3.26 \(\int \frac {\sqrt {b x+\sqrt {a+b^2 x^2}}}{\sqrt {a+b^2 x^2}} \, dx\)

Optimal. Leaf size=26 \[ \frac {2 \sqrt {\sqrt {a+b^2 x^2}+b x}}{b} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 26, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.057, Rules used = {2122, 30} \[ \frac {2 \sqrt {\sqrt {a+b^2 x^2}+b x}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x + Sqrt[a + b^2*x^2]])/b

Rule 30

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

Rule 2122

Int[((g_) + (i_.)*(x_)^2)^(m_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^(n_.), x_Symbol] :> Dis
t[(1*(i/c)^m)/(2^(2*m + 1)*e*f^(2*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*} \int \frac {\sqrt {b x+\sqrt {a+b^2 x^2}}}{\sqrt {a+b^2 x^2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\sqrt {x}} \, dx,x,b x+\sqrt {a+b^2 x^2}\right )}{b}\\ &=\frac {2 \sqrt {b x+\sqrt {a+b^2 x^2}}}{b}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 26, normalized size = 1.00 \[ \frac {2 \sqrt {\sqrt {a+b^2 x^2}+b x}}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b*x + Sqrt[a + b^2*x^2]])/b

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fricas [A]  time = 0.99, size = 22, normalized size = 0.85 \[ \frac {2 \, \sqrt {b x + \sqrt {b^{2} x^{2} + a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x + sqrt(b^2*x^2 + a))/b

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giac [A]  time = 0.98, size = 22, normalized size = 0.85 \[ \frac {2 \, \sqrt {b x + \sqrt {b^{2} x^{2} + a}}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*sqrt(b*x + sqrt(b^2*x^2 + a))/b

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x +\sqrt {b^{2} x^{2}+a}}}{\sqrt {b^{2} x^{2}+a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {b x + \sqrt {b^{2} x^{2} + a}}}{\sqrt {b^{2} x^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.51, size = 22, normalized size = 0.85 \[ \frac {2\,\sqrt {\sqrt {b^2\,x^2+a}+b\,x}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.07, size = 27, normalized size = 1.04 \[ \begin {cases} \frac {2 \sqrt {b x + \sqrt {a + b^{2} x^{2}}}}{b} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt [4]{a}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*sqrt(b*x + sqrt(a + b**2*x**2))/b, Ne(b, 0)), (x/a**(1/4), True))

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