3.2104 \(\int \frac{1}{(a+\frac{b}{x^4})^{5/2} x^2} \, dx\)

Optimal. Leaf size=131 \[ -\frac{5 \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{24 a^{9/4} \sqrt [4]{b} \sqrt{a+\frac{b}{x^4}}}-\frac{5}{12 a^2 x \sqrt{a+\frac{b}{x^4}}}-\frac{1}{6 a x \left (a+\frac{b}{x^4}\right )^{3/2}} \]

[Out]

-1/(6*a*(a + b/x^4)^(3/2)*x) - 5/(12*a^2*Sqrt[a + b/x^4]*x) - (5*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])/(24*a^(9/4)*b^(1/4)*Sqrt[a + b/x^4])

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Rubi [A]  time = 0.0555304, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {335, 199, 220} \[ -\frac{5}{12 a^2 x \sqrt{a+\frac{b}{x^4}}}-\frac{5 \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{24 a^{9/4} \sqrt [4]{b} \sqrt{a+\frac{b}{x^4}}}-\frac{1}{6 a x \left (a+\frac{b}{x^4}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^4)^(5/2)*x^2),x]

[Out]

-1/(6*a*(a + b/x^4)^(3/2)*x) - 5/(12*a^2*Sqrt[a + b/x^4]*x) - (5*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])/(24*a^(9/4)*b^(1/4)*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 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || 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 \frac{1}{\left (a+\frac{b}{x^4}\right )^{5/2} x^2} \, dx &=-\operatorname{Subst}\left (\int \frac{1}{\left (a+b x^4\right )^{5/2}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2} x}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^4\right )^{3/2}} \, dx,x,\frac{1}{x}\right )}{6 a}\\ &=-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2} x}-\frac{5}{12 a^2 \sqrt{a+\frac{b}{x^4}} x}-\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )}{12 a^2}\\ &=-\frac{1}{6 a \left (a+\frac{b}{x^4}\right )^{3/2} x}-\frac{5}{12 a^2 \sqrt{a+\frac{b}{x^4}} x}-\frac{5 \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 )}{24 a^{9/4} \sqrt [4]{b} \sqrt{a+\frac{b}{x^4}}}\\ \end{align*}

Mathematica [C]  time = 0.0400305, size = 82, normalized size = 0.63 \[ \frac{5 \left (a x^4+b\right ) \sqrt{\frac{a x^4}{b}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{a x^4}{b}\right )-7 a x^4-5 b}{12 a^2 x \sqrt{a+\frac{b}{x^4}} \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^4)^(5/2)*x^2),x]

[Out]

(-5*b - 7*a*x^4 + 5*(b + a*x^4)*Sqrt[1 + (a*x^4)/b]*Hypergeometric2F1[1/4, 1/2, 5/4, -((a*x^4)/b)])/(12*a^2*Sq
rt[a + b/x^4]*x*(b + a*x^4))

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Maple [C]  time = 0.018, size = 279, normalized size = 2.1 \begin{align*} -{\frac{1}{12\,{a}^{2}{x}^{10}} \left ( -5\,\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}^{8}{a}^{2}+7\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{x}^{9}{a}^{2}-10\,\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}^{4}ab+12\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{x}^{5}ab-5\,\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 ){b}^{2}+5\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}x{b}^{2} \right ) \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^4)^(5/2)/x^2,x)

[Out]

-1/12*(-5*(-(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^8*a^2+7*(I*a^(1/2)/b^(1/2))^(1/2)*x^9*a^2-10*(-(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^4*a*b+12*(I*a^(1/2)/b^(1/2)
)^(1/2)*x^5*a*b-5*(-(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)*b^2+5*(I*a^(1/2)/b^(1/2))^(1/2)*x*b^2)/a^2/((a*x^4+b)/x^4)^(5/2)/x^10/(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 \frac{1}{{\left (a + \frac{b}{x^{4}}\right )}^{\frac{5}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^4)^(5/2)/x^2,x, algorithm="maxima")

[Out]

integrate(1/((a + b/x^4)^(5/2)*x^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{10} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{a^{3} x^{12} + 3 \, a^{2} b x^{8} + 3 \, a b^{2} x^{4} + b^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^4)^(5/2)/x^2,x, algorithm="fricas")

[Out]

integral(x^10*sqrt((a*x^4 + b)/x^4)/(a^3*x^12 + 3*a^2*b*x^8 + 3*a*b^2*x^4 + b^3), x)

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Sympy [C]  time = 2.17315, size = 37, normalized size = 0.28 \begin{align*} - \frac{\Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{4}, \frac{5}{2} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac{5}{2}} x \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**4)**(5/2)/x**2,x)

[Out]

-gamma(1/4)*hyper((1/4, 5/2), (5/4,), b*exp_polar(I*pi)/(a*x**4))/(4*a**(5/2)*x*gamma(5/4))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a + \frac{b}{x^{4}}\right )}^{\frac{5}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^4)^(5/2)/x^2,x, algorithm="giac")

[Out]

integrate(1/((a + b/x^4)^(5/2)*x^2), x)