3.53.63 \(\int \frac {(-16-2 x) \log ^2(x)+(16 x+2 x^2+2 x^3) \log (\frac {8+x}{4})+\log (x) (16+2 x+2 x^2+(-16 x-2 x^2) \log (\frac {8+x}{4}))}{8 x^3+x^4} \, dx\)

Optimal. Leaf size=20 \[ \left (-\log \left (2+\frac {x}{4}\right )-\frac {\log (x)}{x}\right )^2 \]

________________________________________________________________________________________

Rubi [B]  time = 1.23, antiderivative size = 66, normalized size of antiderivative = 3.30, number of steps used = 35, number of rules used = 19, integrand size = 77, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.247, Rules used = {1593, 6688, 12, 6742, 893, 2304, 2418, 2395, 36, 31, 29, 2390, 2301, 2376, 2392, 2391, 2357, 2317, 2305} \begin {gather*} \frac {\log ^2(x)}{x^2}+\log ^2\left (\frac {x+8}{4}\right )-\frac {1}{4} \log \left (\frac {x}{8}+1\right ) \log (x)+\frac {2 \log \left (\frac {x}{4}+2\right ) \log (x)}{x}+\frac {1}{4} \log (x+8) \log (x)-\frac {1}{4} \log (8) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-16 - 2*x)*Log[x]^2 + (16*x + 2*x^2 + 2*x^3)*Log[(8 + x)/4] + Log[x]*(16 + 2*x + 2*x^2 + (-16*x - 2*x^2)
*Log[(8 + x)/4]))/(8*x^3 + x^4),x]

[Out]

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

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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

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

Rule 2317

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(Log[1 + (e*x)/d]*(a +
b*Log[c*x^n])^p)/e, x] - Dist[(b*n*p)/e, Int[(Log[1 + (e*x)/d]*(a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2376

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] &
& RationalQ[q])) && NeQ[q, -1]

Rule 2390

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2392

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + (e*x)/d]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 2418

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

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 \frac {(-16-2 x) \log ^2(x)+\left (16 x+2 x^2+2 x^3\right ) \log \left (\frac {8+x}{4}\right )+\log (x) \left (16+2 x+2 x^2+\left (-16 x-2 x^2\right ) \log \left (\frac {8+x}{4}\right )\right )}{x^3 (8+x)} \, dx\\ &=\int \frac {2 \left (8+x+x^2-(8+x) \log (x)\right ) \left (\log (x)+x \log \left (\frac {8+x}{4}\right )\right )}{x^3 (8+x)} \, dx\\ &=2 \int \frac {\left (8+x+x^2-(8+x) \log (x)\right ) \left (\log (x)+x \log \left (\frac {8+x}{4}\right )\right )}{x^3 (8+x)} \, dx\\ &=2 \int \left (\frac {\log \left (2+\frac {x}{4}\right ) \left (8+x+x^2-8 \log (x)-x \log (x)\right )}{x^2 (8+x)}+\frac {\log (x) \left (8+x+x^2-8 \log (x)-x \log (x)\right )}{x^3 (8+x)}\right ) \, dx\\ &=2 \int \frac {\log \left (2+\frac {x}{4}\right ) \left (8+x+x^2-8 \log (x)-x \log (x)\right )}{x^2 (8+x)} \, dx+2 \int \frac {\log (x) \left (8+x+x^2-8 \log (x)-x \log (x)\right )}{x^3 (8+x)} \, dx\\ &=2 \int \left (\frac {\left (8+x+x^2\right ) \log \left (2+\frac {x}{4}\right )}{x^2 (8+x)}-\frac {\log \left (2+\frac {x}{4}\right ) \log (x)}{x^2}\right ) \, dx+2 \int \left (\frac {\left (8+x+x^2\right ) \log (x)}{x^3 (8+x)}-\frac {\log ^2(x)}{x^3}\right ) \, dx\\ &=2 \int \frac {\left (8+x+x^2\right ) \log \left (2+\frac {x}{4}\right )}{x^2 (8+x)} \, dx+2 \int \frac {\left (8+x+x^2\right ) \log (x)}{x^3 (8+x)} \, dx-2 \int \frac {\log \left (2+\frac {x}{4}\right ) \log (x)}{x^2} \, dx-2 \int \frac {\log ^2(x)}{x^3} \, dx\\ &=\frac {2 \log \left (2+\frac {x}{4}\right ) \log (x)}{x}-\frac {\log ^2(x)}{4}+\frac {\log ^2(x)}{x^2}+\frac {1}{4} \log (x) \log (8+x)+2 \int \left (\frac {\log \left (2+\frac {x}{4}\right )}{x^2}+\frac {\log \left (2+\frac {x}{4}\right )}{8+x}\right ) \, dx-2 \int \frac {\log (x)}{x^3} \, dx+2 \int \left (\frac {\log (x)}{x^3}+\frac {\log (x)}{8 x}-\frac {\log (x)}{8 (8+x)}\right ) \, dx+2 \int \left (-\frac {\log \left (2+\frac {x}{4}\right )}{x^2}+\frac {\log (x)}{8 x}-\frac {\log (8+x)}{8 x}\right ) \, dx\\ &=\frac {1}{2 x^2}+\frac {\log (x)}{x^2}+\frac {2 \log \left (2+\frac {x}{4}\right ) \log (x)}{x}-\frac {\log ^2(x)}{4}+\frac {\log ^2(x)}{x^2}+\frac {1}{4} \log (x) \log (8+x)+2 \left (\frac {1}{4} \int \frac {\log (x)}{x} \, dx\right )-\frac {1}{4} \int \frac {\log (x)}{8+x} \, dx-\frac {1}{4} \int \frac {\log (8+x)}{x} \, dx+2 \int \frac {\log \left (2+\frac {x}{4}\right )}{8+x} \, dx+2 \int \frac {\log (x)}{x^3} \, dx\\ &=-\frac {1}{4} \log (8) \log (x)-\frac {1}{4} \log \left (1+\frac {x}{8}\right ) \log (x)+\frac {2 \log \left (2+\frac {x}{4}\right ) \log (x)}{x}+\frac {\log ^2(x)}{x^2}+\frac {1}{4} \log (x) \log (8+x)+8 \operatorname {Subst}\left (\int \frac {\log (x)}{4 x} \, dx,x,2+\frac {x}{4}\right )\\ &=-\frac {1}{4} \log (8) \log (x)-\frac {1}{4} \log \left (1+\frac {x}{8}\right ) \log (x)+\frac {2 \log \left (2+\frac {x}{4}\right ) \log (x)}{x}+\frac {\log ^2(x)}{x^2}+\frac {1}{4} \log (x) \log (8+x)+2 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,2+\frac {x}{4}\right )\\ &=-\frac {1}{4} \log (8) \log (x)-\frac {1}{4} \log \left (1+\frac {x}{8}\right ) \log (x)+\frac {2 \log \left (2+\frac {x}{4}\right ) \log (x)}{x}+\frac {\log ^2(x)}{x^2}+\log ^2\left (\frac {8+x}{4}\right )+\frac {1}{4} \log (x) \log (8+x)\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [B]  time = 0.12, size = 42, normalized size = 2.10 \begin {gather*} 2 \left (\frac {\log ^2(x)}{2 x^2}+\frac {\log (x) \log \left (\frac {8+x}{4}\right )}{x}+\frac {1}{2} \log ^2\left (\frac {8+x}{4}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((-16 - 2*x)*Log[x]^2 + (16*x + 2*x^2 + 2*x^3)*Log[(8 + x)/4] + Log[x]*(16 + 2*x + 2*x^2 + (-16*x -
2*x^2)*Log[(8 + x)/4]))/(8*x^3 + x^4),x]

