3.493 \(\int \frac{1}{x^3 \sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=50 \[ \frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}-\frac{\sqrt{a+b x^2}}{2 a x^2} \]

[Out]

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

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Rubi [A]  time = 0.0261624, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ \frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}-\frac{\sqrt{a+b x^2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[a + b*x^2]),x]

[Out]

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{1}{x^3 \sqrt{a+b x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=-\frac{\sqrt{a+b x^2}}{2 a x^2}-\frac{b \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{\sqrt{a+b x^2}}{2 a x^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{2 a}\\ &=-\frac{\sqrt{a+b x^2}}{2 a x^2}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.0474586, size = 61, normalized size = 1.22 \[ \frac{b \sqrt{a+b x^2} \left (\frac{\tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )}{2 \sqrt{\frac{b x^2}{a}+1}}-\frac{a}{2 b x^2}\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[a + b*x^2]),x]

[Out]

(b*Sqrt[a + b*x^2]*(-a/(2*b*x^2) + ArcTanh[Sqrt[1 + (b*x^2)/a]]/(2*Sqrt[1 + (b*x^2)/a])))/a^2

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Maple [A]  time = 0.005, size = 48, normalized size = 1. \begin{align*} -{\frac{1}{2\,a{x}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{b}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.35134, size = 263, normalized size = 5.26 \begin{align*} \left [\frac{\sqrt{a} b x^{2} \log \left (-\frac{b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \, \sqrt{b x^{2} + a} a}{4 \, a^{2} x^{2}}, -\frac{\sqrt{-a} b x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) + \sqrt{b x^{2} + a} a}{2 \, a^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 2.26539, size = 42, normalized size = 0.84 \begin{align*} - \frac{\sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{2 a x} + \frac{b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(b*x**2+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 2.88439, size = 65, normalized size = 1.3 \begin{align*} -\frac{1}{2} \, b{\left (\frac{\arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{\sqrt{b x^{2} + a}}{a b x^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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