3.1899 \(\int (a+\frac{b}{x^2})^{3/2} x^3 \, dx\)

Optimal. Leaf size=68 \[ \frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 \sqrt{a}}+\frac{1}{4} x^4 \left (a+\frac{b}{x^2}\right )^{3/2}+\frac{3}{8} b x^2 \sqrt{a+\frac{b}{x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0346493, antiderivative size = 68, 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, 47, 63, 208} \[ \frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+\frac{b}{x^2}}}{\sqrt{a}}\right )}{8 \sqrt{a}}+\frac{1}{4} x^4 \left (a+\frac{b}{x^2}\right )^{3/2}+\frac{3}{8} b x^2 \sqrt{a+\frac{b}{x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)^(3/2)*x^3,x]

[Out]

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

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

Mathematica [A]  time = 0.110525, size = 66, normalized size = 0.97 \[ \frac{1}{8} x \sqrt{a+\frac{b}{x^2}} \left (\frac{3 b^{3/2} \sinh ^{-1}\left (\frac{\sqrt{a} x}{\sqrt{b}}\right )}{\sqrt{a} \sqrt{\frac{a x^2}{b}+1}}+2 a x^3+5 b x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)^(3/2)*x^3,x]

[Out]

(Sqrt[a + b/x^2]*x*(5*b*x + 2*a*x^3 + (3*b^(3/2)*ArcSinh[(Sqrt[a]*x)/Sqrt[b]])/(Sqrt[a]*Sqrt[1 + (a*x^2)/b])))
/8

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Maple [A]  time = 0.004, size = 84, normalized size = 1.2 \begin{align*}{\frac{{x}^{3}}{8} \left ({\frac{a{x}^{2}+b}{{x}^{2}}} \right ) ^{{\frac{3}{2}}} \left ( 2\,x \left ( a{x}^{2}+b \right ) ^{3/2}\sqrt{a}+3\,\sqrt{a}\sqrt{a{x}^{2}+b}xb+3\,\ln \left ( x\sqrt{a}+\sqrt{a{x}^{2}+b} \right ){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^3,x)

[Out]

1/8*((a*x^2+b)/x^2)^(3/2)*x^3*(2*x*(a*x^2+b)^(3/2)*a^(1/2)+3*a^(1/2)*(a*x^2+b)^(1/2)*x*b+3*ln(x*a^(1/2)+(a*x^2
+b)^(1/2))*b^2)/(a*x^2+b)^(3/2)/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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.60095, size = 354, normalized size = 5.21 \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} + 5 \, a b x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{16 \, a}, -\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} + 5 \, a b x^{2}\right )} \sqrt{\frac{a x^{2} + b}{x^{2}}}}{8 \, a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)^(3/2)*x^3,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 + 5*a*b*x^2)*sqrt(
(a*x^2 + b)/x^2))/a, -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
+ 5*a*b*x^2)*sqrt((a*x^2 + b)/x^2))/a]

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Sympy [A]  time = 3.52827, size = 70, normalized size = 1.03 \begin{align*} \frac{a \sqrt{b} x^{3} \sqrt{\frac{a x^{2}}{b} + 1}}{4} + \frac{5 b^{\frac{3}{2}} x \sqrt{\frac{a x^{2}}{b} + 1}}{8} + \frac{3 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a} x}{\sqrt{b}} \right )}}{8 \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20645, size = 92, normalized size = 1.35 \begin{align*} -\frac{3 \, b^{2} \log \left ({\left | -\sqrt{a} x + \sqrt{a x^{2} + b} \right |}\right ) \mathrm{sgn}\left (x\right )}{8 \, \sqrt{a}} + \frac{3 \, b^{2} \log \left ({\left | b \right |}\right ) \mathrm{sgn}\left (x\right )}{16 \, \sqrt{a}} + \frac{1}{8} \,{\left (2 \, a x^{2} \mathrm{sgn}\left (x\right ) + 5 \, b \mathrm{sgn}\left (x\right )\right )} \sqrt{a x^{2} + b} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8*b^2*log(abs(-sqrt(a)*x + sqrt(a*x^2 + b)))*sgn(x)/sqrt(a) + 3/16*b^2*log(abs(b))*sgn(x)/sqrt(a) + 1/8*(2*
a*x^2*sgn(x) + 5*b*sgn(x))*sqrt(a*x^2 + b)*x