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

Optimal. Leaf size=261 \[ \frac {7 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 b^{11/4} \sqrt {a+b x^4}}-\frac {7 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^4}}+\frac {7 a^2 x \sqrt {a+b x^4}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {7 a x^3 \sqrt {a+b x^4}}{45 b^2}+\frac {x^7 \sqrt {a+b x^4}}{9 b} \]

[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^2*x*(b*x^4+a)^(1/2)/b^(5/2)/(a^(1/2)+x^2*b^(1
/2))-7/15*a^(9/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticE(sin(2*
arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2)^(1/2)/b^(11/
4)/(b*x^4+a)^(1/2)+7/30*a^(9/4)*(cos(2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*El
lipticF(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x^2*b^(1/2))*((b*x^4+a)/(a^(1/2)+x^2*b^(1/2))^2
)^(1/2)/b^(11/4)/(b*x^4+a)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 261, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {321, 305, 220, 1196} \[ \frac {7 a^2 x \sqrt {a+b x^4}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {7 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 b^{11/4} \sqrt {a+b x^4}}-\frac {7 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^4}}-\frac {7 a x^3 \sqrt {a+b x^4}}{45 b^2}+\frac {x^7 \sqrt {a+b x^4}}{9 b} \]

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^2*x*Sqrt[a + b*x^4])/(15*b^(5/2)*(Sqr
t[a] + Sqrt[b]*x^2)) - (7*a^(9/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Elliptic
E[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(15*b^(11/4)*Sqrt[a + b*x^4]) + (7*a^(9/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt
[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(30*b^(11/4)*Sqrt[a + b
*x^4])

Rule 220

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

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, 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] && PosQ[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 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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 {7 a^2 x \sqrt {a+b x^4}}{15 b^{5/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {7 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{15 b^{11/4} \sqrt {a+b x^4}}+\frac {7 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 b^{11/4} \sqrt {a+b x^4}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 80, normalized size = 0.31 \[ \frac {x^3 \left (7 a^2 \sqrt {\frac {b x^4}{a}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {b x^4}{a}\right )-7 a^2-2 a b x^4+5 b^2 x^8\right )}{45 b^2 \sqrt {a+b x^4}} \]

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 = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{10}}{\sqrt {b x^{4} + a}}, x\right ) \]

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(x^10/sqrt(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 [C]  time = 0.01, size = 133, normalized size = 0.51 \[ \frac {\sqrt {b \,x^{4}+a}\, x^{7}}{9 b}-\frac {7 \sqrt {b \,x^{4}+a}\, a \,x^{3}}{45 b^{2}}+\frac {7 i \sqrt {-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \sqrt {\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}+1}\, \left (-\EllipticE \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )+\EllipticF \left (\sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, x , i\right )\right ) a^{\frac {5}{2}}}{15 \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, b^{\frac {5}{2}}} \]

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*I*a^(5/2)/b^(5/2)/(I/a^(1/2)*b^(1/2))^(1/2)*(-I/
a^(1/2)*b^(1/2)*x^2+1)^(1/2)*(I/a^(1/2)*b^(1/2)*x^2+1)^(1/2)/(b*x^4+a)^(1/2)*(EllipticF((I/a^(1/2)*b^(1/2))^(1
/2)*x,I)-EllipticE((I/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 = -0.00 \[ \int \frac {x^{10}}{\sqrt {b\,x^4+a}} \,d x \]

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 [C]  time = 1.38, size = 37, normalized size = 0.14 \[ \frac {x^{11} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {15}{4}\right )} \]

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(I*pi)/a)/(4*sqrt(a)*gamma(15/4))

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