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

Optimal. Leaf size=69 \[ -\frac{3 b}{2 a^2 \sqrt{a+b x^2}}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{5/2}}-\frac{1}{2 a x^2 \sqrt{a+b x^2}} \]

[Out]

(-3*b)/(2*a^2*Sqrt[a + b*x^2]) - 1/(2*a*x^2*Sqrt[a + b*x^2]) + (3*b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(5/
2))

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Rubi [A]  time = 0.0390216, antiderivative size = 68, normalized size of antiderivative = 0.99, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac{3 \sqrt{a+b x^2}}{2 a^2 x^2}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{5/2}}+\frac{1}{a x^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*x^2)^(3/2)),x]

[Out]

1/(a*x^2*Sqrt[a + b*x^2]) - (3*Sqrt[a + b*x^2])/(2*a^2*x^2) + (3*b*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/(2*a^(5/2
))

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 \left (a+b x^2\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)^{3/2}} \, dx,x,x^2\right )\\ &=\frac{1}{a x^2 \sqrt{a+b x^2}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{2 a}\\ &=\frac{1}{a x^2 \sqrt{a+b x^2}}-\frac{3 \sqrt{a+b x^2}}{2 a^2 x^2}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )}{4 a^2}\\ &=\frac{1}{a x^2 \sqrt{a+b x^2}}-\frac{3 \sqrt{a+b x^2}}{2 a^2 x^2}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{2 a^2}\\ &=\frac{1}{a x^2 \sqrt{a+b x^2}}-\frac{3 \sqrt{a+b x^2}}{2 a^2 x^2}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.007585, size = 35, normalized size = 0.51 \[ -\frac{b \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b x^2}{a}+1\right )}{a^2 \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*x^2)^(3/2)),x]

[Out]

-((b*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (b*x^2)/a])/(a^2*Sqrt[a + b*x^2]))

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Maple [A]  time = 0.004, size = 63, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,a{x}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{3\,b}{2\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{3\,b}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a/x^2/(b*x^2+a)^(1/2)-3/2*b/a^2/(b*x^2+a)^(1/2)+3/2*b/a^(5/2)*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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.35542, size = 382, normalized size = 5.54 \begin{align*} \left [\frac{3 \,{\left (b^{2} x^{4} + a b x^{2}\right )} \sqrt{a} \log \left (-\frac{b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (3 \, a b x^{2} + a^{2}\right )} \sqrt{b x^{2} + a}}{4 \,{\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}, -\frac{3 \,{\left (b^{2} x^{4} + a b x^{2}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (3 \, a b x^{2} + a^{2}\right )} \sqrt{b x^{2} + a}}{2 \,{\left (a^{3} b x^{4} + a^{4} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(3*(b^2*x^4 + a*b*x^2)*sqrt(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(3*a*b*x^2 + a^2)*
sqrt(b*x^2 + a))/(a^3*b*x^4 + a^4*x^2), -1/2*(3*(b^2*x^4 + a*b*x^2)*sqrt(-a)*arctan(sqrt(-a)/sqrt(b*x^2 + a))
+ (3*a*b*x^2 + a^2)*sqrt(b*x^2 + a))/(a^3*b*x^4 + a^4*x^2)]

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Sympy [A]  time = 3.19302, size = 73, normalized size = 1.06 \begin{align*} - \frac{1}{2 a \sqrt{b} x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{3 \sqrt{b}}{2 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{3 b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(2*a*sqrt(b)*x**3*sqrt(a/(b*x**2) + 1)) - 3*sqrt(b)/(2*a**2*x*sqrt(a/(b*x**2) + 1)) + 3*b*asinh(sqrt(a)/(sq
rt(b)*x))/(2*a**(5/2))

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Giac [A]  time = 1.72575, size = 89, normalized size = 1.29 \begin{align*} -\frac{1}{2} \, b{\left (\frac{3 \, \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \, b x^{2} + a}{{\left ({\left (b x^{2} + a\right )}^{\frac{3}{2}} - \sqrt{b x^{2} + a} a\right )} a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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