3.17.67 \(\int \frac {-64-48 x-16 x^2-32 x \log (x)}{-64 x+48 x^2-12 x^3+x^4} \, dx\)

Optimal. Leaf size=21 \[ 10+\frac {x^2 (x+\log (x))}{\left (x-\frac {x^2}{4}\right )^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 33, normalized size of antiderivative = 1.57, number of steps used = 10, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {6688, 12, 6742, 44, 37, 2319} \begin {gather*} \frac {2 x^2}{(4-x)^2}+\frac {32}{(4-x)^2}+\frac {16 \log (x)}{(4-x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-64 - 48*x - 16*x^2 - 32*x*Log[x])/(-64*x + 48*x^2 - 12*x^3 + x^4),x]

[Out]

32/(4 - x)^2 + (2*x^2)/(4 - x)^2 + (16*Log[x])/(4 - 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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 2319

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1
)*(a + b*Log[c*x^n])^p)/(e*(q + 1)), x] - Dist[(b*n*p)/(e*(q + 1)), Int[((d + e*x)^(q + 1)*(a + b*Log[c*x^n])^
(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, n, p, q}, x] && GtQ[p, 0] && NeQ[q, -1] && (EqQ[p, 1] || (Integers
Q[2*p, 2*q] &&  !IGtQ[q, 0]) || (EqQ[p, 2] && NeQ[q, 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 {16 \left (4+3 x+x^2+2 x \log (x)\right )}{(4-x)^3 x} \, dx\\ &=16 \int \frac {4+3 x+x^2+2 x \log (x)}{(4-x)^3 x} \, dx\\ &=16 \int \left (-\frac {3}{(-4+x)^3}-\frac {4}{(-4+x)^3 x}-\frac {x}{(-4+x)^3}-\frac {2 \log (x)}{(-4+x)^3}\right ) \, dx\\ &=\frac {24}{(4-x)^2}-16 \int \frac {x}{(-4+x)^3} \, dx-32 \int \frac {\log (x)}{(-4+x)^3} \, dx-64 \int \frac {1}{(-4+x)^3 x} \, dx\\ &=\frac {24}{(4-x)^2}+\frac {2 x^2}{(4-x)^2}+\frac {16 \log (x)}{(4-x)^2}-16 \int \frac {1}{(-4+x)^2 x} \, dx-64 \int \left (\frac {1}{4 (-4+x)^3}-\frac {1}{16 (-4+x)^2}+\frac {1}{64 (-4+x)}-\frac {1}{64 x}\right ) \, dx\\ &=\frac {32}{(4-x)^2}+\frac {4}{4-x}+\frac {2 x^2}{(4-x)^2}-\log (4-x)+\log (x)+\frac {16 \log (x)}{(4-x)^2}-16 \int \left (\frac {1}{4 (-4+x)^2}-\frac {1}{16 (-4+x)}+\frac {1}{16 x}\right ) \, dx\\ &=\frac {32}{(4-x)^2}+\frac {2 x^2}{(4-x)^2}+\frac {16 \log (x)}{(4-x)^2}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 11, normalized size = 0.52 \begin {gather*} \frac {16 (x+\log (x))}{(-4+x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-64 - 48*x - 16*x^2 - 32*x*Log[x])/(-64*x + 48*x^2 - 12*x^3 + x^4),x]

[Out]

(16*(x + Log[x]))/(-4 + x)^2

________________________________________________________________________________________

fricas [A]  time = 0.75, size = 16, normalized size = 0.76 \begin {gather*} \frac {16 \, {\left (x + \log \relax (x)\right )}}{x^{2} - 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*x*log(x)-16*x^2-48*x-64)/(x^4-12*x^3+48*x^2-64*x),x, algorithm="fricas")

[Out]

16*(x + log(x))/(x^2 - 8*x + 16)

________________________________________________________________________________________

giac [A]  time = 0.21, size = 28, normalized size = 1.33 \begin {gather*} \frac {16 \, x}{x^{2} - 8 \, x + 16} + \frac {16 \, \log \relax (x)}{x^{2} - 8 \, x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*x*log(x)-16*x^2-48*x-64)/(x^4-12*x^3+48*x^2-64*x),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.04, size = 15, normalized size = 0.71

method	result	size
norman	\(\frac {16 x +16 \ln \relax (x )}{\left (x -4\right )^{2}}\)	\(15\)
risch	\(\frac {16 \ln \relax (x )}{x^{2}-8 x +16}+\frac {16 x}{x^{2}-8 x +16}\)	\(29\)
default	\(\frac {16}{x -4}-\frac {\ln \relax (x ) x \left (-8+x \right )}{\left (x -4\right )^{2}}+\frac {64}{\left (x -4\right )^{2}}+\ln \relax (x )\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-32*x*ln(x)-16*x^2-48*x-64)/(x^4-12*x^3+48*x^2-64*x),x,method=_RETURNVERBOSE)

[Out]

(16*x+16*ln(x))/(x-4)^2

________________________________________________________________________________________

maxima [B]  time = 0.77, size = 64, normalized size = 3.05 \begin {gather*} \frac {16 \, {\left (x - 2\right )}}{x^{2} - 8 \, x + 16} - \frac {4 \, {\left (x - 6\right )}}{x^{2} - 8 \, x + 16} + \frac {16 \, \log \relax (x)}{x^{2} - 8 \, x + 16} + \frac {24}{x^{2} - 8 \, x + 16} + \frac {4}{x - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*x*log(x)-16*x^2-48*x-64)/(x^4-12*x^3+48*x^2-64*x),x, algorithm="maxima")

[Out]

16*(x - 2)/(x^2 - 8*x + 16) - 4*(x - 6)/(x^2 - 8*x + 16) + 16*log(x)/(x^2 - 8*x + 16) + 24/(x^2 - 8*x + 16) +
4/(x - 4)

________________________________________________________________________________________

mupad [B]  time = 1.13, size = 11, normalized size = 0.52 \begin {gather*} \frac {16\,\left (x+\ln \relax (x)\right )}{{\left (x-4\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((48*x + 32*x*log(x) + 16*x^2 + 64)/(64*x - 48*x^2 + 12*x^3 - x^4),x)

[Out]

(16*(x + log(x)))/(x - 4)^2

________________________________________________________________________________________

sympy [A]  time = 0.14, size = 24, normalized size = 1.14 \begin {gather*} \frac {16 x}{x^{2} - 8 x + 16} + \frac {16 \log {\relax (x )}}{x^{2} - 8 x + 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-32*x*ln(x)-16*x**2-48*x-64)/(x**4-12*x**3+48*x**2-64*x),x)

[Out]

16*x/(x**2 - 8*x + 16) + 16*log(x)/(x**2 - 8*x + 16)

________________________________________________________________________________________