3.846 \(\int \frac {x^{10}}{\sqrt {a-b x^4}} \, dx\)

Optimal. Leaf size=158 \[ -\frac {7 a^{11/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{15 b^{11/4} \sqrt {a-b x^4}}+\frac {7 a^{11/4} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{15 b^{11/4} \sqrt {a-b x^4}}-\frac {7 a x^3 \sqrt {a-b x^4}}{45 b^2}-\frac {x^7 \sqrt {a-b x^4}}{9 b} \]

[Out]

-7/45*a*x^3*(-b*x^4+a)^(1/2)/b^2-1/9*x^7*(-b*x^4+a)^(1/2)/b+7/15*a^(11/4)*EllipticE(b^(1/4)*x/a^(1/4),I)*(1-b*
x^4/a)^(1/2)/b^(11/4)/(-b*x^4+a)^(1/2)-7/15*a^(11/4)*EllipticF(b^(1/4)*x/a^(1/4),I)*(1-b*x^4/a)^(1/2)/b^(11/4)
/(-b*x^4+a)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {321, 307, 224, 221, 1200, 1199, 424} \[ -\frac {7 a^{11/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{15 b^{11/4} \sqrt {a-b x^4}}+\frac {7 a^{11/4} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{15 b^{11/4} \sqrt {a-b x^4}}-\frac {7 a x^3 \sqrt {a-b x^4}}{45 b^2}-\frac {x^7 \sqrt {a-b x^4}}{9 b} \]

Antiderivative was successfully verified.

[In]

Int[x^10/Sqrt[a - b*x^4],x]

[Out]

(-7*a*x^3*Sqrt[a - b*x^4])/(45*b^2) - (x^7*Sqrt[a - b*x^4])/(9*b) + (7*a^(11/4)*Sqrt[1 - (b*x^4)/a]*EllipticE[
ArcSin[(b^(1/4)*x)/a^(1/4)], -1])/(15*b^(11/4)*Sqrt[a - b*x^4]) - (7*a^(11/4)*Sqrt[1 - (b*x^4)/a]*EllipticF[Ar
cSin[(b^(1/4)*x)/a^(1/4)], -1])/(15*b^(11/4)*Sqrt[a - b*x^4])

Rule 221

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 224

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 307

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

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 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 1199

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[d/Sqrt[a], Int[Sqrt[1 + (e*x^2)/d]/Sqrt
[1 - (e*x^2)/d], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] && GtQ[a, 0]

