3.859 \(\int \frac{x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx\)

Optimal. Leaf size=151 \[ \frac{15 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{28 b^{13/4} \sqrt{a+b x^4}}-\frac{15 a x \sqrt{a+b x^4}}{14 b^3}+\frac{9 x^5 \sqrt{a+b x^4}}{14 b^2}-\frac{x^9}{2 b \sqrt{a+b x^4}} \]

[Out]

-x^9/(2*b*Sqrt[a + b*x^4]) - (15*a*x*Sqrt[a + b*x^4])/(14*b^3) + (9*x^5*Sqrt[a +
b*x^4])/(14*b^2) + (15*a^(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*b^(13/4)*
Sqrt[a + b*x^4])

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Rubi [A]  time = 0.140961, antiderivative size = 151, 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 \[ \frac{15 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{28 b^{13/4} \sqrt{a+b x^4}}-\frac{15 a x \sqrt{a+b x^4}}{14 b^3}+\frac{9 x^5 \sqrt{a+b x^4}}{14 b^2}-\frac{x^9}{2 b \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]  Int[x^12/(a + b*x^4)^(3/2),x]

[Out]

-x^9/(2*b*Sqrt[a + b*x^4]) - (15*a*x*Sqrt[a + b*x^4])/(14*b^3) + (9*x^5*Sqrt[a +
b*x^4])/(14*b^2) + (15*a^(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*b^(13/4)*
Sqrt[a + b*x^4])

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Rubi in Sympy [A]  time = 15.7999, size = 138, normalized size = 0.91 \[ \frac{15 a^{\frac{7}{4}} \sqrt{\frac{a + b x^{4}}{\left (\sqrt{a} + \sqrt{b} x^{2}\right )^{2}}} \left (\sqrt{a} + \sqrt{b} x^{2}\right ) F\left (2 \operatorname{atan}{\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}} \right )}\middle | \frac{1}{2}\right )}{28 b^{\frac{13}{4}} \sqrt{a + b x^{4}}} - \frac{15 a x \sqrt{a + b x^{4}}}{14 b^{3}} - \frac{x^{9}}{2 b \sqrt{a + b x^{4}}} + \frac{9 x^{5} \sqrt{a + b x^{4}}}{14 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  rubi_integrate(x**12/(b*x**4+a)**(3/2),x)

[Out]

15*a**(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*b**(13/4)*sqrt(a + b*x**4))
- 15*a*x*sqrt(a + b*x**4)/(14*b**3) - x**9/(2*b*sqrt(a + b*x**4)) + 9*x**5*sqrt
(a + b*x**4)/(14*b**2)

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Mathematica [C]  time = 0.263205, size = 106, normalized size = 0.7 \[ \frac{-\frac{15 i a^2 \sqrt{\frac{b x^4}{a}+1} F\left (\left .i \sinh ^{-1}\left (\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}} x\right )\right |-1\right )}{\sqrt{\frac{i \sqrt{b}}{\sqrt{a}}}}-15 a^2 x-6 a b x^5+2 b^2 x^9}{14 b^3 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]  Integrate[x^12/(a + b*x^4)^(3/2),x]

[Out]

(-15*a^2*x - 6*a*b*x^5 + 2*b^2*x^9 - ((15*I)*a^2*Sqrt[1 + (b*x^4)/a]*EllipticF[I
*ArcSinh[Sqrt[(I*Sqrt[b])/Sqrt[a]]*x], -1])/Sqrt[(I*Sqrt[b])/Sqrt[a]])/(14*b^3*S
qrt[a + b*x^4])

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Maple [C]  time = 0.024, size = 133, normalized size = 0.9 \[ -{\frac{x{a}^{2}}{2\,{b}^{3}}{\frac{1}{\sqrt{ \left ({x}^{4}+{\frac{a}{b}} \right ) b}}}}+{\frac{{x}^{5}}{7\,{b}^{2}}\sqrt{b{x}^{4}+a}}-{\frac{4\,ax}{7\,{b}^{3}}\sqrt{b{x}^{4}+a}}+{\frac{15\,{a}^{2}}{14\,{b}^{3}}\sqrt{1-{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}\sqrt{1+{i{x}^{2}\sqrt{b}{\frac{1}{\sqrt{a}}}}}{\it EllipticF} \left ( x\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}},i \right ){\frac{1}{\sqrt{{i\sqrt{b}{\frac{1}{\sqrt{a}}}}}}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  int(x^12/(b*x^4+a)^(3/2),x)

[Out]

-1/2/b^3*a^2*x/((x^4+a/b)*b)^(1/2)+1/7*x^5*(b*x^4+a)^(1/2)/b^2-4/7*a*x*(b*x^4+a)
^(1/2)/b^3+15/14*a^2/b^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)*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. \[ \int \frac{x^{12}}{{\left (b x^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^12/(b*x^4 + a)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^12/(b*x^4 + a)^(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^12/(b*x^4 + a)^(3/2),x, algorithm="fricas")

[Out]

integral(x^12/(b*x^4 + a)^(3/2), x)

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Sympy [A]  time = 6.99806, size = 37, normalized size = 0.25 \[ \frac{x^{13} \Gamma \left (\frac{13}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{3}{2}, \frac{13}{4} \\ \frac{17}{4} \end{matrix}\middle |{\frac{b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac{3}{2}} \Gamma \left (\frac{17}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x**12/(b*x**4+a)**(3/2),x)

[Out]

x**13*gamma(13/4)*hyper((3/2, 13/4), (17/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**(3
/2)*gamma(17/4))

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GIAC/XCAS [F]  time = 0., size = 0, normalized size = 0. \[ \int \frac{x^{12}}{{\left (b x^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]  integrate(x^12/(b*x^4 + a)^(3/2),x, algorithm="giac")

[Out]

integrate(x^12/(b*x^4 + a)^(3/2), x)