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

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

________________________________________________________________________________________

Rubi [B]  time = 0.18, antiderivative size = 34, normalized size of antiderivative = 2.27, number of steps used = 10, number of rules used = 8, integrand size = 54, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {6742, 2411, 12, 2346, 2301, 2295, 2389, 2296} \begin {gather*} x-4 \left (x+\frac {1}{\log (4)}\right ) \log ^2\left (x+\frac {1}{\log (4)}\right )+\frac {4 \log ^2\left (x+\frac {1}{\log (4)}\right )}{\log (4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - 4*(x + Log[4]^(-1))*Log[x + Log[4]^(-1)]^2 + (4*Log[x + Log[4]^(-1)]^2)/Log[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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2296

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*Log[c*x^n])^p, x] - Dist[b*n*p, In
t[(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{a, b, c, n}, x] && GtQ[p, 0] && IntegerQ[2*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 2346

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.))/(x_), x_Symbol] :> Dist[d, Int[((d
+ e*x)^(q - 1)*(a + b*Log[c*x^n])^p)/x, x], x] + Dist[e, Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /
; FreeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]

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 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 (1-\frac {8 x \log (4) \log \left (x+\frac {1}{\log (4)}\right )}{1+x \log (4)}-4 \log ^2\left (x+\frac {1}{\log (4)}\right )\right ) \, dx\\ &=x-4 \int \log ^2\left (x+\frac {1}{\log (4)}\right ) \, dx-(8 \log (4)) \int \frac {x \log \left (x+\frac {1}{\log (4)}\right )}{1+x \log (4)} \, dx\\ &=x-4 \operatorname {Subst}\left (\int \log ^2(x) \, dx,x,x+\frac {1}{\log (4)}\right )-(8 \log (4)) \operatorname {Subst}\left (\int \frac {\left (x-\frac {1}{\log (4)}\right ) \log (x)}{x \log (4)} \, dx,x,x+\frac {1}{\log (4)}\right )\\ &=x-4 \left (x+\frac {1}{\log (4)}\right ) \log ^2\left (x+\frac {1}{\log (4)}\right )+8 \operatorname {Subst}\left (\int \log (x) \, dx,x,x+\frac {1}{\log (4)}\right )-8 \operatorname {Subst}\left (\int \frac {\left (x-\frac {1}{\log (4)}\right ) \log (x)}{x} \, dx,x,x+\frac {1}{\log (4)}\right )\\ &=-7 x+8 \left (x+\frac {1}{\log (4)}\right ) \log \left (x+\frac {1}{\log (4)}\right )-4 \left (x+\frac {1}{\log (4)}\right ) \log ^2\left (x+\frac {1}{\log (4)}\right )-8 \operatorname {Subst}\left (\int \log (x) \, dx,x,x+\frac {1}{\log (4)}\right )+\frac {8 \operatorname {Subst}\left (\int \frac {\log (x)}{x} \, dx,x,x+\frac {1}{\log (4)}\right )}{\log (4)}\\ &=x-4 \left (x+\frac {1}{\log (4)}\right ) \log ^2\left (x+\frac {1}{\log (4)}\right )+\frac {4 \log ^2\left (x+\frac {1}{\log (4)}\right )}{\log (4)}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.06, size = 14, normalized size = 0.93 \begin {gather*} x-4 x \log ^2\left (x+\frac {1}{\log (4)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

x - 4*x*Log[x + Log[4]^(-1)]^2

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 21, normalized size = 1.40 \begin {gather*} -4 \, x \log \left (\frac {2 \, x \log \relax (2) + 1}{2 \, \log \relax (2)}\right )^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.48, size = 56, normalized size = 3.73 \begin {gather*} 8 \, x {\left (\log \relax (2) + \log \left (\log \relax (2)\right )\right )} \log \left (2 \, x \log \relax (2) + 1\right ) - 4 \, x \log \left (2 \, x \log \relax (2) + 1\right )^{2} - {\left (4 \, \log \relax (2)^{2} + 8 \, \log \relax (2) \log \left (\log \relax (2)\right ) + 4 \, \log \left (\log \relax (2)\right )^{2} - 1\right )} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8*x*(log(2) + log(log(2)))*log(2*x*log(2) + 1) - 4*x*log(2*x*log(2) + 1)^2 - (4*log(2)^2 + 8*log(2)*log(log(2)
) + 4*log(log(2))^2 - 1)*x