[Out]

2*(Log[x]^2/(2*x^2) + (Log[x]*Log[(8 + x)/4])/x + Log[(8 + x)/4]^2/2)

________________________________________________________________________________________

fricas [B]  time = 0.61, size = 32, normalized size = 1.60 \begin {gather*} \frac {x^{2} \log \left (\frac {1}{4} \, x + 2\right )^{2} + 2 \, x \log \relax (x) \log \left (\frac {1}{4} \, x + 2\right ) + \log \relax (x)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-16)*log(x)^2+((-2*x^2-16*x)*log(2+1/4*x)+2*x^2+2*x+16)*log(x)+(2*x^3+2*x^2+16*x)*log(2+1/4*x)
)/(x^4+8*x^3),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.16, size = 51, normalized size = 2.55 \begin {gather*} 2 \, {\left (\frac {\log \relax (x)}{x} + \log \left (x + 8\right )\right )} \log \left (x + 8\right ) - 4 \, \log \relax (2) \log \left (x + 8\right ) - \log \left (x + 8\right )^{2} - \frac {4 \, \log \relax (2) \log \relax (x)}{x} + \frac {\log \relax (x)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-16)*log(x)^2+((-2*x^2-16*x)*log(2+1/4*x)+2*x^2+2*x+16)*log(x)+(2*x^3+2*x^2+16*x)*log(2+1/4*x)
)/(x^4+8*x^3),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.19, size = 31, normalized size = 1.55

method	result	size
risch	\(\ln \left (2+\frac {x}{4}\right )^{2}+\frac {2 \ln \relax (x ) \ln \left (2+\frac {x}{4}\right )}{x}+\frac {\ln \relax (x )^{2}}{x^{2}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.50, size = 45, normalized size = 2.25 \begin {gather*} \frac {x^{2} \log \left (x + 8\right )^{2} - 4 \, x \log \relax (2) \log \relax (x) - 2 \, {\left (2 \, x^{2} \log \relax (2) - x \log \relax (x)\right )} \log \left (x + 8\right ) + \log \relax (x)^{2}}{x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-16)*log(x)^2+((-2*x^2-16*x)*log(2+1/4*x)+2*x^2+2*x+16)*log(x)+(2*x^3+2*x^2+16*x)*log(2+1/4*x)
)/(x^4+8*x^3),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.70, size = 17, normalized size = 0.85 \begin {gather*} \frac {{\left (\ln \relax (x)+x\,\ln \left (\frac {x}{4}+2\right )\right )}^2}{x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(2*x - log(x/4 + 2)*(16*x + 2*x^2) + 2*x^2 + 16) + log(x/4 + 2)*(16*x + 2*x^2 + 2*x^3) - log(x)^2*
(2*x + 16))/(8*x^3 + x^4),x)

[Out]

(log(x) + x*log(x/4 + 2))^2/x^2

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-16)*ln(x)**2+((-2*x**2-16*x)*ln(2+1/4*x)+2*x**2+2*x+16)*ln(x)+(2*x**3+2*x**2+16*x)*ln(2+1/4*x
))/(x**4+8*x**3),x)

[Out]

Timed out

________________________________________________________________________________________