3.33.39 $\int \frac{9x^2-15x^3+9x^4+(-6x+13x^2-9x^3)\log (\frac{x^3}{-1+x})+(-2x+2x^2){\log }^2(\frac{x^3}{-1+x})}{(-2+2x){\log }^2(5)}dx$

Optimal. Leaf size=31 $$\frac{x^2{(\frac{3x}{2}-\log \left(\frac{x^3}{-1+x}\right))}^2}{2{\log }^2(5)}$$

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Rubi [A]  time = 0.46, antiderivative size = 62, normalized size of antiderivative = 2.00, number of steps used = 42, number of rules used = 16, integrand size = 75, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.213, Rules used = {12, 6742, 893, 2514, 2487, 29, 8, 2494, 2390, 2301, 2394, 2315, 2495, 43, 30, 2498} $$\begin{array}{c}\frac{9x^4}{8{\log }^2(5)}-\frac{3x^3\log (-\frac{x^3}{1-x})}{2{\log }^2(5)}+\frac{x^2{\log }^2(-\frac{x^3}{1-x})}{2{\log }^2(5)}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(9*x^2 - 15*x^3 + 9*x^4 + (-6*x + 13*x^2 - 9*x^3)*Log[x^3/(-1 + x)] + (-2*x + 2*x^2)*Log[x^3/(-1 + x)]^2)/
((-2 + 2*x)*Log[5]^2),x]

[Out]

(9*x^4)/(8*Log[5]^2) - (3*x^3*Log[-(x^3/(1 - x))])/(2*Log[5]^2) + (x^2*Log[-(x^3/(1 - x))]^2)/(2*Log[5]^2)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 30

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

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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

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 2394

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[(Log[(e*(f +
g*x))/(e*f - d*g)]*(a + b*Log[c*(d + e*x)^n]))/g, x] - Dist[(b*e*n)/g, Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2487

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.), x_Symbol] :> Simp[((
a + b*x)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/b, x] + (Dist[(q*r*s*(b*c - a*d))/b, Int[Log[e*(f*(a + b*x)^p
*(c + d*x)^q)^r]^(s - 1)/(c + d*x), x], x] - Dist[r*s*(p + q), Int[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1
), x], x]) /; FreeQ[{a, b, c, d, e, f, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && NeQ[p + q, 0] && IGtQ[s, 0] &&
LtQ[s, 4]

Rule 2494

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]/((g_.) + (h_.)*(x_)), x_Sym
bol] :> Simp[(Log[g + h*x]*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/h, x] + (-Dist[(b*p*r)/h, Int[Log[g + h*x]/(a
+ b*x), x], x] - Dist[(d*q*r)/h, Int[Log[g + h*x]/(c + d*x), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, p, q,
r}, x] && NeQ[b*c - a*d, 0]

Rule 2495

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]*((g_.) + (h_.)*(x_))^(m_.),
x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r])/(h*(m + 1)), x] + (-Dist[(b*p*r)/(
h*(m + 1)), Int[(g + h*x)^(m + 1)/(a + b*x), x], x] - Dist[(d*q*r)/(h*(m + 1)), Int[(g + h*x)^(m + 1)/(c + d*x
), x], x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1]

Rule 2498

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_)*((g_.) + (h_.)*(x_))^(
m_.), x_Symbol] :> Simp[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s)/(h*(m + 1)), x] + (-Dist[(b
*p*r*s)/(h*(m + 1)), Int[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/(a + b*x), x], x] -
Dist[(d*q*r*s)/(h*(m + 1)), Int[((g + h*x)^(m + 1)*Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^(s - 1))/(c + d*x), x]
, x]) /; FreeQ[{a, b, c, d, e, f, g, h, m, p, q, r, s}, x] && NeQ[b*c - a*d, 0] && IGtQ[s, 0] && NeQ[m, -1]

