\(\int \frac {x^4}{(-2+3 x^2)^{3/4}} \, dx\) [906]

Optimal result

Integrand size = 15, antiderivative size = 120 \[ \int \frac {x^4}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {8}{63} x \sqrt [4]{-2+3 x^2}+\frac {2}{21} x^3 \sqrt [4]{-2+3 x^2}+\frac {8\ 2^{3/4} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{63 \sqrt {3} x} \]

[Out]

8/63*x*(3*x^2-2)^(1/4)+2/21*x^3*(3*x^2-2)^(1/4)+8/189*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 = 0.06 (sec) , antiderivative size = 120, 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 = {327, 240, 226} \[ \int \frac {x^4}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {8\ 2^{3/4} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{63 \sqrt {3} x}+\frac {8}{63} \sqrt [4]{3 x^2-2} x+\frac {2}{21} \sqrt [4]{3 x^2-2} x^3 \]

[In]

Int[x^4/(-2 + 3*x^2)^(3/4),x]

[Out]

(8*x*(-2 + 3*x^2)^(1/4))/63 + (2*x^3*(-2 + 3*x^2)^(1/4))/21 + (8*2^(3/4)*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])/(63*Sqrt[3]*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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {2}{21} x^3 \sqrt [4]{-2+3 x^2}+\frac {4}{7} \int \frac {x^2}{\left (-2+3 x^2\right )^{3/4}} \, dx \\ & = \frac {8}{63} x \sqrt [4]{-2+3 x^2}+\frac {2}{21} x^3 \sqrt [4]{-2+3 x^2}+\frac {16}{63} \int \frac {1}{\left (-2+3 x^2\right )^{3/4}} \, dx \\ & = \frac {8}{63} x \sqrt [4]{-2+3 x^2}+\frac {2}{21} x^3 \sqrt [4]{-2+3 x^2}+\frac {\left (16 \sqrt {\frac {2}{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 )}{63 x} \\ & = \frac {8}{63} x \sqrt [4]{-2+3 x^2}+\frac {2}{21} x^3 \sqrt [4]{-2+3 x^2}+\frac {8\ 2^{3/4} \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 )}{63 \sqrt {3} x} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 5.35 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.52 \[ \int \frac {x^4}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {2 x \left (-8+6 x^2+9 x^4+4 \sqrt [4]{2} \left (2-3 x^2\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {3 x^2}{2}\right )\right )}{63 \left (-2+3 x^2\right )^{3/4}} \]

[In]

Integrate[x^4/(-2 + 3*x^2)^(3/4),x]

[Out]

(2*x*(-8 + 6*x^2 + 9*x^4 + 4*2^(1/4)*(2 - 3*x^2)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, (3*x^2)/2]))/(63*(-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 = 2.19 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.35

method	result	size
meijerg	\(\frac {2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} x^{5} {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {3}{4},\frac {5}{2};\frac {7}{2};\frac {3 x^{2}}{2}\right )}{10 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}}}\)	\(42\)
risch	\(\frac {2 x \left (3 x^{2}+4\right ) \left (3 x^{2}-2\right )^{\frac {1}{4}}}{63}+\frac {8 \,2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} x {}_{2}^{}{\moversetsp {}{\mundersetsp {}{F_{1}^{}}}}\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\frac {3 x^{2}}{2}\right )}{63 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}}}\)	\(60\)

[In]

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

[Out]

1/10*2^(1/4)/signum(-1+3/2*x^2)^(3/4)*(-signum(-1+3/2*x^2))^(3/4)*x^5*hypergeom([3/4,5/2],[7/2],3/2*x^2)

Fricas [F]

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

[In]

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

[Out]

integral(x^4/(3*x^2 - 2)^(3/4), x)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.44 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.26 \[ \int \frac {x^4}{\left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{2} x^{5} e^{- \frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{10} \]

[In]

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

[Out]

2**(1/4)*x**5*exp(-3*I*pi/4)*hyper((3/4, 5/2), (7/2,), 3*x**2/2)/10

Maxima [F]

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

[In]

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

[Out]

integrate(x^4/(3*x^2 - 2)^(3/4), x)

Giac [F]

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

[In]

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

[Out]

integrate(x^4/(3*x^2 - 2)^(3/4), x)

Mupad [F(-1)]

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

[In]

int(x^4/(3*x^2 - 2)^(3/4),x)

[Out]

int(x^4/(3*x^2 - 2)^(3/4), x)