3.1930 \(\int \frac{x^3}{(a+\frac{b}{x^2})^{3/2}} \, dx\)

Optimal. Leaf size=95 \[ -\frac{15 b^2}{8 a^3 \sqrt{a+\frac{b}{x^2}}}+\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{7/2}}-\frac{5 b x^2}{8 a^2 \sqrt{a+\frac{b}{x^2}}}+\frac{x^4}{4 a \sqrt{a+\frac{b}{x^2}}} \]

[Out]

(-15*b^2)/(8*a^3*Sqrt[a + b/x^2]) - (5*b*x^2)/(8*a^2*Sqrt[a + b/x^2]) + x^4/(4*a*Sqrt[a + b/x^2]) + (15*b^2*Ar
cTanh[Sqrt[a + b/x^2]/Sqrt[a]])/(8*a^(7/2))

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Rubi [A]  time = 0.0454288, antiderivative size = 93, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ \frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{7/2}}+\frac{5 x^4 \sqrt{a+\frac{b}{x^2}}}{4 a^2}-\frac{15 b x^2 \sqrt{a+\frac{b}{x^2}}}{8 a^3}-\frac{x^4}{a \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/(a + b/x^2)^(3/2),x]

[Out]

(-15*b*Sqrt[a + b/x^2]*x^2)/(8*a^3) - x^4/(a*Sqrt[a + b/x^2]) + (5*Sqrt[a + b/x^2]*x^4)/(4*a^2) + (15*b^2*ArcT
anh[Sqrt[a + b/x^2]/Sqrt[a]])/(8*a^(7/2))

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}{\left (a+\frac{b}{x^2}\right )^{3/2}} \, dx &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^3 (a+b x)^{3/2}} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=-\frac{x^4}{a \sqrt{a+\frac{b}{x^2}}}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )}{2 a}\\ &=-\frac{x^4}{a \sqrt{a+\frac{b}{x^2}}}+\frac{5 \sqrt{a+\frac{b}{x^2}} x^4}{4 a^2}+\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )}{8 a^2}\\ &=-\frac{15 b \sqrt{a+\frac{b}{x^2}} x^2}{8 a^3}-\frac{x^4}{a \sqrt{a+\frac{b}{x^2}}}+\frac{5 \sqrt{a+\frac{b}{x^2}} x^4}{4 a^2}-\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\frac{1}{x^2}\right )}{16 a^3}\\ &=-\frac{15 b \sqrt{a+\frac{b}{x^2}} x^2}{8 a^3}-\frac{x^4}{a \sqrt{a+\frac{b}{x^2}}}+\frac{5 \sqrt{a+\frac{b}{x^2}} x^4}{4 a^2}-\frac{(15 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^3}\\ &=-\frac{15 b \sqrt{a+\frac{b}{x^2}} x^2}{8 a^3}-\frac{x^4}{a \sqrt{a+\frac{b}{x^2}}}+\frac{5 \sqrt{a+\frac{b}{x^2}} x^4}{4 a^2}+\frac{15 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 a^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.0555257, size = 86, normalized size = 0.91 \[ \frac{\sqrt{a} x \left (2 a^2 x^4-5 a b x^2-15 b^2\right )+15 b^{5/2} \sqrt{\frac{a x^2}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{8 a^{7/2} x \sqrt{a+\frac{b}{x^2}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/(a + b/x^2)^(3/2),x]

[Out]

(Sqrt[a]*x*(-15*b^2 - 5*a*b*x^2 + 2*a^2*x^4) + 15*b^(5/2)*Sqrt[1 + (a*x^2)/b]*ArcSinh[(Sqrt[a]*x)/Sqrt[b]])/(8
*a^(7/2)*Sqrt[a + b/x^2]*x)

