3.1915 \(\int \frac{x^3}{\sqrt{a+\frac{b}{x^2}}} \, dx\)

Optimal. Leaf size=74 \[ \frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{5/2}}-\frac{3 b x^2 \sqrt{a+\frac{b}{x^2}}}{8 a^2}+\frac{x^4 \sqrt{a+\frac{b}{x^2}}}{4 a} \]

[Out]

(-3*b*Sqrt[a + b/x^2]*x^2)/(8*a^2) + (Sqrt[a + b/x^2]*x^4)/(4*a) + (3*b^2*ArcTanh[Sqrt[a + b/x^2]/Sqrt[a]])/(8
*a^(5/2))

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Rubi [A]  time = 0.0366002, antiderivative size = 74, normalized size of antiderivative = 1., 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} \[ \frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{5/2}}-\frac{3 b x^2 \sqrt{a+\frac{b}{x^2}}}{8 a^2}+\frac{x^4 \sqrt{a+\frac{b}{x^2}}}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Sqrt[a + b/x^2],x]

[Out]

(-3*b*Sqrt[a + b/x^2]*x^2)/(8*a^2) + (Sqrt[a + b/x^2]*x^4)/(4*a) + (3*b^2*ArcTanh[Sqrt[a + b/x^2]/Sqrt[a]])/(8
*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{x^3}{\sqrt{a+\frac{b}{x^2}}} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{\sqrt{a+\frac{b}{x^2}} x^4}{4 a}+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )}{8 a}\\ &=-\frac{3 b \sqrt{a+\frac{b}{x^2}} x^2}{8 a^2}+\frac{\sqrt{a+\frac{b}{x^2}} x^4}{4 a}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )}{16 a^2}\\ &=-\frac{3 b \sqrt{a+\frac{b}{x^2}} x^2}{8 a^2}+\frac{\sqrt{a+\frac{b}{x^2}} x^4}{4 a}-\frac{(3 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^2}\\ &=-\frac{3 b \sqrt{a+\frac{b}{x^2}} x^2}{8 a^2}+\frac{\sqrt{a+\frac{b}{x^2}} x^4}{4 a}+\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0340435, size = 87, normalized size = 1.18 \[ \frac{\sqrt{a} x \left (2 a^2 x^4-a b x^2-3 b^2\right )+3 b^2 \sqrt{a x^2+b} \tanh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{a x^2+b}}\right )}{8 a^{5/2} x \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Sqrt[a + b/x^2],x]

[Out]

(Sqrt[a]*x*(-3*b^2 - a*b*x^2 + 2*a^2*x^4) + 3*b^2*Sqrt[b + a*x^2]*ArcTanh[(Sqrt[a]*x)/Sqrt[b + a*x^2]])/(8*a^(
5/2)*Sqrt[a + b/x^2]*x)

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Maple [A]  time = 0.007, size = 87, normalized size = 1.2 \begin{align*}{\frac{1}{8\,x}\sqrt{a{x}^{2}+b} \left ( 2\,{x}^{3}\sqrt{a{x}^{2}+b}{a}^{5/2}-3\,{a}^{3/2}\sqrt{a{x}^{2}+b}xb+3\,\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ) a{b}^{2} \right ){\frac{1}{\sqrt{{\frac{a{x}^{2}+b}{{x}^{2}}}}}}{a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a*x^2+b)^(1/2)*(2*x^3*(a*x^2+b)^(1/2)*a^(5/2)-3*a^(3/2)*(a*x^2+b)^(1/2)*x*b+3*ln(x*a^(1/2)+(a*x^2+b)^(1/2
))*a*b^2)/((a*x^2+b)/x^2)^(1/2)/x/a^(7/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60108, size = 359, normalized size = 4.85 \begin{align*} \left [\frac{3 \, \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} - 3 \, a b x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{16 \, a^{3}}, -\frac{3 \, \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} - 3 \, a b x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{8 \, a^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*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 - 3*a*b*x^2)*sqrt(
(a*x^2 + b)/x^2))/a^3, -1/8*(3*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 - 3*a*b*x^2)*sqrt((a*x^2 + b)/x^2))/a^3]

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Sympy [A]  time = 5.38876, size = 95, normalized size = 1.28 \begin{align*} \frac{x^{5}}{4 \sqrt{b} \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{\sqrt{b} x^{3}}{8 a \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{3 b^{\frac{3}{2}} x}{8 a^{2} \sqrt{\frac{a x^{2}}{b} + 1}} + \frac{3 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} x}{\sqrt{b}} \right )}}{8 a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20663, size = 134, normalized size = 1.81 \begin{align*} -\frac{1}{8} \, b^{2}{\left (\frac{3 \, \arctan \left (\frac{\sqrt{\frac{a x^{2} + b}{x^{2}}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} - \frac{5 \, a \sqrt{\frac{a x^{2} + b}{x^{2}}} - \frac{3 \,{\left (a x^{2} + b\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{x^{2}}}{{\left (a - \frac{a x^{2} + b}{x^{2}}\right )}^{2} a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b^2*(3*arctan(sqrt((a*x^2 + b)/x^2)/sqrt(-a))/(sqrt(-a)*a^2) - (5*a*sqrt((a*x^2 + b)/x^2) - 3*(a*x^2 + b)
*sqrt((a*x^2 + b)/x^2)/x^2)/((a - (a*x^2 + b)/x^2)^2*a^2))