\(\int \frac {1}{x^4 (-2+3 x^2)^{3/4}} \, dx\) [910]

Optimal result

Integrand size = 15, antiderivative size = 122 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{-2+3 x^2}}{6 x^3}+\frac {5 \sqrt [4]{-2+3 x^2}}{8 x}+\frac {5 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{16 \sqrt [4]{2} x} \]

[Out]

1/6*(3*x^2-2)^(1/4)/x^3+5/8*(3*x^2-2)^(1/4)/x+5/32*2^(3/4)*(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 = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {331, 240, 226} \[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\frac {5 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{16 \sqrt [4]{2} x}+\frac {5 \sqrt [4]{3 x^2-2}}{8 x}+\frac {\sqrt [4]{3 x^2-2}}{6 x^3} \]

[In]

Int[1/(x^4*(-2 + 3*x^2)^(3/4)),x]

[Out]

(-2 + 3*x^2)^(1/4)/(6*x^3) + (5*(-2 + 3*x^2)^(1/4))/(8*x) + (5*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])/(16*2^(1/4)*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 240

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Dist[2*(Sqrt[(-b)*(x^2/a)]/(b*x)), Subst[Int[1/Sqrt[1 - x^4/a],
x], x, (a + b*x^2)^(1/4)], x] /; FreeQ[{a, b}, x] && NegQ[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt [4]{-2+3 x^2}}{6 x^3}+\frac {5}{4} \int \frac {1}{x^2 \left (-2+3 x^2\right )^{3/4}} \, dx \\ & = \frac {\sqrt [4]{-2+3 x^2}}{6 x^3}+\frac {5 \sqrt [4]{-2+3 x^2}}{8 x}+\frac {15}{16} \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx \\ & = \frac {\sqrt [4]{-2+3 x^2}}{6 x^3}+\frac {5 \sqrt [4]{-2+3 x^2}}{8 x}+\frac {\left (5 \sqrt {\frac {3}{2}} \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 {\sqrt [4]{-2+3 x^2}}{6 x^3}+\frac {5 \sqrt [4]{-2+3 x^2}}{8 x}+\frac {5 \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 )}{16 \sqrt [4]{2} 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 = 0.39 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=-\frac {\left (1-\frac {3 x^2}{2}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},-\frac {1}{2},\frac {3 x^2}{2}\right )}{3 x^3 \left (-2+3 x^2\right )^{3/4}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 2.20 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.34

method	result	size
meijerg	\(-\frac {2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (-\frac {3}{2},\frac {3}{4};-\frac {1}{2};\frac {3 x^{2}}{2}\right )}{6 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}} x^{3}}\)	\(42\)
risch	\(\frac {45 x^{4}-18 x^{2}-8}{24 x^{3} \left (3 x^{2}-2\right )^{\frac {3}{4}}}+\frac {15 \,2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{32 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}}}\)	\(67\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((3*x^2 - 2)^(1/4)/(3*x^6 - 2*x^4), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.55 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{2} e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{6 x^{3}} \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(x^4*(3*x^2 - 2)^(3/4)),x)

[Out]

int(1/(x^4*(3*x^2 - 2)^(3/4)), x)