\(\int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a^2+x^2}} \, dx\) [25]

Optimal result

Integrand size = 29, antiderivative size = 19 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a^2+x^2}} \, dx=2 \sqrt {x+\sqrt {a^2+x^2}} \]

[Out]

2*(x+(a^2+x^2)^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2147, 30} \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a^2+x^2}} \, dx=2 \sqrt {\sqrt {a^2+x^2}+x} \]

[In]

Int[Sqrt[x + Sqrt[a^2 + x^2]]/Sqrt[a^2 + x^2],x]

[Out]

2*Sqrt[x + Sqrt[a^2 + x^2]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2147

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1/(2^(2*m + 1)*e*f^(2*m)))*(i/c)^m, Subst[Int[x^n*((d^2 + a*f^2 - 2*d*x + x^2)^(2*m + 1)/(-d + x)^(2*(m + 1
))), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, d, e, f, g, i, n}, x] && EqQ[e^2 - c*f^2, 0] && E
qQ[c*g - a*i, 0] && IntegerQ[2*m] && (IntegerQ[m] || GtQ[i/c, 0])

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a^2+x^2}} \, dx=2 \sqrt {x+\sqrt {a^2+x^2}} \]

[In]

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

[Out]

2*Sqrt[x + Sqrt[a^2 + x^2]]

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(2 \sqrt {x +\sqrt {a^{2}+x^{2}}}\)	\(16\)
default	\(2 \sqrt {x +\sqrt {a^{2}+x^{2}}}\)	\(16\)

[In]

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

[Out]

2*(x+(a^2+x^2)^(1/2))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a^2+x^2}} \, dx=2 \, \sqrt {x + \sqrt {a^{2} + x^{2}}} \]

[In]

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

[Out]

2*sqrt(x + sqrt(a^2 + x^2))

Sympy [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a^2+x^2}} \, dx=2 \sqrt {x + \sqrt {a^{2} + x^{2}}} \]

[In]

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

[Out]

2*sqrt(x + sqrt(a**2 + x**2))

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a^2+x^2}} \, dx=2 \, \sqrt {x + \sqrt {a^{2} + x^{2}}} \]

[In]

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

[Out]

2*sqrt(x + sqrt(a^2 + x^2))

Mupad [B] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a^2+x^2}} \, dx=2\,\sqrt {x+\sqrt {a^2+x^2}} \]

[In]

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

[Out]

2*(x + (a^2 + x^2)^(1/2))^(1/2)