3.1196 \(\int x^{12} \sqrt [4]{a-b x^4} \, dx\)

Optimal. Leaf size=156 \[ -\frac{3 a^{7/2} x^3 \left (1-\frac{a}{b x^4}\right )^{3/4} \text{EllipticF}\left (\frac{1}{2} \csc ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right ),2\right )}{112 b^{5/2} \left (a-b x^4\right )^{3/4}}-\frac{3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac{3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b} \]

[Out]

(-3*a^3*x*(a - b*x^4)^(1/4))/(112*b^3) - (3*a^2*x^5*(a - b*x^4)^(1/4))/(280*b^2) - (a*x^9*(a - b*x^4)^(1/4))/(
140*b) + (x^13*(a - b*x^4)^(1/4))/14 - (3*a^(7/2)*(1 - a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCsc[(Sqrt[b]*x^2)/Sqr
t[a]]/2, 2])/(112*b^(5/2)*(a - b*x^4)^(3/4))

________________________________________________________________________________________

Rubi [A]  time = 0.0774046, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {279, 321, 237, 335, 275, 232} \[ -\frac{3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac{3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}-\frac{3 a^{7/2} x^3 \left (1-\frac{a}{b x^4}\right )^{3/4} F\left (\left .\frac{1}{2} \csc ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{112 b^{5/2} \left (a-b x^4\right )^{3/4}}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b} \]

Antiderivative was successfully verified.

[In]

Int[x^12*(a - b*x^4)^(1/4),x]

[Out]

(-3*a^3*x*(a - b*x^4)^(1/4))/(112*b^3) - (3*a^2*x^5*(a - b*x^4)^(1/4))/(280*b^2) - (a*x^9*(a - b*x^4)^(1/4))/(
140*b) + (x^13*(a - b*x^4)^(1/4))/14 - (3*a^(7/2)*(1 - a/(b*x^4))^(3/4)*x^3*EllipticF[ArcCsc[(Sqrt[b]*x^2)/Sqr
t[a]]/2, 2])/(112*b^(5/2)*(a - b*x^4)^(3/4))

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(x^3*(1 + a/(b*x^4))^(3/4))/(a + b*x^4)^(3/4), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 232

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcSin[Rt[-(b/a), 2]*x])/2, 2])/(a^(3/4)*R
t[-(b/a), 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rubi steps

\begin{align*} \int x^{12} \sqrt [4]{a-b x^4} \, dx &=\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}+\frac{1}{14} a \int \frac{x^{12}}{\left (a-b x^4\right )^{3/4}} \, dx\\ &=-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}+\frac{\left (9 a^2\right ) \int \frac{x^8}{\left (a-b x^4\right )^{3/4}} \, dx}{140 b}\\ &=-\frac{3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}+\frac{\left (3 a^3\right ) \int \frac{x^4}{\left (a-b x^4\right )^{3/4}} \, dx}{56 b^2}\\ &=-\frac{3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}-\frac{3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}+\frac{\left (3 a^4\right ) \int \frac{1}{\left (a-b x^4\right )^{3/4}} \, dx}{112 b^3}\\ &=-\frac{3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}-\frac{3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}+\frac{\left (3 a^4 \left (1-\frac{a}{b x^4}\right )^{3/4} x^3\right ) \int \frac{1}{\left (1-\frac{a}{b x^4}\right )^{3/4} x^3} \, dx}{112 b^3 \left (a-b x^4\right )^{3/4}}\\ &=-\frac{3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}-\frac{3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}-\frac{\left (3 a^4 \left (1-\frac{a}{b x^4}\right )^{3/4} x^3\right ) \operatorname{Subst}\left (\int \frac{x}{\left (1-\frac{a x^4}{b}\right )^{3/4}} \, dx,x,\frac{1}{x}\right )}{112 b^3 \left (a-b x^4\right )^{3/4}}\\ &=-\frac{3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}-\frac{3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}-\frac{\left (3 a^4 \left (1-\frac{a}{b x^4}\right )^{3/4} x^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1-\frac{a x^2}{b}\right )^{3/4}} \, dx,x,\frac{1}{x^2}\right )}{224 b^3 \left (a-b x^4\right )^{3/4}}\\ &=-\frac{3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}-\frac{3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac{a x^9 \sqrt [4]{a-b x^4}}{140 b}+\frac{1}{14} x^{13} \sqrt [4]{a-b x^4}-\frac{3 a^{7/2} \left (1-\frac{a}{b x^4}\right )^{3/4} x^3 F\left (\left .\frac{1}{2} \csc ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )\right |2\right )}{112 b^{5/2} \left (a-b x^4\right )^{3/4}}\\ \end{align*}

Mathematica [C]  time = 0.0613502, size = 108, normalized size = 0.69 \[ \frac{x \sqrt [4]{a-b x^4} \left (15 a^3 \, _2F_1\left (-\frac{1}{4},\frac{1}{4};\frac{5}{4};\frac{b x^4}{a}\right )-\sqrt [4]{1-\frac{b x^4}{a}} \left (3 a^2 b x^4+15 a^3+2 a b^2 x^8-20 b^3 x^{12}\right )\right )}{280 b^3 \sqrt [4]{1-\frac{b x^4}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^12*(a - b*x^4)^(1/4),x]

[Out]

(x*(a - b*x^4)^(1/4)*(-((1 - (b*x^4)/a)^(1/4)*(15*a^3 + 3*a^2*b*x^4 + 2*a*b^2*x^8 - 20*b^3*x^12)) + 15*a^3*Hyp
ergeometric2F1[-1/4, 1/4, 5/4, (b*x^4)/a]))/(280*b^3*(1 - (b*x^4)/a)^(1/4))

________________________________________________________________________________________

Maple [F]  time = 0.017, size = 0, normalized size = 0. \begin{align*} \int{x}^{12}\sqrt [4]{-b{x}^{4}+a}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^12*(-b*x^4+a)^(1/4),x)

[Out]

int(x^12*(-b*x^4+a)^(1/4),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-b x^{4} + a\right )}^{\frac{1}{4}} x^{12}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((-b*x^4 + a)^(1/4)*x^12, x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (-b x^{4} + a\right )}^{\frac{1}{4}} x^{12}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(1/4)*x^12, x)

________________________________________________________________________________________

Sympy [C]  time = 2.86644, size = 41, normalized size = 0.26 \begin{align*} \frac{\sqrt [4]{a} x^{13} \Gamma \left (\frac{13}{4}\right ){{}_{2}F_{1}\left (\begin{matrix} - \frac{1}{4}, \frac{13}{4} \\ \frac{17}{4} \end{matrix}\middle |{\frac{b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \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)**(1/4),x)

[Out]

a**(1/4)*x**13*gamma(13/4)*hyper((-1/4, 13/4), (17/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*gamma(17/4))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-b x^{4} + a\right )}^{\frac{1}{4}} x^{12}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^12*(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate((-b*x^4 + a)^(1/4)*x^12, x)