\(\int \frac {-4+10 x^2-2 x^3}{x^2} \, dx\) [6598]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 16, antiderivative size = 22 \[ \int \frac {-4+10 x^2-2 x^3}{x^2} \, dx=-3+x+(8-x) x+\frac {4+x+x^2}{x}+\log (2) \]

[Out]

(x^2+x+4)/x+x-3+x*(8-x)+ln(2)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {14} \[ \int \frac {-4+10 x^2-2 x^3}{x^2} \, dx=-x^2+10 x+\frac {4}{x} \]

[In]

Int[(-4 + 10*x^2 - 2*x^3)/x^2,x]

[Out]

4/x + 10*x - x^2

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]]

Rubi steps \begin{align*} \text {integral}& = \int \left (10-\frac {4}{x^2}-2 x\right ) \, dx \\ & = \frac {4}{x}+10 x-x^2 \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-4+10 x^2-2 x^3}{x^2} \, dx=\frac {4}{x}+10 x-x^2 \]

[In]

Integrate[(-4 + 10*x^2 - 2*x^3)/x^2,x]

[Out]

4/x + 10*x - x^2

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68

method result size
default \(10 x -x^{2}+\frac {4}{x}\) \(15\)
risch \(10 x -x^{2}+\frac {4}{x}\) \(15\)
gosper \(-\frac {x^{3}-10 x^{2}-4}{x}\) \(16\)
parallelrisch \(-\frac {x^{3}-10 x^{2}-4}{x}\) \(16\)
norman \(\frac {-x^{3}+10 x^{2}+4}{x}\) \(17\)

[In]

int((-2*x^3+10*x^2-4)/x^2,x,method=_RETURNVERBOSE)

[Out]

10*x-x^2+4/x

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.68 \[ \int \frac {-4+10 x^2-2 x^3}{x^2} \, dx=-\frac {x^{3} - 10 \, x^{2} - 4}{x} \]

[In]

integrate((-2*x^3+10*x^2-4)/x^2,x, algorithm="fricas")

[Out]

-(x^3 - 10*x^2 - 4)/x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 8, normalized size of antiderivative = 0.36 \[ \int \frac {-4+10 x^2-2 x^3}{x^2} \, dx=- x^{2} + 10 x + \frac {4}{x} \]

[In]

integrate((-2*x**3+10*x**2-4)/x**2,x)

[Out]

-x**2 + 10*x + 4/x

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-4+10 x^2-2 x^3}{x^2} \, dx=-x^{2} + 10 \, x + \frac {4}{x} \]

[In]

integrate((-2*x^3+10*x^2-4)/x^2,x, algorithm="maxima")

[Out]

-x^2 + 10*x + 4/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.64 \[ \int \frac {-4+10 x^2-2 x^3}{x^2} \, dx=-x^{2} + 10 \, x + \frac {4}{x} \]

[In]

integrate((-2*x^3+10*x^2-4)/x^2,x, algorithm="giac")

[Out]

-x^2 + 10*x + 4/x

Mupad [B] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.73 \[ \int \frac {-4+10 x^2-2 x^3}{x^2} \, dx=\frac {-x^3+10\,x^2+4}{x} \]

[In]

int(-(2*x^3 - 10*x^2 + 4)/x^2,x)

[Out]

(10*x^2 - x^3 + 4)/x