\(\int \frac {x^4}{\sqrt {a-b x^4}} \, dx\) [842]

Optimal result

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

[Out]

-1/3*x*(-b*x^4+a)^(1/2)/b+1/3*a^(5/4)*EllipticF(b^(1/4)*x/a^(1/4),I)*(1-b*x^4/a)^(1/2)/b^(5/4)/(-b*x^4+a)^(1/2
)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {327, 230, 227} \[ \int \frac {x^4}{\sqrt {a-b x^4}} \, dx=\frac {a^{5/4} \sqrt {1-\frac {b x^4}{a}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),-1\right )}{3 b^{5/4} \sqrt {a-b x^4}}-\frac {x \sqrt {a-b x^4}}{3 b} \]

[In]

Int[x^4/Sqrt[a - b*x^4],x]

[Out]

-1/3*(x*Sqrt[a - b*x^4])/b + (a^(5/4)*Sqrt[1 - (b*x^4)/a]*EllipticF[ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(3*b^(5/
4)*Sqrt[a - b*x^4])

Rule 227

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

Rule 230

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + b*(x^4/a)]/Sqrt[a + b*x^4], Int[1/Sqrt[1 + b*(x^4/
a)], x], x] /; FreeQ[{a, b}, x] && NegQ[b/a] &&  !GtQ[a, 0]

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 {x \sqrt {a-b x^4}}{3 b}+\frac {a \int \frac {1}{\sqrt {a-b x^4}} \, dx}{3 b} \\ & = -\frac {x \sqrt {a-b x^4}}{3 b}+\frac {\left (a \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1}{\sqrt {1-\frac {b x^4}{a}}} \, dx}{3 b \sqrt {a-b x^4}} \\ & = -\frac {x \sqrt {a-b x^4}}{3 b}+\frac {a^{5/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{3 b^{5/4} \sqrt {a-b x^4}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.83 \[ \int \frac {x^4}{\sqrt {a-b x^4}} \, dx=\frac {x \left (-a+b x^4+a \sqrt {1-\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},\frac {b x^4}{a}\right )\right )}{3 b \sqrt {a-b x^4}} \]

[In]

Integrate[x^4/Sqrt[a - b*x^4],x]

[Out]

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

Maple [A] (verified)

Time = 4.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.12

method	result	size
default	\(-\frac {x \sqrt {-b \,x^{4}+a}}{3 b}+\frac {a \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(86\)
risch	\(-\frac {x \sqrt {-b \,x^{4}+a}}{3 b}+\frac {a \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(86\)
elliptic	\(-\frac {x \sqrt {-b \,x^{4}+a}}{3 b}+\frac {a \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )}{3 b \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(86\)

[In]

int(x^4/(-b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3*x*(-b*x^4+a)^(1/2)/b+1/3*a/b/(1/a^(1/2)*b^(1/2))^(1/2)*(1-x^2*b^(1/2)/a^(1/2))^(1/2)*(1+x^2*b^(1/2)/a^(1/
2))^(1/2)/(-b*x^4+a)^(1/2)*EllipticF(x*(1/a^(1/2)*b^(1/2))^(1/2),I)

Fricas [A] (verification not implemented)

none

Time = 0.09 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.60 \[ \int \frac {x^4}{\sqrt {a-b x^4}} \, dx=\frac {\sqrt {-b} \left (\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - \sqrt {-b x^{4} + a} x}{3 \, b} \]

[In]

integrate(x^4/(-b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

1/3*(sqrt(-b)*(a/b)^(3/4)*elliptic_f(arcsin((a/b)^(1/4)/x), -1) - sqrt(-b*x^4 + a)*x)/b

Sympy [A] (verification not implemented)

Time = 0.48 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.51 \[ \int \frac {x^4}{\sqrt {a-b x^4}} \, dx=\frac {x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {9}{4}\right )} \]

[In]

integrate(x**4/(-b*x**4+a)**(1/2),x)

[Out]

x**5*gamma(5/4)*hyper((1/2, 5/4), (9/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(9/4))

Maxima [F]

\[ \int \frac {x^4}{\sqrt {a-b x^4}} \, dx=\int { \frac {x^{4}}{\sqrt {-b x^{4} + a}} \,d x } \]

[In]

integrate(x^4/(-b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^4/sqrt(-b*x^4 + a), x)

Giac [F]

\[ \int \frac {x^4}{\sqrt {a-b x^4}} \, dx=\int { \frac {x^{4}}{\sqrt {-b x^{4} + a}} \,d x } \]

[In]

integrate(x^4/(-b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^4/sqrt(-b*x^4 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt {a-b x^4}} \, dx=\int \frac {x^4}{\sqrt {a-b\,x^4}} \,d x \]

[In]

int(x^4/(a - b*x^4)^(1/2),x)

[Out]

int(x^4/(a - b*x^4)^(1/2), x)