3.1.26 \(\int \frac {\sqrt {b x+\sqrt {a+b^2 x^2}}}{\sqrt {a+b^2 x^2}} \, dx\) [26]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {2147, 30} \begin {gather*} \frac {2 \sqrt {\sqrt {a+b^2 x^2}+b x}}{b} \end {gather*}

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 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*} \int \frac {\sqrt {b x+\sqrt {a+b^2 x^2}}}{\sqrt {a+b^2 x^2}} \, dx &=\frac {\text {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 = 0.04, size = 26, normalized size = 1.00 \begin {gather*} \frac {2 \sqrt {b x+\sqrt {a+b^2 x^2}}}{b} \end {gather*}

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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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 2 in optimal.
time = 2.13, size = 33, normalized size = 1.27 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {2 \sqrt {b x+\sqrt {a+b^2 x^2}}}{b},b\text {!=}0\right \}\right \},\frac {x}{a^{\frac {1}{4}}}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Sqrt[b*x + Sqrt[a + b^2*x^2]]/Sqrt[a + b^2*x^2],x]')

[Out]

Piecewise[{{2 Sqrt[b x + Sqrt[a + b ^ 2 x ^ 2]] / b, b != 0}}, x / a ^ (1 / 4)]

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

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.60, size = 22, normalized size = 0.85 \begin {gather*} \frac {2 \, \sqrt {b x + \sqrt {b^{2} x^{2} + a}}}{b} \end {gather*}

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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Sympy [A]
time = 0.51, size = 27, normalized size = 1.04 \begin {gather*} \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} \end {gather*}

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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Giac [A]
time = 0.00, size = 25, normalized size = 0.96 \begin {gather*} \frac {2 \sqrt {b x+\sqrt {a+b^{2} x^{2}}}}{b} \end {gather*}

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]

2*sqrt(b*x + sqrt(b^2*x^2 + a))/b

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Mupad [B]
time = 0.51, size = 22, normalized size = 0.85 \begin {gather*} \frac {2\,\sqrt {\sqrt {b^2\,x^2+a}+b\,x}}{b} \end {gather*}

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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