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

Optimal. Leaf size=22 $$(x+\log (x)(x+\frac{{\log }^2(x)}{4}))(x+\log (3x))$$

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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} $$\begin{array}{c}-\frac{x^2}{2}+(x^2+x)\log (x)+\frac{1}{6}(3x+1{)}^2-x+\frac{1}{4}x{\log }^3(x)+\frac{1}{4}\log (3x){\log }^3(x)-x\log (x)+x\log (3x)\log (x)+x\log (3x)\end{array}$$

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{array}{c}\begin{array}{rl}\text{integral}& =\frac{1}{4}\int \frac{4x+12x^2+(4x+8x^2)\log (x)+3x{\log }^2(x)+(1+x){\log }^3(x)+(8x+4x\log (x)+3{\log }^2(x))\log (3x)}{x}dx\\ & =\frac{1}{4}\int (\frac{4x+12x^2+4x\log (x)+8x^2\log (x)+3x{\log }^2(x)+{\log }^3(x)+x{\log }^3(x)}{x}+\frac{(8x+4x\log (x)+3{\log }^2(x))\log (3x)}{x})dx\\ & =\frac{1}{4}\int \frac{4x+12x^2+4x\log (x)+8x^2\log (x)+3x{\log }^2(x)+{\log }^3(x)+x{\log }^3(x)}{x}dx+\frac{1}{4}\int \frac{(8x+4x\log (x)+3{\log }^2(x))\log (3x)}{x}dx\\ & =\frac{1}{4}\int (4(1+3x)+4(1+2x)\log (x)+3{\log }^2(x)+\frac{(1+x){\log }^3(x)}{x})dx+\frac{1}{4}\int (8\log (3x)+4\log (x)\log (3x)+\frac{3{\log }^2(x)\log (3x)}{x})dx\\ & =\frac{1}{6}(1+3x{)}^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 (3x)}{x}dx+2\int \log (3x)dx+\int (1+2x)\log (x)dx+\int \log (x)\log (3x)dx\\ & =-2x+\frac{1}{6}(1+3x{)}^2+(x+x^2)\log (x)+\frac{3}{4}x{\log }^2(x)+x\log (3x)+x\log (x)\log (3x)+\frac{1}{4}{\log }^3(x)\log (3x)+\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)}{3x}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+3x{)}^2-\frac{3}{2}x\log (x)+(x+x^2)\log (x)+\frac{3}{4}x{\log }^2(x)+\frac{1}{4}x{\log }^3(x)+x\log (3x)+x\log (x)\log (3x)+\frac{1}{4}{\log }^3(x)\log (3x)-\frac{1}{4}\int \frac{{\log }^3(x)}{x}dx+\frac{1}{4}\mathrm{Subst}(\int x^{3}dx,x,\log (x))-\frac{3}{4}\int {\log }^2(x)dx-\int \log (x)dx\\ & =\frac{x}{2}-\frac{x^2}{2}+\frac{1}{6}(1+3x{)}^2-\frac{5}{2}x\log (x)+(x+x^2)\log (x)+\frac{1}{4}x{\log }^3(x)+\frac{{\log }^4(x)}{16}+x\log (3x)+x\log (x)\log (3x)+\frac{1}{4}{\log }^3(x)\log (3x)-\frac{1}{4}\mathrm{Subst}(\int x^{3}dx,x,\log (x))+\frac{3}{2}\int \log (x)dx\\ & =-x-\frac{x^2}{2}+\frac{1}{6}(1+3x{)}^2-x\log (x)+(x+x^2)\log (x)+\frac{1}{4}x{\log }^3(x)+x\log (3x)+x\log (x)\log (3x)+\frac{1}{4}{\log }^3(x)\log (3x)\end{array}\end{array}$$

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

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 $$\begin{array}{c}\frac{1}{4}(x+\log (3))\log (x{)}^3+\frac{1}{4}\log (x{)}^4+x\log (x{)}^2+x^2+x\log (3)+(x^2+x\log (3)+x)\log (x)\end{array}$$

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 $$\begin{array}{c}\frac{1}{4}(x+\log (3))\log (x{)}^3+\frac{1}{4}\log (x{)}^4+x\log (x{)}^2+x^2+x\log (3)+(x^2+x(\log (3)+1))\log (x)\end{array}$$

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	$\frac{\ln (x{)}^4}{4}+\frac{(\ln (3)+x)\ln (x{)}^3}{4}+x\ln (x{)}^2+\frac{(4x\ln (3)+4x^2+4x)\ln (x)}{4}+x\ln (3)+x^2$	$49$
default	$\frac{x\ln (x{)}^3}{4}+x\ln (x{)}^2+x\ln (x)+\frac{\ln (x{)}^4}{4}+\frac{\ln (3)\ln (x{)}^3}{4}+\ln (3)(x\ln (x)-x)+x^2\ln (x)+x^2+2x\ln (3)$	$58$

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 $$\begin{array}{c}\frac{1}{16}\log {\left(3x\right)}^4-\frac{1}{4}\log {\left(3x\right)}^3\log (x)+\frac{3}{8}\log {\left(3x\right)}^2\log (x{)}^2+\frac{1}{16}\log (x{)}^4+x^2\log (x)+\frac{1}{4}(\log (x{)}^3-3\log (x{)}^2+6\log (x)-6)x+\frac{3}{4}(\log (x{)}^2-2\log (x)+2)x+x^2-x(\log (3)-2)+2x\log \left(3x\right)+(x\log \left(3x\right)-x)\log (x)-2x\end{array}$$

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 $$\begin{array}{c}\frac{(x+\ln (3)+\ln (x))({\ln (x)}^3+4x\ln (x)+4x)}{4}\end{array}$$

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 $$\begin{array}{c}{x}^2+x\log {(x)}^2+x\log (3)+(\frac{x}{4}+\frac{\log (3)}{4})\log {(x)}^3+(x^2+x+x\log (3))\log (x)+\frac{\log {(x)}^4}{4}\end{array}$$

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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