Rule 2514

Int[Log[(e_.)*((f_.)*((a_.) + (b_.)*(x_))^(p_.)*((c_.) + (d_.)*(x_))^(q_.))^(r_.)]^(s_.)*(RFx_), x_Symbol] :>
With[{u = ExpandIntegrand[Log[e*(f*(a + b*x)^p*(c + d*x)^q)^r]^s, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a,
b, c, d, e, f, p, q, r, s}, x] && RationalFunctionQ[RFx, x] && IGtQ[s, 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{\int \frac{9x^2-15x^3+9x^4+(-6x+13x^2-9x^3)\log \left(\frac{x^3}{-1+x}\right)+(-2x+2x^2){\log }^2\left(\frac{x^3}{-1+x}\right)}{-2+2x}dx}{{\log }^2(5)}\\ & =\frac{\int (\frac{3x^2(3-5x+3x^2)}{2(-1+x)}-\frac{x(6-13x+9x^2)\log \left(\frac{x^3}{-1+x}\right)}{2(-1+x)}+x{\log }^2\left(\frac{x^3}{-1+x}\right))dx}{{\log }^2(5)}\\ & =-\frac{\int \frac{x(6-13x+9x^2)\log \left(\frac{x^3}{-1+x}\right)}{-1+x}dx}{2{\log }^2(5)}+\frac{\int x{\log }^2\left(\frac{x^3}{-1+x}\right)dx}{{\log }^2(5)}+\frac{3\int \frac{x^2(3-5x+3x^2)}{-1+x}dx}{2{\log }^2(5)}\\ & =\frac{x^2{\log }^2(-\frac{x^3}{1-x})}{2{\log }^2(5)}-\frac{\int (2\log \left(\frac{x^3}{-1+x}\right)+\frac{2\log \left(\frac{x^3}{-1+x}\right)}{-1+x}-4x\log \left(\frac{x^3}{-1+x}\right)+9x^2\log \left(\frac{x^3}{-1+x}\right))dx}{2{\log }^2(5)}+\frac{\int \frac{x^2\log \left(\frac{x^3}{-1+x}\right)}{-1+x}dx}{{\log }^2(5)}+\frac{3\int (1+\frac{1}{-1+x}+x-2x^2+3x^3)dx}{2{\log }^2(5)}-\frac{3\int x\log \left(\frac{x^3}{-1+x}\right)dx}{{\log }^2(5)}\\ & =\frac{3x}{2{\log }^2(5)}+\frac{3x^2}{4{\log }^2(5)}-\frac{x^3}{{\log }^2(5)}+\frac{9x^4}{8{\log }^2(5)}+\frac{3\log (1-x)}{2{\log }^2(5)}-\frac{3x^2\log (-\frac{x^3}{1-x})}{2{\log }^2(5)}+\frac{x^2{\log }^2(-\frac{x^3}{1-x})}{2{\log }^2(5)}-\frac{\int \log \left(\frac{x^3}{-1+x}\right)dx}{{\log }^2(5)}-\frac{\int \frac{\log \left(\frac{x^3}{-1+x}\right)}{-1+x}dx}{{\log }^2(5)}+\frac{\int (\log \left(\frac{x^3}{-1+x}\right)+\frac{\log \left(\frac{x^3}{-1+x}\right)}{-1+x}+x\log \left(\frac{x^3}{-1+x}\right))dx}{{\log }^2(5)}-\frac{3\int \frac{x^2}{-1+x}dx}{2{\log }^2(5)}+\frac{2\int x\log \left(\frac{x^3}{-1+x}\right)dx}{{\log }^2(5)}+\frac{9\int xdx}{2{\log }^2(5)}-\frac{9\int x^2\log \left(\frac{x^3}{-1+x}\right)dx}{2{\log }^2(5)}\\ & =\frac{3x}{2{\log }^2(5)}+\frac{3x^2}{{\log }^2(5)}-\frac{x^3}{{\log }^2(5)}+\frac{9x^4}{8{\log }^2(5)}+\frac{3\log (1-x)}{2{\log }^2(5)}+\frac{(1-x)\log (-\frac{x^3}{1-x})}{{\log }^2(5)}-\frac{x^2\log (-\frac{x^3}{1-x})}{2{\log }^2(5)}-\frac{3x^3\log (-\frac{x^3}{1-x})}{2{\log }^2(5)}-\frac{\log (-1+x)\log (-\frac{x^3}{1-x})}{{\log }^2(5)}+\frac{x^2{\log }^2(-\frac{x^3}{1-x})}{2{\log }^2(5)}+\frac{\int \frac{x^2}{-1+x}dx}{{\log }^2(5)}-\frac{\int \frac{\log (-1+x)}{-1+x}dx}{{\log }^2(5)}+\frac{\int \log \left(\frac{x^3}{-1+x}\right)dx}{{\log }^2(5)}+\frac{\int \frac{\log \left(\frac{x^3}{-1+x}\right)}{-1+x}dx}{{\log }^2(5)}+\frac{\int x\log \left(\frac{x^3}{-1+x}\right)dx}{{\log }^2(5)}-\frac{3\int \frac{x^3}{-1+x}dx}{2{\log }^2(5)}-\frac{3\int (1+\frac{1}{-1+x}+x)dx}{2{\log }^2(5)}+\frac{2\int 1dx}{{\log }^2(5)}-\frac{3\int \frac{1}{x}dx}{{\log }^2(5)}-\frac{3\int xdx}{{\log }^2(5)}+\frac{3\int \frac{\log (-1+x)}{x}dx}{{\log }^2(5)}+\frac{9\int x^{2}dx}{2{\log }^2(5)}\\ & =\frac{2x}{{\log }^2(5)}+\frac{3x^2}{4{\log }^2(5)}+\frac{x^3}{2{\log }^2(5)}+\frac{9x^4}{8{\log }^2(5)}-\frac{3\log (x)}{{\log }^2(5)}+\frac{3\log (-1+x)\log (x)}{{\log }^2(5)}-\frac{3x^3\log (-\frac{x^3}{1-x})}{2{\log }^2(5)}+\frac{x^2{\log }^2(-\frac{x^3}{1-x})}{2{\log }^2(5)}+\frac{\int \frac{x^2}{-1+x}dx}{2{\log }^2(5)}+\frac{\int (1+\frac{1}{-1+x}+x)dx}{{\log }^2(5)}+\frac{\int \frac{\log (-1+x)}{-1+x}dx}{{\log }^2(5)}-\frac{\mathrm{Subst}(\int \frac{\log (x)}{x}dx,x,-1+x)}{{\log }^2(5)}-\frac{3\int xdx}{2{\log }^2(5)}-\frac{3\int (1+\frac{1}{-1+x}+x+x^2)dx}{2{\log }^2(5)}-\frac{2\int 1dx}{{\log }^2(5)}+\frac{3\int \frac{1}{x}dx}{{\log }^2(5)}-\frac{3\int \frac{\log (-1+x)}{x}dx}{{\log }^2(5)}-\frac{3\int \frac{\log (x)}{-1+x}dx}{{\log }^2(5)}\\ & =-\frac{x}{2{\log }^2(5)}-\frac{x^2}{4{\log }^2(5)}+\frac{9x^4}{8{\log }^2(5)}-\frac{\log (1-x)}{2{\log }^2(5)}-\frac{{\log }^2(-1+x)}{2{\log }^2(5)}-\frac{3x^3\log (-\frac{x^3}{1-x})}{2{\log }^2(5)}+\frac{x^2{\log }^2(-\frac{x^3}{1-x})}{2{\log }^2(5)}+\frac{3{\text{Li}}_2(1-x)}{{\log }^2(5)}+\frac{\int (1+\frac{1}{-1+x}+x)dx}{2{\log }^2(5)}+\frac{\mathrm{Subst}(\int \frac{\log (x)}{x}dx,x,-1+x)}{{\log }^2(5)}+\frac{3\int \frac{\log (x)}{-1+x}dx}{{\log }^2(5)}\\ & =\frac{9x^4}{8{\log }^2(5)}-\frac{3x^3\log (-\frac{x^3}{1-x})}{2{\log }^2(5)}+\frac{x^2{\log }^2(-\frac{x^3}{1-x})}{2{\log }^2(5)}\end{array}\end{array}$$

