3.2103 \(\int \frac {1}{(a+\frac {b}{x^4})^{5/2}} \, dx\)

Optimal. Leaf size=277 \[ -\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{8 a^{11/4} \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{11/4} \sqrt {a+\frac {b}{x^4}}}+\frac {7 x \sqrt {a+\frac {b}{x^4}}}{4 a^3}-\frac {7 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{4 a^3 x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}} \]

[Out]

-1/6*x/a/(a+b/x^4)^(3/2)-7/12*x/a^2/(a+b/x^4)^(1/2)+7/4*x*(a+b/x^4)^(1/2)/a^3-7/4*b^(1/2)*(a+b/x^4)^(1/2)/a^3/
x/(a^(1/2)+b^(1/2)/x^2)+7/4*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^(11/4)/(a+b/x^4)^(1/2)-7/8*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^(11/4)/(a+b/x^4)^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 277, 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, 290, 325, 305, 220, 1196} \[ \frac {7 x \sqrt {a+\frac {b}{x^4}}}{4 a^3}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}-\frac {7 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{4 a^3 x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{8 a^{11/4} \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt [4]{b} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{11/4} \sqrt {a+\frac {b}{x^4}}}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(-5/2),x]

[Out]

(-7*Sqrt[b]*Sqrt[a + b/x^4])/(4*a^3*(Sqrt[a] + Sqrt[b]/x^2)*x) - x/(6*a*(a + b/x^4)^(3/2)) - (7*x)/(12*a^2*Sqr
t[a + b/x^4]) + (7*Sqrt[a + b/x^4]*x)/(4*a^3) + (7*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])/(4*a^(11/4)*Sqrt[a + b/x^4]) - (7*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])
/(8*a^(11/4)*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 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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 325

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

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 {1}{\left (a+\frac {b}{x^4}\right )^{5/2}} \, dx &=-\operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b x^4\right )^{5/2}} \, dx,x,\frac {1}{x}\right )\\ &=-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b x^4\right )^{3/2}} \, dx,x,\frac {1}{x}\right )}{6 a}\\ &=-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}-\frac {7 \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{4 a^2}\\ &=-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt {a+\frac {b}{x^4}} x}{4 a^3}-\frac {(7 b) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{4 a^3}\\ &=-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt {a+\frac {b}{x^4}} x}{4 a^3}-\frac {\left (7 \sqrt {b}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{4 a^{5/2}}+\frac {\left (7 \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 )}{4 a^{5/2}}\\ &=-\frac {7 \sqrt {b} \sqrt {a+\frac {b}{x^4}}}{4 a^3 \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) x}-\frac {x}{6 a \left (a+\frac {b}{x^4}\right )^{3/2}}-\frac {7 x}{12 a^2 \sqrt {a+\frac {b}{x^4}}}+\frac {7 \sqrt {a+\frac {b}{x^4}} x}{4 a^3}+\frac {7 \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 )}{4 a^{11/4} \sqrt {a+\frac {b}{x^4}}}-\frac {7 \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 )}{8 a^{11/4} \sqrt {a+\frac {b}{x^4}}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 81, normalized size = 0.29 \[ \frac {-7 x \left (a x^4+b\right ) \sqrt {\frac {a x^4}{b}+1} \, _2F_1\left (\frac {3}{4},\frac {5}{2};\frac {7}{4};-\frac {a x^4}{b}\right )+3 a x^5+7 b x}{3 a^2 \sqrt {a+\frac {b}{x^4}} \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(-5/2),x]

[Out]

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

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fricas [F]  time = 0.97, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{12} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a^{3} x^{12} + 3 \, a^{2} b x^{8} + 3 \, a b^{2} x^{4} + b^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^4)^(5/2),x, algorithm="fricas")

[Out]

integral(x^12*sqrt((a*x^4 + b)/x^4)/(a^3*x^12 + 3*a^2*b*x^8 + 3*a*b^2*x^4 + b^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 = 503, normalized size = 1.82 \[ -\frac {9 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{\frac {9}{2}} x^{11}+21 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{4} \sqrt {b}\, x^{8} \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-21 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{4} \sqrt {b}\, x^{8} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )+16 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{\frac {7}{2}} b \,x^{7}+42 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{3} b^{\frac {3}{2}} x^{4} \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-42 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{3} b^{\frac {3}{2}} x^{4} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )+7 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{\frac {5}{2}} b^{2} x^{3}+21 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{2} b^{\frac {5}{2}} \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-21 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{2} b^{\frac {5}{2}} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )}{12 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{\frac {9}{2}} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b/x^4)^(5/2),x)

[Out]

-1/12*(9*(I*a^(1/2)/b^(1/2))^(1/2)*a^(9/2)*x^11-21*I*(-(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)*b^(1/2)*x^8*a^4+21*I*(-(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)*EllipticE((I*a^(1/2)/b^(1/2))^(1/2)*x,I)*b^(1/2)*x^8*
a^4+16*(I*a^(1/2)/b^(1/2))^(1/2)*a^(7/2)*x^7*b-42*I*(-(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)*b^(3/2)*x^4*a^3+42*I*(-(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)*EllipticE((I*a^(1/2)/b^(1/2))^(1/2)*x,I)*b^(3/2)*x^4*a
^3+7*(I*a^(1/2)/b^(1/2))^(1/2)*a^(5/2)*x^3*b^2-21*I*(-(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)*b^(5/2)*a^2+21*I*(-(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)*EllipticE((I*a^(1/2)/b^(1/2))^(1/2)*x,I)*b^(5/2)*a^2)/a^(9
/2)/((a*x^4+b)/x^4)^(5/2)/x^10/(I*a^(1/2)/b^(1/2))^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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 = 1.30, size = 43, normalized size = 0.16 \[ \frac {x\,{\left (\frac {a\,x^4}{b}+1\right )}^{5/2}\,\sqrt {x^{20}}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{2},\frac {11}{4};\ \frac {15}{4};\ -\frac {a\,x^4}{b}\right )}{11\,{\left (a\,x^4+b\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b/x^4)^(5/2),x)

[Out]

(x*((a*x^4)/b + 1)^(5/2)*(x^20)^(1/2)*hypergeom([5/2, 11/4], 15/4, -(a*x^4)/b))/(11*(b + a*x^4)^(5/2))

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sympy [C]  time = 1.55, size = 41, normalized size = 0.15 \[ - \frac {x \Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {5}{2}} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**4)**(5/2),x)

[Out]

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

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