3.2 \(\int \frac {1}{x^4 (1+x^2)} \, dx\)

Optimal. Leaf size=13 \[ -\frac {1}{3 x^3}+\frac {1}{x}+\tan ^{-1}(x) \]

[Out]

-1/3/x^3+1/x+arctan(x)

________________________________________________________________________________________

Rubi [A]  time = 0.00, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {325, 203} \[ -\frac {1}{3 x^3}+\frac {1}{x}+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(1 + x^2)),x]

[Out]

-1/(3*x^3) + x^(-1) + ArcTan[x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.00, size = 13, normalized size = 1.00 \[ -\frac {1}{3 x^3}+\frac {1}{x}+\tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(1 + x^2)),x]

[Out]

-1/3*1/x^3 + x^(-1) + ArcTan[x]

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 19, normalized size = 1.46 \[ \frac {3 \, x^{3} \arctan \relax (x) + 3 \, x^{2} - 1}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*x^3*arctan(x) + 3*x^2 - 1)/x^3

________________________________________________________________________________________

giac [A]  time = 0.80, size = 15, normalized size = 1.15 \[ \frac {3 \, x^{2} - 1}{3 \, x^{3}} + \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*x^2 - 1)/x^3 + arctan(x)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 12, normalized size = 0.92 \[ \arctan \relax (x )+\frac {1}{x}-\frac {1}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(x^2+1),x)

[Out]

-1/3/x^3+1/x+arctan(x)

________________________________________________________________________________________

maxima [A]  time = 1.29, size = 15, normalized size = 1.15 \[ \frac {3 \, x^{2} - 1}{3 \, x^{3}} + \arctan \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*x^2 - 1)/x^3 + arctan(x)

________________________________________________________________________________________

mupad [B]  time = 0.16, size = 12, normalized size = 0.92 \[ \mathrm {atan}\relax (x)+\frac {x^2-\frac {1}{3}}{x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(x^2 + 1)),x)

[Out]

atan(x) + (x^2 - 1/3)/x^3

________________________________________________________________________________________

sympy [A]  time = 0.11, size = 14, normalized size = 1.08 \[ \operatorname {atan}{\relax (x )} + \frac {3 x^{2} - 1}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

atan(x) + (3*x**2 - 1)/(3*x**3)

________________________________________________________________________________________