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

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

[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)/(Sqr
t[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 = 0.13909, 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 \[ \frac{15 b^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{28 a^{13/4} \sqrt{a+b x^4}}+\frac{15 b \sqrt{a+b x^4}}{14 a^3 x^3}-\frac{9 \sqrt{a+b x^4}}{14 a^2 x^7}+\frac{1}{2 a x^7 \sqrt{a+b x^4}} \]

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)/(Sqr
t[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 in Sympy [A]  time = 15.0845, size = 141, normalized size = 0.92 \[ \frac{1}{2 a x^{7} \sqrt{a + b x^{4}}} - \frac{9 \sqrt{a + b x^{4}}}{14 a^{2} x^{7}} + \frac{15 b \sqrt{a + b x^{4}}}{14 a^{3} x^{3}} + \frac{15 b^{\frac{7}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{28 a^{\frac{13}{4}} \sqrt{a + b x^{4}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(1/x**8/(b*x**4+a)**(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 + b*x**4)/(sqrt(a) + sqrt(b)*x**2
)**2)*(sqrt(a) + sqrt(b)*x**2)*elliptic_f(2*atan(b**(1/4)*x/a**(1/4)), 1/2)/(28*
a**(13/4)*sqrt(a + b*x**4))

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Mathematica [C]  time = 0.301633, size = 106, normalized size = 0.69 \[ \frac{-\frac{2 a^2}{x^7}-\frac{15 i b^2 \sqrt{\frac{b x^4}{a}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}}}+\frac{6 a b}{x^3}+15 b^2 x}{14 a^3 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

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

[Out]

((-2*a^2)/x^7 + (6*a*b)/x^3 + 15*b^2*x - ((15*I)*b^2*Sqrt[1 + (b*x^4)/a]*Ellipti
cF[I*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/Sqrt[(I*Sqrt[b])/Sqrt[a]])/(14*a
^3*Sqrt[a + b*x^4])

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Maple [C]  time = 0.032, size = 135, normalized size = 0.9 \[{\frac{{b}^{2}x}{2\,{a}^{3}}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}-{\frac{1}{7\,{a}^{2}{x}^{7}}\sqrt{b{x}^{4}+a}}+{\frac{4\,b}{7\,{a}^{3}{x}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{15\,{b}^{2}}{14\,{a}^{3}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(1/x^8/(b*x^4+a)^(3/2),x)

[Out]

1/2*b^2/a^3*x/((x^4+a/b)*b)^(1/2)-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+15/14/a^3*b^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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(1/((b*x^4 + a)^(3/2)*x^8),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. \[{\rm integral}\left (\frac{1}{{\left (b x^{12} + a x^{8}\right )} \sqrt{b x^{4} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

integral(1/((b*x^12 + a*x^8)*sqrt(b*x^4 + a)), x)

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Sympy [A]  time = 7.0108, size = 44, normalized size = 0.29 \[ \frac{\Gamma \left (- \frac{7}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{7}{4}, \frac{3}{2} \\ - \frac{3}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} x^{7} \Gamma \left (- \frac{3}{4}\right )} \]

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

Verification of antiderivative is not currently implemented for this CAS.

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

[Out]

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