\(\int \frac {x^6}{\sqrt {a+b x^4}} \, dx\) [827]

Optimal result

Integrand size = 15, antiderivative size = 237 \[ \int \frac {x^6}{\sqrt {a+b x^4}} \, dx=\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {3 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {3 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{10 b^{7/4} \sqrt {a+b x^4}} \]

[Out]

1/5*x^3*(b*x^4+a)^(1/2)/b-3/5*a*x*(b*x^4+a)^(1/2)/b^(3/2)/(a^(1/2)+x^2*b^(1/2))+3/5*a^(5/4)*(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^(7/4)/(b*x^4+a)^(1/2)-3/10*a^(5/4)*(cos(
2*arctan(b^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x/a^(1/4)))*EllipticF(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^(7/4)/(b*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {327, 311, 226, 1210} \[ \int \frac {x^6}{\sqrt {a+b x^4}} \, dx=-\frac {3 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{10 b^{7/4} \sqrt {a+b x^4}}+\frac {3 a^{5/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {3 a x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {x^3 \sqrt {a+b x^4}}{5 b} \]

[In]

Int[x^6/Sqrt[a + b*x^4],x]

[Out]

(x^3*Sqrt[a + b*x^4])/(5*b) - (3*a*x*Sqrt[a + b*x^4])/(5*b^(3/2)*(Sqrt[a] + Sqrt[b]*x^2)) + (3*a^(5/4)*(Sqrt[a
] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(5
*b^(7/4)*Sqrt[a + b*x^4]) - (3*a^(5/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*Ell
ipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(10*b^(7/4)*Sqrt[a + b*x^4])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

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 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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = \frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {(3 a) \int \frac {x^2}{\sqrt {a+b x^4}} \, dx}{5 b} \\ & = \frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {\left (3 a^{3/2}\right ) \int \frac {1}{\sqrt {a+b x^4}} \, dx}{5 b^{3/2}}+\frac {\left (3 a^{3/2}\right ) \int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx}{5 b^{3/2}} \\ & = \frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a x \sqrt {a+b x^4}}{5 b^{3/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}+\frac {3 a^{5/4} \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 )}{5 b^{7/4} \sqrt {a+b x^4}}-\frac {3 a^{5/4} \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 )}{10 b^{7/4} \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 = 10.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.27 \[ \int \frac {x^6}{\sqrt {a+b x^4}} \, dx=\frac {x^3 \left (a+b x^4-a \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{5 b \sqrt {a+b x^4}} \]

[In]

Integrate[x^6/Sqrt[a + b*x^4],x]

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 4.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.49

method	result	size
default	\(\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(115\)
risch	\(\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(115\)
elliptic	\(\frac {x^{3} \sqrt {b \,x^{4}+a}}{5 b}-\frac {3 i a^{\frac {3}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{5 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\)	\(115\)

[In]

int(x^6/(b*x^4+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.38 \[ \int \frac {x^6}{\sqrt {a+b x^4}} \, dx=-\frac {3 \, a \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 3 \, a \sqrt {b} x \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - \sqrt {b x^{4} + a} {\left (b x^{4} - 3 \, a\right )}}{5 \, b^{2} x} \]

[In]

integrate(x^6/(b*x^4+a)^(1/2),x, algorithm="fricas")

[Out]

-1/5*(3*a*sqrt(b)*x*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b)^(1/4)/x), -1) - 3*a*sqrt(b)*x*(-a/b)^(3/4)*elliptic_
f(arcsin((-a/b)^(1/4)/x), -1) - sqrt(b*x^4 + a)*(b*x^4 - 3*a))/(b^2*x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.48 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.16 \[ \int \frac {x^6}{\sqrt {a+b x^4}} \, dx=\frac {x^{7} \Gamma \left (\frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {7}{4} \\ \frac {11}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {11}{4}\right )} \]

[In]

integrate(x**6/(b*x**4+a)**(1/2),x)

[Out]

x**7*gamma(7/4)*hyper((1/2, 7/4), (11/4,), b*x**4*exp_polar(I*pi)/a)/(4*sqrt(a)*gamma(11/4))

Maxima [F]

\[ \int \frac {x^6}{\sqrt {a+b x^4}} \, dx=\int { \frac {x^{6}}{\sqrt {b x^{4} + a}} \,d x } \]

[In]

integrate(x^6/(b*x^4+a)^(1/2),x, algorithm="maxima")

[Out]

integrate(x^6/sqrt(b*x^4 + a), x)

Giac [F]

\[ \int \frac {x^6}{\sqrt {a+b x^4}} \, dx=\int { \frac {x^{6}}{\sqrt {b x^{4} + a}} \,d x } \]

[In]

integrate(x^6/(b*x^4+a)^(1/2),x, algorithm="giac")

[Out]

integrate(x^6/sqrt(b*x^4 + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{\sqrt {a+b x^4}} \, dx=\int \frac {x^6}{\sqrt {b\,x^4+a}} \,d x \]

[In]

int(x^6/(a + b*x^4)^(1/2),x)

[Out]

int(x^6/(a + b*x^4)^(1/2), x)