3.1891 \(\int \sqrt{a+\frac{b}{x^2}} x^3 \, dx\)

Optimal. Leaf size=71 \[ -\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{3/2}}+\frac{1}{4} x^4 \sqrt{a+\frac{b}{x^2}}+\frac{b x^2 \sqrt{a+\frac{b}{x^2}}}{8 a} \]

[Out]

(b*Sqrt[a + b/x^2]*x^2)/(8*a) + (Sqrt[a + b/x^2]*x^4)/4 - (b^2*ArcTanh[Sqrt[a + b/x^2]/Sqrt[a]])/(8*a^(3/2))

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Rubi [A]  time = 0.0357065, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {266, 47, 51, 63, 208} \[ -\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{3/2}}+\frac{1}{4} x^4 \sqrt{a+\frac{b}{x^2}}+\frac{b x^2 \sqrt{a+\frac{b}{x^2}}}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b/x^2]*x^3,x]

[Out]

(b*Sqrt[a + b/x^2]*x^2)/(8*a) + (Sqrt[a + b/x^2]*x^4)/4 - (b^2*ArcTanh[Sqrt[a + b/x^2]/Sqrt[a]])/(8*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 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 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 \sqrt{a+\frac{b}{x^2}} x^3 \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x^3} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{1}{4} \sqrt{a+\frac{b}{x^2}} x^4-\frac{1}{8} b \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )\\ &=\frac{b \sqrt{a+\frac{b}{x^2}} x^2}{8 a}+\frac{1}{4} \sqrt{a+\frac{b}{x^2}} x^4+\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )}{16 a}\\ &=\frac{b \sqrt{a+\frac{b}{x^2}} x^2}{8 a}+\frac{1}{4} \sqrt{a+\frac{b}{x^2}} x^4+\frac{b \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+\frac{b}{x^2}}\right )}{8 a}\\ &=\frac{b \sqrt{a+\frac{b}{x^2}} x^2}{8 a}+\frac{1}{4} \sqrt{a+\frac{b}{x^2}} x^4-\frac{b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.039887, size = 88, normalized size = 1.24 \[ x \sqrt{a+\frac{b}{x^2}} \left (\frac{b x}{8 a}+\frac{x^3}{4}\right )-\frac{b^2 x \sqrt{a+\frac{b}{x^2}} \log \left (\sqrt{a} \sqrt{a x^2+b}+a x\right )}{8 a^{3/2} \sqrt{a x^2+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b/x^2]*x^3,x]

[Out]

Sqrt[a + b/x^2]*x*((b*x)/(8*a) + x^3/4) - (b^2*Sqrt[a + b/x^2]*x*Log[a*x + Sqrt[a]*Sqrt[b + a*x^2]])/(8*a^(3/2
)*Sqrt[b + a*x^2])

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Maple [A]  time = 0.01, size = 82, normalized size = 1.2 \begin{align*}{\frac{x}{8}\sqrt{{\frac{a{x}^{2}+b}{{x}^{2}}}} \left ( 2\,x \left ( a{x}^{2}+b \right ) ^{3/2}\sqrt{a}-\sqrt{a}\sqrt{a{x}^{2}+b}xb-\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ){b}^{2} \right ){\frac{1}{\sqrt{a{x}^{2}+b}}}{a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*((a*x^2+b)/x^2)^(1/2)*x*(2*x*(a*x^2+b)^(3/2)*a^(1/2)-a^(1/2)*(a*x^2+b)^(1/2)*x*b-ln(x*a^(1/2)+(a*x^2+b)^(1
/2))*b^2)/(a*x^2+b)^(1/2)/a^(3/2)

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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)^(1/2)*x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.54799, size = 347, normalized size = 4.89 \begin{align*} \left [\frac{\sqrt{a} b^{2} \log \left (-2 \, a x^{2} + 2 \, \sqrt{a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}} - b\right ) + 2 \,{\left (2 \, a^{2} x^{4} + a b x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{16 \, a^{2}}, \frac{\sqrt{-a} b^{2} \arctan \left (\frac{\sqrt{-a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) +{\left (2 \, a^{2} x^{4} + a b x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{8 \, a^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(sqrt(a)*b^2*log(-2*a*x^2 + 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) + 2*(2*a^2*x^4 + a*b*x^2)*sqrt((a*x
^2 + b)/x^2))/a^2, 1/8*(sqrt(-a)*b^2*arctan(sqrt(-a)*x^2*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) + (2*a^2*x^4 + a*b
*x^2)*sqrt((a*x^2 + b)/x^2))/a^2]

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Sympy [A]  time = 4.43026, size = 92, normalized size = 1.3 \begin{align*} \frac{a x^{5}}{4 \sqrt{b} \sqrt{\frac{a x^{2}}{b} + 1}} + \frac{3 \sqrt{b} x^{3}}{8 \sqrt{\frac{a x^{2}}{b} + 1}} + \frac{b^{\frac{3}{2}} x}{8 a \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} x}{\sqrt{b}} \right )}}{8 a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x**5/(4*sqrt(b)*sqrt(a*x**2/b + 1)) + 3*sqrt(b)*x**3/(8*sqrt(a*x**2/b + 1)) + b**(3/2)*x/(8*a*sqrt(a*x**2/b
+ 1)) - b**2*asinh(sqrt(a)*x/sqrt(b))/(8*a**(3/2))

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Giac [A]  time = 1.21975, size = 93, normalized size = 1.31 \begin{align*} \frac{1}{8} \, \sqrt{a x^{2} + b}{\left (2 \, x^{2} \mathrm{sgn}\left (x\right ) + \frac{b \mathrm{sgn}\left (x\right )}{a}\right )} x + \frac{b^{2} \log \left ({\left | -\sqrt{a} x + \sqrt{a x^{2} + b} \right |}\right ) \mathrm{sgn}\left (x\right )}{8 \, a^{\frac{3}{2}}} - \frac{b^{2} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (x\right )}{16 \, a^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(a*x^2 + b)*(2*x^2*sgn(x) + b*sgn(x)/a)*x + 1/8*b^2*log(abs(-sqrt(a)*x + sqrt(a*x^2 + b)))*sgn(x)/a^(3
/2) - 1/16*b^2*log(abs(b))*sgn(x)/a^(3/2)