3.866 \(\int \frac{1}{x^8 (a+b x^4)^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ \frac{15 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{28 a^{13/4} \sqrt{a+b x^4}}+\frac{15 b \sqrt{a+b x^4}}{14 a^3 x^3}-\frac{9 \sqrt{a+b x^4}}{14 a^2 x^7}+\frac{1}{2 a x^7 \sqrt{a+b x^4}} \]

[Out]

1/(2*a*x^7*Sqrt[a + b*x^4]) - (9*Sqrt[a + b*x^4])/(14*a^2*x^7) + (15*b*Sqrt[a + b*x^4])/(14*a^3*x^3) + (15*b^(
7/4)*(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])/(28*a^(13/4)*Sqrt[a + b*x^4])

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Rubi [A]  time = 0.0503625, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {290, 325, 220} \[ \frac{15 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{28 a^{13/4} \sqrt{a+b x^4}}+\frac{15 b \sqrt{a+b x^4}}{14 a^3 x^3}-\frac{9 \sqrt{a+b x^4}}{14 a^2 x^7}+\frac{1}{2 a x^7 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^8*(a + b*x^4)^(3/2)),x]

[Out]

1/(2*a*x^7*Sqrt[a + b*x^4]) - (9*Sqrt[a + b*x^4])/(14*a^2*x^7) + (15*b*Sqrt[a + b*x^4])/(14*a^3*x^3) + (15*b^(
7/4)*(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])/(28*a^(13/4)*Sqrt[a + b*x^4])

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 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 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]

Rubi steps

\begin{align*} \int \frac{1}{x^8 \left (a+b x^4\right )^{3/2}} \, dx &=\frac{1}{2 a x^7 \sqrt{a+b x^4}}+\frac{9 \int \frac{1}{x^8 \sqrt{a+b x^4}} \, dx}{2 a}\\ &=\frac{1}{2 a x^7 \sqrt{a+b x^4}}-\frac{9 \sqrt{a+b x^4}}{14 a^2 x^7}-\frac{(45 b) \int \frac{1}{x^4 \sqrt{a+b x^4}} \, dx}{14 a^2}\\ &=\frac{1}{2 a x^7 \sqrt{a+b x^4}}-\frac{9 \sqrt{a+b x^4}}{14 a^2 x^7}+\frac{15 b \sqrt{a+b x^4}}{14 a^3 x^3}+\frac{\left (15 b^2\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx}{14 a^3}\\ &=\frac{1}{2 a x^7 \sqrt{a+b x^4}}-\frac{9 \sqrt{a+b x^4}}{14 a^2 x^7}+\frac{15 b \sqrt{a+b x^4}}{14 a^3 x^3}+\frac{15 b^{7/4} \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 )}{28 a^{13/4} \sqrt{a+b x^4}}\\ \end{align*}

Mathematica [C]  time = 0.0087661, size = 54, normalized size = 0.35 \[ -\frac{\sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (-\frac{7}{4},\frac{3}{2};-\frac{3}{4};-\frac{b x^4}{a}\right )}{7 a x^7 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^8*(a + b*x^4)^(3/2)),x]

[Out]

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

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Maple [C]  time = 0.02, size = 135, normalized size = 0.9 \begin{align*} -{\frac{1}{7\,{a}^{2}{x}^{7}}\sqrt{b{x}^{4}+a}}+{\frac{4\,b}{7\,{a}^{3}{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{{b}^{2}x}{2\,{a}^{3}}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}+{\frac{15\,{b}^{2}}{14\,{a}^{3}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\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^8/(b*x^4+a)^(3/2),x)

[Out]

-1/7*(b*x^4+a)^(1/2)/a^2/x^7+4/7*b*(b*x^4+a)^(1/2)/a^3/x^3+1/2*b^2/a^3*x/((x^4+1/b*a)*b)^(1/2)+15/14*b^2/a^3/(
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)*Ellip
ticF(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}{{\left (b x^{4} + a\right )}^{\frac{3}{2}} x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^8), 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^{2} x^{16} + 2 \, a b x^{12} + a^{2} x^{8}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^4 + a)/(b^2*x^16 + 2*a*b*x^12 + a^2*x^8), x)

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Sympy [C]  time = 2.26804, size = 44, normalized size = 0.29 \begin{align*} \frac{\Gamma \left (- \frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{4}, \frac{3}{2} \\ - \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} x^{7} \Gamma \left (- \frac{3}{4}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**8/(b*x**4+a)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{4} + a\right )}^{\frac{3}{2}} x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^8/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^4 + a)^(3/2)*x^8), x)