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

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

[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])

________________________________________________________________________________________

Rubi [A]  time = 0.0515274, 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, Rules used = {288, 321, 220} \[ \frac{15 a^{7/4} \left (\sqrt{a}+\sqrt{b} x^2\right ) \sqrt{\frac{a+b x^4}{\left (\sqrt{a}+\sqrt{b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{28 b^{13/4} \sqrt{a+b x^4}}+\frac{9 x^5 \sqrt{a+b x^4}}{14 b^2}-\frac{15 a x \sqrt{a+b x^4}}{14 b^3}-\frac{x^9}{2 b \sqrt{a+b x^4}} \]

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])

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && 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{x^{12}}{\left (a+b x^4\right )^{3/2}} \, dx &=-\frac{x^9}{2 b \sqrt{a+b x^4}}+\frac{9 \int \frac{x^8}{\sqrt{a+b x^4}} \, dx}{2 b}\\ &=-\frac{x^9}{2 b \sqrt{a+b x^4}}+\frac{9 x^5 \sqrt{a+b x^4}}{14 b^2}-\frac{(45 a) \int \frac{x^4}{\sqrt{a+b x^4}} \, dx}{14 b^2}\\ &=-\frac{x^9}{2 b \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{\left (15 a^2\right ) \int \frac{1}{\sqrt{a+b x^4}} \, dx}{14 b^3}\\ &=-\frac{x^9}{2 b \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{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}}\\ \end{align*}

Mathematica [C]  time = 0.0213767, size = 79, normalized size = 0.52 \[ \frac{15 a^2 x \sqrt{\frac{b x^4}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^4}{a}\right )-15 a^2 x-6 a b x^5+2 b^2 x^9}{14 b^3 \sqrt{a+b x^4}} \]

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*a^2*x*Sqrt[1 + (b*x^4)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^4)/a
)])/(14*b^3*Sqrt[a + b*x^4])

________________________________________________________________________________________

Maple [C]  time = 0.015, size = 133, normalized size = 0.9 \begin{align*} -{\frac{{a}^{2}x}{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}}}} \end{align*}

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*x*a^2/((x^4+1/b*a)*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)*Ellipti
cF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{12}}{{\left (b x^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{4} + a} x^{12}}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}, x\right ) \end{align*}

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

________________________________________________________________________________________

Sympy [C]  time = 2.05152, size = 37, normalized size = 0.25 \begin{align*} \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 )} \end{align*}

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))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{12}}{{\left (b x^{4} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

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)