3.103.82 \(\int \frac {(32-4 x) \log ^2(-8+x)+(-32+4 x) \log ^2(-8+x) \log (2 x)+(-15 x^3+(-120 x^2+15 x^3) \log (-8+x)+(8 x^2-x^3) \log ^2(-8+x)+(15 x^2+(120 x-15 x^2) \log (-8+x)+(-8 x+x^2) \log ^2(-8+x)) \log (2 x)) \log ^2(\frac {x-\log (2 x)}{x})}{((8 x^2-x^3) \log ^2(-8+x)+(-8 x+x^2) \log ^2(-8+x) \log (2 x)) \log ^2(\frac {x-\log (2 x)}{x})} \, dx\)

Optimal. Leaf size=27 \[ x-\frac {15 x}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \]

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Rubi [A]  time = 0.76, antiderivative size = 39, normalized size of antiderivative = 1.44, number of steps used = 14, number of rules used = 10, integrand size = 179, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {6688, 2411, 2353, 2297, 2298, 2302, 30, 2389, 6742, 6686} \begin {gather*} x+\frac {15 (8-x)}{\log (x-8)}-\frac {120}{\log (x-8)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((32 - 4*x)*Log[-8 + x]^2 + (-32 + 4*x)*Log[-8 + x]^2*Log[2*x] + (-15*x^3 + (-120*x^2 + 15*x^3)*Log[-8 + x
] + (8*x^2 - x^3)*Log[-8 + x]^2 + (15*x^2 + (120*x - 15*x^2)*Log[-8 + x] + (-8*x + x^2)*Log[-8 + x]^2)*Log[2*x
])*Log[(x - Log[2*x])/x]^2)/(((8*x^2 - x^3)*Log[-8 + x]^2 + (-8*x + x^2)*Log[-8 + x]^2*Log[2*x])*Log[(x - Log[
2*x])/x]^2),x]

[Out]

x - 120/Log[-8 + x] + (15*(8 - x))/Log[-8 + x] + 4/Log[1 - Log[2*x]/x]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2298

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

Rule 2302

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 2353

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 2389

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

Rule 2411

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 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 6742

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\frac {15 x}{(-8+x) \log ^2(-8+x)}-\frac {15}{\log (-8+x)}+\frac {4+x^2 \log ^2\left (1-\frac {\log (2 x)}{x}\right )-\log (2 x) \left (4+x \log ^2\left (1-\frac {\log (2 x)}{x}\right )\right )}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )}\right ) \, dx\\ &=15 \int \frac {x}{(-8+x) \log ^2(-8+x)} \, dx-15 \int \frac {1}{\log (-8+x)} \, dx+\int \frac {4+x^2 \log ^2\left (1-\frac {\log (2 x)}{x}\right )-\log (2 x) \left (4+x \log ^2\left (1-\frac {\log (2 x)}{x}\right )\right )}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )} \, dx\\ &=15 \operatorname {Subst}\left (\int \frac {8+x}{x \log ^2(x)} \, dx,x,-8+x\right )-15 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-8+x\right )+\int \left (1-\frac {4 (-1+\log (2 x))}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )}\right ) \, dx\\ &=x-15 \text {li}(-8+x)-4 \int \frac {-1+\log (2 x)}{x (x-\log (2 x)) \log ^2\left (1-\frac {\log (2 x)}{x}\right )} \, dx+15 \operatorname {Subst}\left (\int \left (\frac {1}{\log ^2(x)}+\frac {8}{x \log ^2(x)}\right ) \, dx,x,-8+x\right )\\ &=x+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )}-15 \text {li}(-8+x)+15 \operatorname {Subst}\left (\int \frac {1}{\log ^2(x)} \, dx,x,-8+x\right )+120 \operatorname {Subst}\left (\int \frac {1}{x \log ^2(x)} \, dx,x,-8+x\right )\\ &=x+\frac {15 (8-x)}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )}-15 \text {li}(-8+x)+15 \operatorname {Subst}\left (\int \frac {1}{\log (x)} \, dx,x,-8+x\right )+120 \operatorname {Subst}\left (\int \frac {1}{x^2} \, dx,x,\log (-8+x)\right )\\ &=x-\frac {120}{\log (-8+x)}+\frac {15 (8-x)}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.24, size = 27, normalized size = 1.00 \begin {gather*} x-\frac {15 x}{\log (-8+x)}+\frac {4}{\log \left (1-\frac {\log (2 x)}{x}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((32 - 4*x)*Log[-8 + x]^2 + (-32 + 4*x)*Log[-8 + x]^2*Log[2*x] + (-15*x^3 + (-120*x^2 + 15*x^3)*Log[
-8 + x] + (8*x^2 - x^3)*Log[-8 + x]^2 + (15*x^2 + (120*x - 15*x^2)*Log[-8 + x] + (-8*x + x^2)*Log[-8 + x]^2)*L
og[2*x])*Log[(x - Log[2*x])/x]^2)/(((8*x^2 - x^3)*Log[-8 + x]^2 + (-8*x + x^2)*Log[-8 + x]^2*Log[2*x])*Log[(x
- Log[2*x])/x]^2),x]