Rule 1200

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> Dist[Sqrt[1 + (c*x^4)/a]/Sqrt[a + c*x^4], In
t[(d + e*x^2)/Sqrt[1 + (c*x^4)/a], x], x] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && EqQ[c*d^2 + a*e^2, 0] &&
!GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {x^{10}}{\sqrt {a-b x^4}} \, dx &=-\frac {x^7 \sqrt {a-b x^4}}{9 b}+\frac {(7 a) \int \frac {x^6}{\sqrt {a-b x^4}} \, dx}{9 b}\\ &=-\frac {7 a x^3 \sqrt {a-b x^4}}{45 b^2}-\frac {x^7 \sqrt {a-b x^4}}{9 b}+\frac {\left (7 a^2\right ) \int \frac {x^2}{\sqrt {a-b x^4}} \, dx}{15 b^2}\\ &=-\frac {7 a x^3 \sqrt {a-b x^4}}{45 b^2}-\frac {x^7 \sqrt {a-b x^4}}{9 b}-\frac {\left (7 a^{5/2}\right ) \int \frac {1}{\sqrt {a-b x^4}} \, dx}{15 b^{5/2}}+\frac {\left (7 a^{5/2}\right ) \int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a-b x^4}} \, dx}{15 b^{5/2}}\\ &=-\frac {7 a x^3 \sqrt {a-b x^4}}{45 b^2}-\frac {x^7 \sqrt {a-b x^4}}{9 b}-\frac {\left (7 a^{5/2} \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1}{\sqrt {1-\frac {b x^4}{a}}} \, dx}{15 b^{5/2} \sqrt {a-b x^4}}+\frac {\left (7 a^{5/2} \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {1+\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {1-\frac {b x^4}{a}}} \, dx}{15 b^{5/2} \sqrt {a-b x^4}}\\ &=-\frac {7 a x^3 \sqrt {a-b x^4}}{45 b^2}-\frac {x^7 \sqrt {a-b x^4}}{9 b}-\frac {7 a^{11/4} \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 )}{15 b^{11/4} \sqrt {a-b x^4}}+\frac {\left (7 a^{5/2} \sqrt {1-\frac {b x^4}{a}}\right ) \int \frac {\sqrt {1+\frac {\sqrt {b} x^2}{\sqrt {a}}}}{\sqrt {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}} \, dx}{15 b^{5/2} \sqrt {a-b x^4}}\\ &=-\frac {7 a x^3 \sqrt {a-b x^4}}{45 b^2}-\frac {x^7 \sqrt {a-b x^4}}{9 b}+\frac {7 a^{11/4} \sqrt {1-\frac {b x^4}{a}} E\left (\left .\sin ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )\right |-1\right )}{15 b^{11/4} \sqrt {a-b x^4}}-\frac {7 a^{11/4} \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 )}{15 b^{11/4} \sqrt {a-b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.04, size = 81, normalized size = 0.51 \[ \frac {x^3 \left (7 a^2 \sqrt {1-\frac {b x^4}{a}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};\frac {b x^4}{a}\right )-7 a^2+2 a b x^4+5 b^2 x^8\right )}{45 b^2 \sqrt {a-b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^10/Sqrt[a - b*x^4],x]

[Out]

(x^3*(-7*a^2 + 2*a*b*x^4 + 5*b^2*x^8 + 7*a^2*Sqrt[1 - (b*x^4)/a]*Hypergeometric2F1[1/2, 3/4, 7/4, (b*x^4)/a]))
/(45*b^2*Sqrt[a - b*x^4])

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-b x^{4} + a} x^{10}}{b x^{4} - a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-b*x^4 + a)*x^10/(b*x^4 - a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{\sqrt {-b x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^10/sqrt(-b*x^4 + a), x)

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maple [A]  time = 0.01, size = 126, normalized size = 0.80 \[ -\frac {\sqrt {-b \,x^{4}+a}\, x^{7}}{9 b}-\frac {7 \sqrt {-b \,x^{4}+a}\, a \,x^{3}}{45 b^{2}}-\frac {7 \sqrt {-\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {\sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, x , i\right )\right ) a^{\frac {5}{2}}}{15 \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}\, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(-b*x^4+a)^(1/2),x)

[Out]

-1/9*x^7*(-b*x^4+a)^(1/2)/b-7/45*a*x^3*(-b*x^4+a)^(1/2)/b^2-7/15*a^(5/2)/b^(5/2)/(1/a^(1/2)*b^(1/2))^(1/2)*(-1
/a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(1/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(-b*x^4+a)^(1/2)*(EllipticF((1/a^(1/2)*b^(1/2))^
(1/2)*x,I)-EllipticE((1/a^(1/2)*b^(1/2))^(1/2)*x,I))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{10}}{\sqrt {-b x^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10/(-b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^10/sqrt(-b*x^4 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^{10}}{\sqrt {a-b\,x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10/(a - b*x^4)^(1/2),x)

[Out]

int(x^10/(a - b*x^4)^(1/2), x)

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sympy [A]  time = 2.25, size = 39, normalized size = 0.25 \[ \frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10/(-b*x**4+a)**(1/2),x)

[Out]

x**11*gamma(11/4)*hyper((1/2, 11/4), (15/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*sqrt(a)*gamma(15/4))

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