3.33.30 $\int \frac{3590-5760x+3264x^2-768x^3+64x^4+(-3600+3264x^2-1536x^3+192x^4)\log (x)}{x^2}dx$

Optimal. Leaf size=25 $$2+\frac{10+16{(-3+2(3-x{)}^2)}^2\log (x)}{x}$$

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Rubi [A]  time = 0.10, antiderivative size = 36, normalized size of antiderivative = 1.44, number of steps used = 10, number of rules used = 4, integrand size = 44, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.091, Rules used = {14, 2357, 2295, 2304} $$\begin{array}{c}64x^3\log (x)-768x^2\log (x)+\frac{10}{x}+3264x\log (x)-5760\log (x)+\frac{3600\log (x)}{x}\end{array}$$

Antiderivative was successfully verified.

[In]

Int[(3590 - 5760*x + 3264*x^2 - 768*x^3 + 64*x^4 + (-3600 + 3264*x^2 - 1536*x^3 + 192*x^4)*Log[x])/x^2,x]

[Out]

10/x - 5760*Log[x] + (3600*Log[x])/x + 3264*x*Log[x] - 768*x^2*Log[x] + 64*x^3*Log[x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 2304

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

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]

Rubi steps

$$\begin{array}{c}\begin{array}{rl}\text{integral}& =\int (\frac{2(1795-2880x+1632x^2-384x^3+32x^4)}{x^2}+\frac{48(15-12x+2x^2)(-5-4x+2x^2)\log (x)}{x^2})dx\\ & =2\int \frac{1795-2880x+1632x^2-384x^3+32x^4}{x^2}dx+48\int \frac{(15-12x+2x^2)(-5-4x+2x^2)\log (x)}{x^2}dx\\ & =2\int (1632+\frac{1795}{x^2}-\frac{2880}{x}-384x+32x^2)dx+48\int (68\log (x)-\frac{75\log (x)}{x^2}-32x\log (x)+4x^2\log (x))dx\\ & =-\frac{3590}{x}+3264x-384x^2+\frac{64x^3}{3}-5760\log (x)+192\int x^2\log (x)dx-1536\int x\log (x)dx+3264\int \log (x)dx-3600\int \frac{\log (x)}{x^2}dx\\ & =\frac{10}{x}-5760\log (x)+\frac{3600\log (x)}{x}+3264x\log (x)-768x^2\log (x)+64x^3\log (x)\end{array}\end{array}$$

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Mathematica [A]  time = 0.01, size = 36, normalized size = 1.44 $$\begin{array}{c}\frac{10}{x}-5760\log (x)+\frac{3600\log (x)}{x}+3264x\log (x)-768x^2\log (x)+64x^3\log (x)\end{array}$$

Antiderivative was successfully verified.

[In]

Integrate[(3590 - 5760*x + 3264*x^2 - 768*x^3 + 64*x^4 + (-3600 + 3264*x^2 - 1536*x^3 + 192*x^4)*Log[x])/x^2,x
]

[Out]

10/x - 5760*Log[x] + (3600*Log[x])/x + 3264*x*Log[x] - 768*x^2*Log[x] + 64*x^3*Log[x]

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fricas [A]  time = 0.69, size = 31, normalized size = 1.24 $$\begin{array}{c}\frac{2(8(4x^4-48x^3+204x^2-360x+225)\log (x)+5)}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((192*x^4-1536*x^3+3264*x^2-3600)*log(x)+64*x^4-768*x^3+3264*x^2-5760*x+3590)/x^2,x, algorithm="fric
as")

[Out]

2*(8*(4*x^4 - 48*x^3 + 204*x^2 - 360*x + 225)*log(x) + 5)/x

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giac [A]  time = 0.23, size = 33, normalized size = 1.32 $$\begin{array}{c}16(4x^3-48x^2+204x+\frac{225}{x})\log (x)+\frac{10}{x}-5760\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((192*x^4-1536*x^3+3264*x^2-3600)*log(x)+64*x^4-768*x^3+3264*x^2-5760*x+3590)/x^2,x, algorithm="giac
")

[Out]

16*(4*x^3 - 48*x^2 + 204*x + 225/x)*log(x) + 10/x - 5760*log(x)

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maple [A]  time = 0.02, size = 37, normalized size = 1.48

method	result	size
default	$64x^3\ln (x)-768x^2\ln (x)+3264x\ln (x)+\frac{3600\ln (x)}{x}+\frac{10}{x}-5760\ln (x)$	$37$
norman	$\frac{10-5760x\ln (x)+3264x^2\ln (x)-768x^3\ln (x)+64x^4\ln (x)+3600\ln (x)}{x}$	$37$
risch	$\frac{16(4x^4-48x^3+204x^2+225)\ln (x)}{x}-\frac{10(576x\ln (x)-1)}{x}$	$38$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((192*x^4-1536*x^3+3264*x^2-3600)*ln(x)+64*x^4-768*x^3+3264*x^2-5760*x+3590)/x^2,x,method=_RETURNVERBOSE)

[Out]

64*x^3*ln(x)-768*x^2*ln(x)+3264*x*ln(x)+3600*ln(x)/x+10/x-5760*ln(x)

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maxima [A]  time = 0.41, size = 36, normalized size = 1.44 $$\begin{array}{c}64x^3\log (x)-768x^2\log (x)+3264x\log (x)+\frac{3600\log (x)}{x}+\frac{10}{x}-5760\log (x)\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((192*x^4-1536*x^3+3264*x^2-3600)*log(x)+64*x^4-768*x^3+3264*x^2-5760*x+3590)/x^2,x, algorithm="maxi
ma")

[Out]

64*x^3*log(x) - 768*x^2*log(x) + 3264*x*log(x) + 3600*log(x)/x + 10/x - 5760*log(x)

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mupad [B]  time = 1.95, size = 34, normalized size = 1.36 $$\begin{array}{c}64x^3\ln (x)-768x^2\ln (x)-5760\ln (x)+3264x\ln (x)+\frac{3600\ln (x)+10}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(3264*x^2 - 1536*x^3 + 192*x^4 - 3600) - 5760*x + 3264*x^2 - 768*x^3 + 64*x^4 + 3590)/x^2,x)

[Out]

64*x^3*log(x) - 768*x^2*log(x) - 5760*log(x) + 3264*x*log(x) + (3600*log(x) + 10)/x

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sympy [A]  time = 0.18, size = 29, normalized size = 1.16 $$\begin{array}{c}-5760\log (x)+\frac{(64x^4-768x^3+3264x^2+3600)\log (x)}{x}+\frac{10}{x}\end{array}$$

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((192*x**4-1536*x**3+3264*x**2-3600)*ln(x)+64*x**4-768*x**3+3264*x**2-5760*x+3590)/x**2,x)

[Out]

-5760*log(x) + (64*x**4 - 768*x**3 + 3264*x**2 + 3600)*log(x)/x + 10/x

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