3.2072 \(\int \frac {(a+\frac {b}{x^4})^{3/2}}{x^2} \, dx\)

Optimal. Leaf size=126 \[ -\frac {2 a^{7/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {2 a \sqrt {a+\frac {b}{x^4}}}{7 x}-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{7 x} \]

[Out]

-1/7*(a+b/x^4)^(3/2)/x-2/7*a*(a+b/x^4)^(1/2)/x-2/7*a^(7/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*ar
ccot(a^(1/4)*x/b^(1/4)))*EllipticF(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a^(1/2)+b^(1/2)/x^2)*((a+b/x
^4)/(a^(1/2)+b^(1/2)/x^2)^2)^(1/2)/b^(1/4)/(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.05, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {335, 195, 220} \[ -\frac {2 a^{7/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}-\frac {2 a \sqrt {a+\frac {b}{x^4}}}{7 x}-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{7 x} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(3/2)/x^2,x]

[Out]

(-2*a*Sqrt[a + b/x^4])/(7*x) - (a + b/x^4)^(3/2)/(7*x) - (2*a^(7/4)*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])/(7*b^(1/4)*Sqrt[a + b/x^4])

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]

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]

Rubi steps

\begin {align*} \int \frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{x^2} \, dx &=-\operatorname {Subst}\left (\int \left (a+b x^4\right )^{3/2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{7 x}-\frac {1}{7} (6 a) \operatorname {Subst}\left (\int \sqrt {a+b x^4} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {2 a \sqrt {a+\frac {b}{x^4}}}{7 x}-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{7 x}-\frac {1}{7} \left (4 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {2 a \sqrt {a+\frac {b}{x^4}}}{7 x}-\frac {\left (a+\frac {b}{x^4}\right )^{3/2}}{7 x}-\frac {2 a^{7/4} \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 )}{7 \sqrt [4]{b} \sqrt {a+\frac {b}{x^4}}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 52, normalized size = 0.41 \[ -\frac {b \sqrt {a+\frac {b}{x^4}} \, _2F_1\left (-\frac {7}{4},-\frac {3}{2};-\frac {3}{4};-\frac {a x^4}{b}\right )}{7 x^5 \sqrt {\frac {a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(3/2)/x^2,x]

[Out]

-1/7*(b*Sqrt[a + b/x^4]*Hypergeometric2F1[-7/4, -3/2, -3/4, -((a*x^4)/b)])/(x^5*Sqrt[1 + (a*x^4)/b])

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a x^{4} + b\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{x^{6}}, x\right ) \]

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^6, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}}{x^{2}}\,{d x} \]

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)

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maple [C]  time = 0.02, size = 158, normalized size = 1.25 \[ \frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {3}{2}} \left (-3 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} x^{8}+4 \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{2} x^{7} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-4 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a b \,x^{4}-\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{2}\right )}{7 \left (a \,x^{4}+b \right )^{2} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*((a*x^4+b)/x^4)^(3/2)*(4*a^2*(-(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((I*a^(1/2)/b^(1/2))^(1/2)*x,I)*x^7-3*(I*a^(1/2)/b^(1/2))^(1/2)*a^2*x^8-4*(I*a^(1/2)/b^(1/2))^(1/
2)*a*b*x^4-(I*a^(1/2)/b^(1/2))^(1/2)*b^2)/x/(a*x^4+b)^2/(I*a^(1/2)/b^(1/2))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {3}{2}}}{x^{2}}\,{d x} \]

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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mupad [B]  time = 1.57, size = 39, normalized size = 0.31 \[ -\frac {{\left (a\,x^4+b\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{4};\ \frac {5}{4};\ -\frac {b}{a\,x^4}\right )}{x\,{\left (\frac {b}{a}+x^4\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^4)^(3/2)/x^2,x)

[Out]

-((b + a*x^4)^(3/2)*hypergeom([-3/2, 1/4], 5/4, -b/(a*x^4)))/(x*(b/a + x^4)^(3/2))

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sympy [C]  time = 1.42, size = 39, normalized size = 0.31 \[ - \frac {a^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 x \Gamma \left (\frac {5}{4}\right )} \]

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)*gamma(1/4)*hyper((-3/2, 1/4), (5/4,), b*exp_polar(I*pi)/(a*x**4))/(4*x*gamma(5/4))

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