3.152 \(\int \sqrt {5+x^2} \, dx\)

Optimal. Leaf size=27 \[ \frac {1}{2} \sqrt {x^2+5} x+\frac {5}{2} \sinh ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \]

[Out]

5/2*arcsinh(1/5*x*5^(1/2))+1/2*x*(x^2+5)^(1/2)

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Rubi [A]  time = 0.00, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {195, 215} \[ \frac {1}{2} \sqrt {x^2+5} x+\frac {5}{2} \sinh ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[5 + x^2],x]

[Out]

(x*Sqrt[5 + x^2])/2 + (5*ArcSinh[x/Sqrt[5]])/2

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 215

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

Rubi steps

\begin {align*} \int \sqrt {5+x^2} \, dx &=\frac {1}{2} x \sqrt {5+x^2}+\frac {5}{2} \int \frac {1}{\sqrt {5+x^2}} \, dx\\ &=\frac {1}{2} x \sqrt {5+x^2}+\frac {5}{2} \sinh ^{-1}\left (\frac {x}{\sqrt {5}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {1}{2} \sqrt {x^2+5} x+\frac {5}{2} \sinh ^{-1}\left (\frac {x}{\sqrt {5}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[5 + x^2],x]

[Out]

(x*Sqrt[5 + x^2])/2 + (5*ArcSinh[x/Sqrt[5]])/2

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fricas [A]  time = 0.39, size = 25, normalized size = 0.93 \[ \frac {1}{2} \, \sqrt {x^{2} + 5} x - \frac {5}{2} \, \log \left (-x + \sqrt {x^{2} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+5)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(x^2 + 5)*x - 5/2*log(-x + sqrt(x^2 + 5))

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giac [A]  time = 0.01, size = 25, normalized size = 0.93 \[ \frac {1}{2} \, \sqrt {x^{2} + 5} x - \frac {5}{2} \, \log \left (-x + \sqrt {x^{2} + 5}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+5)^(1/2),x, algorithm="giac")

[Out]

1/2*sqrt(x^2 + 5)*x - 5/2*log(-x + sqrt(x^2 + 5))

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maple [A]  time = 0.00, size = 21, normalized size = 0.78 \[ \frac {\sqrt {x^{2}+5}\, x}{2}+\frac {5 \arcsinh \left (\frac {\sqrt {5}\, x}{5}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2+5)^(1/2),x)

[Out]

5/2*arcsinh(1/5*x*5^(1/2))+1/2*x*(x^2+5)^(1/2)

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maxima [A]  time = 1.27, size = 20, normalized size = 0.74 \[ \frac {1}{2} \, \sqrt {x^{2} + 5} x + \frac {5}{2} \, \operatorname {arsinh}\left (\frac {1}{5} \, \sqrt {5} x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2+5)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(x^2 + 5)*x + 5/2*arcsinh(1/5*sqrt(5)*x)

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mupad [B]  time = 0.09, size = 20, normalized size = 0.74 \[ \frac {5\,\mathrm {asinh}\left (\frac {\sqrt {5}\,x}{5}\right )}{2}+\frac {x\,\sqrt {x^2+5}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2 + 5)^(1/2),x)

[Out]

(5*asinh((5^(1/2)*x)/5))/2 + (x*(x^2 + 5)^(1/2))/2

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sympy [A]  time = 0.21, size = 24, normalized size = 0.89 \[ \frac {x \sqrt {x^{2} + 5}}{2} + \frac {5 \operatorname {asinh}{\left (\frac {\sqrt {5} x}{5} \right )}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2+5)**(1/2),x)

[Out]

x*sqrt(x**2 + 5)/2 + 5*asinh(sqrt(5)*x/5)/2

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