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

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

[Out]

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

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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 = {331, 209} \begin {gather*} -\frac {1}{3 x^3}+\frac {1}{x}+\tan ^{-1}(x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(1 + x^2)),x]

[Out]

-1/3*1/x^3 + x^(-1) + ArcTan[x]

Rule 209

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

Rule 331

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*}

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

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(1 + x^2)),x]

[Out]

-1/3*1/x^3 + x^(-1) + ArcTan[x]

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Mathics [A]
time = 1.78, size = 11, normalized size = 0.85 \begin {gather*} -\frac {1}{3 x^3}+\frac {1}{x}+\text {ArcTan}\left [x\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

mathics('Integrate[1/(x^4*(1 + x^2)),x]')

[Out]

-1 / (3 x ^ 3) + 1 / x + ArcTan[x]

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Maple [A]
time = 0.03, size = 12, normalized size = 0.92

method	result	size
default	\(-\frac {1}{3 x^{3}}+\frac {1}{x}+\arctan \left (x \right )\)	\(12\)
meijerg	\(-\frac {1}{3 x^{3}}+\frac {1}{x}+\arctan \left (x \right )\)	\(12\)
risch	\(\frac {x^{2}-\frac {1}{3}}{x^{3}}+\arctan \left (x \right )\)	\(13\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.38, size = 15, normalized size = 1.15 \begin {gather*} \frac {3 \, x^{2} - 1}{3 \, x^{3}} + \arctan \left (x\right ) \end {gather*}

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)

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Fricas [A]
time = 0.30, size = 19, normalized size = 1.46 \begin {gather*} \frac {3 \, x^{3} \arctan \left (x\right ) + 3 \, x^{2} - 1}{3 \, x^{3}} \end {gather*}

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

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Sympy [A]
time = 0.05, size = 14, normalized size = 1.08 \begin {gather*} \operatorname {atan}{\left (x \right )} + \frac {3 x^{2} - 1}{3 x^{3}} \end {gather*}

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)

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Giac [A]
time = 0.00, size = 17, normalized size = 1.31 \begin {gather*} \frac {3 x^{2}-1}{3 x^{3}}+\arctan x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*x^2 - 1)/x^3 + arctan(x)

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Mupad [B]
time = 0.16, size = 12, normalized size = 0.92 \begin {gather*} \mathrm {atan}\left (x\right )+\frac {x^2-\frac {1}{3}}{x^3} \end {gather*}

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

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