3.64.77 \(\int \frac {8 \log (\frac {4+x}{4 x})}{4 x+x^2} \, dx\)

Optimal. Leaf size=20 \[ 5+\log (4)-\log ^2\left (\frac {1}{4} \left (1+\frac {4}{x}\right )\right ) \]

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Rubi [C]  time = 0.20, antiderivative size = 45, normalized size of antiderivative = 2.25, number of steps used = 15, number of rules used = 12, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {12, 1593, 2469, 2466, 2454, 2392, 2391, 2462, 260, 2416, 2390, 2301} \begin {gather*} 2 \text {Li}_2\left (-\frac {4}{x}\right )+2 \text {Li}_2\left (-\frac {x}{4}\right )+\log ^2(x+4)-2 \log \left (\frac {1}{x}+\frac {1}{4}\right ) \log (x+4)-4 \log (4) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-4*Log[4]*Log[x] - 2*Log[1/4 + x^(-1)]*Log[4 + x] + Log[4 + x]^2 + 2*PolyLog[2, -4/x] + 2*PolyLog[2, -1/4*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 260

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

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 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 2416

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((h_.)*(x_))^(m_.)*((f_) + (g_.)*(x_)^(r_.))^(q
_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x)^n])^p, (h*x)^m*(f + g*x^r)^q, x], x] /; FreeQ[{a,
b, c, d, e, f, g, h, m, n, p, q, r}, x] && IntegerQ[m] && IntegerQ[q]

Rule 2454

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 2462

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

Rule 2466

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_.) + (g_.)*(x_))^(r_.), x_S
ymbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e,
f, g, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]

Rule 2469

Int[((a_.) + Log[(c_.)*(v_)^(p_.)]*(b_.))^(q_.)*(u_)^(r_.)*((h_.)*(x_))^(m_.), x_Symbol] :> Int[(h*x)^m*Expand
ToSum[u, x]^r*(a + b*Log[c*ExpandToSum[v, x]^p])^q, x] /; FreeQ[{a, b, c, h, m, p, q, r}, x] && LinearQ[u, x]
&& BinomialQ[v, x] &&  !(LinearMatchQ[u, x] && BinomialMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=8 \int \frac {\log \left (\frac {4+x}{4 x}\right )}{4 x+x^2} \, dx\\ &=8 \int \frac {\log \left (\frac {4+x}{4 x}\right )}{x (4+x)} \, dx\\ &=8 \int \frac {\log \left (\frac {1}{4}+\frac {1}{x}\right )}{x (4+x)} \, dx\\ &=8 \int \left (\frac {\log \left (\frac {1}{4}+\frac {1}{x}\right )}{4 x}-\frac {\log \left (\frac {1}{4}+\frac {1}{x}\right )}{4 (4+x)}\right ) \, dx\\ &=2 \int \frac {\log \left (\frac {1}{4}+\frac {1}{x}\right )}{x} \, dx-2 \int \frac {\log \left (\frac {1}{4}+\frac {1}{x}\right )}{4+x} \, dx\\ &=-2 \log \left (\frac {1}{4}+\frac {1}{x}\right ) \log (4+x)-2 \int \frac {\log (4+x)}{\left (\frac {1}{4}+\frac {1}{x}\right ) x^2} \, dx-2 \operatorname {Subst}\left (\int \frac {\log \left (\frac {1}{4}+x\right )}{x} \, dx,x,\frac {1}{x}\right )\\ &=-2 \log (4) \log (x)-2 \log \left (\frac {1}{4}+\frac {1}{x}\right ) \log (4+x)-2 \int \left (\frac {\log (4+x)}{-4-x}+\frac {\log (4+x)}{x}\right ) \, dx-2 \operatorname {Subst}\left (\int \frac {\log (1+4 x)}{x} \, dx,x,\frac {1}{x}\right )\\ &=-2 \log (4) \log (x)-2 \log \left (\frac {1}{4}+\frac {1}{x}\right ) \log (4+x)+2 \text {Li}_2\left (-\frac {4}{x}\right )-2 \int \frac {\log (4+x)}{-4-x} \, dx-2 \int \frac {\log (4+x)}{x} \, dx\\ &=-4 \log (4) \log (x)-2 \log \left (\frac {1}{4}+\frac {1}{x}\right ) \log (4+x)+2 \text {Li}_2\left (-\frac {4}{x}\right )-2 \int \frac {\log \left (1+\frac {x}{4}\right )}{x} \, dx+2 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,4+x\right )\\ &=-4 \log (4) \log (x)-2 \log \left (\frac {1}{4}+\frac {1}{x}\right ) \log (4+x)+\log ^2(4+x)+2 \text {Li}_2\left (-\frac {4}{x}\right )+2 \text {Li}_2\left (-\frac {x}{4}\right )\\ \end {aligned} \end {gather*}

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Mathematica [C]  time = 0.02, size = 85, normalized size = 4.25 \begin {gather*} 8 \left (-\frac {1}{4} \log (4) \log (x)+\frac {1}{8} \log ^2(4+x)-\frac {1}{4} \log \left (-\frac {4}{x}\right ) \log \left (\frac {4+x}{4 x}\right )-\frac {1}{4} \log (4+x) \log \left (\frac {4+x}{4 x}\right )+\frac {\text {Li}_2\left (-\frac {x}{4}\right )}{4}-\frac {1}{4} \text {Li}_2\left (\frac {4+x}{x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 13, normalized size = 0.65 \begin {gather*} -\log \left (\frac {x + 4}{4 \, x}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(1/4*(x + 4)/x)^2

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giac [A]  time = 0.20, size = 13, normalized size = 0.65 \begin {gather*} -\log \left (\frac {x + 4}{4 \, x}\right )^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log(1/4*(x + 4)/x)^2

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maple [A]  time = 0.38, size = 11, normalized size = 0.55

method	result	size
derivativedivides	\(-\ln \left (\frac {1}{4}+\frac {1}{x}\right )^{2}\)	\(11\)
default	\(-\ln \left (\frac {1}{4}+\frac {1}{x}\right )^{2}\)	\(11\)
risch	\(-\ln \left (\frac {1}{4}+\frac {1}{x}\right )^{2}\)	\(11\)
norman	\(-\ln \left (\frac {4+x}{4 x}\right )^{2}\)	\(14\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.36, size = 39, normalized size = 1.95 \begin {gather*} \log \left (x + 4\right )^{2} - 2 \, \log \left (x + 4\right ) \log \relax (x) + \log \relax (x)^{2} - 2 \, {\left (\log \left (x + 4\right ) - \log \relax (x)\right )} \log \left (\frac {x + 4}{4 \, x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.17, size = 13, normalized size = 0.65 \begin {gather*} -{\ln \left (\frac {x+4}{4\,x}\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*log((x/4 + 1)/x))/(4*x + x^2),x)

[Out]

-log((x + 4)/(4*x))^2

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sympy [A]  time = 0.11, size = 10, normalized size = 0.50 \begin {gather*} - \log {\left (\frac {\frac {x}{4} + 1}{x} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log((x/4 + 1)/x)**2

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