3.2081 \(\int \frac {(a+\frac {b}{x^4})^{5/2}}{x^4} \, dx\)

Optimal. Leaf size=278 \[ -\frac {4 a^{13/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{39 b^{3/4} \sqrt {a+\frac {b}{x^4}}}+\frac {8 a^{13/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{39 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {8 a^3 \sqrt {a+\frac {b}{x^4}}}{39 \sqrt {b} x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {4 a^2 \sqrt {a+\frac {b}{x^4}}}{39 x^3}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{117 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{13 x^3} \]

[Out]

-10/117*a*(a+b/x^4)^(3/2)/x^3-1/13*(a+b/x^4)^(5/2)/x^3-4/39*a^2*(a+b/x^4)^(1/2)/x^3-8/39*a^3*(a+b/x^4)^(1/2)/x
/b^(1/2)/(a^(1/2)+b^(1/2)/x^2)+8/39*a^(13/4)*(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)/b^(3/4)/(a+b/x^4)^(1/2)-4/39*a^(13/4)*(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^(3/4)/(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {4 a^{13/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{39 b^{3/4} \sqrt {a+\frac {b}{x^4}}}+\frac {8 a^{13/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{39 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {8 a^3 \sqrt {a+\frac {b}{x^4}}}{39 \sqrt {b} x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {4 a^2 \sqrt {a+\frac {b}{x^4}}}{39 x^3}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{117 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{13 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(5/2)/x^4,x]

[Out]

(-4*a^2*Sqrt[a + b/x^4])/(39*x^3) - (10*a*(a + b/x^4)^(3/2))/(117*x^3) - (a + b/x^4)^(5/2)/(13*x^3) - (8*a^3*S
qrt[a + b/x^4])/(39*Sqrt[b]*(Sqrt[a] + Sqrt[b]/x^2)*x) + (8*a^(13/4)*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])/(39*b^(3/4)*Sqrt[a + b/x^4]) - (4*a^
(13/4)*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])/(39*b^(3/4)*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 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 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 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 )^{5/2}}{x^4} \, dx &=-\operatorname {Subst}\left (\int x^2 \left (a+b x^4\right )^{5/2} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{13 x^3}-\frac {1}{13} (10 a) \operatorname {Subst}\left (\int x^2 \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}}{117 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{13 x^3}-\frac {1}{39} \left (20 a^2\right ) \operatorname {Subst}\left (\int x^2 \sqrt {a+b x^4} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {4 a^2 \sqrt {a+\frac {b}{x^4}}}{39 x^3}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{117 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{13 x^3}-\frac {1}{39} \left (8 a^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {4 a^2 \sqrt {a+\frac {b}{x^4}}}{39 x^3}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{117 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{13 x^3}-\frac {\left (8 a^{7/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{39 \sqrt {b}}+\frac {\left (8 a^{7/2}\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 )}{39 \sqrt {b}}\\ &=-\frac {4 a^2 \sqrt {a+\frac {b}{x^4}}}{39 x^3}-\frac {10 a \left (a+\frac {b}{x^4}\right )^{3/2}}{117 x^3}-\frac {\left (a+\frac {b}{x^4}\right )^{5/2}}{13 x^3}-\frac {8 a^3 \sqrt {a+\frac {b}{x^4}}}{39 \sqrt {b} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}+\frac {8 a^{13/4} \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 )}{39 b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {4 a^{13/4} \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 )}{39 b^{3/4} \sqrt {a+\frac {b}{x^4}}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.19 \[ -\frac {b^2 \sqrt {a+\frac {b}{x^4}} \, _2F_1\left (-\frac {13}{4},-\frac {5}{2};-\frac {9}{4};-\frac {a x^4}{b}\right )}{13 x^{11} \sqrt {\frac {a x^4}{b}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(5/2)/x^4,x]

[Out]

-1/13*(b^2*Sqrt[a + b/x^4]*Hypergeometric2F1[-13/4, -5/2, -9/4, -((a*x^4)/b)])/(x^11*Sqrt[1 + (a*x^4)/b])

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fricas [F]  time = 1.10, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{2} x^{8} + 2 \, a b x^{4} + b^{2}\right )} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{x^{12}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x^4)^(5/2)/x^4, x)

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maple [C]  time = 0.03, size = 279, normalized size = 1.00 \[ \frac {\left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \left (-24 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{4} \sqrt {b}\, x^{16}-24 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{\frac {7}{2}} b \,x^{13} \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )+24 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{\frac {7}{2}} b \,x^{13} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-55 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{3} b^{\frac {3}{2}} x^{12}-59 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{2} b^{\frac {5}{2}} x^{8}-37 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,b^{\frac {7}{2}} x^{4}-9 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{\frac {9}{2}}\right )}{117 \left (a \,x^{4}+b \right )^{3} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b^{\frac {3}{2}} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/117*((a*x^4+b)/x^4)^(5/2)*(24*I*a^(7/2)*(-(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^13*b*EllipticF((I*a^(1/2)/b^(1/2))^(1/2)*x,I)-24*I*a^(7/2)*(-(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^13*b*EllipticE((I*a^(1/2)/b^(1/2))^(1/2)*x,I)-24*b^(1/2)*(I*a^(
1/2)/b^(1/2))^(1/2)*x^16*a^4-55*b^(3/2)*(I*a^(1/2)/b^(1/2))^(1/2)*x^12*a^3-59*b^(5/2)*(I*a^(1/2)/b^(1/2))^(1/2
)*x^8*a^2-37*b^(7/2)*(I*a^(1/2)/b^(1/2))^(1/2)*x^4*a-9*b^(9/2)*(I*a^(1/2)/b^(1/2))^(1/2))/x^3/(a*x^4+b)^3/b^(3
/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 \[ \int \frac {{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a + b/x^4)^(5/2)/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {b}{x^4}\right )}^{5/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x^4)^(5/2)/x^4,x)

[Out]

int((a + b/x^4)^(5/2)/x^4, x)

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sympy [C]  time = 1.89, size = 41, normalized size = 0.15 \[ - \frac {a^{\frac {5}{2}} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{2}, \frac {3}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 x^{3} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**(5/2)*gamma(3/4)*hyper((-5/2, 3/4), (7/4,), b*exp_polar(I*pi)/(a*x**4))/(4*x**3*gamma(7/4))

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