3.22.73 \(\int \frac {(3-4 x+x^2)^2}{x^6} \, dx\) [2173]

Optimal. Leaf size=30 \[ -\frac {9}{5 x^5}+\frac {6}{x^4}-\frac {22}{3 x^3}+\frac {4}{x^2}-\frac {1}{x} \]

[Out]

-9/5/x^5+6/x^4-22/3/x^3+4/x^2-1/x

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {712} \begin {gather*} -\frac {9}{5 x^5}+\frac {6}{x^4}-\frac {22}{3 x^3}+\frac {4}{x^2}-\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(3 - 4*x + x^2)^2/x^6,x]

[Out]

-9/(5*x^5) + 6/x^4 - 22/(3*x^3) + 4/x^2 - x^(-1)

Rule 712

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d +
e*x)^m*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*
e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {\left (3-4 x+x^2\right )^2}{x^6} \, dx &=\int \left (\frac {9}{x^6}-\frac {24}{x^5}+\frac {22}{x^4}-\frac {8}{x^3}+\frac {1}{x^2}\right ) \, dx\\ &=-\frac {9}{5 x^5}+\frac {6}{x^4}-\frac {22}{3 x^3}+\frac {4}{x^2}-\frac {1}{x}\\ \end {align*}

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Mathematica [A]
time = 0.00, size = 30, normalized size = 1.00 \begin {gather*} -\frac {9}{5 x^5}+\frac {6}{x^4}-\frac {22}{3 x^3}+\frac {4}{x^2}-\frac {1}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(3 - 4*x + x^2)^2/x^6,x]

[Out]

-9/(5*x^5) + 6/x^4 - 22/(3*x^3) + 4/x^2 - x^(-1)

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Maple [A]
time = 0.66, size = 27, normalized size = 0.90

method	result	size
norman	\(\frac {-x^{4}+4 x^{3}-\frac {22}{3} x^{2}+6 x -\frac {9}{5}}{x^{5}}\)	\(25\)
risch	\(\frac {-x^{4}+4 x^{3}-\frac {22}{3} x^{2}+6 x -\frac {9}{5}}{x^{5}}\)	\(25\)
gosper	\(-\frac {15 x^{4}-60 x^{3}+110 x^{2}-90 x +27}{15 x^{5}}\)	\(26\)
default	\(-\frac {9}{5 x^{5}}+\frac {6}{x^{4}}-\frac {22}{3 x^{3}}+\frac {4}{x^{2}}-\frac {1}{x}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-4*x+3)^2/x^6,x,method=_RETURNVERBOSE)

[Out]

-9/5/x^5+6/x^4-22/3/x^3+4/x^2-1/x

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Maxima [A]
time = 0.28, size = 25, normalized size = 0.83 \begin {gather*} -\frac {15 \, x^{4} - 60 \, x^{3} + 110 \, x^{2} - 90 \, x + 27}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-4*x+3)^2/x^6,x, algorithm="maxima")

[Out]

-1/15*(15*x^4 - 60*x^3 + 110*x^2 - 90*x + 27)/x^5

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Fricas [A]
time = 2.50, size = 25, normalized size = 0.83 \begin {gather*} -\frac {15 \, x^{4} - 60 \, x^{3} + 110 \, x^{2} - 90 \, x + 27}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-4*x+3)^2/x^6,x, algorithm="fricas")

[Out]

-1/15*(15*x^4 - 60*x^3 + 110*x^2 - 90*x + 27)/x^5

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Sympy [A]
time = 0.04, size = 24, normalized size = 0.80 \begin {gather*} \frac {- 15 x^{4} + 60 x^{3} - 110 x^{2} + 90 x - 27}{15 x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-4*x+3)**2/x**6,x)

[Out]

(-15*x**4 + 60*x**3 - 110*x**2 + 90*x - 27)/(15*x**5)

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Giac [A]
time = 1.09, size = 25, normalized size = 0.83 \begin {gather*} -\frac {15 \, x^{4} - 60 \, x^{3} + 110 \, x^{2} - 90 \, x + 27}{15 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-4*x+3)^2/x^6,x, algorithm="giac")

[Out]

-1/15*(15*x^4 - 60*x^3 + 110*x^2 - 90*x + 27)/x^5

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Mupad [B]
time = 0.02, size = 23, normalized size = 0.77 \begin {gather*} -\frac {x^4-4\,x^3+\frac {22\,x^2}{3}-6\,x+\frac {9}{5}}{x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 - 4*x + 3)^2/x^6,x)

[Out]

-((22*x^2)/3 - 6*x - 4*x^3 + x^4 + 9/5)/x^5

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