Integrand size = 15, antiderivative size = 122 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{-2+3 x^2}}{6 x^3}+\frac {5 \sqrt [4]{-2+3 x^2}}{8 x}+\frac {5 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {2}+\sqrt {-2+3 x^2}\right )^2}} \left (\sqrt {2}+\sqrt {-2+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{-2+3 x^2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{16 \sqrt [4]{2} x} \] Output:
1/6*(3*x^2-2)^(1/4)/x^3+5/8*(3*x^2-2)^(1/4)/x+5/32*2^(3/4)*(x^2/(2^(1/2)+( 3*x^2-2)^(1/2))^2)^(1/2)*(2^(1/2)+(3*x^2-2)^(1/2))*InverseJacobiAM(2*arcta n(1/2*(3*x^2-2)^(1/4)*2^(3/4)),1/2*2^(1/2))*3^(1/2)/x
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 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=-\frac {\left (1-\frac {3 x^2}{2}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},\frac {3}{4},-\frac {1}{2},\frac {3 x^2}{2}\right )}{3 x^3 \left (-2+3 x^2\right )^{3/4}} \] Input:
Integrate[1/(x^4*(-2 + 3*x^2)^(3/4)),x]
Output:
-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))
Time = 0.21 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.04, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {264, 264, 232, 761}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{x^4 \left (3 x^2-2\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5}{4} \int \frac {1}{x^2 \left (3 x^2-2\right )^{3/4}}dx+\frac {\sqrt [4]{3 x^2-2}}{6 x^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {5}{4} \left (\frac {3}{4} \int \frac {1}{\left (3 x^2-2\right )^{3/4}}dx+\frac {\sqrt [4]{3 x^2-2}}{2 x}\right )+\frac {\sqrt [4]{3 x^2-2}}{6 x^3}\) |
\(\Big \downarrow \) 232 |
\(\displaystyle \frac {5}{4} \left (\frac {\sqrt {\frac {3}{2}} \sqrt {x^2} \int \frac {1}{\sqrt {\frac {1}{2} \left (3 x^2-2\right )+1}}d\sqrt [4]{3 x^2-2}}{2 x}+\frac {\sqrt [4]{3 x^2-2}}{2 x}\right )+\frac {\sqrt [4]{3 x^2-2}}{6 x^3}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {5}{4} \left (\frac {\sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-2}+\sqrt {2}\right )^2}} \left (\sqrt {3 x^2-2}+\sqrt {2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{3 x^2-2}}{\sqrt [4]{2}}\right ),\frac {1}{2}\right )}{4 \sqrt [4]{2} x}+\frac {\sqrt [4]{3 x^2-2}}{2 x}\right )+\frac {\sqrt [4]{3 x^2-2}}{6 x^3}\) |
Input:
Int[1/(x^4*(-2 + 3*x^2)^(3/4)),x]
Output:
(-2 + 3*x^2)^(1/4)/(6*x^3) + (5*((-2 + 3*x^2)^(1/4)/(2*x) + (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])/(4*2^(1/4)*x)))/4
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[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]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
Time = 0.37 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.34
method | result | size |
meijerg | \(-\frac {2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} \operatorname {hypergeom}\left (\left [-\frac {3}{2}, \frac {3}{4}\right ], \left [-\frac {1}{2}\right ], \frac {3 x^{2}}{2}\right )}{6 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}} x^{3}}\) | \(42\) |
risch | \(\frac {45 x^{4}-18 x^{2}-8}{24 x^{3} \left (3 x^{2}-2\right )^{\frac {3}{4}}}+\frac {15 \,2^{\frac {1}{4}} {\left (-\operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )\right )}^{\frac {3}{4}} x \operatorname {hypergeom}\left (\left [\frac {1}{2}, \frac {3}{4}\right ], \left [\frac {3}{2}\right ], \frac {3 x^{2}}{2}\right )}{32 \operatorname {signum}\left (-1+\frac {3 x^{2}}{2}\right )^{\frac {3}{4}}}\) | \(67\) |
Input:
int(1/x^4/(3*x^2-2)^(3/4),x,method=_RETURNVERBOSE)
Output:
-1/6*2^(1/4)/signum(-1+3/2*x^2)^(3/4)*(-signum(-1+3/2*x^2))^(3/4)/x^3*hype rgeom([-3/2,3/4],[-1/2],3/2*x^2)
\[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(3*x^2-2)^(3/4),x, algorithm="fricas")
Output:
integral((3*x^2 - 2)^(1/4)/(3*x^6 - 2*x^4), x)
Result contains complex when optimal does not.
Time = 0.51 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.26 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\frac {\sqrt [4]{2} e^{\frac {i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, \frac {3}{4} \\ - \frac {1}{2} \end {matrix}\middle | {\frac {3 x^{2}}{2}} \right )}}{6 x^{3}} \] Input:
integrate(1/x**4/(3*x**2-2)**(3/4),x)
Output:
2**(1/4)*exp(I*pi/4)*hyper((-3/2, 3/4), (-1/2,), 3*x**2/2)/(6*x**3)
\[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(3*x^2-2)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((3*x^2 - 2)^(3/4)*x^4), x)
\[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (3 \, x^{2} - 2\right )}^{\frac {3}{4}} x^{4}} \,d x } \] Input:
integrate(1/x^4/(3*x^2-2)^(3/4),x, algorithm="giac")
Output:
integrate(1/((3*x^2 - 2)^(3/4)*x^4), x)
Timed out. \[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\int \frac {1}{x^4\,{\left (3\,x^2-2\right )}^{3/4}} \,d x \] Input:
int(1/(x^4*(3*x^2 - 2)^(3/4)),x)
Output:
int(1/(x^4*(3*x^2 - 2)^(3/4)), x)
\[ \int \frac {1}{x^4 \left (-2+3 x^2\right )^{3/4}} \, dx=\int \frac {1}{\left (3 x^{2}-2\right )^{\frac {3}{4}} x^{4}}d x \] Input:
int(1/x^4/(3*x^2-2)^(3/4),x)
Output:
int(1/((3*x**2 - 2)**(3/4)*x**4),x)