3.8.26 \(\int \frac {-16 x^2+16 x^3+(24 x^2-12 x^3) \log (\frac {25}{4 x^2-4 x^3+x^4})}{-2+x} \, dx\)

Optimal. Leaf size=20 \[ 4-4 x^3 \log \left (\frac {25}{(2-x)^2 x^2}\right ) \]

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 10, number of rules used = 7, integrand size = 48, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {6688, 12, 6742, 77, 2495, 43, 30} \begin {gather*} -4 x^3 \log \left (\frac {25}{(2-x)^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-16*x^2 + 16*x^3 + (24*x^2 - 12*x^3)*Log[25/(4*x^2 - 4*x^3 + x^4)])/(-2 + x),x]

[Out]

-4*x^3*Log[25/((2 - x)^2*x^2)]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 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 \frac {4 x^2 \left (4-4 x+3 (-2+x) \log \left (\frac {25}{(-2+x)^2 x^2}\right )\right )}{2-x} \, dx\\ &=4 \int \frac {x^2 \left (4-4 x+3 (-2+x) \log \left (\frac {25}{(-2+x)^2 x^2}\right )\right )}{2-x} \, dx\\ &=4 \int \left (\frac {4 (-1+x) x^2}{-2+x}-3 x^2 \log \left (\frac {25}{(-2+x)^2 x^2}\right )\right ) \, dx\\ &=-\left (12 \int x^2 \log \left (\frac {25}{(-2+x)^2 x^2}\right ) \, dx\right )+16 \int \frac {(-1+x) x^2}{-2+x} \, dx\\ &=-4 x^3 \log \left (\frac {25}{(2-x)^2 x^2}\right )-8 \int x^2 \, dx-8 \int \frac {x^3}{-2+x} \, dx+16 \int \left (2+\frac {4}{-2+x}+x+x^2\right ) \, dx\\ &=32 x+8 x^2+\frac {8 x^3}{3}+64 \log (2-x)-4 x^3 \log \left (\frac {25}{(2-x)^2 x^2}\right )-8 \int \left (4+\frac {8}{-2+x}+2 x+x^2\right ) \, dx\\ &=-4 x^3 \log \left (\frac {25}{(2-x)^2 x^2}\right )\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 16, normalized size = 0.80 \begin {gather*} -4 x^3 \log \left (\frac {25}{(-2+x)^2 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-16*x^2 + 16*x^3 + (24*x^2 - 12*x^3)*Log[25/(4*x^2 - 4*x^3 + x^4)])/(-2 + x),x]

[Out]

-4*x^3*Log[25/((-2 + x)^2*x^2)]

________________________________________________________________________________________

fricas [A]  time = 1.03, size = 24, normalized size = 1.20 \begin {gather*} -4 \, x^{3} \log \left (\frac {25}{x^{4} - 4 \, x^{3} + 4 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+24*x^2)*log(25/(x^4-4*x^3+4*x^2))+16*x^3-16*x^2)/(x-2),x, algorithm="fricas")

[Out]

-4*x^3*log(25/(x^4 - 4*x^3 + 4*x^2))

________________________________________________________________________________________

giac [A]  time = 0.26, size = 24, normalized size = 1.20 \begin {gather*} -4 \, x^{3} \log \left (\frac {25}{x^{4} - 4 \, x^{3} + 4 \, x^{2}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+24*x^2)*log(25/(x^4-4*x^3+4*x^2))+16*x^3-16*x^2)/(x-2),x, algorithm="giac")

[Out]

-4*x^3*log(25/(x^4 - 4*x^3 + 4*x^2))

________________________________________________________________________________________

maple [A]  time = 0.17, size = 25, normalized size = 1.25

method	result	size
norman	\(-4 x^{3} \ln \left (\frac {25}{x^{4}-4 x^{3}+4 x^{2}}\right )\)	\(25\)
risch	\(-4 x^{3} \ln \left (\frac {25}{x^{4}-4 x^{3}+4 x^{2}}\right )\)	\(25\)
default	\(-8 x^{3} \ln \relax (5)-4 x^{3} \ln \left (\frac {1}{x^{2} \left (x^{2}-4 x +4\right )}\right )\)	\(29\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-12*x^3+24*x^2)*ln(25/(x^4-4*x^3+4*x^2))+16*x^3-16*x^2)/(x-2),x,method=_RETURNVERBOSE)

[Out]

-4*x^3*ln(25/(x^4-4*x^3+4*x^2))

________________________________________________________________________________________

maxima [B]  time = 1.00, size = 41, normalized size = 2.05 \begin {gather*} -\frac {8}{3} \, x^{3} {\left (3 \, \log \relax (5) + 2\right )} + 8 \, x^{3} \log \relax (x) + \frac {16}{3} \, x^{3} + 8 \, {\left (x^{3} - 8\right )} \log \left (x - 2\right ) + 64 \, \log \left (x - 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x^3+24*x^2)*log(25/(x^4-4*x^3+4*x^2))+16*x^3-16*x^2)/(x-2),x, algorithm="maxima")

[Out]

-8/3*x^3*(3*log(5) + 2) + 8*x^3*log(x) + 16/3*x^3 + 8*(x^3 - 8)*log(x - 2) + 64*log(x - 2)

________________________________________________________________________________________

mupad [B]  time = 0.73, size = 27, normalized size = 1.35 \begin {gather*} -4\,x^3\,\left (2\,\ln \relax (5)+\ln \left (\frac {1}{x^4-4\,x^3+4\,x^2}\right )\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((16*x^3 - 16*x^2 + log(25/(4*x^2 - 4*x^3 + x^4))*(24*x^2 - 12*x^3))/(x - 2),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.15, size = 22, normalized size = 1.10 \begin {gather*} - 4 x^{3} \log {\left (\frac {25}{x^{4} - 4 x^{3} + 4 x^{2}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-12*x**3+24*x**2)*ln(25/(x**4-4*x**3+4*x**2))+16*x**3-16*x**2)/(x-2),x)

[Out]

-4*x**3*log(25/(x**4 - 4*x**3 + 4*x**2))

________________________________________________________________________________________