3.85.99 $\int \frac{12x^2+4x^3-6x^4-12x\log (16-48x+36x^2)+(-2+3x){\log }^2(16-48x+36x^2)}{-2x^2+3x^3}dx$

Optimal. Leaf size=31 $$-x^2+\frac{x+x^2-{(x-\log ((4-6x{)}^2))}^2}{x}$$

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Rubi [A]  time = 0.41, antiderivative size = 54, normalized size of antiderivative = 1.74, number of steps used = 13, number of rules used = 11, integrand size = 63, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.175, Rules used = {1593, 6742, 698, 2411, 2344, 2301, 2317, 2391, 2397, 2394, 2315} $$\begin{array}{c}-x^2-\frac{3}{2}{\log }^2(4(2-3x{)}^2)-\frac{(2-3x){\log }^2(4(2-3x{)}^2)}{2x}+4\log (2-3x)\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(12*x^2 + 4*x^3 - 6*x^4 - 12*x*Log[16 - 48*x + 36*x^2] + (-2 + 3*x)*Log[16 - 48*x + 36*x^2]^2)/(-2*x^2 + 3
*x^3),x]

[Out]

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

Rule 698

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - 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 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 2317

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

Rule 2344

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

Rule 2391

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

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 2397

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

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{12x^2+4x^3-6x^4-12x\log (16-48x+36x^2)+(-2+3x){\log }^2(16-48x+36x^2)}{x^2(-2+3x)}dx\\ & =\int (-\frac{2(-6-2x+3x^2)}{-2+3x}-\frac{12\log (4(2-3x{)}^2)}{x(-2+3x)}+\frac{{\log }^2(4(2-3x{)}^2)}{x^2})dx\\ & =-(2\int \frac{-6-2x+3x^2}{-2+3x}dx)-12\int \frac{\log (4(2-3x{)}^2)}{x(-2+3x)}dx+\int \frac{{\log }^2(4(2-3x{)}^2)}{x^2}dx\\ & =-\frac{(2-3x){\log }^2(4(2-3x{)}^2)}{2x}-2\int (x-\frac{6}{-2+3x})dx-4\mathrm{Subst}(\int \frac{\log \left(4x^2\right)}{(\frac{2}{3}-\frac{x}{3})x}dx,x,2-3x)-6\int \frac{\log (4(2-3x{)}^2)}{x}dx\\ & =-x^2+4\log (2-3x)-\frac{(2-3x){\log }^2(4(2-3x{)}^2)}{2x}-6\log (4(2-3x{)}^2)\log \left(\frac{3x}{2}\right)-2\mathrm{Subst}(\int \frac{\log \left(4x^2\right)}{\frac{2}{3}-\frac{x}{3}}dx,x,2-3x)-6\mathrm{Subst}(\int \frac{\log \left(4x^2\right)}{x}dx,x,2-3x)-36\int \frac{\log \left(\frac{3x}{2}\right)}{2-3x}dx\\ & =-x^2+4\log (2-3x)-\frac{3}{2}{\log }^2(4(2-3x{)}^2)-\frac{(2-3x){\log }^2(4(2-3x{)}^2)}{2x}-12{\text{Li}}_2(1-\frac{3x}{2})-12\mathrm{Subst}(\int \frac{\log (1-\frac{x}{2})}{x}dx,x,2-3x)\\ & =-x^2+4\log (2-3x)-\frac{3}{2}{\log }^2(4(2-3x{)}^2)-\frac{(2-3x){\log }^2(4(2-3x{)}^2)}{2x}\end{array}\end{array}$$

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Mathematica [A]  time = 0.08, size = 35, normalized size = 1.13 $$\begin{array}{c}-x^2+2\log (4(2-3x{)}^2)-\frac{{\log }^2(4(2-3x{)}^2)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(12*x^2 + 4*x^3 - 6*x^4 - 12*x*Log[16 - 48*x + 36*x^2] + (-2 + 3*x)*Log[16 - 48*x + 36*x^2]^2)/(-2*x
^2 + 3*x^3),x]

[Out]

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

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fricas [A]  time = 0.62, size = 36, normalized size = 1.16 $$\begin{array}{c}-\frac{x^3-2x\log (36x^2-48x+16)+\log {(36x^2-48x+16)}^2}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-2)*log(36*x^2-48*x+16)^2-12*x*log(36*x^2-48*x+16)-6*x^4+4*x^3+12*x^2)/(3*x^3-2*x^2),x, algorit
hm="fricas")

[Out]

-(x^3 - 2*x*log(36*x^2 - 48*x + 16) + log(36*x^2 - 48*x + 16)^2)/x

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giac [A]  time = 0.22, size = 32, normalized size = 1.03 $$\begin{array}{c}-x^2-\frac{\log {(36x^2-48x+16)}^2}{x}+4\log (3x-2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-2)*log(36*x^2-48*x+16)^2-12*x*log(36*x^2-48*x+16)-6*x^4+4*x^3+12*x^2)/(3*x^3-2*x^2),x, algorit
hm="giac")

[Out]

-x^2 - log(36*x^2 - 48*x + 16)^2/x + 4*log(3*x - 2)

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maple [A]  time = 0.38, size = 33, normalized size = 1.06

method	result	size
risch	$-\frac{\ln {(36x^2-48x+16)}^2}{x}-x^2+4\ln (3x-2)$	$33$
norman	$\frac{-x^3-\ln {(36x^2-48x+16)}^2}{x}+4\ln (3x-2)$	$35$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x-2)*ln(36*x^2-48*x+16)^2-12*x*ln(36*x^2-48*x+16)-6*x^4+4*x^3+12*x^2)/(3*x^3-2*x^2),x,method=_RETURNVE
RBOSE)

[Out]

-1/x*ln(36*x^2-48*x+16)^2-x^2+4*ln(3*x-2)

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maxima [A]  time = 0.50, size = 42, normalized size = 1.35 $$\begin{array}{c}-x^2-\frac{4(\log (2{)}^2+2\log (2)\log (3x-2)+\log {(3x-2)}^2)}{x}+4\log (3x-2)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-2)*log(36*x^2-48*x+16)^2-12*x*log(36*x^2-48*x+16)-6*x^4+4*x^3+12*x^2)/(3*x^3-2*x^2),x, algorit
hm="maxima")

[Out]

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

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mupad [B]  time = 0.18, size = 34, normalized size = 1.10 $$\begin{array}{c}2\ln \left({(3x-2)}^2\right)-\frac{{\ln (36x^2-48x+16)}^2}{x}-x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(36*x^2 - 48*x + 16)^2*(3*x - 2) - 12*x*log(36*x^2 - 48*x + 16) + 12*x^2 + 4*x^3 - 6*x^4)/(2*x^2 - 3*
x^3),x)

[Out]

2*log((3*x - 2)^2) - log(36*x^2 - 48*x + 16)^2/x - x^2

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sympy [A]  time = 0.17, size = 26, normalized size = 0.84 $$\begin{array}{c}-x^2+4\log (3x-2)-\frac{\log {(36x^2-48x+16)}^2}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x-2)*ln(36*x**2-48*x+16)**2-12*x*ln(36*x**2-48*x+16)-6*x**4+4*x**3+12*x**2)/(3*x**3-2*x**2),x)

[Out]

-x**2 + 4*log(3*x - 2) - log(36*x**2 - 48*x + 16)**2/x

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