3.1901 \(\int (a+\frac{b}{x^2})^{3/2} x \, dx\)

Optimal. Leaf size=63 \[ \frac{1}{2} x^2 \left (a+\frac{b}{x^2}\right )^{3/2}-\frac{3}{2} b \sqrt{a+\frac{b}{x^2}}+\frac{3}{2} \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.030937, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {266, 47, 50, 63, 208} \[ \frac{1}{2} x^2 \left (a+\frac{b}{x^2}\right )^{3/2}-\frac{3}{2} b \sqrt{a+\frac{b}{x^2}}+\frac{3}{2} \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(3/2)*x,x]

[Out]

(-3*b*Sqrt[a + b/x^2])/2 + ((a + b/x^2)^(3/2)*x^2)/2 + (3*Sqrt[a]*b*ArcTanh[Sqrt[a + b/x^2]/Sqrt[a]])/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 47

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \left (a+\frac{b}{x^2}\right )^{3/2} x \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{x^2} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{1}{2} \left (a+\frac{b}{x^2}\right )^{3/2} x^2-\frac{1}{4} (3 b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{3}{2} b \sqrt{a+\frac{b}{x^2}}+\frac{1}{2} \left (a+\frac{b}{x^2}\right )^{3/2} x^2-\frac{1}{4} (3 a b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{3}{2} b \sqrt{a+\frac{b}{x^2}}+\frac{1}{2} \left (a+\frac{b}{x^2}\right )^{3/2} x^2-\frac{1}{2} (3 a) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x^2}}\right )\\ &=-\frac{3}{2} b \sqrt{a+\frac{b}{x^2}}+\frac{1}{2} \left (a+\frac{b}{x^2}\right )^{3/2} x^2+\frac{3}{2} \sqrt{a} b \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )\\ \end{align*}

Mathematica [C]  time = 0.011747, size = 47, normalized size = 0.75 \[ -\frac{b \sqrt{a+\frac{b}{x^2}} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{a x^2}{b}\right )}{\sqrt{\frac{a x^2}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(3/2)*x,x]

[Out]

-((b*Sqrt[a + b/x^2]*Hypergeometric2F1[-3/2, -1/2, 1/2, -((a*x^2)/b)])/Sqrt[1 + (a*x^2)/b])

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Maple [B]  time = 0.005, size = 107, normalized size = 1.7 \begin{align*}{\frac{{x}^{2}}{2\,b} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 2\,{a}^{3/2} \left ( a{x}^{2}+b \right ) ^{3/2}{x}^{2}+3\,{a}^{3/2}\sqrt{a{x}^{2}+b}{x}^{2}b-2\, \left ( a{x}^{2}+b \right ) ^{5/2}\sqrt{a}+3\,\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ) xa{b}^{2} \right ) \left ( a{x}^{2}+b \right ) ^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a*x^2+b)/x^2)^(3/2)*x^2*(2*a^(3/2)*(a*x^2+b)^(3/2)*x^2+3*a^(3/2)*(a*x^2+b)^(1/2)*x^2*b-2*(a*x^2+b)^(5/2)
*a^(1/2)+3*ln(x*a^(1/2)+(a*x^2+b)^(1/2))*x*a*b^2)/(a*x^2+b)^(3/2)/b/a^(1/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60251, size = 312, normalized size = 4.95 \begin{align*} \left [\frac{3}{4} \, \sqrt{a} b \log \left (-2 \, a x^{2} - 2 \, \sqrt{a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}} - b\right ) + \frac{1}{2} \,{\left (a x^{2} - 2 \, b\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}, -\frac{3}{2} \, \sqrt{-a} b \arctan \left (\frac{\sqrt{-a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) + \frac{1}{2} \,{\left (a x^{2} - 2 \, b\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.64258, size = 88, normalized size = 1.4 \begin{align*} \frac{3 \sqrt{a} b \operatorname{asinh}{\left (\frac{\sqrt{a} x}{\sqrt{b}} \right )}}{2} + \frac{a^{2} x^{3}}{2 \sqrt{b} \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{a \sqrt{b} x}{2 \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{b^{\frac{3}{2}}}{x \sqrt{\frac{a x^{2}}{b} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27286, size = 107, normalized size = 1.7 \begin{align*} \frac{1}{2} \, \sqrt{a x^{2} + b} a x \mathrm{sgn}\left (x\right ) - \frac{3}{4} \, \sqrt{a} b \log \left ({\left (\sqrt{a} x - \sqrt{a x^{2} + b}\right )}^{2}\right ) \mathrm{sgn}\left (x\right ) + \frac{2 \, \sqrt{a} b^{2} \mathrm{sgn}\left (x\right )}{{\left (\sqrt{a} x - \sqrt{a x^{2} + b}\right )}^{2} - b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(a*x^2 + b)*a*x*sgn(x) - 3/4*sqrt(a)*b*log((sqrt(a)*x - sqrt(a*x^2 + b))^2)*sgn(x) + 2*sqrt(a)*b^2*sgn
(x)/((sqrt(a)*x - sqrt(a*x^2 + b))^2 - b)