3.13 \(\int \frac{5+2 x^2}{-1+x^4} \, dx\)

Optimal. Leaf size=13 \[ -\frac{3}{2} \tan ^{-1}(x)-\frac{7}{2} \tanh ^{-1}(x) \]

[Out]

(-3*ArcTan[x])/2 - (7*ArcTanh[x])/2

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Rubi [A]  time = 0.0062671, antiderivative size = 13, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1167, 207, 203} \[ -\frac{3}{2} \tan ^{-1}(x)-\frac{7}{2} \tanh ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(5 + 2*x^2)/(-1 + x^4),x]

[Out]

(-3*ArcTan[x])/2 - (7*ArcTanh[x])/2

Rule 1167

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q)
, Int[1/(-q + c*x^2), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x^2), x], x]] /; FreeQ[{a, c, d, e}, x] &&
NeQ[c*d^2 - a*e^2, 0] && PosQ[-(a*c)]

Rule 207

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

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])

Rubi steps

\begin{align*} \int \frac{5+2 x^2}{-1+x^4} \, dx &=-\left (\frac{3}{2} \int \frac{1}{1+x^2} \, dx\right )+\frac{7}{2} \int \frac{1}{-1+x^2} \, dx\\ &=-\frac{3}{2} \tan ^{-1}(x)-\frac{7}{2} \tanh ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.00597, size = 25, normalized size = 1.92 \[ \frac{7}{4} \log (1-x)-\frac{7}{4} \log (x+1)-\frac{3}{2} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Integrate[(5 + 2*x^2)/(-1 + x^4),x]

[Out]

(-3*ArcTan[x])/2 + (7*Log[1 - x])/4 - (7*Log[1 + x])/4

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Maple [A]  time = 0.044, size = 18, normalized size = 1.4 \begin{align*}{\frac{7\,\ln \left ( -1+x \right ) }{4}}-{\frac{7\,\ln \left ( 1+x \right ) }{4}}-{\frac{3\,\arctan \left ( x \right ) }{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x^2+5)/(x^4-1),x)

[Out]

7/4*ln(-1+x)-7/4*ln(1+x)-3/2*arctan(x)

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Maxima [A]  time = 1.48577, size = 23, normalized size = 1.77 \begin{align*} -\frac{3}{2} \, \arctan \left (x\right ) - \frac{7}{4} \, \log \left (x + 1\right ) + \frac{7}{4} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+5)/(x^4-1),x, algorithm="maxima")

[Out]

-3/2*arctan(x) - 7/4*log(x + 1) + 7/4*log(x - 1)

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Fricas [A]  time = 1.34957, size = 68, normalized size = 5.23 \begin{align*} -\frac{3}{2} \, \arctan \left (x\right ) - \frac{7}{4} \, \log \left (x + 1\right ) + \frac{7}{4} \, \log \left (x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+5)/(x^4-1),x, algorithm="fricas")

[Out]

-3/2*arctan(x) - 7/4*log(x + 1) + 7/4*log(x - 1)

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Sympy [A]  time = 0.132927, size = 22, normalized size = 1.69 \begin{align*} \frac{7 \log{\left (x - 1 \right )}}{4} - \frac{7 \log{\left (x + 1 \right )}}{4} - \frac{3 \operatorname{atan}{\left (x \right )}}{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x**2+5)/(x**4-1),x)

[Out]

7*log(x - 1)/4 - 7*log(x + 1)/4 - 3*atan(x)/2

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Giac [B]  time = 1.12775, size = 26, normalized size = 2. \begin{align*} -\frac{3}{2} \, \arctan \left (x\right ) - \frac{7}{4} \, \log \left ({\left | x + 1 \right |}\right ) + \frac{7}{4} \, \log \left ({\left | x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2*x^2+5)/(x^4-1),x, algorithm="giac")

[Out]

-3/2*arctan(x) - 7/4*log(abs(x + 1)) + 7/4*log(abs(x - 1))