\(\int \frac {(3-4 x+x^2)^2}{x^4} \, dx\) [2171]

Optimal result

Integrand size = 14, antiderivative size = 21 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^4} \, dx=-\frac {3}{x^3}+\frac {12}{x^2}-\frac {22}{x}+x-8 \log (x) \]

[Out]

-3/x^3+12/x^2-22/x+x-8*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {712} \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^4} \, dx=-\frac {3}{x^3}+\frac {12}{x^2}+x-\frac {22}{x}-8 \log (x) \]

[In]

Int[(3 - 4*x + x^2)^2/x^4,x]

[Out]

-3/x^3 + 12/x^2 - 22/x + x - 8*Log[x]

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*} \text {integral}& = \int \left (1+\frac {9}{x^4}-\frac {24}{x^3}+\frac {22}{x^2}-\frac {8}{x}\right ) \, dx \\ & = -\frac {3}{x^3}+\frac {12}{x^2}-\frac {22}{x}+x-8 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^4} \, dx=-\frac {3}{x^3}+\frac {12}{x^2}-\frac {22}{x}+x-8 \log (x) \]

[In]

Integrate[(3 - 4*x + x^2)^2/x^4,x]

[Out]

-3/x^3 + 12/x^2 - 22/x + x - 8*Log[x]

Maple [A] (verified)

Time = 16.64 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00

method	result	size
risch	\(x +\frac {-22 x^{2}+12 x -3}{x^{3}}-8 \ln \left (x \right )\)	\(21\)
default	\(-\frac {3}{x^{3}}+\frac {12}{x^{2}}-\frac {22}{x}+x -8 \ln \left (x \right )\)	\(22\)
norman	\(\frac {x^{4}-22 x^{2}+12 x -3}{x^{3}}-8 \ln \left (x \right )\)	\(23\)
parallelrisch	\(-\frac {8 \ln \left (x \right ) x^{3}-x^{4}+3+22 x^{2}-12 x}{x^{3}}\)	\(28\)

[In]

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

[Out]

x+(-22*x^2+12*x-3)/x^3-8*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^4} \, dx=\frac {x^{4} - 8 \, x^{3} \log \left (x\right ) - 22 \, x^{2} + 12 \, x - 3}{x^{3}} \]

[In]

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

[Out]

(x^4 - 8*x^3*log(x) - 22*x^2 + 12*x - 3)/x^3

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^4} \, dx=x - 8 \log {\left (x \right )} + \frac {- 22 x^{2} + 12 x - 3}{x^{3}} \]

[In]

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

[Out]

x - 8*log(x) + (-22*x**2 + 12*x - 3)/x**3

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^4} \, dx=x - \frac {22 \, x^{2} - 12 \, x + 3}{x^{3}} - 8 \, \log \left (x\right ) \]

[In]

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

[Out]

x - (22*x^2 - 12*x + 3)/x^3 - 8*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^4} \, dx=x - \frac {22 \, x^{2} - 12 \, x + 3}{x^{3}} - 8 \, \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

x - (22*x^2 - 12*x + 3)/x^3 - 8*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^4} \, dx=x-8\,\ln \left (x\right )-\frac {22\,x^2-12\,x+3}{x^3} \]

[In]

int((x^2 - 4*x + 3)^2/x^4,x)

[Out]

x - 8*log(x) - (22*x^2 - 12*x + 3)/x^3