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

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

[Out]

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

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Rubi [A]  time = 0.0135447, antiderivative size = 48, normalized size of antiderivative = 1., 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} \[ \frac{3 \tanh ^{-1}\left (\frac{x}{\sqrt{2} \sqrt{x^2+1}}\right )}{4 \sqrt{2}}-\frac{x \sqrt{x^2+1}}{4 \left (x^2+2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 + x^2]*(2 + x^2)^2),x]

[Out]

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

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]

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 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])

Rubi steps

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

Mathematica [A]  time = 0.24494, size = 74, normalized size = 1.54 \[ \frac{\sqrt{x^2+1} \left (3 \sqrt{2} \sqrt{\frac{x^2}{x^2+1}} \left (x^2+2\right ) \tanh ^{-1}\left (\sqrt{\frac{x^2}{2 x^2+2}}\right )-2 x^2\right )}{8 x \left (x^2+2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.014, size = 46, normalized size = 1. \begin{align*}{\frac{x}{4}{\frac{1}{\sqrt{{x}^{2}+1}}} \left ({\frac{{x}^{2}}{{x}^{2}+1}}-2 \right ) ^{-1}}+{\frac{3\,\sqrt{2}}{8}{\it Artanh} \left ({\frac{x\sqrt{2}}{2}{\frac{1}{\sqrt{{x}^{2}+1}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{2} + 2\right )}^{2} \sqrt{x^{2} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.82398, size = 215, normalized size = 4.48 \begin{align*} \frac{3 \, \sqrt{2}{\left (x^{2} + 2\right )} \log \left (\frac{9 \, x^{2} + 2 \, \sqrt{2}{\left (3 \, x^{2} + 2\right )} + 2 \, \sqrt{x^{2} + 1}{\left (3 \, \sqrt{2} x + 4 \, x\right )} + 6}{x^{2} + 2}\right ) - 4 \, x^{2} - 4 \, \sqrt{x^{2} + 1} x - 8}{16 \,{\left (x^{2} + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*sqrt(2)*(x^2 + 2)*log((9*x^2 + 2*sqrt(2)*(3*x^2 + 2) + 2*sqrt(x^2 + 1)*(3*sqrt(2)*x + 4*x) + 6)/(x^2 +
2)) - 4*x^2 - 4*sqrt(x^2 + 1)*x - 8)/(x^2 + 2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{x^{2} + 1} \left (x^{2} + 2\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.07245, size = 136, normalized size = 2.83 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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