\(\int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx\) [448]

Optimal result

Integrand size = 15, antiderivative size = 35 \[ \int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\sqrt {x+\sqrt {1+x^2}} \]

[Out]

-1/3/(x+(x^2+1)^(1/2))^(3/2)+(x+(x^2+1)^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2142, 14} \[ \int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\sqrt {\sqrt {x^2+1}+x}-\frac {1}{3 \left (\sqrt {x^2+1}+x\right )^{3/2}} \]

[In]

Int[1/Sqrt[x + Sqrt[1 + x^2]],x]

[Out]

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2142

Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_))^(p_.), x_Symbol] :> Dist[1/(2*
e), Subst[Int[(g + h*x^n)^p*((d^2 + a*f^2 - 2*d*x + x^2)/(d - x)^2), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /
; FreeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1+x^2}{x^{5/2}} \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \left (\frac {1}{x^{5/2}}+\frac {1}{\sqrt {x}}\right ) \, dx,x,x+\sqrt {1+x^2}\right ) \\ & = -\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\sqrt {x+\sqrt {1+x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {1}{3 \left (x+\sqrt {1+x^2}\right )^{3/2}}+\sqrt {x+\sqrt {1+x^2}} \]

[In]

Integrate[1/Sqrt[x + Sqrt[1 + x^2]],x]

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 3 vs. order 2.

Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.77

method	result	size
meijerg	\(-\frac {-\frac {32 \sqrt {\pi }\, \sqrt {2}\, \cosh \left (\frac {3 \,\operatorname {arcsinh}\left (\frac {1}{x}\right )}{2}\right )}{3 x^{\frac {3}{2}}}-\frac {8 \sqrt {\pi }\, \sqrt {2}\, x^{\frac {3}{2}} \left (-\frac {4}{3 x^{4}}-\frac {2}{3 x^{2}}+\frac {2}{3}\right ) \sinh \left (\frac {3 \,\operatorname {arcsinh}\left (\frac {1}{x}\right )}{2}\right )}{\sqrt {1+\frac {1}{x^{2}}}}}{8 \sqrt {\pi }}\)	\(62\)

[In]

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

[Out]

-1/8/Pi^(1/2)*(-32/3*Pi^(1/2)*2^(1/2)/x^(3/2)*cosh(3/2*arcsinh(1/x))-8*Pi^(1/2)*2^(1/2)*x^(3/2)*(-4/3/x^4-2/3/
x^2+2/3)*sinh(3/2*arcsinh(1/x))/(1+1/x^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx=-\frac {2}{3} \, {\left (x^{2} - \sqrt {x^{2} + 1} x - 1\right )} \sqrt {x + \sqrt {x^{2} + 1}} \]

[In]

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

[Out]

-2/3*(x^2 - sqrt(x^2 + 1)*x - 1)*sqrt(x + sqrt(x^2 + 1))

Sympy [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\frac {4 x}{3 \sqrt {x + \sqrt {x^{2} + 1}}} + \frac {2 \sqrt {x^{2} + 1}}{3 \sqrt {x + \sqrt {x^{2} + 1}}} \]

[In]

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

[Out]

4*x/(3*sqrt(x + sqrt(x**2 + 1))) + 2*sqrt(x**2 + 1)/(3*sqrt(x + sqrt(x**2 + 1)))

Maxima [F]

\[ \int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\int { \frac {1}{\sqrt {x + \sqrt {x^{2} + 1}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\int { \frac {1}{\sqrt {x + \sqrt {x^{2} + 1}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {x+\sqrt {1+x^2}}} \, dx=\int \frac {1}{\sqrt {x+\sqrt {x^2+1}}} \,d x \]

[In]

int(1/(x + (x^2 + 1)^(1/2))^(1/2),x)

[Out]

int(1/(x + (x^2 + 1)^(1/2))^(1/2), x)