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

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

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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} $$\begin{array}{c}\frac{2x^2}{(4-x{)}^2}+\frac{32}{(4-x{)}^2}+\frac{16\log (x)}{(4-x{)}^2}\end{array}$$

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

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int \frac{16(4+3x+x^2+2x\log (x))}{(4-x{)}^{3}x}dx\\ & =16\int \frac{4+3x+x^2+2x\log (x)}{(4-x{)}^{3}x}dx\\ & =16\int (-\frac{3}{(-4+x{)}^3}-\frac{4}{(-4+x{)}^{3}x}-\frac{x}{(-4+x{)}^3}-\frac{2\log (x)}{(-4+x{)}^3})dx\\ & =\frac{24}{(4-x{)}^2}-16\int \frac{x}{(-4+x{)}^3}dx-32\int \frac{\log (x)}{(-4+x{)}^3}dx-64\int \frac{1}{(-4+x{)}^{3}x}dx\\ & =\frac{24}{(4-x{)}^2}+\frac{2x^2}{(4-x{)}^2}+\frac{16\log (x)}{(4-x{)}^2}-16\int \frac{1}{(-4+x{)}^{2}x}dx-64\int (\frac{1}{4(-4+x{)}^3}-\frac{1}{16(-4+x{)}^2}+\frac{1}{64(-4+x)}-\frac{1}{64x})dx\\ & =\frac{32}{(4-x{)}^2}+\frac{4}{4-x}+\frac{2x^2}{(4-x{)}^2}-\log (4-x)+\log (x)+\frac{16\log (x)}{(4-x{)}^2}-16\int (\frac{1}{4(-4+x{)}^2}-\frac{1}{16(-4+x)}+\frac{1}{16x})dx\\ & =\frac{32}{(4-x{)}^2}+\frac{2x^2}{(4-x{)}^2}+\frac{16\log (x)}{(4-x{)}^2}\end{array}\end{array}$$

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

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

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fricas [A]  time = 0.75, size = 16, normalized size = 0.76 $$\begin{array}{c}\frac{16(x+\log (x))}{x^2-8x+16}\end{array}$$

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)

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

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)

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

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

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

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maxima [B]  time = 0.77, size = 64, normalized size = 3.05 $$\begin{array}{c}\frac{16(x-2)}{x^2-8x+16}-\frac{4(x-6)}{x^2-8x+16}+\frac{16\log (x)}{x^2-8x+16}+\frac{24}{x^2-8x+16}+\frac{4}{x-4}\end{array}$$

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)

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mupad [B]  time = 1.13, size = 11, normalized size = 0.52 $$\begin{array}{c}\frac{16(x+\ln (x))}{{(x-4)}^2}\end{array}$$

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

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

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)

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