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

Optimal. Leaf size=82 \[ \frac{4 x}{3 \left (1-3 x^2\right )}-\frac{2 \sqrt{x^2+1}}{3 \left (1-3 x^2\right )}+\frac{\tanh ^{-1}\left (\frac{1}{2} \sqrt{3} \sqrt{x^2+1}\right )}{3 \sqrt{3}}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0758864, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {6742, 199, 207, 444, 47, 63} \[ \frac{4 x}{3 \left (1-3 x^2\right )}-\frac{2 \sqrt{x^2+1}}{3 \left (1-3 x^2\right )}+\frac{\tanh ^{-1}\left (\frac{1}{2} \sqrt{3} \sqrt{x^2+1}\right )}{3 \sqrt{3}}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}} \]

Antiderivative was successfully verified.

[In]

Int[(2*x + Sqrt[1 + x^2])^(-2),x]

[Out]

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

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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 444

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

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rubi steps

\begin{align*} \int \frac{1}{\left (2 x+\sqrt{1+x^2}\right )^2} \, dx &=\int \left (\frac{8}{3 \left (-1+3 x^2\right )^2}-\frac{4 x \sqrt{1+x^2}}{\left (-1+3 x^2\right )^2}+\frac{5}{3 \left (-1+3 x^2\right )}\right ) \, dx\\ &=\frac{5}{3} \int \frac{1}{-1+3 x^2} \, dx+\frac{8}{3} \int \frac{1}{\left (-1+3 x^2\right )^2} \, dx-4 \int \frac{x \sqrt{1+x^2}}{\left (-1+3 x^2\right )^2} \, dx\\ &=\frac{4 x}{3 \left (1-3 x^2\right )}-\frac{5 \tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}}-\frac{4}{3} \int \frac{1}{-1+3 x^2} \, dx-2 \operatorname{Subst}\left (\int \frac{\sqrt{1+x}}{(-1+3 x)^2} \, dx,x,x^2\right )\\ &=\frac{4 x}{3 \left (1-3 x^2\right )}-\frac{2 \sqrt{1+x^2}}{3 \left (1-3 x^2\right )}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}}-\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+x} (-1+3 x)} \, dx,x,x^2\right )\\ &=\frac{4 x}{3 \left (1-3 x^2\right )}-\frac{2 \sqrt{1+x^2}}{3 \left (1-3 x^2\right )}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}}-\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{-4+3 x^2} \, dx,x,\sqrt{1+x^2}\right )\\ &=\frac{4 x}{3 \left (1-3 x^2\right )}-\frac{2 \sqrt{1+x^2}}{3 \left (1-3 x^2\right )}-\frac{\tanh ^{-1}\left (\sqrt{3} x\right )}{3 \sqrt{3}}+\frac{\tanh ^{-1}\left (\frac{1}{2} \sqrt{3} \sqrt{1+x^2}\right )}{3 \sqrt{3}}\\ \end{align*}

Mathematica [A]  time = 0.107544, size = 69, normalized size = 0.84 \[ \frac{1}{9} \left (\frac{6 \left (\sqrt{x^2+1}-2 x\right )}{3 x^2-1}+\sqrt{3} \tanh ^{-1}\left (\frac{1}{2} \sqrt{3} \sqrt{x^2+1}\right )-\sqrt{3} \tanh ^{-1}\left (\sqrt{3} x\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*(-2*x + Sqrt[1 + x^2]))/(-1 + 3*x^2) - Sqrt[3]*ArcTanh[Sqrt[3]*x] + Sqrt[3]*ArcTanh[(Sqrt[3]*Sqrt[1 + x^2]
)/2])/9

