3.2073 \(\int \frac{(a+\frac{b}{x^4})^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=257 \[ -\frac{2 a^{9/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) \text{EllipticF}\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+\frac{b}{x^4}}}+\frac{4 a^{9/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+\frac{b}{x^4}}}-\frac{4 a^2 \sqrt{a+\frac{b}{x^4}}}{15 \sqrt{b} x \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )}-\frac{2 a \sqrt{a+\frac{b}{x^4}}}{15 x^3}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{9 x^3} \]

[Out]

(-2*a*Sqrt[a + b/x^4])/(15*x^3) - (a + b/x^4)^(3/2)/(9*x^3) - (4*a^2*Sqrt[a + b/x^4])/(15*Sqrt[b]*(Sqrt[a] + S
qrt[b]/x^2)*x) + (4*a^(9/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*Ar
cCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a + b/x^4]) - (2*a^(9/4)*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])/(15*b^(3/4)*Sqrt[a + b/x^4])

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Rubi [A]  time = 0.135677, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {335, 279, 305, 220, 1196} \[ -\frac{2 a^{9/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+\frac{b}{x^4}}}+\frac{4 a^{9/4} \sqrt{\frac{a+\frac{b}{x^4}}{\left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )^2}} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+\frac{b}{x^4}}}-\frac{4 a^2 \sqrt{a+\frac{b}{x^4}}}{15 \sqrt{b} x \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right )}-\frac{2 a \sqrt{a+\frac{b}{x^4}}}{15 x^3}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(3/2)/x^4,x]

[Out]

(-2*a*Sqrt[a + b/x^4])/(15*x^3) - (a + b/x^4)^(3/2)/(9*x^3) - (4*a^2*Sqrt[a + b/x^4])/(15*Sqrt[b]*(Sqrt[a] + S
qrt[b]/x^2)*x) + (4*a^(9/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*Ar
cCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(15*b^(3/4)*Sqrt[a + b/x^4]) - (2*a^(9/4)*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])/(15*b^(3/4)*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 279

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

Rule 305

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

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 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{x^4} \, dx &=-\operatorname{Subst}\left (\int x^2 \left (a+b x^4\right )^{3/2} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{9 x^3}-\frac{1}{3} (2 a) \operatorname{Subst}\left (\int x^2 \sqrt{a+b x^4} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 a \sqrt{a+\frac{b}{x^4}}}{15 x^3}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{9 x^3}-\frac{1}{15} \left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{2 a \sqrt{a+\frac{b}{x^4}}}{15 x^3}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{9 x^3}-\frac{\left (4 a^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )}{15 \sqrt{b}}+\frac{\left (4 a^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx,x,\frac{1}{x}\right )}{15 \sqrt{b}}\\ &=-\frac{2 a \sqrt{a+\frac{b}{x^4}}}{15 x^3}-\frac{\left (a+\frac{b}{x^4}\right )^{3/2}}{9 x^3}-\frac{4 a^2 \sqrt{a+\frac{b}{x^4}}}{15 \sqrt{b} \left (\sqrt{a}+\frac{\sqrt{b}}{x^2}\right ) x}+\frac{4 a^{9/4} \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 ) E\left (2 \cot ^{-1}\left (\frac{\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac{1}{2}\right )}{15 b^{3/4} \sqrt{a+\frac{b}{x^4}}}-\frac{2 a^{9/4} \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 )}{15 b^{3/4} \sqrt{a+\frac{b}{x^4}}}\\ \end{align*}

Mathematica [C]  time = 0.0125343, size = 52, normalized size = 0.2 \[ -\frac{b \sqrt{a+\frac{b}{x^4}} \, _2F_1\left (-\frac{9}{4},-\frac{3}{2};-\frac{5}{4};-\frac{a x^4}{b}\right )}{9 x^7 \sqrt{\frac{a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(3/2)/x^4,x]

[Out]

-(b*Sqrt[a + b/x^4]*Hypergeometric2F1[-9/4, -3/2, -5/4, -((a*x^4)/b)])/(9*x^7*Sqrt[1 + (a*x^4)/b])

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Maple [C]  time = 0.017, size = 257, normalized size = 1. \begin{align*} -{\frac{1}{45\,{x}^{3} \left ( a{x}^{4}+b \right ) ^{2}} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{{\frac{3}{2}}} \left ( -12\,i{a}^{{\frac{5}{2}}}\sqrt{-{ \left ( i\sqrt{a}{x}^{2}-\sqrt{b} \right ){\frac{1}{\sqrt{b}}}}}\sqrt{{ \left ( i\sqrt{a}{x}^{2}+\sqrt{b} \right ){\frac{1}{\sqrt{b}}}}}{x}^{9}b{\it EllipticF} \left ( x\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}},i \right ) +12\,i{a}^{{\frac{5}{2}}}\sqrt{-{ \left ( i\sqrt{a}{x}^{2}-\sqrt{b} \right ){\frac{1}{\sqrt{b}}}}}\sqrt{{ \left ( i\sqrt{a}{x}^{2}+\sqrt{b} \right ){\frac{1}{\sqrt{b}}}}}{x}^{9}b{\it EllipticE} \left ( x\sqrt{{i\sqrt{a}{\frac{1}{\sqrt{b}}}}},i \right ) +12\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}\sqrt{b}{x}^{12}{a}^{3}+23\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{b}^{3/2}{x}^{8}{a}^{2}+16\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{b}^{5/2}{x}^{4}a+5\,\sqrt{{\frac{i\sqrt{a}}{\sqrt{b}}}}{b}^{7/2} \right ){b}^{-{\frac{3}{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((a+b/x^4)^(3/2)/x^4,x)

[Out]

-1/45*((a*x^4+b)/x^4)^(3/2)*(-12*I*a^(5/2)*(-(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)*x^9*b*EllipticF(x*(I*a^(1/2)/b^(1/2))^(1/2),I)+12*I*a^(5/2)*(-(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)*x^9*b*EllipticE(x*(I*a^(1/2)/b^(1/2))^(1/2),I)+12*(I*a^(1/2)/b^(1
/2))^(1/2)*b^(1/2)*x^12*a^3+23*(I*a^(1/2)/b^(1/2))^(1/2)*b^(3/2)*x^8*a^2+16*(I*a^(1/2)/b^(1/2))^(1/2)*b^(5/2)*
x^4*a+5*(I*a^(1/2)/b^(1/2))^(1/2)*b^(7/2))/x^3/(a*x^4+b)^2/b^(3/2)/(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 \frac{{\left (a + \frac{b}{x^{4}}\right )}^{\frac{3}{2}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x^4)^(3/2)/x^4, 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^{8}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 2.09188, size = 41, normalized size = 0.16 \begin{align*} - \frac{a^{\frac{3}{2}} \Gamma \left (\frac{3}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{3}{2}, \frac{3}{4} \\ \frac{7}{4} \end{matrix}\middle |{\frac{b e^{i \pi }}{a x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac{7}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(3/2)*gamma(3/4)*hyper((-3/2, 3/4), (7/4,), b*exp_polar(I*pi)/(a*x**4))/(4*x**3*gamma(7/4))

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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{3}{2}}}{x^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x^4)^(3/2)/x^4, x)