3.2105 \(\int \frac {1}{(a+\frac {b}{x^4})^{5/2} x^4} \, dx\)

Optimal. Leaf size=262 \[ \frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{8 a^{7/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{7/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {1}{4 a^2 x^3 \sqrt {a+\frac {b}{x^4}}}+\frac {\sqrt {a+\frac {b}{x^4}}}{4 a^2 \sqrt {b} x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {1}{6 a x^3 \left (a+\frac {b}{x^4}\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 262, normalized size of antiderivative = 1.00, 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, 290, 305, 220, 1196} \[ \frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{8 a^{7/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}-\frac {\sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) E\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{4 a^{7/4} b^{3/4} \sqrt {a+\frac {b}{x^4}}}+\frac {\sqrt {a+\frac {b}{x^4}}}{4 a^2 \sqrt {b} x \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )}-\frac {1}{4 a^2 x^3 \sqrt {a+\frac {b}{x^4}}}-\frac {1}{6 a x^3 \left (a+\frac {b}{x^4}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^4)^(5/2)*x^4),x]

[Out]

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

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Mathematica [C]  time = 0.06, size = 77, normalized size = 0.29 \[ \frac {x \left (a x^4+b\right ) \sqrt {\frac {a x^4}{b}+1} \, _2F_1\left (\frac {3}{4},\frac {5}{2};\frac {7}{4};-\frac {a x^4}{b}\right )-b x}{3 a b \sqrt {a+\frac {b}{x^4}} \left (a x^4+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{8} \sqrt {\frac {a x^{4} + b}{x^{4}}}}{a^{3} x^{12} + 3 \, a^{2} b x^{8} + 3 \, a b^{2} x^{4} + b^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(x^8*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 \[ \int \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.02, size = 503, normalized size = 1.92 \[ -\frac {-3 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{\frac {7}{2}} \sqrt {b}\, x^{11}-3 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{3} b \,x^{8} \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )+3 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{3} b \,x^{8} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-4 \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{\frac {5}{2}} b^{\frac {3}{2}} x^{7}-6 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{2} b^{2} x^{4} \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )+6 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a^{2} b^{2} x^{4} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )-\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{\frac {3}{2}} b^{\frac {5}{2}} x^{3}-3 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a \,b^{3} \EllipticE \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )+3 i \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, a \,b^{3} \EllipticF \left (\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, x , i\right )}{12 \left (\frac {a \,x^{4}+b}{x^{4}}\right )^{\frac {5}{2}} \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a^{\frac {5}{2}} b^{\frac {3}{2}} x^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(-3*a^(7/2)*b^(1/2)*(I*a^(1/2)/b^(1/2))^(1/2)*x^11+3*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)*x^8*a^3*b-3*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)*x^8*a^3*b-4
*a^(5/2)*b^(3/2)*(I*a^(1/2)/b^(1/2))^(1/2)*x^7+6*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)*x^4*a^2*b^2-6*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)*x^4*a^2*b^2-a^(3/2)
*b^(5/2)*(I*a^(1/2)/b^(1/2))^(1/2)*x^3+3*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)*a*b^3-3*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)*a*b^3)/a^(5/2)/((a*x^4+b)/x^4)^(5
/2)/x^10/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 \[ \int \frac {1}{{\left (a + \frac {b}{x^{4}}\right )}^{\frac {5}{2}} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b/x^4)^(5/2)),x)

[Out]

int(1/(x^4*(a + b/x^4)^(5/2)), x)

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sympy [C]  time = 2.11, size = 39, normalized size = 0.15 \[ - \frac {\Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 a^{\frac {5}{2}} x^{3} \Gamma \left (\frac {7}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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