3.2079 \(\int (a+\frac {b}{x^4})^{5/2} \, dx\)

Optimal. Leaf size=272 \[ -\frac {4 a^{9/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3 \sqrt {a+\frac {b}{x^4}}}+\frac {8 a^{9/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3 \sqrt {a+\frac {b}{x^4}}}-\frac {8 a^2 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{3 x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}+x \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}-\frac {4 a b \sqrt {a+\frac {b}{x^4}}}{3 x^3} \]

[Out]

-10/9*b*(a+b/x^4)^(3/2)/x^3+(a+b/x^4)^(5/2)*x-4/3*a*b*(a+b/x^4)^(1/2)/x^3-8/3*a^2*b^(1/2)*(a+b/x^4)^(1/2)/x/(a
^(1/2)+b^(1/2)/x^2)+8/3*a^(9/4)*b^(1/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/4)*x/b^(1
/4)))*EllipticE(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)/(a+b/x^4)^(1/2)-4/3*a^(9/4)*b^(1/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(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)/(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 272, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {242, 277, 279, 305, 220, 1196} \[ -\frac {8 a^2 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{3 x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {4 a^{9/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3 \sqrt {a+\frac {b}{x^4}}}+\frac {8 a^{9/4} \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{3 \sqrt {a+\frac {b}{x^4}}}+x \left (a+\frac {b}{x^4}\right )^{5/2}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}-\frac {4 a b \sqrt {a+\frac {b}{x^4}}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(5/2),x]

[Out]

(-4*a*b*Sqrt[a + b/x^4])/(3*x^3) - (10*b*(a + b/x^4)^(3/2))/(9*x^3) - (8*a^2*Sqrt[b]*Sqrt[a + b/x^4])/(3*(Sqrt
[a] + Sqrt[b]/x^2)*x) + (a + b/x^4)^(5/2)*x + (8*a^(9/4)*b^(1/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(
Sqrt[a] + Sqrt[b]/x^2)*EllipticE[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(3*Sqrt[a + b/x^4]) - (4*a^(9/4)*b^(1/4)
*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 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 242

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

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 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 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 \left (a+\frac {b}{x^4}\right )^{5/2} \, dx &=-\operatorname {Subst}\left (\int \frac {\left (a+b x^4\right )^{5/2}}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\left (a+\frac {b}{x^4}\right )^{5/2} x-(10 b) \operatorname {Subst}\left (\int x^2 \left (a+b x^4\right )^{3/2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}+\left (a+\frac {b}{x^4}\right )^{5/2} x-\frac {1}{3} (20 a b) \operatorname {Subst}\left (\int x^2 \sqrt {a+b x^4} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {4 a b \sqrt {a+\frac {b}{x^4}}}{3 x^3}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}+\left (a+\frac {b}{x^4}\right )^{5/2} x-\frac {1}{3} \left (8 a^2 b\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {4 a b \sqrt {a+\frac {b}{x^4}}}{3 x^3}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}+\left (a+\frac {b}{x^4}\right )^{5/2} x-\frac {1}{3} \left (8 a^{5/2} \sqrt {b}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )+\frac {1}{3} \left (8 a^{5/2} \sqrt {b}\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 )\\ &=-\frac {4 a b \sqrt {a+\frac {b}{x^4}}}{3 x^3}-\frac {10 b \left (a+\frac {b}{x^4}\right )^{3/2}}{9 x^3}-\frac {8 a^2 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{3 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+\left (a+\frac {b}{x^4}\right )^{5/2} x+\frac {8 a^{9/4} \sqrt [4]{b} \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 )}{3 \sqrt {a+\frac {b}{x^4}}}-\frac {4 a^{9/4} \sqrt [4]{b} \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*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.20 \[ -\frac {b^2 \sqrt {a+\frac {b}{x^4}} \, _2F_1\left (-\frac {5}{2},-\frac {9}{4};-\frac {5}{4};-\frac {a x^4}{b}\right )}{9 x^7 \sqrt {\frac {a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(5/2),x]

[Out]

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

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fricas [F]  time = 0.72, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{x^{8}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x^4)^(5/2), x)

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maple [C]  time = 0.02, size = 251, normalized size = 0.92 \[ \frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (-15 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{3} x^{12}-24 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{\frac {5}{2}} \sqrt {b}\, x^{9} \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )+24 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{\frac {5}{2}} \sqrt {b}\, x^{9} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-19 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} b \,x^{8}-5 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,b^{2} x^{4}-\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{3}\right ) x}{9 \left (a \,x^{4}+b \right )^{3} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*((a*x^4+b)/x^4)^(5/2)*x*(24*I*a^(5/2)*b^(1/2)*(-(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*EllipticF((I*a^(1/2)/b^(1/2))^(1/2)*x,I)-24*I*a^(5/2)*b^(1/2)*(-(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*EllipticE((I*a^(1/2)/b^(1/2))^(1/2)*x,I)-15*(I*a
^(1/2)/b^(1/2))^(1/2)*a^3*x^12-19*(I*a^(1/2)/b^(1/2))^(1/2)*a^2*b*x^8-5*(I*a^(1/2)/b^(1/2))^(1/2)*a*b^2*x^4-(I
*a^(1/2)/b^(1/2))^(1/2)*b^3)/(a*x^4+b)^3/(I*a^(1/2)/b^(1/2))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x^4)^(5/2), x)

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mupad [B]  time = 1.71, size = 38, normalized size = 0.14 \[ -\frac {x\,{\left (a+\frac {b}{x^4}\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},-\frac {9}{4};\ -\frac {5}{4};\ -\frac {a\,x^4}{b}\right )}{9\,{\left (\frac {a\,x^4}{b}+1\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^4)^(5/2),x)

[Out]

-(x*(a + b/x^4)^(5/2)*hypergeom([-5/2, -9/4], -5/4, -(a*x^4)/b))/(9*((a*x^4)/b + 1)^(5/2))

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sympy [C]  time = 1.98, size = 42, normalized size = 0.15 \[ - \frac {a^{\frac {5}{2}} x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, - \frac {1}{4} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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