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Mathematica [C]  time = 0.12, size = 142, normalized size = 4.58 $$\begin{array}{c}\frac{\frac{9x^4}{4}-3\log (1-x)+{\log }^2(1-x)+3\log (-1+x)-{\log }^2(-1+x)+6\log (-1+x)\log (x)-3x^3\log (-\frac{x^3}{1-x})+2\log (1-x)\log (-\frac{x^3}{1-x})+x^2{\log }^2(-\frac{x^3}{1-x})-2\log (-1+x)\log \left(\frac{x^3}{-1+x}\right)+6{\text{Li}}_2(1-x)+6{\text{Li}}_2(x)}{2{\log }^2(5)}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(9*x^2 - 15*x^3 + 9*x^4 + (-6*x + 13*x^2 - 9*x^3)*Log[x^3/(-1 + x)] + (-2*x + 2*x^2)*Log[x^3/(-1 + x
)]^2)/((-2 + 2*x)*Log[5]^2),x]

[Out]

((9*x^4)/4 - 3*Log[1 - x] + Log[1 - x]^2 + 3*Log[-1 + x] - Log[-1 + x]^2 + 6*Log[-1 + x]*Log[x] - 3*x^3*Log[-(
x^3/(1 - x))] + 2*Log[1 - x]*Log[-(x^3/(1 - x))] + x^2*Log[-(x^3/(1 - x))]^2 - 2*Log[-1 + x]*Log[x^3/(-1 + x)]
+ 6*PolyLog[2, 1 - x] + 6*PolyLog[2, x])/(2*Log[5]^2)

