3.12.84 $\int \frac{-6x+4x^2+(-3+2x)\log (\frac{1}{2}(3x-2x^2))+\log (x)(-3+4x+(-3+2x)\log (\frac{1}{2}(3x-2x^2)))}{-3+2x}dx$

Optimal. Leaf size=19 $$x(x+\log (x)\log (\frac{3x}{2}-x^2))$$

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Rubi [A]  time = 0.32, antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 11, integrand size = 65, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.169, Rules used = {6688, 2487, 29, 8, 6742, 43, 2357, 2295, 2316, 2315, 2556} $$\begin{array}{c}{x}^2+x\log (x)\log (\frac{1}{2}(3-2x)x)\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

x^2 + x*Log[x]*Log[((3 - 2*x)*x)/2]

Rule 8

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

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 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{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 2316

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

Rule 2357

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 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 2556

Int[Log[v_]*Log[w_], x_Symbol] :> Simp[x*Log[v]*Log[w], x] + (-Int[SimplifyIntegrand[(x*Log[w]*D[v, x])/v, x],
x] - Int[SimplifyIntegrand[(x*Log[v]*D[w, x])/w, x], x]) /; InverseFunctionFreeQ[v, x] && InverseFunctionFree
Q[w, x]

Rule 6688

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

Rule 6742

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

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (2x+\log (\frac{1}{2}(3-2x)x)+\frac{\log (x)(-3+4x+(-3+2x)\log (\frac{1}{2}(3-2x)x))}{-3+2x})dx\\ & =x^2+\int \log (\frac{1}{2}(3-2x)x)dx+\int \frac{\log (x)(-3+4x+(-3+2x)\log (\frac{1}{2}(3-2x)x))}{-3+2x}dx\\ & =x^2-\frac{1}{2}(3-2x)\log (\frac{1}{2}(3-2x)x)+\frac{3}{2}\int \frac{1}{x}dx-2\int 1dx+\int (\frac{(-3+4x)\log (x)}{-3+2x}+\log (x)\log (\frac{1}{2}(3-2x)x))dx\\ & =-2x+x^2+\frac{3\log (x)}{2}-\frac{1}{2}(3-2x)\log (\frac{1}{2}(3-2x)x)+\int \frac{(-3+4x)\log (x)}{-3+2x}dx+\int \log (x)\log (\frac{1}{2}(3-2x)x)dx\\ & =-2x+x^2+\frac{3\log (x)}{2}-\frac{1}{2}(3-2x)\log (\frac{1}{2}(3-2x)x)+x\log (x)\log (\frac{1}{2}(3-2x)x)-\int \frac{(3-4x)\log (x)}{3-2x}dx+\int (2\log (x)+\frac{3\log (x)}{-3+2x})dx-\int \log (\frac{1}{2}(3-2x)x)dx\\ & =-2x+x^2+\frac{3\log (x)}{2}+x\log (x)\log (\frac{1}{2}(3-2x)x)-\frac{3}{2}\int \frac{1}{x}dx+2\int 1dx+2\int \log (x)dx+3\int \frac{\log (x)}{-3+2x}dx-\int (2\log (x)+\frac{3\log (x)}{-3+2x})dx\\ & =-2x+x^2+2x\log (x)+x\log (x)\log (\frac{1}{2}(3-2x)x)+\frac{3}{2}\log \left(\frac{3}{2}\right)\log (-3+2x)-2\int \log (x)dx+3\int \frac{\log \left(\frac{2x}{3}\right)}{-3+2x}dx-3\int \frac{\log (x)}{-3+2x}dx\\ & =x^2+x\log (x)\log (\frac{1}{2}(3-2x)x)-\frac{3}{2}{\text{Li}}_2(1-\frac{2x}{3})-3\int \frac{\log \left(\frac{2x}{3}\right)}{-3+2x}dx\\ & =x^2+x\log (x)\log (\frac{1}{2}(3-2x)x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.07, size = 18, normalized size = 0.95 $$\begin{array}{c}x(x+\log (x)\log (\frac{1}{2}(3-2x)x))\end{array}$$

Antiderivative was successfully verified.

