3.1.48 \(\int \sqrt {a+b x^2} \, dx\) [48]

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

[Out]

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

________________________________________________________________________________________

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 = {201, 223, 212} \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 201

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 212

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

Rule 223

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 \text {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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 48, normalized size = 1.04 \begin {gather*} \frac {1}{2} x \sqrt {a+b x^2}-\frac {a \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\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[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*Sqrt[b])

________________________________________________________________________________________

Maple [A]
time = 0.05, size = 36, normalized size = 0.78

method	result	size
default	\(\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\)	\(36\)
risch	\(\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\)	\(36\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.28, 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)

________________________________________________________________________________________

Fricas [A]
time = 1.28, 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]

________________________________________________________________________________________

Sympy [A]
time = 0.89, 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))

________________________________________________________________________________________

Giac [A]
time = 0.66, 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)

________________________________________________________________________________________

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))

________________________________________________________________________________________