3.55.57 \(\int \frac {5+x+2 \log (4)+2 \log (\frac {1}{4} (5+x+2 \log (4)))}{5+x+2 \log (4)} \, dx\)

Optimal. Leaf size=28 \[ -4+x+\log ^2\left (1+\frac {1-x}{4}+x+\frac {1}{2} (-x+\log (4))\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 20, normalized size of antiderivative = 0.71, number of steps used = 7, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {6688, 6742, 43, 2390, 2301} \begin {gather*} x+\log ^2(x+5+\log (16))-\log (16) \log (x+5+\log (16)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - Log[16]*Log[5 + x + Log[16]] + Log[5 + x + Log[16]]^2

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 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 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 {5+x+2 \log (5+x+\log (16))}{5+x+\log (16)} \, dx\\ &=\int \left (\frac {5+x}{5+x+\log (16)}+\frac {2 \log (5+x+\log (16))}{5+x+\log (16)}\right ) \, dx\\ &=2 \int \frac {\log (5+x+\log (16))}{5+x+\log (16)} \, dx+\int \frac {5+x}{5+x+\log (16)} \, dx\\ &=2 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,5+x+\log (16)\right )+\int \left (1-\frac {\log (16)}{5+x+\log (16)}\right ) \, dx\\ &=x-\log (16) \log (5+x+\log (16))+\log ^2(5+x+\log (16))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 21, normalized size = 0.75 \begin {gather*} x+\frac {1}{4} (-\log (16)+2 \log (5+x+\log (16)))^2 \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x + (-Log[16] + 2*Log[5 + x + Log[16]])^2/4

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fricas [A]  time = 1.25, size = 12, normalized size = 0.43 \begin {gather*} \log \left (\frac {1}{4} \, x + \log \relax (2) + \frac {5}{4}\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

log(1/4*x + log(2) + 5/4)^2 + x

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giac [A]  time = 0.17, size = 24, normalized size = 0.86 \begin {gather*} -4 \, \log \relax (2) \log \left (x + 4 \, \log \relax (2) + 5\right ) + \log \left (x + 4 \, \log \relax (2) + 5\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*log(2)*log(x + 4*log(2) + 5) + log(x + 4*log(2) + 5)^2 + x

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maple [A]  time = 0.21, size = 13, normalized size = 0.46

method	result	size
norman	\(x +\ln \left (\ln \relax (2)+\frac {5}{4}+\frac {x}{4}\right )^{2}\)	\(13\)
risch	\(x +\ln \left (\ln \relax (2)+\frac {5}{4}+\frac {x}{4}\right )^{2}\)	\(13\)
derivativedivides	\(\ln \left (\ln \relax (2)+\frac {5}{4}+\frac {x}{4}\right )^{2}+4 \ln \relax (2)+5+x\)	\(18\)
default	\(\ln \left (\ln \relax (2)+\frac {5}{4}+\frac {x}{4}\right )^{2}+4 \ln \relax (2)+5+x\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x+ln(ln(2)+5/4+1/4*x)^2

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maxima [B]  time = 0.48, size = 38, normalized size = 1.36 \begin {gather*} -{\left (4 \, \log \relax (2) + 5\right )} \log \left (x + 4 \, \log \relax (2) + 5\right ) + \log \left (x + 4 \, \log \relax (2) + 5\right )^{2} + x + 5 \, \log \left (x + 4 \, \log \relax (2) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(4*log(2) + 5)*log(x + 4*log(2) + 5) + log(x + 4*log(2) + 5)^2 + x + 5*log(x + 4*log(2) + 5)

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mupad [B]  time = 3.64, size = 12, normalized size = 0.43 \begin {gather*} {\ln \left (\frac {x}{4}+\ln \relax (2)+\frac {5}{4}\right )}^2+x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + log(x/4 + log(2) + 5/4)^2

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sympy [A]  time = 0.11, size = 14, normalized size = 0.50 \begin {gather*} x + \log {\left (\frac {x}{4} + \log {\relax (2 )} + \frac {5}{4} \right )}^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x + log(x/4 + log(2) + 5/4)**2

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