\(\int \frac {1+2 x+x^2}{x^4} \, dx\) [2181]

Optimal result

Integrand size = 12, antiderivative size = 18 \[ \int \frac {1+2 x+x^2}{x^4} \, dx=-\frac {1}{3 x^3}-\frac {1}{x^2}-\frac {1}{x} \]

[Out]

-1/3/x^3-1/x^2-1/x

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {14} \[ \int \frac {1+2 x+x^2}{x^4} \, dx=-\frac {1}{3 x^3}-\frac {1}{x^2}-\frac {1}{x} \]

[In]

Int[(1 + 2*x + x^2)/x^4,x]

[Out]

-1/3*1/x^3 - x^(-2) - x^(-1)

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 (\frac {1}{x^4}+\frac {2}{x^3}+\frac {1}{x^2}\right ) \, dx \\ & = -\frac {1}{3 x^3}-\frac {1}{x^2}-\frac {1}{x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x+x^2}{x^4} \, dx=-\frac {1}{3 x^3}-\frac {1}{x^2}-\frac {1}{x} \]

[In]

Integrate[(1 + 2*x + x^2)/x^4,x]

[Out]

-1/3*1/x^3 - x^(-2) - x^(-1)

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83

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

[In]

int((x^2+2*x+1)/x^4,x,method=_RETURNVERBOSE)

[Out]

(-x^2-x-1/3)/x^3

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x+x^2}{x^4} \, dx=-\frac {3 \, x^{2} + 3 \, x + 1}{3 \, x^{3}} \]

[In]

integrate((x^2+2*x+1)/x^4,x, algorithm="fricas")

[Out]

-1/3*(3*x^2 + 3*x + 1)/x^3

Sympy [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x+x^2}{x^4} \, dx=\frac {- 3 x^{2} - 3 x - 1}{3 x^{3}} \]

[In]

integrate((x**2+2*x+1)/x**4,x)

[Out]

(-3*x**2 - 3*x - 1)/(3*x**3)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x+x^2}{x^4} \, dx=-\frac {3 \, x^{2} + 3 \, x + 1}{3 \, x^{3}} \]

[In]

integrate((x^2+2*x+1)/x^4,x, algorithm="maxima")

[Out]

-1/3*(3*x^2 + 3*x + 1)/x^3

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int \frac {1+2 x+x^2}{x^4} \, dx=-\frac {3 \, x^{2} + 3 \, x + 1}{3 \, x^{3}} \]

[In]

integrate((x^2+2*x+1)/x^4,x, algorithm="giac")

[Out]

-1/3*(3*x^2 + 3*x + 1)/x^3

Mupad [B] (verification not implemented)

Time = 0.02 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.61 \[ \int \frac {1+2 x+x^2}{x^4} \, dx=-\frac {x^2+x+\frac {1}{3}}{x^3} \]

[In]

int((2*x + x^2 + 1)/x^4,x)

[Out]

-(x + x^2 + 1/3)/x^3