\(\int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx\) [896]

Optimal result

Integrand size = 15, antiderivative size = 242 \[ \int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx=\frac {\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac {9 x \sqrt [4]{-2+3 x^2}}{8 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4\ 2^{3/4} x}-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{8\ 2^{3/4} x} \]

[Out]

1/6*(3*x^2-2)^(3/4)/x^3+3/8*(3*x^2-2)^(3/4)/x-9/8*x*(3*x^2-2)^(1/4)/(2^(1/2)+(3*x^2-2)^(1/2))+3/8*2^(1/4)*(cos
(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))^2)^(1/2)/cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))*EllipticE(sin(2*ar
ctan(1/2*(3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1/2))*(2^(1/2)+(3*x^2-2)^(1/2))*(x^2/(2^(1/2)+(3*x^2-2)^(1/2))^2)^(1
/2)/x*3^(1/2)-3/16*2^(1/4)*(cos(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4)))^2)^(1/2)/cos(2*arctan(1/2*(3*x^2-2)^(1/
4)*2^(3/4)))*EllipticF(sin(2*arctan(1/2*(3*x^2-2)^(1/4)*2^(3/4))),1/2*2^(1/2))*(2^(1/2)+(3*x^2-2)^(1/2))*(x^2/
(2^(1/2)+(3*x^2-2)^(1/2))^2)^(1/2)/x*3^(1/2)

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {331, 236, 311, 226, 1210} \[ \int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx=-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{8\ 2^{3/4} x}+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) E\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4\ 2^{3/4} x}-\frac {9 \sqrt [4]{3 x^2-2} x}{8 \left (\sqrt {3 x^2-2}+\sqrt {2}\right )}+\frac {3 \left (3 x^2-2\right )^{3/4}}{8 x}+\frac {\left (3 x^2-2\right )^{3/4}}{6 x^3} \]

[In]

Int[1/(x^4*(-2 + 3*x^2)^(1/4)),x]

[Out]

(-2 + 3*x^2)^(3/4)/(6*x^3) + (3*(-2 + 3*x^2)^(3/4))/(8*x) - (9*x*(-2 + 3*x^2)^(1/4))/(8*(Sqrt[2] + Sqrt[-2 + 3
*x^2])) + (3*Sqrt[3]*Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2] + Sqrt[-2 + 3*x^2])*EllipticE[2*ArcTan[
(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(4*2^(3/4)*x) - (3*Sqrt[3]*Sqrt[x^2/(Sqrt[2] + Sqrt[-2 + 3*x^2])^2]*(Sqrt[2
] + Sqrt[-2 + 3*x^2])*EllipticF[2*ArcTan[(-2 + 3*x^2)^(1/4)/2^(1/4)], 1/2])/(8*2^(3/4)*x)

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 236

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[2*(Sqrt[(-b)*(x^2/a)]/(b*x)), Subst[Int[x^2/Sqrt[1 - x^4/a
], x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[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 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 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 {\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3}{4} \int \frac {1}{x^2 \sqrt [4]{-2+3 x^2}} \, dx \\ & = \frac {\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac {9}{16} \int \frac {1}{\sqrt [4]{-2+3 x^2}} \, dx \\ & = \frac {\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac {\left (3 \sqrt {\frac {3}{2}} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{8 x} \\ & = \frac {\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{8 x}+\frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1-\frac {x^2}{\sqrt {2}}}{\sqrt {1+\frac {x^4}{2}}} \, dx,x,\sqrt [4]{-2+3 x^2}\right )}{8 x} \\ & = \frac {\left (-2+3 x^2\right )^{3/4}}{6 x^3}+\frac {3 \left (-2+3 x^2\right )^{3/4}}{8 x}-\frac {9 x \sqrt [4]{-2+3 x^2}}{8 \left (\sqrt {2}+\sqrt {-2+3 x^2}\right )}+\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{4\ 2^{3/4} x}-\frac {3 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right )|\frac {1}{2}\right )}{8\ 2^{3/4} x} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.20 \[ \int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx=-\frac {\sqrt [4]{1-\frac {3 x^2}{2}} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {1}{4},-\frac {1}{2},\frac {3 x^2}{2}\right )}{3 x^3 \sqrt [4]{-2+3 x^2}} \]

[In]

Integrate[1/(x^4*(-2 + 3*x^2)^(1/4)),x]

[Out]

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

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 2.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.17

method	result	size
meijerg	\(-\frac {2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{2},\frac {1}{4};-\frac {1}{2};\frac {3 x^{2}}{2}\right )}{6 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}} x^{3}}\)	\(42\)
risch	\(\frac {27 x^{4}-6 x^{2}-8}{24 x^{3} \left (3 x^{2}-2\right )^{\frac {1}{4}}}-\frac {9 \,2^{\frac {3}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {1}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{32 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {1}{4}}}\)	\(67\)

[In]

int(1/x^4/(3*x^2-2)^(1/4),x,method=_RETURNVERBOSE)

[Out]

-1/6*2^(3/4)/signum(-1+3/2*x^2)^(1/4)*(-signum(-1+3/2*x^2))^(1/4)/x^3*hypergeom([-3/2,1/4],[-1/2],3/2*x^2)

Fricas [F]

\[ \int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{4}} \,d x } \]

[In]

integrate(1/x^4/(3*x^2-2)^(1/4),x, algorithm="fricas")

[Out]

integral((3*x^2 - 2)^(3/4)/(3*x^6 - 2*x^4), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.52 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.14 \[ \int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx=\frac {2^{\frac {3}{4}} e^{\frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {1}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{6 x^{3}} \]

[In]

integrate(1/x**4/(3*x**2-2)**(1/4),x)

[Out]

2**(3/4)*exp(3*I*pi/4)*hyper((-3/2, 1/4), (-1/2,), 3*x**2/2)/(6*x**3)

Maxima [F]

\[ \int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{4}} \,d x } \]

[In]

integrate(1/x^4/(3*x^2-2)^(1/4),x, algorithm="maxima")

[Out]

integrate(1/((3*x^2 - 2)^(1/4)*x^4), x)

Giac [F]

\[ \int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {1}{4}} x^{4}} \,d x } \]

[In]

integrate(1/x^4/(3*x^2-2)^(1/4),x, algorithm="giac")

[Out]

integrate(1/((3*x^2 - 2)^(1/4)*x^4), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^4 \sqrt [4]{-2+3 x^2}} \, dx=\int \frac {1}{x^4\,{\left (3\,x^2-2\right )}^{1/4}} \,d x \]

[In]

int(1/(x^4*(3*x^2 - 2)^(1/4)),x)

[Out]

int(1/(x^4*(3*x^2 - 2)^(1/4)), x)