3.1213 \(\int \frac {x^9}{\sqrt [4]{a-b x^4}} \, dx\)

Optimal. Leaf size=108 \[ \frac {4 a^{5/2} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a-b x^4}}-\frac {2 a x^2 \left (a-b x^4\right )^{3/4}}{15 b^2}-\frac {x^6 \left (a-b x^4\right )^{3/4}}{9 b} \]

[Out]

-2/15*a*x^2*(-b*x^4+a)^(3/4)/b^2-1/9*x^6*(-b*x^4+a)^(3/4)/b+4/15*a^(5/2)*(1-b*x^4/a)^(1/4)*(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)))*EllipticE(sin(1/2*arcsin(x^2*b^(1/2)/a^(1/2))
),2^(1/2))/b^(5/2)/(-b*x^4+a)^(1/4)

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Rubi [A]  time = 0.07, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {275, 321, 229, 228} \[ \frac {4 a^{5/2} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a-b x^4}}-\frac {2 a x^2 \left (a-b x^4\right )^{3/4}}{15 b^2}-\frac {x^6 \left (a-b x^4\right )^{3/4}}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(a - b*x^4)^(1/4),x]

[Out]

(-2*a*x^2*(a - b*x^4)^(3/4))/(15*b^2) - (x^6*(a - b*x^4)^(3/4))/(9*b) + (4*a^(5/2)*(1 - (b*x^4)/a)^(1/4)*Ellip
ticE[ArcSin[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(15*b^(5/2)*(a - b*x^4)^(1/4))

Rule 228

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

Rule 229

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

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 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]

Rubi steps

\begin {align*} \int \frac {x^9}{\sqrt [4]{a-b x^4}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x^4}{\sqrt [4]{a-b x^2}} \, dx,x,x^2\right )\\ &=-\frac {x^6 \left (a-b x^4\right )^{3/4}}{9 b}+\frac {a \operatorname {Subst}\left (\int \frac {x^2}{\sqrt [4]{a-b x^2}} \, dx,x,x^2\right )}{3 b}\\ &=-\frac {2 a x^2 \left (a-b x^4\right )^{3/4}}{15 b^2}-\frac {x^6 \left (a-b x^4\right )^{3/4}}{9 b}+\frac {\left (2 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{a-b x^2}} \, dx,x,x^2\right )}{15 b^2}\\ &=-\frac {2 a x^2 \left (a-b x^4\right )^{3/4}}{15 b^2}-\frac {x^6 \left (a-b x^4\right )^{3/4}}{9 b}+\frac {\left (2 a^2 \sqrt [4]{1-\frac {b x^4}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}} \, dx,x,x^2\right )}{15 b^2 \sqrt [4]{a-b x^4}}\\ &=-\frac {2 a x^2 \left (a-b x^4\right )^{3/4}}{15 b^2}-\frac {x^6 \left (a-b x^4\right )^{3/4}}{9 b}+\frac {4 a^{5/2} \sqrt [4]{1-\frac {b x^4}{a}} E\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{15 b^{5/2} \sqrt [4]{a-b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 80, normalized size = 0.74 \[ \frac {x^2 \left (6 a^2 \sqrt [4]{1-\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {b x^4}{a}\right )-6 a^2+a b x^4+5 b^2 x^8\right )}{45 b^2 \sqrt [4]{a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a - b*x^4)^(1/4),x]

[Out]

(x^2*(-6*a^2 + a*b*x^4 + 5*b^2*x^8 + 6*a^2*(1 - (b*x^4)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (b*x^4)/a]))
/(45*b^2*(a - b*x^4)^(1/4))

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}} x^{9}}{b x^{4} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(-b*x^4 + a)^(3/4)*x^9/(b*x^4 - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{9}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^9/(-b*x^4 + a)^(1/4), x)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {x^{9}}{\left (-b \,x^{4}+a \right )^{\frac {1}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{9}}{{\left (-b x^{4} + a\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^9/(-b*x^4 + a)^(1/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^9}{{\left (a-b\,x^4\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(a - b*x^4)^(1/4),x)

[Out]

int(x^9/(a - b*x^4)^(1/4), x)

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sympy [C]  time = 1.54, size = 29, normalized size = 0.27 \[ \frac {x^{10} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{10 \sqrt [4]{a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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