3.2070 \(\int (a+\frac{b}{x^4})^{3/2} x^2 \, dx\)

Optimal. Leaf size=126 \[ -\frac{2 a^{3/4} b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{3 \sqrt{a+\frac{b}{x^4}}}+\frac{1}{3} x^3 \left (a+\frac{b}{x^4}\right )^{3/2}-\frac{2 b \sqrt{a+\frac{b}{x^4}}}{3 x} \]

[Out]

(-2*b*Sqrt[a + b/x^4])/(3*x) + ((a + b/x^4)^(3/2)*x^3)/3 - (2*a^(3/4)*b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt
[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(3*Sqrt[a + b/x^4])

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Rubi [A]  time = 0.0590829, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {335, 277, 195, 220} \[ -\frac{2 a^{3/4} b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{3 \sqrt{a+\frac{b}{x^4}}}+\frac{1}{3} x^3 \left (a+\frac{b}{x^4}\right )^{3/2}-\frac{2 b \sqrt{a+\frac{b}{x^4}}}{3 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(3/2)*x^2,x]

[Out]

(-2*b*Sqrt[a + b/x^4])/(3*x) + ((a + b/x^4)^(3/2)*x^3)/3 - (2*a^(3/4)*b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt
[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(3*Sqrt[a + b/x^4])

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rubi steps

\begin{align*} \int \left (a+\frac{b}{x^4}\right )^{3/2} x^2 \, dx &=-\operatorname{Subst}\left (\int \frac{\left (a+b x^4\right )^{3/2}}{x^4} \, dx,x,\frac{1}{x}\right )\\ &=\frac{1}{3} \left (a+\frac{b}{x^4}\right )^{3/2} x^3-(2 b) \operatorname{Subst}\left (\int \sqrt{a+b x^4} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 b \sqrt{a+\frac{b}{x^4}}}{3 x}+\frac{1}{3} \left (a+\frac{b}{x^4}\right )^{3/2} x^3-\frac{1}{3} (4 a b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 b \sqrt{a+\frac{b}{x^4}}}{3 x}+\frac{1}{3} \left (a+\frac{b}{x^4}\right )^{3/2} x^3-\frac{2 a^{3/4} b^{3/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{3 \sqrt{a+\frac{b}{x^4}}}\\ \end{align*}

Mathematica [C]  time = 0.0123281, size = 52, normalized size = 0.41 \[ -\frac{b \sqrt{a+\frac{b}{x^4}} \, _2F_1\left (-\frac{3}{2},-\frac{3}{4};\frac{1}{4};-\frac{a x^4}{b}\right )}{3 x \sqrt{\frac{a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(3/2)*x^2,x]

[Out]

-(b*Sqrt[a + b/x^4]*Hypergeometric2F1[-3/2, -3/4, 1/4, -((a*x^4)/b)])/(3*x*Sqrt[1 + (a*x^4)/b])

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Maple [C]  time = 0.013, size = 138, normalized size = 1.1 \begin{align*}{\frac{{x}^{3}}{3\, \left ( a{x}^{4}+b \right ) ^{2}} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{{\frac{3}{2}}} \left ( \sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}{x}^{8}{a}^{2}+4\,ab\sqrt{-{\frac{i\sqrt{a}{x}^{2}-\sqrt{b}}{\sqrt{b}}}}\sqrt{{\frac{i\sqrt{a}{x}^{2}+\sqrt{b}}{\sqrt{b}}}}{\it EllipticF} \left ( x\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}},i \right ){x}^{3}-\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}{b}^{2} \right ){\frac{1}{\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^4)^(3/2)*x^2,x)

[Out]

1/3*((a*x^4+b)/x^4)^(3/2)*x^3*((I*a^(1/2)/b^(1/2))^(1/2)*x^8*a^2+4*a*b*(-(I*a^(1/2)*x^2-b^(1/2))/b^(1/2))^(1/2
)*((I*a^(1/2)*x^2+b^(1/2))/b^(1/2))^(1/2)*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)*x^3-(I*a^(1/2)/b^(1/2))^(1/
2)*b^2)/(a*x^4+b)^2/(I*a^(1/2)/b^(1/2))^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a + \frac{b}{x^{4}}\right )}^{\frac{3}{2}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^4)^(3/2)*x^2,x, algorithm="maxima")

[Out]

integrate((a + b/x^4)^(3/2)*x^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (a x^{4} + b\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^4)^(3/2)*x^2,x, algorithm="fricas")

[Out]

integral((a*x^4 + b)*sqrt((a*x^4 + b)/x^4)/x^2, x)

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Sympy [C]  time = 1.88431, size = 44, normalized size = 0.35 \begin{align*} - \frac{a^{\frac{3}{2}} x^{3} \Gamma \left (- \frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, - \frac{3}{4} \\ \frac{1}{4} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**4)**(3/2)*x**2,x)

[Out]

-a**(3/2)*x**3*gamma(-3/4)*hyper((-3/2, -3/4), (1/4,), b*exp_polar(I*pi)/(a*x**4))/(4*gamma(1/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a + \frac{b}{x^{4}}\right )}^{\frac{3}{2}} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^4)^(3/2)*x^2,x, algorithm="giac")

[Out]

integrate((a + b/x^4)^(3/2)*x^2, x)