\(\int x^9 \sqrt [4]{a-b x^4} \, dx\) [1183]

Optimal result

Integrand size = 16, antiderivative size = 130 \[ \int x^9 \sqrt [4]{a-b x^4} \, dx=-\frac {2 a^2 x^2 \sqrt [4]{a-b x^4}}{77 b^2}-\frac {a x^6 \sqrt [4]{a-b x^4}}{77 b}+\frac {1}{11} x^{10} \sqrt [4]{a-b x^4}+\frac {4 a^{7/2} \left (1-\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{77 b^{5/2} \left (a-b x^4\right )^{3/4}} \]

[Out]

-2/77*a^2*x^2*(-b*x^4+a)^(1/4)/b^2-1/77*a*x^6*(-b*x^4+a)^(1/4)/b+1/11*x^10*(-b*x^4+a)^(1/4)+4/77*a^(7/2)*(1-b*
x^4/a)^(3/4)*(cos(1/2*arcsin(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arcsin(x^2*b^(1/2)/a^(1/2)))*EllipticF(sin
(1/2*arcsin(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 = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {281, 285, 327, 239, 238} \[ \int x^9 \sqrt [4]{a-b x^4} \, dx=\frac {4 a^{7/2} \left (1-\frac {b x^4}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right ),2\right )}{77 b^{5/2} \left (a-b x^4\right )^{3/4}}-\frac {2 a^2 x^2 \sqrt [4]{a-b x^4}}{77 b^2}+\frac {1}{11} x^{10} \sqrt [4]{a-b x^4}-\frac {a x^6 \sqrt [4]{a-b x^4}}{77 b} \]

[In]

Int[x^9*(a - b*x^4)^(1/4),x]

[Out]

(-2*a^2*x^2*(a - b*x^4)^(1/4))/(77*b^2) - (a*x^6*(a - b*x^4)^(1/4))/(77*b) + (x^10*(a - b*x^4)^(1/4))/11 + (4*
a^(7/2)*(1 - (b*x^4)/a)^(3/4)*EllipticF[ArcSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(77*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 239

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[(1 + b*(x^2/a))^(3/4)/(a + b*x^2)^(3/4), Int[1/(1 + b*(x^2
/a))^(3/4), x], x] /; FreeQ[{a, b}, x] && PosQ[a]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^4 \sqrt [4]{a-b x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{11} x^{10} \sqrt [4]{a-b x^4}+\frac {1}{22} a \text {Subst}\left (\int \frac {x^4}{\left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right ) \\ & = -\frac {a x^6 \sqrt [4]{a-b x^4}}{77 b}+\frac {1}{11} x^{10} \sqrt [4]{a-b x^4}+\frac {\left (3 a^2\right ) \text {Subst}\left (\int \frac {x^2}{\left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{77 b} \\ & = -\frac {2 a^2 x^2 \sqrt [4]{a-b x^4}}{77 b^2}-\frac {a x^6 \sqrt [4]{a-b x^4}}{77 b}+\frac {1}{11} x^{10} \sqrt [4]{a-b x^4}+\frac {\left (2 a^3\right ) \text {Subst}\left (\int \frac {1}{\left (a-b x^2\right )^{3/4}} \, dx,x,x^2\right )}{77 b^2} \\ & = -\frac {2 a^2 x^2 \sqrt [4]{a-b x^4}}{77 b^2}-\frac {a x^6 \sqrt [4]{a-b x^4}}{77 b}+\frac {1}{11} x^{10} \sqrt [4]{a-b x^4}+\frac {\left (2 a^3 \left (1-\frac {b x^4}{a}\right )^{3/4}\right ) \text {Subst}\left (\int \frac {1}{\left (1-\frac {b x^2}{a}\right )^{3/4}} \, dx,x,x^2\right )}{77 b^2 \left (a-b x^4\right )^{3/4}} \\ & = -\frac {2 a^2 x^2 \sqrt [4]{a-b x^4}}{77 b^2}-\frac {a x^6 \sqrt [4]{a-b x^4}}{77 b}+\frac {1}{11} x^{10} \sqrt [4]{a-b x^4}+\frac {4 a^{7/2} \left (1-\frac {b x^4}{a}\right )^{3/4} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{77 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 = 7.46 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.75 \[ \int x^9 \sqrt [4]{a-b x^4} \, dx=\frac {x^2 \sqrt [4]{a-b x^4} \left (-\sqrt [4]{1-\frac {b x^4}{a}} \left (6 a^2+a b x^4-7 b^2 x^8\right )+6 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {b x^4}{a}\right )\right )}{77 b^2 \sqrt [4]{1-\frac {b x^4}{a}}} \]

[In]

Integrate[x^9*(a - b*x^4)^(1/4),x]

[Out]

(x^2*(a - b*x^4)^(1/4)*(-((1 - (b*x^4)/a)^(1/4)*(6*a^2 + a*b*x^4 - 7*b^2*x^8)) + 6*a^2*Hypergeometric2F1[-1/4,
1/2, 3/2, (b*x^4)/a]))/(77*b^2*(1 - (b*x^4)/a)^(1/4))

Maple [F]

\[\int x^{9} \left (-b \,x^{4}+a \right )^{\frac {1}{4}}d x\]

[In]

int(x^9*(-b*x^4+a)^(1/4),x)

[Out]

int(x^9*(-b*x^4+a)^(1/4),x)

Fricas [F]

\[ \int x^9 \sqrt [4]{a-b x^4} \, dx=\int { {\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{9} \,d x } \]

[In]

integrate(x^9*(-b*x^4+a)^(1/4),x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(1/4)*x^9, x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.66 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.24 \[ \int x^9 \sqrt [4]{a-b x^4} \, dx=\frac {\sqrt [4]{a} x^{10} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{10} \]

[In]

integrate(x**9*(-b*x**4+a)**(1/4),x)

[Out]

a**(1/4)*x**10*hyper((-1/4, 5/2), (7/2,), b*x**4*exp_polar(2*I*pi)/a)/10

Maxima [F]

\[ \int x^9 \sqrt [4]{a-b x^4} \, dx=\int { {\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{9} \,d x } \]

[In]

integrate(x^9*(-b*x^4+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((-b*x^4 + a)^(1/4)*x^9, x)

Giac [F]

\[ \int x^9 \sqrt [4]{a-b x^4} \, dx=\int { {\left (-b x^{4} + a\right )}^{\frac {1}{4}} x^{9} \,d x } \]

[In]

integrate(x^9*(-b*x^4+a)^(1/4),x, algorithm="giac")

[Out]

integrate((-b*x^4 + a)^(1/4)*x^9, x)

Mupad [F(-1)]

Timed out. \[ \int x^9 \sqrt [4]{a-b x^4} \, dx=\int x^9\,{\left (a-b\,x^4\right )}^{1/4} \,d x \]

[In]

int(x^9*(a - b*x^4)^(1/4),x)

[Out]

int(x^9*(a - b*x^4)^(1/4), x)