3.830 \(\int \frac{1}{x^6 \sqrt{a+b x^4}} \, dx\)

Optimal. Leaf size=261 \[ -\frac{3 b^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{10 a^{7/4} \sqrt{a+b x^4}}-\frac{3 b^{3/2} x \sqrt{a+b x^4}}{5 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{3 b^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{7/4} \sqrt{a+b x^4}}+\frac{3 b \sqrt{a+b x^4}}{5 a^2 x}-\frac{\sqrt{a+b x^4}}{5 a x^5} \]

[Out]

-Sqrt[a + b*x^4]/(5*a*x^5) + (3*b*Sqrt[a + b*x^4])/(5*a^2*x) - (3*b^(3/2)*x*Sqrt[a + b*x^4])/(5*a^2*(Sqrt[a] +
Sqrt[b]*x^2)) + (3*b^(5/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*Ar
cTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(5*a^(7/4)*Sqrt[a + b*x^4]) - (3*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])/(10*a^(7/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.0921261, antiderivative size = 261, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {325, 305, 220, 1196} \[ -\frac{3 b^{3/2} x \sqrt{a+b x^4}}{5 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}-\frac{3 b^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{10 a^{7/4} \sqrt{a+b x^4}}+\frac{3 b^{5/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{5 a^{7/4} \sqrt{a+b x^4}}+\frac{3 b \sqrt{a+b x^4}}{5 a^2 x}-\frac{\sqrt{a+b x^4}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*Sqrt[a + b*x^4]),x]

[Out]

-Sqrt[a + b*x^4]/(5*a*x^5) + (3*b*Sqrt[a + b*x^4])/(5*a^2*x) - (3*b^(3/2)*x*Sqrt[a + b*x^4])/(5*a^2*(Sqrt[a] +
Sqrt[b]*x^2)) + (3*b^(5/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*Ar
cTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(5*a^(7/4)*Sqrt[a + b*x^4]) - (3*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])/(10*a^(7/4)*Sqrt[a + b*x^4])

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 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 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 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 \sqrt{a+b x^4}} \, dx &=-\frac{\sqrt{a+b x^4}}{5 a x^5}-\frac{(3 b) \int \frac{1}{x^2 \sqrt{a+b x^4}} \, dx}{5 a}\\ &=-\frac{\sqrt{a+b x^4}}{5 a x^5}+\frac{3 b \sqrt{a+b x^4}}{5 a^2 x}-\frac{\left (3 b^2\right ) \int \frac{x^2}{\sqrt{a+b x^4}} \, dx}{5 a^2}\\ &=-\frac{\sqrt{a+b x^4}}{5 a x^5}+\frac{3 b \sqrt{a+b x^4}}{5 a^2 x}-\frac{\left (3 b^{3/2}\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx}{5 a^{3/2}}+\frac{\left (3 b^{3/2}\right ) \int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a}}}{\sqrt{a+b x^4}} \, dx}{5 a^{3/2}}\\ &=-\frac{\sqrt{a+b x^4}}{5 a x^5}+\frac{3 b \sqrt{a+b x^4}}{5 a^2 x}-\frac{3 b^{3/2} x \sqrt{a+b x^4}}{5 a^2 \left (\sqrt{a}+\sqrt{b} x^2\right )}+\frac{3 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 )}{5 a^{7/4} \sqrt{a+b x^4}}-\frac{3 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 )}{10 a^{7/4} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0087659, size = 51, normalized size = 0.2 \[ -\frac{\sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{5}{4},\frac{1}{2};-\frac{1}{4};-\frac{b x^4}{a}\right )}{5 x^5 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^6*Sqrt[a + b*x^4]),x]

[Out]

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

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Maple [C]  time = 0.011, size = 133, normalized size = 0.5 \begin{align*} -{\frac{1}{5\,a{x}^{5}}\sqrt{b{x}^{4}+a}}+{\frac{3\,b}{5\,{a}^{2}x}\sqrt{b{x}^{4}+a}}-{{\frac{3\,i}{5}}{b}^{{\frac{3}{2}}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}} \left ({\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) -{\it EllipticE} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ) \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{4} + a} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^4 + a)*x^6), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{4} + a}}{b x^{10} + a x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C]  time = 1.39197, size = 44, normalized size = 0.17 \begin{align*} \frac{\Gamma \left (- \frac{5}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{5}{4}, \frac{1}{2} \\ - \frac{1}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt{a} x^{5} \Gamma \left (- \frac{1}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b x^{4} + a} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/(sqrt(b*x^4 + a)*x^6), x)