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

Optimal. Leaf size=14 \[ -\frac {1}{4 x^4 \left (1+x^2\right )^2} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {457, 75} \begin {gather*} -\frac {1}{4 x^4 \left (x^2+1\right )^2} \end {gather*}

Antiderivative was successfully verified.

[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*} \int \frac {1+2 x^2}{x^5 \left (1+x^2\right )^3} \, dx &=\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*}

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Mathematica [A]
time = 0.01, size = 14, normalized size = 1.00 \begin {gather*} -\frac {1}{4 x^4 \left (1+x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*1/(x^4*(1 + x^2)^2)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(29\) vs. \(2(12)=24\).
time = 0.11, size = 30, normalized size = 2.14

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	\(-\frac {1}{4 x^{4} \left (x^{2}+1\right )^{2}}\)	\(13\)
default	\(-\frac {1}{4 x^{4}}+\frac {1}{2 x^{2}}-\frac {1}{4 \left (x^{2}+1\right )^{2}}-\frac {1}{2 \left (x^{2}+1\right )}\)	\(30\)
meijerg	\(-\frac {x^{2} \left (7 x^{2}+8\right )}{4 \left (x^{2}+1\right )^{2}}-\frac {3}{4}-\frac {1}{4 x^{4}}+\frac {1}{2 x^{2}}+\frac {x^{2} \left (5 x^{2}+6\right )}{2 \left (x^{2}+1\right )^{2}}\)	\(51\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 16, normalized size = 1.14 \begin {gather*} -\frac {1}{4 \, {\left (x^{8} + 2 \, x^{6} + x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]
time = 0.89, size = 16, normalized size = 1.14 \begin {gather*} -\frac {1}{4 \, {\left (x^{8} + 2 \, x^{6} + x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [A]
time = 0.04, size = 17, normalized size = 1.21 \begin {gather*} - \frac {1}{4 x^{8} + 8 x^{6} + 4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(4*x**8 + 8*x**6 + 4*x**4)

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Giac [A]
time = 1.68, size = 11, normalized size = 0.79 \begin {gather*} -\frac {1}{4 \, {\left (x^{4} + x^{2}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/(x^4 + x^2)^2

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Mupad [B]
time = 0.02, size = 20, normalized size = 1.43 \begin {gather*} -\frac {1}{4\,x^8+8\,x^6+4\,x^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2 + 1)/(x^5*(x^2 + 1)^3),x)

[Out]

-1/(4*x^4 + 8*x^6 + 4*x^8)

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