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

Optimal. Leaf size=19 \[ x \left (x+\log (x) \log \left (\frac {3 x}{2}-x^2\right )\right ) \]

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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 {gather*} x^2+x \log (x) \log \left (\frac {1}{2} (3-2 x) x\right ) \end {gather*}

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 {gather*} \begin {aligned} \text {integral} &=\int \left (2 x+\log \left (\frac {1}{2} (3-2 x) x\right )+\frac {\log (x) \left (-3+4 x+(-3+2 x) \log \left (\frac {1}{2} (3-2 x) x\right )\right )}{-3+2 x}\right ) \, dx\\ &=x^2+\int \log \left (\frac {1}{2} (3-2 x) x\right ) \, dx+\int \frac {\log (x) \left (-3+4 x+(-3+2 x) \log \left (\frac {1}{2} (3-2 x) x\right )\right )}{-3+2 x} \, dx\\ &=x^2-\frac {1}{2} (3-2 x) \log \left (\frac {1}{2} (3-2 x) x\right )+\frac {3}{2} \int \frac {1}{x} \, dx-2 \int 1 \, dx+\int \left (\frac {(-3+4 x) \log (x)}{-3+2 x}+\log (x) \log \left (\frac {1}{2} (3-2 x) x\right )\right ) \, dx\\ &=-2 x+x^2+\frac {3 \log (x)}{2}-\frac {1}{2} (3-2 x) \log \left (\frac {1}{2} (3-2 x) x\right )+\int \frac {(-3+4 x) \log (x)}{-3+2 x} \, dx+\int \log (x) \log \left (\frac {1}{2} (3-2 x) x\right ) \, dx\\ &=-2 x+x^2+\frac {3 \log (x)}{2}-\frac {1}{2} (3-2 x) \log \left (\frac {1}{2} (3-2 x) x\right )+x \log (x) \log \left (\frac {1}{2} (3-2 x) x\right )-\int \frac {(3-4 x) \log (x)}{3-2 x} \, dx+\int \left (2 \log (x)+\frac {3 \log (x)}{-3+2 x}\right ) \, dx-\int \log \left (\frac {1}{2} (3-2 x) x\right ) \, dx\\ &=-2 x+x^2+\frac {3 \log (x)}{2}+x \log (x) \log \left (\frac {1}{2} (3-2 x) x\right )-\frac {3}{2} \int \frac {1}{x} \, dx+2 \int 1 \, dx+2 \int \log (x) \, dx+3 \int \frac {\log (x)}{-3+2 x} \, dx-\int \left (2 \log (x)+\frac {3 \log (x)}{-3+2 x}\right ) \, dx\\ &=-2 x+x^2+2 x \log (x)+x \log (x) \log \left (\frac {1}{2} (3-2 x) x\right )+\frac {3}{2} \log \left (\frac {3}{2}\right ) \log (-3+2 x)-2 \int \log (x) \, dx+3 \int \frac {\log \left (\frac {2 x}{3}\right )}{-3+2 x} \, dx-3 \int \frac {\log (x)}{-3+2 x} \, dx\\ &=x^2+x \log (x) \log \left (\frac {1}{2} (3-2 x) x\right )-\frac {3}{2} \text {Li}_2\left (1-\frac {2 x}{3}\right )-3 \int \frac {\log \left (\frac {2 x}{3}\right )}{-3+2 x} \, dx\\ &=x^2+x \log (x) \log \left (\frac {1}{2} (3-2 x) x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 18, normalized size = 0.95 \begin {gather*} x \left (x+\log (x) \log \left (\frac {1}{2} (3-2 x) x\right )\right ) \end {gather*}

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 {gather*} x \log \left (-x^{2} + \frac {3}{2} \, x\right ) \log \relax (x) + x^{2} \end {gather*}

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 {gather*} -x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2} + x \log \relax (x) \log \left (-2 \, x + 3\right ) + x^{2} \end {gather*}

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	\(x^{2}+\ln \relax (x ) \ln \left (-x^{2}+\frac {3}{2} x \right ) x\)	\(19\)
default	\(-x \ln \relax (2) \ln \relax (x )+x \ln \relax (x ) \ln \left (-2 x^{2}+3 x \right )+x^{2}\)	\(26\)
risch	\(\ln \relax (x ) x \ln \left (x -\frac {3}{2}\right )+x \ln \relax (x )^{2}-\frac {i \ln \relax (x ) x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i \left (x -\frac {3}{2}\right )\right ) \mathrm {csgn}\left (i x \left (x -\frac {3}{2}\right )\right )}{2}+\frac {i \ln \relax (x ) x \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x \left (x -\frac {3}{2}\right )\right )^{2}}{2}-i \ln \relax (x ) x \pi \mathrm {csgn}\left (i x \left (x -\frac {3}{2}\right )\right )^{2}+\frac {i \ln \relax (x ) x \pi \,\mathrm {csgn}\left (i \left (x -\frac {3}{2}\right )\right ) \mathrm {csgn}\left (i x \left (x -\frac {3}{2}\right )\right )^{2}}{2}+\frac {i \ln \relax (x ) x \pi \mathrm {csgn}\left (i x \left (x -\frac {3}{2}\right )\right )^{3}}{2}+i \ln \relax (x ) x \pi -x \ln \relax (2) \ln \relax (x )+x^{2}\)	\(140\)

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 {gather*} -x \log \relax (2) \log \relax (x) + x \log \relax (x)^{2} + x \log \relax (x) \log \left (-2 \, x + 3\right ) + x^{2} \end {gather*}

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 {gather*} x^2+x\,\ln \left (\frac {3\,x}{2}-x^2\right )\,\ln \relax (x) \end {gather*}

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 {gather*} x^{2} + \left (x \log {\relax (x )} - \frac {1}{24}\right ) \log {\left (- x^{2} + \frac {3 x}{2} \right )} + \frac {\log {\left (2 x^{2} - 3 x \right )}}{24} \end {gather*}

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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