3.7 \(\int \frac {1}{\sqrt {-1+x^2} (1+x^2)^2} \, dx\)

Optimal. Leaf size=48 \[ \frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right )}{4 \sqrt {2}}-\frac {x \sqrt {x^2-1}}{4 \left (x^2+1\right )} \]

[Out]

3/8*arctanh(x*2^(1/2)/(x^2-1)^(1/2))*2^(1/2)-1/4*x*(x^2-1)^(1/2)/(x^2+1)

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Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {382, 377, 206} \[ \frac {3 \tanh ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^2-1}}\right )}{4 \sqrt {2}}-\frac {x \sqrt {x^2-1}}{4 \left (x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[-1 + x^2]*(1 + x^2)^2),x]

[Out]

-(x*Sqrt[-1 + x^2])/(4*(1 + x^2)) + (3*ArcTanh[(Sqrt[2]*x)/Sqrt[-1 + x^2]])/(4*Sqrt[2])

Rule 206

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

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 75, normalized size = 1.56 \[ \frac {\sqrt {x^2-1} \left (3 \sqrt {2} \sqrt {\frac {x^2}{x^2-1}} \left (x^2+1\right ) \tanh ^{-1}\left (\sqrt {2} \sqrt {\frac {x^2}{x^2-1}}\right )-2 x^2\right )}{8 \left (x^3+x\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[-1 + x^2]*(1 + x^2)^2),x]

[Out]

(Sqrt[-1 + x^2]*(-2*x^2 + 3*Sqrt[2]*Sqrt[x^2/(-1 + x^2)]*(1 + x^2)*ArcTanh[Sqrt[2]*Sqrt[x^2/(-1 + x^2)]]))/(8*
(x + x^3))

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fricas [B]  time = 1.01, size = 83, normalized size = 1.73 \[ \frac {3 \, \sqrt {2} {\left (x^{2} + 1\right )} \log \left (\frac {9 \, x^{2} + 2 \, \sqrt {2} {\left (3 \, x^{2} - 1\right )} + 2 \, \sqrt {x^{2} - 1} {\left (3 \, \sqrt {2} x + 4 \, x\right )} - 3}{x^{2} + 1}\right ) - 4 \, x^{2} - 4 \, \sqrt {x^{2} - 1} x - 4}{16 \, {\left (x^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*sqrt(2)*(x^2 + 1)*log((9*x^2 + 2*sqrt(2)*(3*x^2 - 1) + 2*sqrt(x^2 - 1)*(3*sqrt(2)*x + 4*x) - 3)/(x^2 +
1)) - 4*x^2 - 4*sqrt(x^2 - 1)*x - 4)/(x^2 + 1)

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giac [B]  time = 0.96, size = 101, normalized size = 2.10 \[ -\frac {3}{16} \, \sqrt {2} \log \left (\frac {{\left (x - \sqrt {x^{2} - 1}\right )}^{2} - 2 \, \sqrt {2} + 3}{{\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 2 \, \sqrt {2} + 3}\right ) - \frac {3 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 1}{2 \, {\left ({\left (x - \sqrt {x^{2} - 1}\right )}^{4} + 6 \, {\left (x - \sqrt {x^{2} - 1}\right )}^{2} + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/16*sqrt(2)*log(((x - sqrt(x^2 - 1))^2 - 2*sqrt(2) + 3)/((x - sqrt(x^2 - 1))^2 + 2*sqrt(2) + 3)) - 1/2*(3*(x
- sqrt(x^2 - 1))^2 + 1)/((x - sqrt(x^2 - 1))^4 + 6*(x - sqrt(x^2 - 1))^2 + 1)

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maple [A]  time = 0.02, size = 45, normalized size = 0.94 \[ -\frac {x}{8 \sqrt {x^{2}-1}\, \left (\frac {x^{2}}{x^{2}-1}-\frac {1}{2}\right )}+\frac {3 \sqrt {2}\, \arctanh \left (\frac {\sqrt {2}\, x}{\sqrt {x^{2}-1}}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*x/(x^2-1)^(1/2)/(x^2/(x^2-1)-1/2)+3/8*arctanh(x*2^(1/2)/(x^2-1)^(1/2))*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{2} + 1\right )}^{2} \sqrt {x^{2} - 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((x^2 + 1)^2*sqrt(x^2 - 1)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{\sqrt {x^2-1}\,{\left (x^2+1\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {\left (x - 1\right ) \left (x + 1\right )} \left (x^{2} + 1\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(sqrt((x - 1)*(x + 1))*(x**2 + 1)**2), x)

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