[In]

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

[Out]

x*(x + Log[x]*Log[((3 - 2*x)*x)/2])

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fricas [A]  time = 1.78, size = 18, normalized size = 0.95 $$\begin{array}{c}x\log (-x^2+\frac{3}{2}x)\log (x)+x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-3)*log(-x^2+3/2*x)+4*x-3)*log(x)+(2*x-3)*log(-x^2+3/2*x)+4*x^2-6*x)/(2*x-3),x, algorithm="fri
cas")

[Out]

x*log(-x^2 + 3/2*x)*log(x) + x^2

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giac [A]  time = 0.38, size = 27, normalized size = 1.42 $$\begin{array}{c}-x\log (2)\log (x)+x\log (x{)}^2+x\log (x)\log (-2x+3)+x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-3)*log(-x^2+3/2*x)+4*x-3)*log(x)+(2*x-3)*log(-x^2+3/2*x)+4*x^2-6*x)/(2*x-3),x, algorithm="gia
c")

[Out]

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

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maple [A]  time = 0.26, size = 19, normalized size = 1.00

method	result	size
norman	$x^2+\ln (x)\ln (-x^2+\frac{3}{2}x)x$	$19$
default	$-x\ln (2)\ln (x)+x\ln (x)\ln (-2x^2+3x)+x^2$	$26$
risch	$\ln (x)x\ln (x-\frac{3}{2})+x\ln (x{)}^2-\frac{i\ln (x)x\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}\left(i(x-\frac{3}{2})\right)\mathrm{csgn}\left(ix(x-\frac{3}{2})\right)}{2}+\frac{i\ln (x)x\pi \mathrm{csgn}\left(ix\right)\mathrm{csgn}{\left(ix(x-\frac{3}{2})\right)}^2}{2}-i\ln (x)x\pi \mathrm{csgn}{\left(ix(x-\frac{3}{2})\right)}^2+\frac{i\ln (x)x\pi \mathrm{csgn}\left(i(x-\frac{3}{2})\right)\mathrm{csgn}{\left(ix(x-\frac{3}{2})\right)}^2}{2}+\frac{i\ln (x)x\pi \mathrm{csgn}{\left(ix(x-\frac{3}{2})\right)}^3}{2}+i\ln (x)x\pi -x\ln (2)\ln (x)+x^2$	$140$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x-3)*ln(-x^2+3/2*x)+4*x-3)*ln(x)+(2*x-3)*ln(-x^2+3/2*x)+4*x^2-6*x)/(2*x-3),x,method=_RETURNVERBOSE)

[Out]

x^2+ln(x)*ln(-x^2+3/2*x)*x

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maxima [A]  time = 0.54, size = 27, normalized size = 1.42 $$\begin{array}{c}-x\log (2)\log (x)+x\log (x{)}^2+x\log (x)\log (-2x+3)+x^2\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-3)*log(-x^2+3/2*x)+4*x-3)*log(x)+(2*x-3)*log(-x^2+3/2*x)+4*x^2-6*x)/(2*x-3),x, algorithm="max
ima")

[Out]

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

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mupad [B]  time = 0.95, size = 18, normalized size = 0.95 $$\begin{array}{c}{x}^2+x\ln (\frac{3x}{2}-x^2)\ln (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log((3*x)/2 - x^2)*(2*x - 3) - 6*x + 4*x^2 + log(x)*(4*x + log((3*x)/2 - x^2)*(2*x - 3) - 3))/(2*x - 3),x
)

[Out]

x^2 + x*log((3*x)/2 - x^2)*log(x)

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sympy [B]  time = 0.95, size = 32, normalized size = 1.68 $$\begin{array}{c}{x}^2+(x\log (x)-\frac{1}{24})\log (-x^2+\frac{3x}{2})+\frac{\log (2x^2-3x)}{24}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x-3)*ln(-x**2+3/2*x)+4*x-3)*ln(x)+(2*x-3)*ln(-x**2+3/2*x)+4*x**2-6*x)/(2*x-3),x)

[Out]

x**2 + (x*log(x) - 1/24)*log(-x**2 + 3*x/2) + log(2*x**2 - 3*x)/24

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