3.872 \(\int \frac {1}{x^6 (a+b x^4)^{3/2}} \, dx\)

Optimal. Leaf size=282 \[ -\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{20 a^{11/4} \sqrt {a+b x^4}}+\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 a^{11/4} \sqrt {a+b x^4}}-\frac {21 b^{3/2} x \sqrt {a+b x^4}}{10 a^3 \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {21 b \sqrt {a+b x^4}}{10 a^3 x}-\frac {7 \sqrt {a+b x^4}}{10 a^2 x^5}+\frac {1}{2 a x^5 \sqrt {a+b x^4}} \]

[Out]

1/2/a/x^5/(b*x^4+a)^(1/2)-7/10*(b*x^4+a)^(1/2)/a^2/x^5+21/10*b*(b*x^4+a)^(1/2)/a^3/x-21/10*b^(3/2)*x*(b*x^4+a)
^(1/2)/a^3/(a^(1/2)+x^2*b^(1/2))+21/10*b^(5/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)
*x/a^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)
+x^2*b^(1/2))^2)^(1/2)/a^(11/4)/(b*x^4+a)^(1/2)-21/20*b^(5/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2
*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*
x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/a^(11/4)/(b*x^4+a)^(1/2)

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Rubi [A]  time = 0.12, antiderivative size = 282, 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 = {290, 325, 305, 220, 1196} \[ -\frac {21 b^{3/2} x \sqrt {a+b x^4}}{10 a^3 \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{20 a^{11/4} \sqrt {a+b x^4}}+\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 a^{11/4} \sqrt {a+b x^4}}+\frac {21 b \sqrt {a+b x^4}}{10 a^3 x}-\frac {7 \sqrt {a+b x^4}}{10 a^2 x^5}+\frac {1}{2 a x^5 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(a + b*x^4)^(3/2)),x]

[Out]

1/(2*a*x^5*Sqrt[a + b*x^4]) - (7*Sqrt[a + b*x^4])/(10*a^2*x^5) + (21*b*Sqrt[a + b*x^4])/(10*a^3*x) - (21*b^(3/
2)*x*Sqrt[a + b*x^4])/(10*a^3*(Sqrt[a] + Sqrt[b]*x^2)) + (21*b^(5/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/
(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(10*a^(11/4)*Sqrt[a + b*x^4]) - (21*
b^(5/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(
1/4)], 1/2])/(20*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 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}{x^6 \left (a+b x^4\right )^{3/2}} \, dx &=\frac {1}{2 a x^5 \sqrt {a+b x^4}}+\frac {7 \int \frac {1}{x^6 \sqrt {a+b x^4}} \, dx}{2 a}\\ &=\frac {1}{2 a x^5 \sqrt {a+b x^4}}-\frac {7 \sqrt {a+b x^4}}{10 a^2 x^5}-\frac {(21 b) \int \frac {1}{x^2 \sqrt {a+b x^4}} \, dx}{10 a^2}\\ &=\frac {1}{2 a x^5 \sqrt {a+b x^4}}-\frac {7 \sqrt {a+b x^4}}{10 a^2 x^5}+\frac {21 b \sqrt {a+b x^4}}{10 a^3 x}-\frac {\left (21 b^2\right ) \int \frac {x^2}{\sqrt {a+b x^4}} \, dx}{10 a^3}\\ &=\frac {1}{2 a x^5 \sqrt {a+b x^4}}-\frac {7 \sqrt {a+b x^4}}{10 a^2 x^5}+\frac {21 b \sqrt {a+b x^4}}{10 a^3 x}-\frac {\left (21 b^{3/2}\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{10 a^{5/2}}+\frac {\left (21 b^{3/2}\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{10 a^{5/2}}\\ &=\frac {1}{2 a x^5 \sqrt {a+b x^4}}-\frac {7 \sqrt {a+b x^4}}{10 a^2 x^5}+\frac {21 b \sqrt {a+b x^4}}{10 a^3 x}-\frac {21 b^{3/2} x \sqrt {a+b x^4}}{10 a^3 \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{10 a^{11/4} \sqrt {a+b x^4}}-\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{20 a^{11/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 54, normalized size = 0.19 \[ -\frac {\sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (-\frac {5}{4},\frac {3}{2};-\frac {1}{4};-\frac {b x^4}{a}\right )}{5 a x^5 \sqrt {a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*(a + b*x^4)^(3/2)),x]

[Out]

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

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fricas [F]  time = 0.87, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{4} + a}}{b^{2} x^{14} + 2 \, a b x^{10} + a^{2} x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)/(b^2*x^14 + 2*a*b*x^10 + a^2*x^6), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^6), x)

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maple [C]  time = 0.02, size = 157, normalized size = 0.56 \[ \frac {b^{2} x^{3}}{2 \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}\, a^{3}}-\frac {21 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )\right ) b^{\frac {3}{2}}}{10 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, a^{\frac {5}{2}}}+\frac {8 \sqrt {b \,x^{4}+a}\, b}{5 a^{3} x}-\frac {\sqrt {b \,x^{4}+a}}{5 a^{2} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(b*x^4+a)^(3/2),x)

[Out]

-1/5*(b*x^4+a)^(1/2)/a^2/x^5+8/5*b*(b*x^4+a)^(1/2)/a^3/x+1/2*b^2/a^3*x^3/((x^4+a/b)*b)^(1/2)-21/10*I/a^(5/2)*b
^(3/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(-I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4+a)^(1
/2)*(EllipticF((I/a^(1/2)*b^(1/2))^(1/2)*x,I)-EllipticE((I/a^(1/2)*b^(1/2))^(1/2)*x,I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{4} + a\right )}^{\frac {3}{2}} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^6), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{x^6\,{\left (b\,x^4+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(a + b*x^4)^(3/2)),x)

[Out]

int(1/(x^6*(a + b*x^4)^(3/2)), x)

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sympy [C]  time = 1.80, size = 44, normalized size = 0.16 \[ \frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} x^{5} \Gamma \left (- \frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(b*x**4+a)**(3/2),x)

[Out]

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

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