3.17.54 \(\int \frac {4 x+12 x^2+(4 x+8 x^2) \log (x)+3 x \log ^2(x)+(1+x) \log ^3(x)+(8 x+4 x \log (x)+3 \log ^2(x)) \log (3 x)}{4 x} \, dx\)

Optimal. Leaf size=22 \[ \left (x+\log (x) \left (x+\frac {\log ^2(x)}{4}\right )\right ) (x+\log (3 x)) \]

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Rubi [B]  time = 0.29, antiderivative size = 70, normalized size of antiderivative = 3.18, number of steps used = 25, number of rules used = 11, integrand size = 63, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.175, Rules used = {12, 14, 2313, 2296, 2295, 2346, 2302, 30, 6742, 2361, 2366} \begin {gather*} -\frac {x^2}{2}+\left (x^2+x\right ) \log (x)+\frac {1}{6} (3 x+1)^2-x+\frac {1}{4} x \log ^3(x)+\frac {1}{4} \log (3 x) \log ^3(x)-x \log (x)+x \log (3 x) \log (x)+x \log (3 x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(4*x + 12*x^2 + (4*x + 8*x^2)*Log[x] + 3*x*Log[x]^2 + (1 + x)*Log[x]^3 + (8*x + 4*x*Log[x] + 3*Log[x]^2)*L
og[3*x])/(4*x),x]

[Out]

-x - x^2/2 + (1 + 3*x)^2/6 - x*Log[x] + (x + x^2)*Log[x] + (x*Log[x]^3)/4 + x*Log[3*x] + x*Log[x]*Log[3*x] + (
Log[x]^3*Log[3*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
 &&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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 2313

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d +
 e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a,
b, c, d, e, n, r}, x] && IGtQ[q, 0]

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 2361

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.)), x_Symbol] :> With[{u =
IntHide[(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[SimplifyIntegrand[u/x, x], x],
 x]] /; FreeQ[{a, b, c, d, e, f, n, p, r}, x]

Rule 2366

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_.)]*(e_.))*((g_.)*(x_))^(m_.), x_Sy
mbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Dist[d + e*Log[f*x^r], u, x] - Dist[e*r, Int[Sim
plifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] &&  !(EqQ[p, 1] && EqQ[a, 0] &&
 NeQ[d, 0])