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fricas [A]  time = 0.52, size = 44, normalized size = 1.42 $$\begin{array}{c}\frac{9x^4-12x^3\log \left(\frac{x^3}{x-1}\right)+4x^2\log {\left(\frac{x^3}{x-1}\right)}^2}{8\log (5{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-2*x)*log(x^3/(x-1))^2+(-9*x^3+13*x^2-6*x)*log(x^3/(x-1))+9*x^4-15*x^3+9*x^2)/(2*x-2)/log(5)^
2,x, algorithm="fricas")

[Out]

1/8*(9*x^4 - 12*x^3*log(x^3/(x - 1)) + 4*x^2*log(x^3/(x - 1))^2)/log(5)^2

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giac [A]  time = 0.32, size = 44, normalized size = 1.42 $$\begin{array}{c}\frac{9x^4-12x^3\log \left(\frac{x^3}{x-1}\right)+4x^2\log {\left(\frac{x^3}{x-1}\right)}^2}{8\log (5{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-2*x)*log(x^3/(x-1))^2+(-9*x^3+13*x^2-6*x)*log(x^3/(x-1))+9*x^4-15*x^3+9*x^2)/(2*x-2)/log(5)^
2,x, algorithm="giac")

[Out]

1/8*(9*x^4 - 12*x^3*log(x^3/(x - 1)) + 4*x^2*log(x^3/(x - 1))^2)/log(5)^2

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maple [A]  time = 0.52, size = 51, normalized size = 1.65

method	result	size
risch	$\frac{\ln {\left(\frac{x^3}{x-1}\right)}^2{x}^2}{2\ln (5{)}^2}-\frac{3\ln \left(\frac{x^3}{x-1}\right)x^3}{2\ln (5{)}^2}+\frac{9x^4}{8\ln (5{)}^2}$	$51$
norman	$\frac{\frac{9x^4}{8\ln (5)}+\frac{x^2\ln {\left(\frac{x^3}{x-1}\right)}^2}{2\ln (5)}-\frac{3x^3\ln \left(\frac{x^3}{x-1}\right)}{2\ln (5)}}{\ln (5)}$	$56$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-2*x)*ln(x^3/(x-1))^2+(-9*x^3+13*x^2-6*x)*ln(x^3/(x-1))+9*x^4-15*x^3+9*x^2)/(2*x-2)/ln(5)^2,x,metho
d=_RETURNVERBOSE)

[Out]

1/2/ln(5)^2*ln(x^3/(x-1))^2*x^2-3/2/ln(5)^2*ln(x^3/(x-1))*x^3+9/8/ln(5)^2*x^4

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maxima [B]  time = 0.88, size = 63, normalized size = 2.03 $$\begin{array}{c}\frac{9x^4+4x^2\log {(x-1)}^2-36x^3\log (x)+36x^2\log (x{)}^2+12(x^3-2x^2\log (x)-1)\log (x-1)+12\log (x-1)}{8\log (5{)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-2*x)*log(x^3/(x-1))^2+(-9*x^3+13*x^2-6*x)*log(x^3/(x-1))+9*x^4-15*x^3+9*x^2)/(2*x-2)/log(5)^
2,x, algorithm="maxima")

[Out]

1/8*(9*x^4 + 4*x^2*log(x - 1)^2 - 36*x^3*log(x) + 36*x^2*log(x)^2 + 12*(x^3 - 2*x^2*log(x) - 1)*log(x - 1) + 1
2*log(x - 1))/log(5)^2

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mupad [B]  time = 2.13, size = 27, normalized size = 0.87 $$\begin{array}{c}\frac{x^2{(3x-2\ln \left(\frac{x^3}{x-1}\right))}^2}{8{\ln (5)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x^3/(x - 1))^2*(2*x - 2*x^2) - 9*x^2 + 15*x^3 - 9*x^4 + log(x^3/(x - 1))*(6*x - 13*x^2 + 9*x^3))/(lo
g(5)^2*(2*x - 2)),x)

[Out]

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

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sympy [B]  time = 0.26, size = 51, normalized size = 1.65 $$\begin{array}{c}\frac{9x^4}{8\log {(5)}^2}-\frac{3x^3\log \left(\frac{x^3}{x-1}\right)}{2\log {(5)}^2}+\frac{x^2\log {\left(\frac{x^3}{x-1}\right)}^2}{2\log {(5)}^2}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**2-2*x)*ln(x**3/(x-1))**2+(-9*x**3+13*x**2-6*x)*ln(x**3/(x-1))+9*x**4-15*x**3+9*x**2)/(2*x-2)/
ln(5)**2,x)

[Out]

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

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