\(\int \frac {x^4}{(-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\) [1088]

Optimal result

Integrand size = 24, antiderivative size = 147 \[ \int \frac {x^4}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {2}{27} x \sqrt [4]{-1+3 x^2}+\frac {1}{9} \sqrt {\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {2 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{27 \sqrt {3} x} \]

[Out]

2/27*x*(3*x^2-1)^(1/4)+1/27*arctan(1/2*x*6^(1/2)/(3*x^2-1)^(1/4))*6^(1/2)-1/27*arctanh(1/2*x*6^(1/2)/(3*x^2-1)
^(1/4))*6^(1/2)+2/81*(cos(2*arctan((3*x^2-1)^(1/4)))^2)^(1/2)/cos(2*arctan((3*x^2-1)^(1/4)))*EllipticF(sin(2*a
rctan((3*x^2-1)^(1/4))),1/2*2^(1/2))*(1+(3*x^2-1)^(1/2))*(x^2/(1+(3*x^2-1)^(1/2))^2)^(1/2)/x*3^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {454, 240, 226, 327, 409, 453} \[ \int \frac {x^4}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {1}{9} \sqrt {\frac {2}{3}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {2 \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right ),\frac {1}{2}\right )}{27 \sqrt {3} x}-\frac {1}{9} \sqrt {\frac {2}{3}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )+\frac {2}{27} \sqrt [4]{3 x^2-1} x \]

[In]

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

[Out]

(2*x*(-1 + 3*x^2)^(1/4))/27 + (Sqrt[2/3]*ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/9 - (Sqrt[2/3]*ArcTanh[(Sqr
t[3/2]*x)/(-1 + 3*x^2)^(1/4)])/9 + (2*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^2]*(1 + Sqrt[-1 + 3*x^2])*EllipticF[2*Ar
cTan[(-1 + 3*x^2)^(1/4)], 1/2])/(27*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]

Rule 409

Int[1/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Dist[1/c, Int[1/(a + b*x^2)^(3/4), x],
x] - Dist[d/c, Int[x^2/((a + b*x^2)^(3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d,
0]

Rule 453

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[(-b/(Sqrt[2]*a*d*Rt[-b^2/a,
4]^3))*ArcTan[(Rt[-b^2/a, 4]*x)/(Sqrt[2]*(a + b*x^2)^(1/4))], x] + Simp[(b/(Sqrt[2]*a*d*Rt[-b^2/a, 4]^3))*ArcT
anh[(Rt[-b^2/a, 4]*x)/(Sqrt[2]*(a + b*x^2)^(1/4))], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && Neg
Q[b^2/a]

Rule 454

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(3/4)*(c + d*x^2)), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a]
|| IntegerQ[m/2])

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{9 \left (-1+3 x^2\right )^{3/4}}+\frac {x^2}{3 \left (-1+3 x^2\right )^{3/4}}+\frac {4}{9 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}}\right ) \, dx \\ & = \frac {2}{9} \int \frac {1}{\left (-1+3 x^2\right )^{3/4}} \, dx+\frac {1}{3} \int \frac {x^2}{\left (-1+3 x^2\right )^{3/4}} \, dx+\frac {4}{9} \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx \\ & = \frac {2}{27} x \sqrt [4]{-1+3 x^2}+\frac {2}{27} \int \frac {1}{\left (-1+3 x^2\right )^{3/4}} \, dx-\frac {2}{9} \int \frac {1}{\left (-1+3 x^2\right )^{3/4}} \, dx+\frac {2}{3} \int \frac {x^2}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx+\frac {\left (4 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{9 \sqrt {3} x} \\ & = \frac {2}{27} x \sqrt [4]{-1+3 x^2}+\frac {1}{9} \sqrt {\frac {2}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {2 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{9 \sqrt {3} x}+\frac {\left (4 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{27 \sqrt {3} x}-\frac {\left (4 \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{9 \sqrt {3} x} \\ & = \frac {2}{27} x \sqrt [4]{-1+3 x^2}+\frac {1}{9} \sqrt {\frac {2}{3}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {1}{9} \sqrt {\frac {2}{3}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )+\frac {2 \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) F\left (2 \tan ^{-1}\left (\sqrt [4]{-1+3 x^2}\right )|\frac {1}{2}\right )}{27 \sqrt {3} x} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 5.82 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.22 \[ \int \frac {x^4}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {2 x \left (-1+3 x^2-2 x^2 \left (1-3 x^2\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},1,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )-\frac {4 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )}{\left (-2+3 x^2\right ) \left (2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},1,\frac {3}{2},3 x^2,\frac {3 x^2}{2}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},2,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )+3 \operatorname {AppellF1}\left (\frac {3}{2},\frac {7}{4},1,\frac {5}{2},3 x^2,\frac {3 x^2}{2}\right )\right )\right )}\right )}{27 \left (-1+3 x^2\right )^{3/4}} \]

[In]

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

[Out]

(2*x*(-1 + 3*x^2 - 2*x^2*(1 - 3*x^2)^(3/4)*AppellF1[3/2, 3/4, 1, 5/2, 3*x^2, (3*x^2)/2] - (4*AppellF1[1/2, 3/4
, 1, 3/2, 3*x^2, (3*x^2)/2])/((-2 + 3*x^2)*(2*AppellF1[1/2, 3/4, 1, 3/2, 3*x^2, (3*x^2)/2] + x^2*(2*AppellF1[3
/2, 3/4, 2, 5/2, 3*x^2, (3*x^2)/2] + 3*AppellF1[3/2, 7/4, 1, 5/2, 3*x^2, (3*x^2)/2])))))/(27*(-1 + 3*x^2)^(3/4
))

Maple [F]

\[\int \frac {x^{4}}{\left (3 x^{2}-2\right ) \left (3 x^{2}-1\right )^{\frac {3}{4}}}d x\]

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((3*x^2 - 1)^(1/4)*x^4/(9*x^4 - 9*x^2 + 2), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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