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

Optimal. Leaf size=63 \[ \frac {4 b x}{3 \sqrt {\sqrt {a x^2+b^2}+b}}+\frac {2 x \sqrt {a x^2+b^2}}{3 \sqrt {\sqrt {a x^2+b^2}+b}} \]

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Rubi [A]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 0.81, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2129} \begin {gather*} \frac {2 b x}{\sqrt {\sqrt {a x^2+b^2}+b}}+\frac {2 a x^3}{3 \left (\sqrt {a x^2+b^2}+b\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[b + Sqrt[b^2 + a*x^2]],x]

[Out]

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

Rule 2129

Int[Sqrt[(a_) + (b_.)*Sqrt[(c_) + (d_.)*(x_)^2]], x_Symbol] :> Simp[(2*b^2*d*x^3)/(3*(a + b*Sqrt[c + d*x^2])^(
3/2)), x] + Simp[(2*a*x)/Sqrt[a + b*Sqrt[c + d*x^2]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2*c, 0]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 41, normalized size = 0.65 \begin {gather*} \frac {2 x \left (\sqrt {a x^2+b^2}+2 b\right )}{3 \sqrt {\sqrt {a x^2+b^2}+b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.12, size = 63, normalized size = 1.00 \begin {gather*} \frac {4 b x}{3 \sqrt {b+\sqrt {b^2+a x^2}}}+\frac {2 x \sqrt {b^2+a x^2}}{3 \sqrt {b+\sqrt {b^2+a x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[b + Sqrt[b^2 + a*x^2]],x]

[Out]

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

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fricas [A]  time = 0.52, size = 47, normalized size = 0.75 \begin {gather*} \frac {2 \, {\left (a x^{2} - b^{2} + \sqrt {a x^{2} + b^{2}} b\right )} \sqrt {b + \sqrt {a x^{2} + b^{2}}}}{3 \, a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {b + \sqrt {a x^{2} + b^{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.05, size = 118, normalized size = 1.87

method	result	size
meijerg	\(-\frac {\left (b^{2}\right )^{\frac {1}{4}} \left (-\frac {32 \sqrt {\pi }\, \sqrt {2}\, x^{3} \sqrt {\frac {a}{b^{2}}}\, a \cosh \left (\frac {3 \arcsinh \left (\frac {x \sqrt {a}}{b}\right )}{2}\right )}{3 b^{2}}-\frac {8 \sqrt {\pi }\, \sqrt {2}\, \sqrt {\frac {a}{b^{2}}}\, \left (-\frac {4 x^{4} a^{2}}{3 b^{4}}-\frac {2 x^{2} a}{3 b^{2}}+\frac {2}{3}\right ) \sinh \left (\frac {3 \arcsinh \left (\frac {x \sqrt {a}}{b}\right )}{2}\right ) b}{\sqrt {a}\, \sqrt {\frac {x^{2} a}{b^{2}}+1}}\right )}{8 \sqrt {\pi }\, \sqrt {\frac {a}{b^{2}}}}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(b^2)^(1/4)/Pi^(1/2)/(a/b^2)^(1/2)*(-32/3*Pi^(1/2)*2^(1/2)*x^3*(a/b^2)^(1/2)*a/b^2*cosh(3/2*arcsinh(x*a^(
1/2)/b))-8*Pi^(1/2)*2^(1/2)*(a/b^2)^(1/2)*(-4/3*x^4*a^2/b^4-2/3*x^2*a/b^2+2/3)*sinh(3/2*arcsinh(x*a^(1/2)/b))/
a^(1/2)*b/(x^2*a/b^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \sqrt {b+\sqrt {b^2+a\,x^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.18, size = 286, normalized size = 4.54 \begin {gather*} - \frac {\sqrt {2} a \sqrt {b} x^{3} \Gamma \left (- \frac {1}{4}\right ) \Gamma \left (\frac {1}{4}\right )}{12 \pi b^{2} \sqrt {\frac {a x^{2}}{b^{2}} + 1} \sqrt {\sqrt {\frac {a x^{2}}{b^{2}} + 1} + 1} + 12 \pi b^{2} \sqrt {\sqrt {\frac {a x^{2}}{b^{2}} + 1} + 1}} - \frac {3 \sqrt {2} b^{\frac {5}{2}} x \sqrt {\frac {a x^{2}}{b^{2}} + 1} \Gamma \left (- \frac {1}{4}\right ) \Gamma \left (\frac {1}{4}\right )}{12 \pi b^{2} \sqrt {\frac {a x^{2}}{b^{2}} + 1} \sqrt {\sqrt {\frac {a x^{2}}{b^{2}} + 1} + 1} + 12 \pi b^{2} \sqrt {\sqrt {\frac {a x^{2}}{b^{2}} + 1} + 1}} - \frac {3 \sqrt {2} b^{\frac {5}{2}} x \Gamma \left (- \frac {1}{4}\right ) \Gamma \left (\frac {1}{4}\right )}{12 \pi b^{2} \sqrt {\frac {a x^{2}}{b^{2}} + 1} \sqrt {\sqrt {\frac {a x^{2}}{b^{2}} + 1} + 1} + 12 \pi b^{2} \sqrt {\sqrt {\frac {a x^{2}}{b^{2}} + 1} + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-sqrt(2)*a*sqrt(b)*x**3*gamma(-1/4)*gamma(1/4)/(12*pi*b**2*sqrt(a*x**2/b**2 + 1)*sqrt(sqrt(a*x**2/b**2 + 1) +
1) + 12*pi*b**2*sqrt(sqrt(a*x**2/b**2 + 1) + 1)) - 3*sqrt(2)*b**(5/2)*x*sqrt(a*x**2/b**2 + 1)*gamma(-1/4)*gamm
a(1/4)/(12*pi*b**2*sqrt(a*x**2/b**2 + 1)*sqrt(sqrt(a*x**2/b**2 + 1) + 1) + 12*pi*b**2*sqrt(sqrt(a*x**2/b**2 +
1) + 1)) - 3*sqrt(2)*b**(5/2)*x*gamma(-1/4)*gamma(1/4)/(12*pi*b**2*sqrt(a*x**2/b**2 + 1)*sqrt(sqrt(a*x**2/b**2
+ 1) + 1) + 12*pi*b**2*sqrt(sqrt(a*x**2/b**2 + 1) + 1))

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