________________________________________________________________________________________

maple [A]  time = 0.18, size = 22, normalized size = 1.47

method	result	size
norman	\(x -4 x \ln \left (\frac {2 x \ln \relax (2)+1}{2 \ln \relax (2)}\right )^{2}\)	\(22\)
risch	\(x -4 x \ln \left (\frac {2 x \ln \relax (2)+1}{2 \ln \relax (2)}\right )^{2}\)	\(22\)
derivativedivides	\(-4 \left (\frac {1}{2 \ln \relax (2)}+x \right ) \ln \left (\frac {1}{2 \ln \relax (2)}+x \right )^{2}+\frac {1}{2 \ln \relax (2)}+x +\frac {2 \ln \left (\frac {1}{2 \ln \relax (2)}+x \right )^{2}}{\ln \relax (2)}\)	\(47\)
default	\(-4 \left (\frac {1}{2 \ln \relax (2)}+x \right ) \ln \left (\frac {1}{2 \ln \relax (2)}+x \right )^{2}+\frac {1}{2 \ln \relax (2)}+x +\frac {2 \ln \left (\frac {1}{2 \ln \relax (2)}+x \right )^{2}}{\ln \relax (2)}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 0.83, size = 187, normalized size = 12.47 \begin {gather*} -4 \, {\left (\frac {2 \, x}{\log \relax (2)} - \frac {\log \left (2 \, x \log \relax (2) + 1\right )}{\log \relax (2)^{2}}\right )} \log \relax (2) \log \left (x + \frac {1}{2 \, \log \relax (2)}\right ) + \frac {1}{2} \, {\left (\frac {2 \, x}{\log \relax (2)} - \frac {\log \left (2 \, x \log \relax (2) + 1\right )}{\log \relax (2)^{2}}\right )} \log \relax (2) - \frac {2 \, \log \left (x + \frac {1}{2 \, \log \relax (2)}\right )^{3}}{3 \, \log \relax (2)} + \frac {2 \, {\left (\log \left (x + \frac {1}{2 \, \log \relax (2)}\right )^{3} - 3 \, {\left (\log \relax (2) \log \left (x + \frac {1}{2 \, \log \relax (2)}\right )^{2} - 2 \, \log \relax (2) \log \left (x + \frac {1}{2 \, \log \relax (2)}\right ) + 2 \, \log \relax (2)\right )} {\left (2 \, x + \frac {1}{\log \relax (2)}\right )}\right )}}{3 \, \log \relax (2)} + \frac {2 \, {\left (4 \, x \log \relax (2) - \log \left (2 \, x \log \relax (2) + 1\right )^{2} - 2 \, \log \left (2 \, x \log \relax (2) + 1\right )\right )}}{\log \relax (2)} + \frac {\log \left (2 \, x \log \relax (2) + 1\right )}{2 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*(2*x/log(2) - log(2*x*log(2) + 1)/log(2)^2)*log(2)*log(x + 1/2/log(2)) + 1/2*(2*x/log(2) - log(2*x*log(2) +
1)/log(2)^2)*log(2) - 2/3*log(x + 1/2/log(2))^3/log(2) + 2/3*(log(x + 1/2/log(2))^3 - 3*(log(2)*log(x + 1/2/l
og(2))^2 - 2*log(2)*log(x + 1/2/log(2)) + 2*log(2))*(2*x + 1/log(2)))/log(2) + 2*(4*x*log(2) - log(2*x*log(2)
+ 1)^2 - 2*log(2*x*log(2) + 1))/log(2) + 1/2*log(2*x*log(2) + 1)/log(2)

________________________________________________________________________________________

mupad [B]  time = 0.53, size = 62, normalized size = 4.13 \begin {gather*} -x\,\left (4\,{\ln \left (\ln \relax (2)\right )}^2-8\,\ln \left (2\,x\,\ln \relax (2)+1\right )\,\ln \left (\ln \relax (2)\right )+4\,{\ln \left (2\,x\,\ln \relax (2)+1\right )}^2+8\,\ln \relax (2)\,\ln \left (\ln \relax (2)\right )+4\,{\ln \relax (2)}^2-8\,\ln \left (2\,x\,\ln \relax (2)+1\right )\,\ln \relax (2)-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x*log(2) - log((x*log(2) + 1/2)/log(2))^2*(8*x*log(2) + 4) - 16*x*log((x*log(2) + 1/2)/log(2))*log(2) +
1)/(2*x*log(2) + 1),x)

[Out]

-x*(4*log(log(2))^2 - 8*log(2*x*log(2) + 1)*log(log(2)) + 4*log(2*x*log(2) + 1)^2 + 8*log(2)*log(log(2)) + 4*l
og(2)^2 - 8*log(2*x*log(2) + 1)*log(2) - 1)

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 19, normalized size = 1.27 \begin {gather*} - 4 x \log {\left (\frac {x \log {\relax (2 )} + \frac {1}{2}}{\log {\relax (2 )}} \right )}^{2} + x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________