3.4.64 \(\int \frac {\sqrt {a+b x^2}}{x^2} \, dx\) [364]

Optimal. Leaf size=42 \[ -\frac {\sqrt {a+b x^2}}{x}+\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 42, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {283, 223, 212} \begin {gather*} \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {\sqrt {a+b x^2}}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/x^2,x]

[Out]

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

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]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 45, normalized size = 1.07 \begin {gather*} -\frac {\sqrt {a+b x^2}}{x}-\sqrt {b} \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*x^2]/x^2,x]

[Out]

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

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Maple [A]
time = 0.03, size = 60, normalized size = 1.43

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}}{x}+\sqrt {b}\, \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )\)	\(36\)
default	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}}}{a x}+\frac {2 b \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{a}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/x*(b*x^2+a)^(3/2)+2*b/a*(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)))

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Maxima [A]
time = 0.28, size = 28, normalized size = 0.67 \begin {gather*} \sqrt {b} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {\sqrt {b x^{2} + a}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.22, size = 88, normalized size = 2.10 \begin {gather*} \left [\frac {\sqrt {b} x \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, \sqrt {b x^{2} + a}}{2 \, x}, -\frac {\sqrt {-b} x \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + \sqrt {b x^{2} + a}}{x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.74, size = 56, normalized size = 1.33 \begin {gather*} - \frac {\sqrt {a}}{x \sqrt {1 + \frac {b x^{2}}{a}}} + \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {b x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 1.53, size = 57, normalized size = 1.36 \begin {gather*} -\frac {1}{2} \, \sqrt {b} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, a \sqrt {b}}{{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.82, size = 56, normalized size = 1.33 \begin {gather*} -\frac {\sqrt {b\,x^2+a}}{x}-\frac {\sqrt {b}\,\mathrm {asin}\left (\frac {\sqrt {b}\,x\,1{}\mathrm {i}}{\sqrt {a}}\right )\,\sqrt {b\,x^2+a}\,1{}\mathrm {i}}{\sqrt {a}\,\sqrt {\frac {b\,x^2}{a}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/2)/x^2,x)

[Out]

- (a + b*x^2)^(1/2)/x - (b^(1/2)*asin((b^(1/2)*x*1i)/a^(1/2))*(a + b*x^2)^(1/2)*1i)/(a^(1/2)*((b*x^2)/a + 1)^(
1/2))

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