\(\int \frac {1+2 x^2}{x^5 (1+x^2)^3} \, dx\) [118]

Optimal result

Integrand size = 18, antiderivative size = 14 \[ \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx=-\frac {1}{4 x^4 \left (1+x^2\right )^2} \]

[Out]

-1/4/x^4/(x^2+1)^2

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {457, 75} \[ \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx=-\frac {1}{4 x^4 \left (x^2+1\right )^2} \]

[In]

Int[(1 + 2*x^2)/(x^5*(1 + x^2)^3),x]

[Out]

-1/4*1/(x^4*(1 + x^2)^2)

Rule 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1+2 x}{x^3 (1+x)^3} \, dx,x,x^2\right ) \\ & = -\frac {1}{4 x^4 \left (1+x^2\right )^2} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.00 \[ \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx=-\frac {1}{4 x^4 \left (1+x^2\right )^2} \]

[In]

Integrate[(1 + 2*x^2)/(x^5*(1 + x^2)^3),x]

[Out]

-1/4*1/(x^4*(1 + x^2)^2)

Maple [A] (verified)

Time = 2.51 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.93

method	result	size
gosper	\(-\frac {1}{4 x^{4} \left (x^{2}+1\right )^{2}}\)	\(13\)
norman	\(-\frac {1}{4 x^{4} \left (x^{2}+1\right )^{2}}\)	\(13\)
risch	\(-\frac {1}{4 x^{4} \left (x^{2}+1\right )^{2}}\)	\(13\)
parallelrisch	\(-\frac {1}{4 x^{4} \left (x^{2}+1\right )^{2}}\)	\(13\)
default	\(-\frac {1}{4 x^{4}}+\frac {1}{2 x^{2}}-\frac {1}{2 \left (x^{2}+1\right )}-\frac {1}{4 \left (x^{2}+1\right )^{2}}\)	\(30\)
meijerg	\(-\frac {1}{4 x^{4}}+\frac {1}{2 x^{2}}-\frac {3}{4}-\frac {x^{2} \left (7 x^{2}+8\right )}{4 \left (x^{2}+1\right )^{2}}+\frac {x^{2} \left (5 x^{2}+6\right )}{2 \left (x^{2}+1\right )^{2}}\)	\(51\)

[In]

int((2*x^2+1)/x^5/(x^2+1)^3,x,method=_RETURNVERBOSE)

[Out]

-1/4/x^4/(x^2+1)^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (x^{8} + 2 \, x^{6} + x^{4}\right )}} \]

[In]

integrate((2*x^2+1)/x^5/(x^2+1)^3,x, algorithm="fricas")

[Out]

-1/4/(x^8 + 2*x^6 + x^4)

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.21 \[ \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx=- \frac {1}{4 x^{8} + 8 x^{6} + 4 x^{4}} \]

[In]

integrate((2*x**2+1)/x**5/(x**2+1)**3,x)

[Out]

-1/(4*x**8 + 8*x**6 + 4*x**4)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.14 \[ \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (x^{8} + 2 \, x^{6} + x^{4}\right )}} \]

[In]

integrate((2*x^2+1)/x^5/(x^2+1)^3,x, algorithm="maxima")

[Out]

-1/4/(x^8 + 2*x^6 + x^4)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.79 \[ \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx=-\frac {1}{4 \, {\left (x^{4} + x^{2}\right )}^{2}} \]

[In]

integrate((2*x^2+1)/x^5/(x^2+1)^3,x, algorithm="giac")

[Out]

-1/4/(x^4 + x^2)^2

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.43 \[ \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx=-\frac {1}{4\,x^8+8\,x^6+4\,x^4} \]

[In]

int((2*x^2 + 1)/(x^5*(x^2 + 1)^3),x)

[Out]

-1/(4*x^4 + 8*x^6 + 4*x^8)