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

Optimal. Leaf size=46 \[ \frac {1}{2} x \sqrt {a+b x^2}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}} \]

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Rubi [A]  time = 0.01, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {195, 217, 206} \begin {gather*} \frac {1}{2} x \sqrt {a+b x^2}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2],x]

[Out]

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

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

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

Rubi steps

\begin {align*} \int \sqrt {a+b x^2} \, dx &=\frac {1}{2} x \sqrt {a+b x^2}+\frac {1}{2} a \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {1}{2} x \sqrt {a+b x^2}+\frac {1}{2} a \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {1}{2} x \sqrt {a+b x^2}+\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 49, normalized size = 1.07 \begin {gather*} \frac {1}{2} x \sqrt {a+b x^2}+\frac {a \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{2 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.00, size = 48, normalized size = 1.04 \begin {gather*} \frac {1}{2} x \sqrt {a+b x^2}-\frac {a \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{2 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + b*x^2],x]

[Out]

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

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fricas [A]  time = 0.75, size = 94, normalized size = 2.04 \begin {gather*} \left [\frac {2 \, \sqrt {b x^{2} + a} b x + a \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right )}{4 \, b}, \frac {\sqrt {b x^{2} + a} b x - a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right )}{2 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*sqrt(b*x^2 + a)*b*x + a*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a))/b, 1/2*(sqrt(b*x^2 +
a)*b*x - a*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)))/b]

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giac [A]  time = 0.58, size = 37, normalized size = 0.80 \begin {gather*} \frac {1}{2} \, \sqrt {b x^{2} + a} x - \frac {a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b*x^2 + a)*x - 1/2*a*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b)

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maple [A]  time = 0.00, size = 36, normalized size = 0.78 \begin {gather*} \frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}+\frac {\sqrt {b \,x^{2}+a}\, x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 28, normalized size = 0.61 \begin {gather*} \frac {1}{2} \, \sqrt {b x^{2} + a} x + \frac {a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.71, size = 35, normalized size = 0.76 \begin {gather*} \frac {x\,\sqrt {b\,x^2+a}}{2}+\frac {a\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{2\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/2),x)

[Out]

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

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sympy [A]  time = 1.86, size = 41, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a} x \sqrt {1 + \frac {b x^{2}}{a}}}{2} + \frac {a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(a)*x*sqrt(1 + b*x**2/a)/2 + a*asinh(sqrt(b)*x/sqrt(a))/(2*sqrt(b))

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