\(\int \frac {(3-4 x+x^2)^2}{x^2} \, dx\) [2169]

Optimal result

Integrand size = 14, antiderivative size = 25 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^2} \, dx=-\frac {9}{x}+22 x-4 x^2+\frac {x^3}{3}-24 \log (x) \]

[Out]

-9/x+22*x-4*x^2+1/3*x^3-24*ln(x)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, 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} \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^2} \, dx=\frac {x^3}{3}-4 x^2+22 x-\frac {9}{x}-24 \log (x) \]

[In]

Int[(3 - 4*x + x^2)^2/x^2,x]

[Out]

-9/x + 22*x - 4*x^2 + x^3/3 - 24*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 (22+\frac {9}{x^2}-\frac {24}{x}-8 x+x^2\right ) \, dx \\ & = -\frac {9}{x}+22 x-4 x^2+\frac {x^3}{3}-24 \log (x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^2} \, dx=-\frac {9}{x}+22 x-4 x^2+\frac {x^3}{3}-24 \log (x) \]

[In]

Integrate[(3 - 4*x + x^2)^2/x^2,x]

[Out]

-9/x + 22*x - 4*x^2 + x^3/3 - 24*Log[x]

Maple [A] (verified)

Time = 16.87 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96

method	result	size
default	\(-\frac {9}{x}+22 x -4 x^{2}+\frac {x^{3}}{3}-24 \ln \left (x \right )\)	\(24\)
risch	\(-\frac {9}{x}+22 x -4 x^{2}+\frac {x^{3}}{3}-24 \ln \left (x \right )\)	\(24\)
norman	\(\frac {-9+22 x^{2}-4 x^{3}+\frac {1}{3} x^{4}}{x}-24 \ln \left (x \right )\)	\(27\)
parallelrisch	\(-\frac {-x^{4}+12 x^{3}+72 \ln \left (x \right ) x -66 x^{2}+27}{3 x}\)	\(28\)

[In]

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

[Out]

-9/x+22*x-4*x^2+1/3*x^3-24*ln(x)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^2} \, dx=\frac {x^{4} - 12 \, x^{3} + 66 \, x^{2} - 72 \, x \log \left (x\right ) - 27}{3 \, x} \]

[In]

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

[Out]

1/3*(x^4 - 12*x^3 + 66*x^2 - 72*x*log(x) - 27)/x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.80 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^2} \, dx=\frac {x^{3}}{3} - 4 x^{2} + 22 x - 24 \log {\left (x \right )} - \frac {9}{x} \]

[In]

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

[Out]

x**3/3 - 4*x**2 + 22*x - 24*log(x) - 9/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^2} \, dx=\frac {1}{3} \, x^{3} - 4 \, x^{2} + 22 \, x - \frac {9}{x} - 24 \, \log \left (x\right ) \]

[In]

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

[Out]

1/3*x^3 - 4*x^2 + 22*x - 9/x - 24*log(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^2} \, dx=\frac {1}{3} \, x^{3} - 4 \, x^{2} + 22 \, x - \frac {9}{x} - 24 \, \log \left ({\left | x \right |}\right ) \]

[In]

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

[Out]

1/3*x^3 - 4*x^2 + 22*x - 9/x - 24*log(abs(x))

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {\left (3-4 x+x^2\right )^2}{x^2} \, dx=22\,x-24\,\ln \left (x\right )-\frac {9}{x}-4\,x^2+\frac {x^3}{3} \]

[In]

int((x^2 - 4*x + 3)^2/x^2,x)

[Out]

22*x - 24*log(x) - 9/x - 4*x^2 + x^3/3