\(\int \frac {1}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx\) [812]

Optimal result

Integrand size = 23, antiderivative size = 61 \[ \int \frac {1}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=-\frac {b^2}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2142, 14} \[ \int \frac {1}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\frac {\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{a}-\frac {b^2}{3 a \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}} \]

[In]

Int[1/Sqrt[a*x + Sqrt[b^2 + a^2*x^2]],x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\frac {2 \left (b^2+3 a x \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}} \]

[In]

Integrate[1/Sqrt[a*x + Sqrt[b^2 + a^2*x^2]],x]

[Out]

(2*(b^2 + 3*a*x*(a*x + Sqrt[b^2 + a^2*x^2])))/(3*a*(a*x + Sqrt[b^2 + a^2*x^2])^(3/2))

Maple [F]

\[\int \frac {1}{\sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}d x\]

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.93 \[ \int \frac {1}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=-\frac {2 \, {\left (a^{2} x^{2} - \sqrt {a^{2} x^{2} + b^{2}} a x - b^{2}\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{3 \, a b^{2}} \]

[In]

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

[Out]

-2/3*(a^2*x^2 - sqrt(a^2*x^2 + b^2)*a*x - b^2)*sqrt(a*x + sqrt(a^2*x^2 + b^2))/(a*b^2)

Sympy [F]

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

[In]

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

[Out]

Integral(1/sqrt(a*x + sqrt(a**2*x**2 + b**2)), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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