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int \frac {4 x+12 x^2+\left (4 x+8 x^2\right ) \log (x)+3 x \log ^2(x)+(1+x) \log ^3(x)+\left (8 x+4 x \log (x)+3 \log ^2(x)\right ) \log (3 x)}{x} \, dx\\ &=\frac {1}{4} \int \left (\frac {4 x+12 x^2+4 x \log (x)+8 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)+x \log ^3(x)}{x}+\frac {\left (8 x+4 x \log (x)+3 \log ^2(x)\right ) \log (3 x)}{x}\right ) \, dx\\ &=\frac {1}{4} \int \frac {4 x+12 x^2+4 x \log (x)+8 x^2 \log (x)+3 x \log ^2(x)+\log ^3(x)+x \log ^3(x)}{x} \, dx+\frac {1}{4} \int \frac {\left (8 x+4 x \log (x)+3 \log ^2(x)\right ) \log (3 x)}{x} \, dx\\ &=\frac {1}{4} \int \left (4 (1+3 x)+4 (1+2 x) \log (x)+3 \log ^2(x)+\frac {(1+x) \log ^3(x)}{x}\right ) \, dx+\frac {1}{4} \int \left (8 \log (3 x)+4 \log (x) \log (3 x)+\frac {3 \log ^2(x) \log (3 x)}{x}\right ) \, dx\\ &=\frac {1}{6} (1+3 x)^2+\frac {1}{4} \int \frac {(1+x) \log ^3(x)}{x} \, dx+\frac {3}{4} \int \log ^2(x) \, dx+\frac {3}{4} \int \frac {\log ^2(x) \log (3 x)}{x} \, dx+2 \int \log (3 x) \, dx+\int (1+2 x) \log (x) \, dx+\int \log (x) \log (3 x) \, dx\\ &=-2 x+\frac {1}{6} (1+3 x)^2+\left (x+x^2\right ) \log (x)+\frac {3}{4} x \log ^2(x)+x \log (3 x)+x \log (x) \log (3 x)+\frac {1}{4} \log ^3(x) \log (3 x)+\frac {1}{4} \int \log ^3(x) \, dx+\frac {1}{4} \int \frac {\log ^3(x)}{x} \, dx-\frac {3}{4} \int \frac {\log ^3(x)}{3 x} \, dx-\frac {3}{2} \int \log (x) \, dx-\int (1+x) \, dx-\int (-1+\log (x)) \, dx\\ &=-\frac {x}{2}-\frac {x^2}{2}+\frac {1}{6} (1+3 x)^2-\frac {3}{2} x \log (x)+\left (x+x^2\right ) \log (x)+\frac {3}{4} x \log ^2(x)+\frac {1}{4} x \log ^3(x)+x \log (3 x)+x \log (x) \log (3 x)+\frac {1}{4} \log ^3(x) \log (3 x)-\frac {1}{4} \int \frac {\log ^3(x)}{x} \, dx+\frac {1}{4} \operatorname {Subst}\left (\int x^3 \, dx,x,\log (x)\right )-\frac {3}{4} \int \log ^2(x) \, dx-\int \log (x) \, dx\\ &=\frac {x}{2}-\frac {x^2}{2}+\frac {1}{6} (1+3 x)^2-\frac {5}{2} x \log (x)+\left (x+x^2\right ) \log (x)+\frac {1}{4} x \log ^3(x)+\frac {\log ^4(x)}{16}+x \log (3 x)+x \log (x) \log (3 x)+\frac {1}{4} \log ^3(x) \log (3 x)-\frac {1}{4} \operatorname {Subst}\left (\int x^3 \, dx,x,\log (x)\right )+\frac {3}{2} \int \log (x) \, dx\\ &=-x-\frac {x^2}{2}+\frac {1}{6} (1+3 x)^2-x \log (x)+\left (x+x^2\right ) \log (x)+\frac {1}{4} x \log ^3(x)+x \log (3 x)+x \log (x) \log (3 x)+\frac {1}{4} \log ^3(x) \log (3 x)\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.02, size = 70, normalized size = 3.18 \begin {gather*} x^2-\frac {1}{4} x \log (81)+\frac {1}{4} x \log (6561)+x \log (x)+x^2 \log (x)+\frac {1}{4} x \log (81) \log (x)+x \log ^2(x)+\frac {1}{4} x \log ^3(x)+\frac {1}{12} \log (27) \log ^3(x)+\frac {\log ^4(x)}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(4*x + 12*x^2 + (4*x + 8*x^2)*Log[x] + 3*x*Log[x]^2 + (1 + x)*Log[x]^3 + (8*x + 4*x*Log[x] + 3*Log[x
]^2)*Log[3*x])/(4*x),x]

[Out]

x^2 - (x*Log[81])/4 + (x*Log[6561])/4 + x*Log[x] + x^2*Log[x] + (x*Log[81]*Log[x])/4 + x*Log[x]^2 + (x*Log[x]^
3)/4 + (Log[27]*Log[x]^3)/12 + Log[x]^4/4

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fricas [A]  time = 0.69, size = 42, normalized size = 1.91 \begin {gather*} \frac {1}{4} \, {\left (x + \log \relax (3)\right )} \log \relax (x)^{3} + \frac {1}{4} \, \log \relax (x)^{4} + x \log \relax (x)^{2} + x^{2} + x \log \relax (3) + {\left (x^{2} + x \log \relax (3) + x\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((3*log(x)^2+4*x*log(x)+8*x)*log(3*x)+(x+1)*log(x)^3+3*x*log(x)^2+(8*x^2+4*x)*log(x)+12*x^2+4*x)
/x,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 43, normalized size = 1.95 \begin {gather*} \frac {1}{4} \, {\left (x + \log \relax (3)\right )} \log \relax (x)^{3} + \frac {1}{4} \, \log \relax (x)^{4} + x \log \relax (x)^{2} + x^{2} + x \log \relax (3) + {\left (x^{2} + x {\left (\log \relax (3) + 1\right )}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((3*log(x)^2+4*x*log(x)+8*x)*log(3*x)+(x+1)*log(x)^3+3*x*log(x)^2+(8*x^2+4*x)*log(x)+12*x^2+4*x)
/x,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.08, size = 49, normalized size = 2.23




method result size



risch \(\frac {\ln \relax (x )^{4}}{4}+\frac {\left (\ln \relax (3)+x \right ) \ln \relax (x )^{3}}{4}+x \ln \relax (x )^{2}+\frac {\left (4 x \ln \relax (3)+4 x^{2}+4 x \right ) \ln \relax (x )}{4}+x \ln \relax (3)+x^{2}\) \(49\)
default \(\frac {x \ln \relax (x )^{3}}{4}+x \ln \relax (x )^{2}+x \ln \relax (x )+\frac {\ln \relax (x )^{4}}{4}+\frac {\ln \relax (3) \ln \relax (x )^{3}}{4}+\ln \relax (3) \left (x \ln \relax (x )-x \right )+x^{2} \ln \relax (x )+x^{2}+2 x \ln \relax (3)\) \(58\)



Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*((3*ln(x)^2+4*x*ln(x)+8*x)*ln(3*x)+(x+1)*ln(x)^3+3*x*ln(x)^2+(8*x^2+4*x)*ln(x)+12*x^2+4*x)/x,x,method=
_RETURNVERBOSE)

[Out]

1/4*ln(x)^4+1/4*(ln(3)+x)*ln(x)^3+x*ln(x)^2+1/4*(4*x*ln(3)+4*x^2+4*x)*ln(x)+x*ln(3)+x^2

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maxima [B]  time = 0.77, size = 108, normalized size = 4.91 \begin {gather*} \frac {1}{16} \, \log \left (3 \, x\right )^{4} - \frac {1}{4} \, \log \left (3 \, x\right )^{3} \log \relax (x) + \frac {3}{8} \, \log \left (3 \, x\right )^{2} \log \relax (x)^{2} + \frac {1}{16} \, \log \relax (x)^{4} + x^{2} \log \relax (x) + \frac {1}{4} \, {\left (\log \relax (x)^{3} - 3 \, \log \relax (x)^{2} + 6 \, \log \relax (x) - 6\right )} x + \frac {3}{4} \, {\left (\log \relax (x)^{2} - 2 \, \log \relax (x) + 2\right )} x + x^{2} - x {\left (\log \relax (3) - 2\right )} + 2 \, x \log \left (3 \, x\right ) + {\left (x \log \left (3 \, x\right ) - x\right )} \log \relax (x) - 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((3*log(x)^2+4*x*log(x)+8*x)*log(3*x)+(x+1)*log(x)^3+3*x*log(x)^2+(8*x^2+4*x)*log(x)+12*x^2+4*x)
/x,x, algorithm="maxima")

[Out]

1/16*log(3*x)^4 - 1/4*log(3*x)^3*log(x) + 3/8*log(3*x)^2*log(x)^2 + 1/16*log(x)^4 + x^2*log(x) + 1/4*(log(x)^3
 - 3*log(x)^2 + 6*log(x) - 6)*x + 3/4*(log(x)^2 - 2*log(x) + 2)*x + x^2 - x*(log(3) - 2) + 2*x*log(3*x) + (x*l
og(3*x) - x)*log(x) - 2*x

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mupad [B]  time = 1.09, size = 21, normalized size = 0.95 \begin {gather*} \frac {\left (x+\ln \relax (3)+\ln \relax (x)\right )\,\left ({\ln \relax (x)}^3+4\,x\,\ln \relax (x)+4\,x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x + (3*x*log(x)^2)/4 + (log(3*x)*(8*x + 3*log(x)^2 + 4*x*log(x)))/4 + (log(x)^3*(x + 1))/4 + (log(x)*(4*x
 + 8*x^2))/4 + 3*x^2)/x,x)

[Out]

((x + log(3) + log(x))*(4*x + log(x)^3 + 4*x*log(x)))/4

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sympy [B]  time = 0.30, size = 48, normalized size = 2.18 \begin {gather*} x^{2} + x \log {\relax (x )}^{2} + x \log {\relax (3 )} + \left (\frac {x}{4} + \frac {\log {\relax (3 )}}{4}\right ) \log {\relax (x )}^{3} + \left (x^{2} + x + x \log {\relax (3 )}\right ) \log {\relax (x )} + \frac {\log {\relax (x )}^{4}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*((3*ln(x)**2+4*x*ln(x)+8*x)*ln(3*x)+(x+1)*ln(x)**3+3*x*ln(x)**2+(8*x**2+4*x)*ln(x)+12*x**2+4*x)/
x,x)

[Out]

x**2 + x*log(x)**2 + x*log(3) + (x/4 + log(3)/4)*log(x)**3 + (x**2 + x + x*log(3))*log(x) + log(x)**4/4

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