∫ 1_____ x∘ _______ x+√ __

Optimal. Leaf size=54 \[ \frac {2}{\sqrt {\sqrt {x^2+1}+x}}+2 \tan ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )-2 \tanh ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right ) \]

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Rubi [A] time = 0.05, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2119, 453, 329, 298, 203, 206} \begin {gather*} \frac {2}{\sqrt {\sqrt {x^2+1}+x}}+2 \tan ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right )-2 \tanh ^{-1}\left (\sqrt {\sqrt {x^2+1}+x}\right ) \end {gather*}

2/Sqrt[x + Sqrt[1 + x^2]] + 2*ArcTan[Sqrt[x + Sqrt[1 + x^2]]] - 2*ArcTanh[Sqrt[x + Sqrt[1 + x^2]]]

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])

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])

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),

2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I

nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]

&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m

+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In

t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||

GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) && !ILtQ[p, -1]

Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dist[1/(2^(

m + 1)*e^(m + 1)), Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*(-(a*f^2*h) + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt

[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[m]

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

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Mathematica [C] time = 0.92, size = 175, normalized size = 3.24 \begin {gather*} \frac {2}{3} \sqrt {\sqrt {x^2+1}+x} \left (-\frac {\sqrt {x^2+1} \left (\sqrt {x^2+1}+x\right )^4 \left (\left (4 x^2+4 \sqrt {x^2+1} x+2\right ) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};\left (x+\sqrt {x^2+1}\right )^2\right )-x^2-\sqrt {x^2+1} x-2\right )}{16 x^6+28 x^4+13 x^2+5 \sqrt {x^2+1} x+16 \sqrt {x^2+1} x^5+20 \sqrt {x^2+1} x^3+1}+\sqrt {x^2+1}-2 x\right ) \end {gather*}

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

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

x^2] + (2 + 4*x^2 + 4*x*Sqrt[1 + x^2])*Hypergeometric2F1[3/4, 1, 7/4, (x + Sqrt[1 + x^2])^2]))/(1 + 13*x^2 +

28*x^4 + 16*x^6 + 5*x*Sqrt[1 + x^2] + 20*x^3*Sqrt[1 + x^2] + 16*x^5*Sqrt[1 + x^2])))/3

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IntegrateAlgebraic [A] time = 0.08, size = 54, normalized size = 1.00 \begin {gather*} \frac {2}{\sqrt {x+\sqrt {1+x^2}}}+2 \tan ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right )-2 \tanh ^{-1}\left (\sqrt {x+\sqrt {1+x^2}}\right ) \end {gather*}

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

2/Sqrt[x + Sqrt[1 + x^2]] + 2*ArcTan[Sqrt[x + Sqrt[1 + x^2]]] - 2*ArcTanh[Sqrt[x + Sqrt[1 + x^2]]]

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fricas [A] time = 0.50, size = 69, normalized size = 1.28 \begin {gather*} -2 \, \sqrt {x + \sqrt {x^{2} + 1}} {\left (x - \sqrt {x^{2} + 1}\right )} + 2 \, \arctan \left (\sqrt {x + \sqrt {x^{2} + 1}}\right ) - \log \left (\sqrt {x + \sqrt {x^{2} + 1}} + 1\right ) + \log \left (\sqrt {x + \sqrt {x^{2} + 1}} - 1\right ) \end {gather*}

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

-2*sqrt(x + sqrt(x^2 + 1))*(x - sqrt(x^2 + 1)) + 2*arctan(sqrt(x + sqrt(x^2 + 1))) - log(sqrt(x + sqrt(x^2 + 1

)) + 1) + log(sqrt(x + sqrt(x^2 + 1)) - 1)

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

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

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

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method	result	size

meijerg	\(-\frac {\sqrt {2}\, \hypergeom \left (\left [\frac {1}{4}, \frac {1}{4}, \frac {3}{4}\right ], \left [\frac {5}{4}, \frac {3}{2}\right ], -\frac {1}{x^{2}}\right )}{\sqrt {x}}\)	\(22\)

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

-2^(1/2)/x^(1/2)*hypergeom([1/4,1/4,3/4],[5/4,3/2],-1/x^2)

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

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

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

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mupad [F] time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{x\,\sqrt {x+\sqrt {x^2+1}}} \,d x \end {gather*}

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sympy [C] time = 1.34, size = 44, normalized size = 0.81 \begin {gather*} - \frac {\Gamma ^{2}\left (\frac {1}{4}\right ) \Gamma \left (\frac {3}{4}\right ) {{}_{3}F_{2}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4}, \frac {3}{4} \\ \frac {5}{4}, \frac {3}{2} \end {matrix}\middle | {\frac {e^{i \pi }}{x^{2}}} \right )}}{4 \pi \sqrt {x} \Gamma \left (\frac {5}{4}\right )} \end {gather*}

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

-gamma(1/4)**2*gamma(3/4)*hyper((1/4, 1/4, 3/4), (5/4, 3/2), exp_polar(I*pi)/x**2)/(4*pi*sqrt(x)*gamma(5/4))

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