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

Optimal. Leaf size=71 \[ -\frac {4 x}{3 \left (3 x^2-1\right )}+\frac {2 \sqrt {x^2+1}}{3 \left (3 x^2-1\right )}-\frac {2 \tanh ^{-1}\left (\frac {x}{\sqrt {3}}-\frac {\sqrt {x^2+1}}{\sqrt {3}}\right )}{3 \sqrt {3}} \]

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Rubi [A]  time = 0.12, antiderivative size = 82, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {6742, 199, 207, 444, 47, 63} \begin {gather*} \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}} \end {gather*}

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 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]

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 6742

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

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*}

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Mathematica [A]  time = 0.14, size = 99, normalized size = 1.39 \begin {gather*} \frac {1}{9} \left (\frac {12 x}{1-3 x^2}-\frac {\frac {6 x^2+6}{1-3 x^2}+\sqrt {3} \sqrt {-x^2-1} \tan ^{-1}\left (\frac {1}{2} \sqrt {3} \sqrt {-x^2-1}\right )}{\sqrt {x^2+1}}-\sqrt {3} \tanh ^{-1}\left (\sqrt {3} x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.42, size = 71, normalized size = 1.00 \begin {gather*} -\frac {4 x}{3 \left (-1+3 x^2\right )}+\frac {2 \sqrt {1+x^2}}{3 \left (-1+3 x^2\right )}-\frac {2 \tanh ^{-1}\left (\frac {x}{\sqrt {3}}-\frac {\sqrt {1+x^2}}{\sqrt {3}}\right )}{3 \sqrt {3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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fricas [A]  time = 0.69, size = 100, normalized size = 1.41 \begin {gather*} \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 {gather*}

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)

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giac [B]  time = 0.22, size = 177, normalized size = 2.49 \begin {gather*} \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 {gather*}

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)

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maple [C]  time = 0.22, size = 72, normalized size = 1.01

method	result	size
trager	\(-\frac {4 x}{3 \left (3 x^{2}-1\right )}+\frac {2 \sqrt {x^{2}+1}}{3 \left (3 x^{2}-1\right )}-\frac {\RootOf \left (\textit {\_Z}^{2}-3\right ) \ln \left (\frac {2 \RootOf \left (\textit {\_Z}^{2}-3\right )-3 \sqrt {x^{2}+1}}{\RootOf \left (\textit {\_Z}^{2}-3\right ) x -1}\right )}{9}\)	\(72\)
default	\(-\frac {x}{2 \left (3 x^{2}-1\right )}-\frac {\sqrt {3}\, \arctanh \left (x \sqrt {3}\right )}{9}-\frac {5 x}{18 \left (x^{2}-\frac {1}{3}\right )}-\sqrt {3}\, \left (-\frac {\left (\left (x -\frac {\sqrt {3}}{3}\right )^{2}+\frac {2 \left (x -\frac {\sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {4}{3}\right )^{\frac {3}{2}}}{12 \left (x -\frac {\sqrt {3}}{3}\right )}+\frac {\sqrt {3}\, \left (\frac {\sqrt {9 \left (x -\frac {\sqrt {3}}{3}\right )^{2}+6 \left (x -\frac {\sqrt {3}}{3}\right ) \sqrt {3}+12}}{3}+\frac {\sqrt {3}\, \arcsinh \relax (x )}{3}-\frac {2 \sqrt {3}\, \arctanh \left (\frac {3 \left (\frac {8}{3}+\frac {2 \left (x -\frac {\sqrt {3}}{3}\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{4 \sqrt {9 \left (x -\frac {\sqrt {3}}{3}\right )^{2}+6 \left (x -\frac {\sqrt {3}}{3}\right ) \sqrt {3}+12}}\right )}{3}\right )}{36}+\frac {x \sqrt {\left (x -\frac {\sqrt {3}}{3}\right )^{2}+\frac {2 \left (x -\frac {\sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {4}{3}}}{12}+\frac {\arcsinh \relax (x )}{12}\right )+\sqrt {3}\, \left (-\frac {\left (\left (x +\frac {\sqrt {3}}{3}\right )^{2}-\frac {2 \left (x +\frac {\sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {4}{3}\right )^{\frac {3}{2}}}{12 \left (x +\frac {\sqrt {3}}{3}\right )}-\frac {\sqrt {3}\, \left (\frac {\sqrt {9 \left (x +\frac {\sqrt {3}}{3}\right )^{2}-6 \left (x +\frac {\sqrt {3}}{3}\right ) \sqrt {3}+12}}{3}-\frac {\sqrt {3}\, \arcsinh \relax (x )}{3}-\frac {2 \sqrt {3}\, \arctanh \left (\frac {3 \left (\frac {8}{3}-\frac {2 \left (x +\frac {\sqrt {3}}{3}\right ) \sqrt {3}}{3}\right ) \sqrt {3}}{4 \sqrt {9 \left (x +\frac {\sqrt {3}}{3}\right )^{2}-6 \left (x +\frac {\sqrt {3}}{3}\right ) \sqrt {3}+12}}\right )}{3}\right )}{36}+\frac {x \sqrt {\left (x +\frac {\sqrt {3}}{3}\right )^{2}-\frac {2 \left (x +\frac {\sqrt {3}}{3}\right ) \sqrt {3}}{3}+\frac {4}{3}}}{12}+\frac {\arcsinh \relax (x )}{12}\right )\)	\(370\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x+(x^2+1)^(1/2))^2,x,method=_RETURNVERBOSE)

