\(\int \frac {-4+5 x^2+x^3}{x^2} \, dx\) [463]

Optimal result

Integrand size = 14, antiderivative size = 16 \[ \int \frac {-4+5 x^2+x^3}{x^2} \, dx=\frac {4}{x}+5 x+\frac {x^2}{2} \]

[Out]

4/x+5*x+1/2*x^2

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, 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 = {14} \[ \int \frac {-4+5 x^2+x^3}{x^2} \, dx=\frac {x^2}{2}+5 x+\frac {4}{x} \]

[In]

Int[(-4 + 5*x^2 + x^3)/x^2,x]

[Out]

4/x + 5*x + x^2/2

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 (5-\frac {4}{x^2}+x\right ) \, dx \\ & = \frac {4}{x}+5 x+\frac {x^2}{2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.00 \[ \int \frac {-4+5 x^2+x^3}{x^2} \, dx=\frac {4}{x}+5 x+\frac {x^2}{2} \]

[In]

Integrate[(-4 + 5*x^2 + x^3)/x^2,x]

[Out]

4/x + 5*x + x^2/2

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {4}{x}+5 x +\frac {x^{2}}{2}\)	\(15\)
risch	\(\frac {4}{x}+5 x +\frac {x^{2}}{2}\)	\(15\)
gosper	\(\frac {x^{3}+10 x^{2}+8}{2 x}\)	\(16\)
parallelrisch	\(\frac {x^{3}+10 x^{2}+8}{2 x}\)	\(16\)
norman	\(\frac {\frac {1}{2} x^{3}+5 x^{2}+4}{x}\)	\(17\)

[In]

int((x^3+5*x^2-4)/x^2,x,method=_RETURNVERBOSE)

[Out]

4/x+5*x+1/2*x^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {-4+5 x^2+x^3}{x^2} \, dx=\frac {x^{3} + 10 \, x^{2} + 8}{2 \, x} \]

[In]

integrate((x^3+5*x^2-4)/x^2,x, algorithm="fricas")

[Out]

1/2*(x^3 + 10*x^2 + 8)/x

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 10, normalized size of antiderivative = 0.62 \[ \int \frac {-4+5 x^2+x^3}{x^2} \, dx=\frac {x^{2}}{2} + 5 x + \frac {4}{x} \]

[In]

integrate((x**3+5*x**2-4)/x**2,x)

[Out]

x**2/2 + 5*x + 4/x

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-4+5 x^2+x^3}{x^2} \, dx=\frac {1}{2} \, x^{2} + 5 \, x + \frac {4}{x} \]

[In]

integrate((x^3+5*x^2-4)/x^2,x, algorithm="maxima")

[Out]

1/2*x^2 + 5*x + 4/x

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.88 \[ \int \frac {-4+5 x^2+x^3}{x^2} \, dx=\frac {1}{2} \, x^{2} + 5 \, x + \frac {4}{x} \]

[In]

integrate((x^3+5*x^2-4)/x^2,x, algorithm="giac")

[Out]

1/2*x^2 + 5*x + 4/x

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.94 \[ \int \frac {-4+5 x^2+x^3}{x^2} \, dx=\frac {x^3+10\,x^2+8}{2\,x} \]

[In]

int((5*x^2 + x^3 - 4)/x^2,x)

[Out]

(10*x^2 + x^3 + 8)/(2*x)