0.77 \begin {gather*} \frac {x^{2} \cdot \left (1 - \log {\left (5 \right )}\right )}{\log {\left (5 \right )}} - 2 x + \left (1 - x\right ) \log {\left (\log {\left (x - 2 \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2-x)*ln(5)*ln(-2+x)*ln(ln(-2+x))+((-2*x**2+2*x+4)*ln(5)+2*x**2-4*x)*ln(-2+x)+(1-x)*ln(5))/(-2+x)/l
n(5)/ln(-2+x),x)

[Out]

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

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Giac [A]
time = 0.42, size = 38, normalized size = 1.23 \begin {gather*} -\frac {x^{2} {\left (\log \left (5\right ) - 1\right )} + x \log \left (5\right ) \log \left (\log \left (x - 2\right )\right ) + 2 \, x \log \left (5\right ) - \log \left (5\right ) \log \left (\log \left (x - 2\right )\right )}{\log \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2-x)*log(5)*log(-2+x)*log(log(-2+x))+((-2*x^2+2*x+4)*log(5)+2*x^2-4*x)*log(-2+x)+(1-x)*log(5))/(-2
+x)/log(5)/log(-2+x),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 4.69, size = 34, normalized size = 1.10 \begin {gather*} -2\,x-\ln \left (\ln \left (x-2\right )\right )-\ln \left (\ln \left (x-2\right )\right )\,\left (x-2\right )-\frac {x^2\,\left (\ln \left (25\right )-2\right )}{2\,\ln \left (5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 2*x - log(log(x - 2)) - log(log(x - 2))*(x - 2) - (x^2*(log(25) - 2))/(2*log(5))

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