3.2080 \(\int \frac{(a+\frac{b}{x^4})^{5/2}}{x^2} \, dx\)

Optimal. Leaf size=147 \[ -\frac{20 a^{11/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{77 \sqrt [4]{b} \sqrt{a+\frac{b}{x^4}}}-\frac{20 a^2 \sqrt{a+\frac{b}{x^4}}}{77 x}-\frac{10 a \left (a+\frac{b}{x^4}\right )^{3/2}}{77 x}-\frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{11 x} \]

[Out]

(-20*a^2*Sqrt[a + b/x^4])/(77*x) - (10*a*(a + b/x^4)^(3/2))/(77*x) - (a + b/x^4)^(5/2)/(11*x) - (20*a^(11/4)*S
qrt[(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])/(77*b^(1/4)*Sqrt[a + b/x^4])

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Rubi [A]  time = 0.0698347, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {335, 195, 220} \[ -\frac{20 a^2 \sqrt{a+\frac{b}{x^4}}}{77 x}-\frac{20 a^{11/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{77 \sqrt [4]{b} \sqrt{a+\frac{b}{x^4}}}-\frac{10 a \left (a+\frac{b}{x^4}\right )^{3/2}}{77 x}-\frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{11 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(5/2)/x^2,x]

[Out]

(-20*a^2*Sqrt[a + b/x^4])/(77*x) - (10*a*(a + b/x^4)^(3/2))/(77*x) - (a + b/x^4)^(5/2)/(11*x) - (20*a^(11/4)*S
qrt[(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])/(77*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 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 \frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{x^2} \, dx &=-\operatorname{Subst}\left (\int \left (a+b x^4\right )^{5/2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{11 x}-\frac{1}{11} (10 a) \operatorname{Subst}\left (\int \left (a+b x^4\right )^{3/2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{10 a \left (a+\frac{b}{x^4}\right )^{3/2}}{77 x}-\frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{11 x}-\frac{1}{77} \left (60 a^2\right ) \operatorname{Subst}\left (\int \sqrt{a+b x^4} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{20 a^2 \sqrt{a+\frac{b}{x^4}}}{77 x}-\frac{10 a \left (a+\frac{b}{x^4}\right )^{3/2}}{77 x}-\frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{11 x}-\frac{1}{77} \left (40 a^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{20 a^2 \sqrt{a+\frac{b}{x^4}}}{77 x}-\frac{10 a \left (a+\frac{b}{x^4}\right )^{3/2}}{77 x}-\frac{\left (a+\frac{b}{x^4}\right )^{5/2}}{11 x}-\frac{20 a^{11/4} \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 )}{77 \sqrt [4]{b} \sqrt{a+\frac{b}{x^4}}}\\ \end{align*}

Mathematica [C]  time = 0.014334, size = 54, normalized size = 0.37 \[ -\frac{b^2 \sqrt{a+\frac{b}{x^4}} \, _2F_1\left (-\frac{11}{4},-\frac{5}{2};-\frac{7}{4};-\frac{a x^4}{b}\right )}{11 x^9 \sqrt{\frac{a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(5/2)/x^2,x]

[Out]

-(b^2*Sqrt[a + b/x^4]*Hypergeometric2F1[-11/4, -5/2, -7/4, -((a*x^4)/b)])/(11*x^9*Sqrt[1 + (a*x^4)/b])

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Maple [C]  time = 0.019, size = 180, normalized size = 1.2 \begin{align*} -{\frac{1}{77\,x \left ( a{x}^{4}+b \right ) ^{3}} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{{\frac{5}{2}}} \left ( -40\,{a}^{3}\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}^{11}+37\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{x}^{12}{a}^{3}+61\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{x}^{8}{a}^{2}b+31\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{x}^{4}a{b}^{2}+7\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{b}^{3} \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)^(5/2)/x^2,x)

[Out]

-1/77*((a*x^4+b)/x^4)^(5/2)*(-40*a^3*(-(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^11+37*(I*a^(1/2)/b^(1/2))^(1/2)*x^12*a^3+61*(I*a^(1/2)/b^(1
/2))^(1/2)*x^8*a^2*b+31*(I*a^(1/2)/b^(1/2))^(1/2)*x^4*a*b^2+7*(I*a^(1/2)/b^(1/2))^(1/2)*b^3)/x/(a*x^4+b)^3/(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{{\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((a+b/x^4)^(5/2)/x^2,x, algorithm="maxima")

[Out]

integrate((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{{\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{x^{10}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 2.28735, size = 39, normalized size = 0.27 \begin{align*} - \frac{a^{\frac{5}{2}} \Gamma \left (\frac{1}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{2}, \frac{1}{4} \\ \frac{5}{4} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{4}}} \right )}}{4 x \Gamma \left (\frac{5}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\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((a+b/x^4)^(5/2)/x^2,x, algorithm="giac")

[Out]

integrate((a + b/x^4)^(5/2)/x^2, x)