[Out]

x - (15*x)/Log[-8 + x] + 4/Log[1 - Log[2*x]/x]

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fricas [A]  time = 0.63, size = 53, normalized size = 1.96 \begin {gather*} \frac {{\left (x \log \left (x - 8\right ) - 15 \, x\right )} \log \left (\frac {x - \log \left (2 \, x\right )}{x}\right ) + 4 \, \log \left (x - 8\right )}{\log \left (x - 8\right ) \log \left (\frac {x - \log \left (2 \, x\right )}{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2-8*x)*log(-8+x)^2+(-15*x^2+120*x)*log(-8+x)+15*x^2)*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2+(15*x^3
-120*x^2)*log(-8+x)-15*x^3)*log((x-log(2*x))/x)^2+(4*x-32)*log(-8+x)^2*log(2*x)+(-4*x+32)*log(-8+x)^2)/((x^2-8
*x)*log(-8+x)^2*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2)/log((x-log(2*x))/x)^2,x, algorithm="fricas")

[Out]

((x*log(x - 8) - 15*x)*log((x - log(2*x))/x) + 4*log(x - 8))/(log(x - 8)*log((x - log(2*x))/x))

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giac [B]  time = 0.76, size = 176, normalized size = 6.52 \begin {gather*} x + \frac {4 \, {\left (x \log \relax (2) - \log \relax (2) \log \left (2 \, x\right ) + x \log \relax (x) - \log \left (2 \, x\right ) \log \relax (x) - x + \log \left (2 \, x\right )\right )}}{x \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) - \log \relax (2) \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) - x \log \left (2 \, x\right ) \log \relax (x) + \log \relax (2) \log \left (2 \, x\right ) \log \relax (x) - \log \left (2 \, x\right ) \log \left (x - \log \left (2 \, x\right )\right ) \log \relax (x) + \log \left (2 \, x\right ) \log \relax (x)^{2} - x \log \left (x - \log \left (2 \, x\right )\right ) + \log \relax (2) \log \left (x - \log \left (2 \, x\right )\right ) + x \log \relax (x) - \log \relax (2) \log \relax (x) + \log \left (x - \log \left (2 \, x\right )\right ) \log \relax (x) - \log \relax (x)^{2}} - \frac {15 \, x}{\log \left (x - 8\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2-8*x)*log(-8+x)^2+(-15*x^2+120*x)*log(-8+x)+15*x^2)*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2+(15*x^3
-120*x^2)*log(-8+x)-15*x^3)*log((x-log(2*x))/x)^2+(4*x-32)*log(-8+x)^2*log(2*x)+(-4*x+32)*log(-8+x)^2)/((x^2-8
*x)*log(-8+x)^2*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2)/log((x-log(2*x))/x)^2,x, algorithm="giac")

[Out]

x + 4*(x*log(2) - log(2)*log(2*x) + x*log(x) - log(2*x)*log(x) - x + log(2*x))/(x*log(2*x)*log(x - log(2*x)) -
log(2)*log(2*x)*log(x - log(2*x)) - x*log(2*x)*log(x) + log(2)*log(2*x)*log(x) - log(2*x)*log(x - log(2*x))*l
og(x) + log(2*x)*log(x)^2 - x*log(x - log(2*x)) + log(2)*log(x - log(2*x)) + x*log(x) - log(2)*log(x) + log(x
- log(2*x))*log(x) - log(x)^2) - 15*x/log(x - 8)

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maple [F]  time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (\left (\left (x^{2}-8 x \right ) \ln \left (-8+x \right )^{2}+\left (-15 x^{2}+120 x \right ) \ln \left (-8+x \right )+15 x^{2}\right ) \ln \left (2 x \right )+\left (-x^{3}+8 x^{2}\right ) \ln \left (-8+x \right )^{2}+\left (15 x^{3}-120 x^{2}\right ) \ln \left (-8+x \right )-15 x^{3}\right ) \ln \left (\frac {x -\ln \left (2 x \right )}{x}\right )^{2}+\left (4 x -32\right ) \ln \left (-8+x \right )^{2} \ln \left (2 x \right )+\left (-4 x +32\right ) \ln \left (-8+x \right )^{2}}{\left (\left (x^{2}-8 x \right ) \ln \left (-8+x \right )^{2} \ln \left (2 x \right )+\left (-x^{3}+8 x^{2}\right ) \ln \left (-8+x \right )^{2}\right ) \ln \left (\frac {x -\ln \left (2 x \right )}{x}\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((((x^2-8*x)*ln(-8+x)^2+(-15*x^2+120*x)*ln(-8+x)+15*x^2)*ln(2*x)+(-x^3+8*x^2)*ln(-8+x)^2+(15*x^3-120*x^2)*
ln(-8+x)-15*x^3)*ln((x-ln(2*x))/x)^2+(4*x-32)*ln(-8+x)^2*ln(2*x)+(-4*x+32)*ln(-8+x)^2)/((x^2-8*x)*ln(-8+x)^2*l
n(2*x)+(-x^3+8*x^2)*ln(-8+x)^2)/ln((x-ln(2*x))/x)^2,x)

