\(\int x^{12} \sqrt [4]{a-b x^4} \, dx\) [1196]

Optimal result

Integrand size = 16, antiderivative size = 156 \[ \int x^{12} \sqrt [4]{a-b x^4} \, dx=-\frac {3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}-\frac {3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}-\frac {a x^9 \sqrt [4]{a-b x^4}}{140 b}+\frac {1}{14} x^{13} \sqrt [4]{a-b x^4}-\frac {3 a^{7/2} \left (1-\frac {a}{b x^4}\right )^{3/4} x^3 \operatorname {EllipticF}\left (\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{112 b^{5/2} \left (a-b x^4\right )^{3/4}} \]

[Out]

-3/112*a^3*x*(-b*x^4+a)^(1/4)/b^3-3/280*a^2*x^5*(-b*x^4+a)^(1/4)/b^2-1/140*a*x^9*(-b*x^4+a)^(1/4)/b+1/14*x^13*
(-b*x^4+a)^(1/4)-3/112*a^(7/2)*(1-a/b/x^4)^(3/4)*x^3*(cos(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*ar
ccsc(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccsc(x^2*b^(1/2)/a^(1/2))),2^(1/2))/b^(5/2)/(-b*x^4+a)^(3/4)

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {285, 327, 243, 342, 281, 238} \[ \int x^{12} \sqrt [4]{a-b x^4} \, dx=-\frac {3 a^{7/2} x^3 \left (1-\frac {a}{b x^4}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{112 b^{5/2} \left (a-b x^4\right )^{3/4}}-\frac {3 a^3 x \sqrt [4]{a-b x^4}}{112 b^3}-\frac {3 a^2 x^5 \sqrt [4]{a-b x^4}}{280 b^2}+\frac {1}{14} x^{13} \sqrt [4]{a-b x^4}-\frac {a x^9 \sqrt [4]{a-b x^4}}{140 b} \]

[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 238

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

Rule 243

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 281

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 285

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 327

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 342

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]

Rubi steps \begin{align*} \text {integral}& = \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 ) \text {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 ) \text {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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 7.67 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.69 \[ \int x^{12} \sqrt [4]{a-b x^4} \, dx=\frac {x \sqrt [4]{a-b x^4} \left (-\sqrt [4]{1-\frac {b x^4}{a}} \left (15 a^3+3 a^2 b x^4+2 a b^2 x^8-20 b^3 x^{12}\right )+15 a^3 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {5}{4},\frac {b x^4}{a}\right )\right )}{280 b^3 \sqrt [4]{1-\frac {b x^4}{a}}} \]

[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]

\[\int x^{12} \left (-b \,x^{4}+a \right )^{\frac {1}{4}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x^{12} \sqrt [4]{a-b x^4} \, dx=\int { {\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{12} \,d x } \]

[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] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.76 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.26 \[ \int x^{12} \sqrt [4]{a-b x^4} \, dx=\frac {\sqrt [4]{a} x^{13} \Gamma \left (\frac {13}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {13}{4} \\ \frac {17}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \Gamma \left (\frac {17}{4}\right )} \]

[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))

Maxima [F]

\[ \int x^{12} \sqrt [4]{a-b x^4} \, dx=\int { {\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{12} \,d x } \]

[In]

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

[Out]

integrate((-b*x^4 + a)^(1/4)*x^12, x)

Giac [F]

\[ \int x^{12} \sqrt [4]{a-b x^4} \, dx=\int { {\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{12} \,d x } \]

[In]

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

[Out]

integrate((-b*x^4 + a)^(1/4)*x^12, x)

Mupad [F(-1)]

Timed out. \[ \int x^{12} \sqrt [4]{a-b x^4} \, dx=\int x^{12}\,{\left (a-b\,x^4\right )}^{1/4} \,d x \]

[In]

int(x^12*(a - b*x^4)^(1/4),x)

[Out]

int(x^12*(a - b*x^4)^(1/4), x)