\(\int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{x} \, dx\) [203]

Optimal result

Integrand size = 21, antiderivative size = 82 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{x} \, dx=2 \sqrt {x+\sqrt {a^2+x^2}}-2 \sqrt {a} \arctan \left (\frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a}}\right )-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a}}\right ) \]

[Out]

-2*arctan((x+(a^2+x^2)^(1/2))^(1/2)/a^(1/2))*a^(1/2)-2*arctanh((x+(a^2+x^2)^(1/2))^(1/2)/a^(1/2))*a^(1/2)+2*(x
+(a^2+x^2)^(1/2))^(1/2)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2144, 470, 335, 218, 212, 209} \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{x} \, dx=-2 \sqrt {a} \arctan \left (\frac {\sqrt {\sqrt {a^2+x^2}+x}}{\sqrt {a}}\right )-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {\sqrt {a^2+x^2}+x}}{\sqrt {a}}\right )+2 \sqrt {\sqrt {a^2+x^2}+x} \]

[In]

Int[Sqrt[x + Sqrt[a^2 + x^2]]/x,x]

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 335

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]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rule 2144

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]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {a^2+x^2}{\sqrt {x} \left (-a^2+x^2\right )} \, dx,x,x+\sqrt {a^2+x^2}\right ) \\ & = 2 \sqrt {x+\sqrt {a^2+x^2}}+\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (-a^2+x^2\right )} \, dx,x,x+\sqrt {a^2+x^2}\right ) \\ & = 2 \sqrt {x+\sqrt {a^2+x^2}}+\left (4 a^2\right ) \text {Subst}\left (\int \frac {1}{-a^2+x^4} \, dx,x,\sqrt {x+\sqrt {a^2+x^2}}\right ) \\ & = 2 \sqrt {x+\sqrt {a^2+x^2}}-(2 a) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\sqrt {x+\sqrt {a^2+x^2}}\right )-(2 a) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,\sqrt {x+\sqrt {a^2+x^2}}\right ) \\ & = 2 \sqrt {x+\sqrt {a^2+x^2}}-2 \sqrt {a} \arctan \left (\frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a}}\right )-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{x} \, dx=2 \sqrt {x+\sqrt {a^2+x^2}}-2 \sqrt {a} \arctan \left (\frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a}}\right )-2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {x+\sqrt {a^2+x^2}}}{\sqrt {a}}\right ) \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3.

Time = 0.07 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.30

method	result	size
meijerg	\(2 \sqrt {2}\, \sqrt {x}\, {}_{3}^{}{\moversetsp {}{\mundersetsp {}{F_{2}^{}}}}\left (-\frac {1}{4},-\frac {1}{4},\frac {1}{4};\frac {1}{2},\frac {3}{4};-\frac {a^{2}}{x^{2}}\right )\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.63 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{x} \, dx=\left [-2 \, \sqrt {a} \arctan \left (\frac {\sqrt {x + \sqrt {a^{2} + x^{2}}}}{\sqrt {a}}\right ) + \sqrt {a} \log \left (\frac {a^{2} + \sqrt {a^{2} + x^{2}} a - {\left ({\left (a - x\right )} \sqrt {a} + \sqrt {a^{2} + x^{2}} \sqrt {a}\right )} \sqrt {x + \sqrt {a^{2} + x^{2}}}}{x}\right ) + 2 \, \sqrt {x + \sqrt {a^{2} + x^{2}}}, 2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {-a} \sqrt {x + \sqrt {a^{2} + x^{2}}}}{a}\right ) + \sqrt {-a} \log \left (-\frac {a^{2} - \sqrt {a^{2} + x^{2}} a + {\left (\sqrt {-a} {\left (a + x\right )} - \sqrt {a^{2} + x^{2}} \sqrt {-a}\right )} \sqrt {x + \sqrt {a^{2} + x^{2}}}}{x}\right ) + 2 \, \sqrt {x + \sqrt {a^{2} + x^{2}}}\right ] \]

[In]

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

[Out]

[-2*sqrt(a)*arctan(sqrt(x + sqrt(a^2 + x^2))/sqrt(a)) + sqrt(a)*log((a^2 + sqrt(a^2 + x^2)*a - ((a - x)*sqrt(a
) + sqrt(a^2 + x^2)*sqrt(a))*sqrt(x + sqrt(a^2 + x^2)))/x) + 2*sqrt(x + sqrt(a^2 + x^2)), 2*sqrt(-a)*arctan(sq
rt(-a)*sqrt(x + sqrt(a^2 + x^2))/a) + sqrt(-a)*log(-(a^2 - sqrt(a^2 + x^2)*a + (sqrt(-a)*(a + x) - sqrt(a^2 +
x^2)*sqrt(-a))*sqrt(x + sqrt(a^2 + x^2)))/x) + 2*sqrt(x + sqrt(a^2 + x^2))]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.83 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {x+\sqrt {a^2+x^2}}}{x} \, dx=\frac {\sqrt {x} \Gamma ^{2}\left (- \frac {1}{4}\right ) \Gamma \left (\frac {1}{4}\right ) {{}_{3}F_{2}\left (\begin {matrix} - \frac {1}{4}, - \frac {1}{4}, \frac {1}{4} \\ \frac {1}{2}, \frac {3}{4} \end {matrix}\middle | {\frac {a^{2} e^{i \pi }}{x^{2}}} \right )}}{8 \pi \Gamma \left (\frac {3}{4}\right )} \]

[In]

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

[Out]

sqrt(x)*gamma(-1/4)**2*gamma(1/4)*hyper((-1/4, -1/4, 1/4), (1/2, 3/4), a**2*exp_polar(I*pi)/x**2)/(8*pi*gamma(
3/4))

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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