3.9.46 \(\int \frac {x^{10}}{\sqrt {a-b x^4}} \, dx\) [846]

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

[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 = 0.07, 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 = {327, 313, 230, 227, 1214, 1213, 435} \begin {gather*} -\frac {7 a^{11/4} \sqrt {1-\frac {b x^4}{a}} F\left (\left .\text {ArcSin}\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 .\text {ArcSin}\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} \end {gather*}

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 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 313

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 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 435

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

Rule 1213

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 1214

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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.04, size = 81, normalized size = 0.51 \begin {gather*} \frac {x^3 \left (-7 a^2+2 a b x^4+5 b^2 x^8+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 )\right )}{45 b^2 \sqrt {a-b x^4}} \end {gather*}

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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Maple [A]
time = 0.15, size = 126, normalized size = 0.80

method	result	size
risch	\(-\frac {x^{3} \left (5 b \,x^{4}+7 a \right ) \sqrt {-b \,x^{4}+a}}{45 b^{2}}-\frac {7 a^{\frac {5}{2}} \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 b^{\frac {5}{2}} \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(117\)
default	\(-\frac {x^{7} \sqrt {-b \,x^{4}+a}}{9 b}-\frac {7 a \,x^{3} \sqrt {-b \,x^{4}+a}}{45 b^{2}}-\frac {7 a^{\frac {5}{2}} \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 b^{\frac {5}{2}} \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(126\)
elliptic	\(-\frac {x^{7} \sqrt {-b \,x^{4}+a}}{9 b}-\frac {7 a \,x^{3} \sqrt {-b \,x^{4}+a}}{45 b^{2}}-\frac {7 a^{\frac {5}{2}} \sqrt {1-\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \sqrt {1+\frac {x^{2} \sqrt {b}}{\sqrt {a}}}\, \left (\EllipticF \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )-\EllipticE \left (x \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}, i\right )\right )}{15 b^{\frac {5}{2}} \sqrt {\frac {\sqrt {b}}{\sqrt {a}}}\, \sqrt {-b \,x^{4}+a}}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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-
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)-EllipticE(x*(1/a^(1/2)*b^(1/2))^(1/2),I))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 0.08, size = 105, normalized size = 0.66 \begin {gather*} -\frac {21 \, a^{2} \sqrt {-b} x \left (\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 21 \, a^{2} \sqrt {-b} x \left (\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (5 \, b^{2} x^{8} + 7 \, a b x^{4} + 21 \, a^{2}\right )} \sqrt {-b x^{4} + a}}{45 \, b^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/45*(21*a^2*sqrt(-b)*x*(a/b)^(3/4)*elliptic_e(arcsin((a/b)^(1/4)/x), -1) - 21*a^2*sqrt(-b)*x*(a/b)^(3/4)*ell
iptic_f(arcsin((a/b)^(1/4)/x), -1) + (5*b^2*x^8 + 7*a*b*x^4 + 21*a^2)*sqrt(-b*x^4 + a))/(b^3*x)

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Sympy [A]
time = 0.55, size = 39, normalized size = 0.25 \begin {gather*} \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 )} \end {gather*}

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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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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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