3.677 \(\int \frac{(1+x^2) \tan ^{-1}(x)}{x^5} \, dx\)

Optimal. Leaf size=31 \[ -\frac{1}{12 x^3}-\frac{\left (x^2+1\right )^2 \tan ^{-1}(x)}{4 x^4}-\frac{1}{4 x} \]

[Out]

-1/(12*x^3) - 1/(4*x) - ((1 + x^2)^2*ArcTan[x])/(4*x^4)

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Rubi [A]  time = 0.0231771, antiderivative size = 31, normalized size of antiderivative = 1., 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 = {4944, 14} \[ -\frac{1}{12 x^3}-\frac{\left (x^2+1\right )^2 \tan ^{-1}(x)}{4 x^4}-\frac{1}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[((1 + x^2)*ArcTan[x])/x^5,x]

[Out]

-1/(12*x^3) - 1/(4*x) - ((1 + x^2)^2*ArcTan[x])/(4*x^4)

Rule 4944

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p)/(d*f*(m + 1)), x] - Dist[(b*c*p)/(f*(m + 1)), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -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*} \int \frac{\left (1+x^2\right ) \tan ^{-1}(x)}{x^5} \, dx &=-\frac{\left (1+x^2\right )^2 \tan ^{-1}(x)}{4 x^4}+\frac{1}{4} \int \frac{1+x^2}{x^4} \, dx\\ &=-\frac{\left (1+x^2\right )^2 \tan ^{-1}(x)}{4 x^4}+\frac{1}{4} \int \left (\frac{1}{x^4}+\frac{1}{x^2}\right ) \, dx\\ &=-\frac{1}{12 x^3}-\frac{1}{4 x}-\frac{\left (1+x^2\right )^2 \tan ^{-1}(x)}{4 x^4}\\ \end{align*}

Mathematica [C]  time = 0.0075305, size = 59, normalized size = 1.9 \[ -\frac{\, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-x^2\right )}{12 x^3}-\frac{\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-x^2\right )}{2 x}-\frac{\tan ^{-1}(x)}{2 x^2}-\frac{\tan ^{-1}(x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[x]/(4*x^4) - ArcTan[x]/(2*x^2) - Hypergeometric2F1[-3/2, 1, -1/2, -x^2]/(12*x^3) - Hypergeometric2F1[-
1/2, 1, 1/2, -x^2]/(2*x)

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Maple [A]  time = 0.008, size = 30, normalized size = 1. \begin{align*} -{\frac{\arctan \left ( x \right ) }{4\,{x}^{4}}}-{\frac{\arctan \left ( x \right ) }{2\,{x}^{2}}}-{\frac{\arctan \left ( x \right ) }{4}}-{\frac{1}{12\,{x}^{3}}}-{\frac{1}{4\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+1)*arctan(x)/x^5,x)

[Out]

-1/4*arctan(x)/x^4-1/2*arctan(x)/x^2-1/4*arctan(x)-1/12/x^3-1/4/x

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Maxima [A]  time = 1.41116, size = 42, normalized size = 1.35 \begin{align*} -\frac{3 \, x^{2} + 1}{12 \, x^{3}} - \frac{{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )}{4 \, x^{4}} - \frac{1}{4} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*x^2 + 1)/x^3 - 1/4*(2*x^2 + 1)*arctan(x)/x^4 - 1/4*arctan(x)

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Fricas [A]  time = 2.43455, size = 74, normalized size = 2.39 \begin{align*} -\frac{3 \, x^{3} + 3 \,{\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (x\right ) + x}{12 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*x^3 + 3*(x^4 + 2*x^2 + 1)*arctan(x) + x)/x^4

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Sympy [A]  time = 0.859894, size = 34, normalized size = 1.1 \begin{align*} - \frac{\operatorname{atan}{\left (x \right )}}{4} - \frac{1}{4 x} - \frac{\operatorname{atan}{\left (x \right )}}{2 x^{2}} - \frac{1}{12 x^{3}} - \frac{\operatorname{atan}{\left (x \right )}}{4 x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-atan(x)/4 - 1/(4*x) - atan(x)/(2*x**2) - 1/(12*x**3) - atan(x)/(4*x**4)

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Giac [A]  time = 1.07741, size = 42, normalized size = 1.35 \begin{align*} -\frac{3 \, x^{2} + 1}{12 \, x^{3}} - \frac{{\left (2 \, x^{2} + 1\right )} \arctan \left (x\right )}{4 \, x^{4}} - \frac{1}{4} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*x^2 + 1)/x^3 - 1/4*(2*x^2 + 1)*arctan(x)/x^4 - 1/4*arctan(x)