3.21.57 \(\int \frac {x+(4+5 x+x^2) \log (\frac {1}{8} (-4-x))}{(4 x+x^2) \log (\frac {1}{8} (-4-x))} \, dx\)

Optimal. Leaf size=19 \[ 1+x+\log \left (\frac {12}{25} x \log \left (\frac {1}{8} (-4-x)\right )\right ) \]

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Rubi [A]  time = 0.35, antiderivative size = 15, normalized size of antiderivative = 0.79, number of steps used = 10, number of rules used = 8, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.186, Rules used = {1593, 6688, 6742, 43, 2390, 12, 2302, 29} \begin {gather*} x+\log (x)+\log \left (\log \left (\frac {1}{8} (-x-4)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x + (4 + 5*x + x^2)*Log[(-4 - x)/8])/((4*x + x^2)*Log[(-4 - x)/8]),x]

[Out]

x + Log[x] + Log[Log[(-4 - x)/8]]

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 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 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 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 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 {x+\left (4+5 x+x^2\right ) \log \left (\frac {1}{8} (-4-x)\right )}{x (4+x) \log \left (\frac {1}{8} (-4-x)\right )} \, dx\\ &=\int \frac {5+\frac {4}{x}+x+\frac {1}{\log \left (\frac {1}{8} (-4-x)\right )}}{4+x} \, dx\\ &=\int \left (\frac {1+x}{x}+\frac {1}{(4+x) \log \left (-\frac {1}{2}-\frac {x}{8}\right )}\right ) \, dx\\ &=\int \frac {1+x}{x} \, dx+\int \frac {1}{(4+x) \log \left (-\frac {1}{2}-\frac {x}{8}\right )} \, dx\\ &=-\left (8 \operatorname {Subst}\left (\int -\frac {1}{8 x \log (x)} \, dx,x,-\frac {1}{2}-\frac {x}{8}\right )\right )+\int \left (1+\frac {1}{x}\right ) \, dx\\ &=x+\log (x)+\operatorname {Subst}\left (\int \frac {1}{x \log (x)} \, dx,x,-\frac {1}{2}-\frac {x}{8}\right )\\ &=x+\log (x)+\operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,\log \left (\frac {1}{8} (-4-x)\right )\right )\\ &=x+\log (x)+\log \left (\log \left (\frac {1}{8} (-4-x)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 15, normalized size = 0.79 \begin {gather*} x+\log (x)+\log \left (\log \left (\frac {1}{8} (-4-x)\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x + (4 + 5*x + x^2)*Log[(-4 - x)/8])/((4*x + x^2)*Log[(-4 - x)/8]),x]

[Out]

x + Log[x] + Log[Log[(-4 - x)/8]]

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fricas [A]  time = 0.61, size = 11, normalized size = 0.58 \begin {gather*} x + \log \relax (x) + \log \left (\log \left (-\frac {1}{8} \, x - \frac {1}{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5*x+4)*log(-1/8*x-1/2)+x)/(x^2+4*x)/log(-1/8*x-1/2),x, algorithm="fricas")

[Out]

x + log(x) + log(log(-1/8*x - 1/2))

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giac [A]  time = 0.26, size = 11, normalized size = 0.58 \begin {gather*} x + \log \relax (x) + \log \left (\log \left (-\frac {1}{8} \, x - \frac {1}{2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5*x+4)*log(-1/8*x-1/2)+x)/(x^2+4*x)/log(-1/8*x-1/2),x, algorithm="giac")

[Out]

x + log(x) + log(log(-1/8*x - 1/2))

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maple [A]  time = 0.35, size = 12, normalized size = 0.63

method	result	size
norman	\(x +\ln \relax (x )+\ln \left (\ln \left (-\frac {x}{8}-\frac {1}{2}\right )\right )\)	\(12\)
risch	\(x +\ln \relax (x )+\ln \left (\ln \left (-\frac {x}{8}-\frac {1}{2}\right )\right )\)	\(12\)
derivativedivides	\(\ln \left (\ln \left (-\frac {x}{8}-\frac {1}{2}\right )\right )+x +4+\ln \left (-\frac {x}{4}\right )\)	\(15\)
default	\(\ln \left (\ln \left (-\frac {x}{8}-\frac {1}{2}\right )\right )+x +4+\ln \left (-\frac {x}{4}\right )\)	\(15\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+5*x+4)*ln(-1/8*x-1/2)+x)/(x^2+4*x)/ln(-1/8*x-1/2),x,method=_RETURNVERBOSE)

[Out]

x+ln(x)+ln(ln(-1/8*x-1/2))

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maxima [B]  time = 0.74, size = 77, normalized size = 4.05 \begin {gather*} 5 \, {\left (3 \, \log \relax (2) - \log \left (-x - 4\right )\right )} \log \left (-3 \, \log \relax (2) + \log \left (-x - 4\right )\right ) + 5 \, \log \left (-\frac {1}{8} \, x - \frac {1}{2}\right ) \log \left (-3 \, \log \relax (2) + \log \left (-x - 4\right )\right ) + x - 5 \, \log \left (x + 4\right ) + \log \relax (x) + 5 \, \log \left (-x - 4\right ) + \log \left (-3 \, \log \relax (2) + \log \left (-x - 4\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+5*x+4)*log(-1/8*x-1/2)+x)/(x^2+4*x)/log(-1/8*x-1/2),x, algorithm="maxima")

[Out]

5*(3*log(2) - log(-x - 4))*log(-3*log(2) + log(-x - 4)) + 5*log(-1/8*x - 1/2)*log(-3*log(2) + log(-x - 4)) + x
- 5*log(x + 4) + log(x) + 5*log(-x - 4) + log(-3*log(2) + log(-x - 4))

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mupad [B]  time = 1.30, size = 11, normalized size = 0.58 \begin {gather*} x+\ln \left (\ln \left (-\frac {x}{8}-\frac {1}{2}\right )\right )+\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + log(- x/8 - 1/2)*(5*x + x^2 + 4))/(log(- x/8 - 1/2)*(4*x + x^2)),x)

[Out]

x + log(log(- x/8 - 1/2)) + log(x)

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sympy [A]  time = 0.13, size = 15, normalized size = 0.79 \begin {gather*} x + \log {\relax (x )} + \log {\left (\log {\left (- \frac {x}{8} - \frac {1}{2} \right )} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+5*x+4)*ln(-1/8*x-1/2)+x)/(x**2+4*x)/ln(-1/8*x-1/2),x)

[Out]

x + log(x) + log(log(-x/8 - 1/2))

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