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Maple [A]  time = 0.009, size = 87, normalized size = 0.9 \begin{align*}{\frac{a{x}^{2}+b}{8\,{x}^{3}} \left ( 2\,{x}^{5}{a}^{7/2}-5\,{a}^{5/2}{x}^{3}b-15\,{a}^{3/2}x{b}^{2}+15\,\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ) \sqrt{a{x}^{2}+b}a{b}^{2} \right ) \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{-{\frac{3}{2}}}{a}^{-{\frac{9}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(a*x^2+b)*(2*x^5*a^(7/2)-5*a^(5/2)*x^3*b-15*a^(3/2)*x*b^2+15*ln(x*a^(1/2)+(a*x^2+b)^(1/2))*(a*x^2+b)^(1/2)
*a*b^2)/((a*x^2+b)/x^2)^(3/2)/x^3/a^(9/2)

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.59922, size = 483, normalized size = 5.08 \begin{align*} \left [\frac{15 \,{\left (a b^{2} x^{2} + b^{3}\right )} \sqrt{a} \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^{3} x^{6} - 5 \, a^{2} b x^{4} - 15 \, a b^{2} x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{16 \,{\left (a^{5} x^{2} + a^{4} b\right )}}, -\frac{15 \,{\left (a b^{2} x^{2} + b^{3}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{-a} x^{2} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{a x^{2} + b}\right ) -{\left (2 \, a^{3} x^{6} - 5 \, a^{2} b x^{4} - 15 \, a b^{2} x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{8 \,{\left (a^{5} x^{2} + a^{4} b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(15*(a*b^2*x^2 + b^3)*sqrt(a)*log(-2*a*x^2 - 2*sqrt(a)*x^2*sqrt((a*x^2 + b)/x^2) - b) + 2*(2*a^3*x^6 - 5
*a^2*b*x^4 - 15*a*b^2*x^2)*sqrt((a*x^2 + b)/x^2))/(a^5*x^2 + a^4*b), -1/8*(15*(a*b^2*x^2 + b^3)*sqrt(-a)*arcta
n(sqrt(-a)*x^2*sqrt((a*x^2 + b)/x^2)/(a*x^2 + b)) - (2*a^3*x^6 - 5*a^2*b*x^4 - 15*a*b^2*x^2)*sqrt((a*x^2 + b)/
x^2))/(a^5*x^2 + a^4*b)]

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Sympy [A]  time = 5.20888, size = 100, normalized size = 1.05 \begin{align*} \frac{x^{5}}{4 a \sqrt{b} \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{5 \sqrt{b} x^{3}}{8 a^{2} \sqrt{\frac{a x^{2}}{b} + 1}} - \frac{15 b^{\frac{3}{2}} x}{8 a^{3} \sqrt{\frac{a x^{2}}{b} + 1}} + \frac{15 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} x}{\sqrt{b}} \right )}}{8 a^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x**5/(4*a*sqrt(b)*sqrt(a*x**2/b + 1)) - 5*sqrt(b)*x**3/(8*a**2*sqrt(a*x**2/b + 1)) - 15*b**(3/2)*x/(8*a**3*sqr
t(a*x**2/b + 1)) + 15*b**2*asinh(sqrt(a)*x/sqrt(b))/(8*a**(7/2))

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Giac [A]  time = 1.25262, size = 158, normalized size = 1.66 \begin{align*} -\frac{1}{8} \, b^{2}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{\frac{a x^{2} + b}{x^{2}}}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{8}{a^{3} \sqrt{\frac{a x^{2} + b}{x^{2}}}} - \frac{9 \, a \sqrt{\frac{a x^{2} + b}{x^{2}}} - \frac{7 \,{\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^{3}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*b^2*(15*arctan(sqrt((a*x^2 + b)/x^2)/sqrt(-a))/(sqrt(-a)*a^3) + 8/(a^3*sqrt((a*x^2 + b)/x^2)) - (9*a*sqrt
((a*x^2 + b)/x^2) - 7*(a*x^2 + b)*sqrt((a*x^2 + b)/x^2)/x^2)/((a - (a*x^2 + b)/x^2)^2*a^3))