[Out]

-4/3*x/(3*x^2-1)+2/3/(3*x^2-1)*(x^2+1)^(1/2)-1/9*RootOf(_Z^2-3)*ln((2*RootOf(_Z^2-3)-3*(x^2+1)^(1/2))/(RootOf(
_Z^2-3)*x-1))

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

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)

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mupad [B]  time = 0.21, size = 204, normalized size = 2.87 \begin {gather*} \frac {\sqrt {3}\,\left (\ln \left (x-\frac {\sqrt {3}}{3}\right )-\ln \left (x+\sqrt {3}+2\,\sqrt {x^2+1}\right )\right )}{18}-\frac {4\,x}{9\,\left (x^2-\frac {1}{3}\right )}+\frac {\sqrt {3}\,\left (\ln \left (x+\frac {\sqrt {3}}{3}\right )-\ln \left (x-\sqrt {3}-2\,\sqrt {x^2+1}\right )\right )}{18}-\frac {\sqrt {3}\,\left (6\,\ln \left (x-\frac {\sqrt {3}}{3}\right )-6\,\ln \left (x+\sqrt {3}+2\,\sqrt {x^2+1}\right )\right )}{54}-\frac {\sqrt {3}\,\left (6\,\ln \left (x+\frac {\sqrt {3}}{3}\right )-6\,\ln \left (x-\sqrt {3}-2\,\sqrt {x^2+1}\right )\right )}{54}+\frac {\sqrt {3}\,\sqrt {x^2+1}}{9\,\left (x-\frac {\sqrt {3}}{3}\right )}-\frac {\sqrt {3}\,\sqrt {x^2+1}}{9\,\left (x+\frac {\sqrt {3}}{3}\right )}+\frac {\sqrt {3}\,\mathrm {atan}\left (\sqrt {3}\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(2*x + (x^2 + 1)^(1/2))^2,x)

[Out]

(3^(1/2)*(log(x - 3^(1/2)/3) - log(x + 3^(1/2) + 2*(x^2 + 1)^(1/2))))/18 + (3^(1/2)*atan(3^(1/2)*x*1i)*1i)/9 -
(4*x)/(9*(x^2 - 1/3)) + (3^(1/2)*(log(x + 3^(1/2)/3) - log(x - 3^(1/2) - 2*(x^2 + 1)^(1/2))))/18 - (3^(1/2)*(
6*log(x - 3^(1/2)/3) - 6*log(x + 3^(1/2) + 2*(x^2 + 1)^(1/2))))/54 - (3^(1/2)*(6*log(x + 3^(1/2)/3) - 6*log(x
- 3^(1/2) - 2*(x^2 + 1)^(1/2))))/54 + (3^(1/2)*(x^2 + 1)^(1/2))/(9*(x - 3^(1/2)/3)) - (3^(1/2)*(x^2 + 1)^(1/2)
)/(9*(x + 3^(1/2)/3))

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

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)

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