________________________________________________________________________________________

Maple [B]  time = 0.041, size = 370, normalized size = 4.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*x/(3*x^2-1)-1/9*arctanh(x*3^(1/2))*3^(1/2)-5/18*x/(x^2-1/3)-3^(1/2)*(-1/12/(x-1/3*3^(1/2))*((x-1/3*3^(1/2
))^2+2/3*3^(1/2)*(x-1/3*3^(1/2))+4/3)^(3/2)+1/36*3^(1/2)*(1/3*(9*(x-1/3*3^(1/2))^2+6*3^(1/2)*(x-1/3*3^(1/2))+1
2)^(1/2)+1/3*3^(1/2)*arcsinh(x)-2/3*3^(1/2)*arctanh(3/4*(8/3+2/3*3^(1/2)*(x-1/3*3^(1/2)))*3^(1/2)/(9*(x-1/3*3^
(1/2))^2+6*3^(1/2)*(x-1/3*3^(1/2))+12)^(1/2)))+1/12*x*((x-1/3*3^(1/2))^2+2/3*3^(1/2)*(x-1/3*3^(1/2))+4/3)^(1/2
)+1/12*arcsinh(x))+3^(1/2)*(-1/12/(x+1/3*3^(1/2))*((x+1/3*3^(1/2))^2-2/3*3^(1/2)*(x+1/3*3^(1/2))+4/3)^(3/2)-1/
36*3^(1/2)*(1/3*(9*(x+1/3*3^(1/2))^2-6*3^(1/2)*(x+1/3*3^(1/2))+12)^(1/2)-1/3*3^(1/2)*arcsinh(x)-2/3*3^(1/2)*ar
ctanh(3/4*(8/3-2/3*3^(1/2)*(x+1/3*3^(1/2)))*3^(1/2)/(9*(x+1/3*3^(1/2))^2-6*3^(1/2)*(x+1/3*3^(1/2))+12)^(1/2)))
+1/12*x*((x+1/3*3^(1/2))^2-2/3*3^(1/2)*(x+1/3*3^(1/2))+4/3)^(1/2)+1/12*arcsinh(x))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.08059, size = 251, normalized size = 3.06 \begin{align*} \frac{\sqrt{3}{\left (3 \, x^{2} - 1\right )} \log \left (\frac{3 \, x^{2} - 2 \, \sqrt{3} x + 1}{3 \, x^{2} - 1}\right ) + \sqrt{3}{\left (3 \, x^{2} - 1\right )} \log \left (\frac{3 \, x^{2} + 4 \, \sqrt{3} \sqrt{x^{2} + 1} + 7}{3 \, x^{2} - 1}\right ) - 24 \, x + 12 \, \sqrt{x^{2} + 1}}{18 \,{\left (3 \, x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(sqrt(3)*(3*x^2 - 1)*log((3*x^2 - 2*sqrt(3)*x + 1)/(3*x^2 - 1)) + sqrt(3)*(3*x^2 - 1)*log((3*x^2 + 4*sqrt
(3)*sqrt(x^2 + 1) + 7)/(3*x^2 - 1)) - 24*x + 12*sqrt(x^2 + 1))/(3*x^2 - 1)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.11115, size = 239, normalized size = 2.91 \begin{align*} \frac{1}{18} \, \sqrt{3} \log \left (\frac{{\left | 6 \, x - 2 \, \sqrt{3} \right |}}{{\left | 6 \, x + 2 \, \sqrt{3} \right |}}\right ) - \frac{1}{18} \, \sqrt{3} \log \left (-\frac{{\left | -6 \, x - 8 \, \sqrt{3} + 6 \, \sqrt{x^{2} + 1} - \frac{6}{x - \sqrt{x^{2} + 1}} \right |}}{2 \,{\left (3 \, x - 4 \, \sqrt{3} - 3 \, \sqrt{x^{2} + 1} + \frac{3}{x - \sqrt{x^{2} + 1}}\right )}}\right ) - \frac{4 \,{\left (x - \sqrt{x^{2} + 1} + \frac{1}{x - \sqrt{x^{2} + 1}}\right )}}{3 \,{\left (3 \,{\left (x - \sqrt{x^{2} + 1} + \frac{1}{x - \sqrt{x^{2} + 1}}\right )}^{2} - 16\right )}} - \frac{4 \, x}{3 \,{\left (3 \, x^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*sqrt(3)*log(abs(6*x - 2*sqrt(3))/abs(6*x + 2*sqrt(3))) - 1/18*sqrt(3)*log(-1/2*abs(-6*x - 8*sqrt(3) + 6*s
qrt(x^2 + 1) - 6/(x - sqrt(x^2 + 1)))/(3*x - 4*sqrt(3) - 3*sqrt(x^2 + 1) + 3/(x - sqrt(x^2 + 1)))) - 4/3*(x -
sqrt(x^2 + 1) + 1/(x - sqrt(x^2 + 1)))/(3*(x - sqrt(x^2 + 1) + 1/(x - sqrt(x^2 + 1)))^2 - 16) - 4/3*x/(3*x^2 -
1)