[Out]

int(((((x^2-8*x)*ln(-8+x)^2+(-15*x^2+120*x)*ln(-8+x)+15*x^2)*ln(2*x)+(-x^3+8*x^2)*ln(-8+x)^2+(15*x^3-120*x^2)*
ln(-8+x)-15*x^3)*ln((x-ln(2*x))/x)^2+(4*x-32)*ln(-8+x)^2*ln(2*x)+(-4*x+32)*ln(-8+x)^2)/((x^2-8*x)*ln(-8+x)^2*l
n(2*x)+(-x^3+8*x^2)*ln(-8+x)^2)/ln((x-ln(2*x))/x)^2,x)

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maxima [B]  time = 0.55, size = 68, normalized size = 2.52 \begin {gather*} \frac {{\left (x \log \left (x - 8\right ) - 15 \, x\right )} \log \left (x - \log \relax (2) - \log \relax (x)\right ) - {\left (x \log \relax (x) - 4\right )} \log \left (x - 8\right ) + 15 \, x \log \relax (x)}{\log \left (x - \log \relax (2) - \log \relax (x)\right ) \log \left (x - 8\right ) - \log \left (x - 8\right ) \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x^2-8*x)*log(-8+x)^2+(-15*x^2+120*x)*log(-8+x)+15*x^2)*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2+(15*x^3
-120*x^2)*log(-8+x)-15*x^3)*log((x-log(2*x))/x)^2+(4*x-32)*log(-8+x)^2*log(2*x)+(-4*x+32)*log(-8+x)^2)/((x^2-8
*x)*log(-8+x)^2*log(2*x)+(-x^3+8*x^2)*log(-8+x)^2)/log((x-log(2*x))/x)^2,x, algorithm="maxima")

[Out]

((x*log(x - 8) - 15*x)*log(x - log(2) - log(x)) - (x*log(x) - 4)*log(x - 8) + 15*x*log(x))/(log(x - log(2) - l
og(x))*log(x - 8) - log(x - 8)*log(x))

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mupad [B]  time = 7.20, size = 44, normalized size = 1.63 \begin {gather*} \frac {4}{\ln \left (\frac {x-\ln \left (2\,x\right )}{x}\right )}-\frac {15\,x-\ln \left (x-8\right )\,\left (15\,x-120\right )}{\ln \left (x-8\right )}-14\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x - 8)^2*(4*x - 32) + log((x - log(2*x))/x)^2*(log(x - 8)*(120*x^2 - 15*x^3) - log(2*x)*(log(x - 8)*
(120*x - 15*x^2) - log(x - 8)^2*(8*x - x^2) + 15*x^2) - log(x - 8)^2*(8*x^2 - x^3) + 15*x^3) - log(2*x)*log(x
- 8)^2*(4*x - 32))/(log((x - log(2*x))/x)^2*(log(x - 8)^2*(8*x^2 - x^3) - log(2*x)*log(x - 8)^2*(8*x - x^2))),
x)

[Out]

4/log((x - log(2*x))/x) - (15*x - log(x - 8)*(15*x - 120))/log(x - 8) - 14*x

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sympy [A]  time = 0.65, size = 20, normalized size = 0.74 \begin {gather*} x - \frac {15 x}{\log {\left (x - 8 \right )}} + \frac {4}{\log {\left (\frac {x - \log {\left (2 x \right )}}{x} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((((x**2-8*x)*ln(-8+x)**2+(-15*x**2+120*x)*ln(-8+x)+15*x**2)*ln(2*x)+(-x**3+8*x**2)*ln(-8+x)**2+(15*
x**3-120*x**2)*ln(-8+x)-15*x**3)*ln((x-ln(2*x))/x)**2+(4*x-32)*ln(-8+x)**2*ln(2*x)+(-4*x+32)*ln(-8+x)**2)/((x*
*2-8*x)*ln(-8+x)**2*ln(2*x)+(-x**3+8*x**2)*ln(-8+x)**2)/ln((x-ln(2*x))/x)**2,x)

[Out]

x - 15*x/log(x - 8) + 4/log((x - log(2*x))/x)

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