\(\int \sqrt {x+\sqrt {1+x^2}} \, dx\) [116]

Optimal result

Integrand size = 15, antiderivative size = 37 \[ \int \sqrt {x+\sqrt {1+x^2}} \, dx=-\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, 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} \[ \int \sqrt {x+\sqrt {1+x^2}} \, dx=\frac {1}{3} \left (\sqrt {x^2+1}+x\right )^{3/2}-\frac {1}{\sqrt {\sqrt {x^2+1}+x}} \]

[In]

Int[Sqrt[x + Sqrt[1 + x^2]],x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \sqrt {x+\sqrt {1+x^2}} \, dx=-\frac {1}{\sqrt {x+\sqrt {1+x^2}}}+\frac {1}{3} \left (x+\sqrt {1+x^2}\right )^{3/2} \]

[In]

Integrate[Sqrt[x + Sqrt[1 + x^2]],x]

[Out]

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

Maple [C] (verified)

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

Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.54

method	result	size
meijerg	\(\frac {\frac {16 \sqrt {\pi }\, \sqrt {2}\, x^{\frac {3}{2}} \left (-\frac {1}{x^{2}}+1\right ) \cosh \left (\frac {\operatorname {arcsinh}\left (\frac {1}{x}\right )}{2}\right )}{3}+\frac {16 \sqrt {\pi }\, \sqrt {2}\, \sqrt {x}\, \sqrt {\frac {1}{x^{2}}+1}\, \sinh \left (\frac {\operatorname {arcsinh}\left (\frac {1}{x}\right )}{2}\right )}{3}}{8 \sqrt {\pi }}\)	\(57\)

[In]

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

[Out]

1/8/Pi^(1/2)*(16/3*Pi^(1/2)*2^(1/2)*x^(3/2)*(-1/x^2+1)*cosh(1/2*arcsinh(1/x))+16/3*Pi^(1/2)*2^(1/2)*x^(1/2)*(1
/x^2+1)^(1/2)*sinh(1/2*arcsinh(1/x)))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.70 \[ \int \sqrt {x+\sqrt {1+x^2}} \, dx=\frac {2}{3} \, {\left (2 \, x - \sqrt {x^{2} + 1}\right )} \sqrt {x + \sqrt {x^{2} + 1}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.14 \[ \int \sqrt {x+\sqrt {1+x^2}} \, dx=\frac {4 x \sqrt {x + \sqrt {x^{2} + 1}}}{3} - \frac {2 \sqrt {x + \sqrt {x^{2} + 1}} \sqrt {x^{2} + 1}}{3} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

integrate(sqrt(x + sqrt(x^2 + 1)), x)

Giac [F]

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

[In]

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

[Out]

integrate(sqrt(x + sqrt(x^2 + 1)), x)

Mupad [F(-1)]

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

[In]

int((x + (x^2 + 1)^(1/2))^(1/2),x)

[Out]

int((x + (x^2 + 1)^(1/2))^(1/2), x)