\(\int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx\) [2086]

Optimal result

Integrand size = 15, antiderivative size = 110 \[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx=\frac {\sqrt {a+\frac {b}{x^4}} x^3}{3 a}+\frac {b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+\frac {b}{x^4}}} \]

[Out]

1/3*x^3*(a+b/x^4)^(1/2)/a+1/6*b^(3/4)*(cos(2*arccot(a^(1/4)*x/b^(1/4)))^2)^(1/2)/cos(2*arccot(a^(1/4)*x/b^(1/4
)))*EllipticF(sin(2*arccot(a^(1/4)*x/b^(1/4))),1/2*2^(1/2))*(a^(1/2)+b^(1/2)/x^2)*((a+b/x^4)/(a^(1/2)+b^(1/2)/
x^2)^2)^(1/2)/a^(5/4)/(a+b/x^4)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {342, 331, 226} \[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx=\frac {b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) \operatorname {EllipticF}\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+\frac {b}{x^4}}}+\frac {x^3 \sqrt {a+\frac {b}{x^4}}}{3 a} \]

[In]

Int[x^2/Sqrt[a + b/x^4],x]

[Out]

(Sqrt[a + b/x^4]*x^3)/(3*a) + (b^(3/4)*Sqrt[(a + b/x^4)/(Sqrt[a] + Sqrt[b]/x^2)^2]*(Sqrt[a] + Sqrt[b]/x^2)*Ell
ipticF[2*ArcCot[(a^(1/4)*x)/b^(1/4)], 1/2])/(6*a^(5/4)*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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{x^4 \sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right ) \\ & = \frac {\sqrt {a+\frac {b}{x^4}} x^3}{3 a}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\frac {1}{x}\right )}{3 a} \\ & = \frac {\sqrt {a+\frac {b}{x^4}} x^3}{3 a}+\frac {b^{3/4} \sqrt {\frac {a+\frac {b}{x^4}}{\left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right )^2}} \left (\sqrt {a}+\frac {\sqrt {b}}{x^2}\right ) F\left (2 \cot ^{-1}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b}}\right )|\frac {1}{2}\right )}{6 a^{5/4} \sqrt {a+\frac {b}{x^4}}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.58 \[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx=\frac {b+a x^4-b \sqrt {1+\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {a x^4}{b}\right )}{3 a \sqrt {a+\frac {b}{x^4}} x} \]

[In]

Integrate[x^2/Sqrt[a + b/x^4],x]

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.01

method	result	size
risch	\(\frac {a \,x^{4}+b}{3 a x \sqrt {\frac {a \,x^{4}+b}{x^{4}}}}-\frac {b \sqrt {1-\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, \sqrt {1+\frac {i \sqrt {a}\, x^{2}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )}{3 a \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2}}\)	\(111\)
default	\(\frac {\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, a \,x^{5}-b \sqrt {-\frac {i \sqrt {a}\, x^{2}-\sqrt {b}}{\sqrt {b}}}\, \sqrt {\frac {i \sqrt {a}\, x^{2}+\sqrt {b}}{\sqrt {b}}}\, F\left (x \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}, i\right )+\sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}\, b x}{3 \sqrt {\frac {a \,x^{4}+b}{x^{4}}}\, x^{2} a \sqrt {\frac {i \sqrt {a}}{\sqrt {b}}}}\)	\(124\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.46 \[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx=\frac {x^{3} \sqrt {\frac {a x^{4} + b}{x^{4}}} - \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {b}{a}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1)}{3 \, a} \]

[In]

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

[Out]

1/3*(x^3*sqrt((a*x^4 + b)/x^4) - sqrt(a)*(-b/a)^(3/4)*elliptic_f(arcsin((-b/a)^(1/4)/x), -1))/a

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.58 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.38 \[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx=- \frac {x^{3} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {1}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b e^{i \pi }}{a x^{4}}} \right )}}{4 \sqrt {a} \Gamma \left (\frac {1}{4}\right )} \]

[In]

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

[Out]

-x**3*gamma(-3/4)*hyper((-3/4, 1/2), (1/4,), b*exp_polar(I*pi)/(a*x**4))/(4*sqrt(a)*gamma(1/4))

Maxima [F]

\[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx=\int { \frac {x^{2}}{\sqrt {a + \frac {b}{x^{4}}}} \,d x } \]

[In]

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

[Out]

integrate(x^2/sqrt(a + b/x^4), x)

Giac [F]

\[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx=\int { \frac {x^{2}}{\sqrt {a + \frac {b}{x^{4}}}} \,d x } \]

[In]

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

[Out]

integrate(x^2/sqrt(a + b/x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \, dx=\int \frac {x^2}{\sqrt {a+\frac {b}{x^4}}} \,d x \]

[In]

int(x^2/(a + b/x^4)^(1/2),x)

[Out]

int(x^2/(a + b/x^4)^(1/2), x)