\(\int \frac {1}{x^4 (-2+3 x^2) (-1+3 x^2)^{3/4}} \, dx\) [1092]

Optimal result

Integrand size = 24, antiderivative size = 165 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=-\frac {\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac {2 \sqrt [4]{-1+3 x^2}}{x}+\frac {3}{8} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {11 \sqrt {3} \sqrt {\frac {x^2}{\left (1+\sqrt {-1+3 x^2}\right )^2}} \left (1+\sqrt {-1+3 x^2}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{-1+3 x^2}\right ),\frac {1}{2}\right )}{8 x} \]

[Out]

-1/6*(3*x^2-1)^(1/4)/x^3-2*(3*x^2-1)^(1/4)/x+3/16*arctan(1/2*x*6^(1/2)/(3*x^2-1)^(1/4))*6^(1/2)-3/16*arctanh(1
/2*x*6^(1/2)/(3*x^2-1)^(1/4))*6^(1/2)-11/8*(cos(2*arctan((3*x^2-1)^(1/4)))^2)^(1/2)/cos(2*arctan((3*x^2-1)^(1/
4)))*EllipticF(sin(2*arctan((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 = 0.11 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {454, 331, 240, 226, 409, 453} \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {3}{8} \sqrt {\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {11 \sqrt {3} \sqrt {\frac {x^2}{\left (\sqrt {3 x^2-1}+1\right )^2}} \left (\sqrt {3 x^2-1}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\sqrt [4]{3 x^2-1}\right ),\frac {1}{2}\right )}{8 x}-\frac {3}{8} \sqrt {\frac {3}{2}} \text {arctanh}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{3 x^2-1}}\right )-\frac {2 \sqrt [4]{3 x^2-1}}{x}-\frac {\sqrt [4]{3 x^2-1}}{6 x^3} \]

[In]

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

[Out]

-1/6*(-1 + 3*x^2)^(1/4)/x^3 - (2*(-1 + 3*x^2)^(1/4))/x + (3*Sqrt[3/2]*ArcTan[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)]
)/8 - (3*Sqrt[3/2]*ArcTanh[(Sqrt[3/2]*x)/(-1 + 3*x^2)^(1/4)])/8 - (11*Sqrt[3]*Sqrt[x^2/(1 + Sqrt[-1 + 3*x^2])^
2]*(1 + Sqrt[-1 + 3*x^2])*EllipticF[2*ArcTan[(-1 + 3*x^2)^(1/4)], 1/2])/(8*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]

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 {1}{2 x^4 \left (-1+3 x^2\right )^{3/4}}-\frac {3}{4 x^2 \left (-1+3 x^2\right )^{3/4}}+\frac {9}{4 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}}\right ) \, dx \\ & = -\left (\frac {1}{2} \int \frac {1}{x^4 \left (-1+3 x^2\right )^{3/4}} \, dx\right )-\frac {3}{4} \int \frac {1}{x^2 \left (-1+3 x^2\right )^{3/4}} \, dx+\frac {9}{4} \int \frac {1}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac {3 \sqrt [4]{-1+3 x^2}}{4 x}-2 \left (\frac {9}{8} \int \frac {1}{\left (-1+3 x^2\right )^{3/4}} \, dx\right )-\frac {5}{4} \int \frac {1}{x^2 \left (-1+3 x^2\right )^{3/4}} \, dx+\frac {27}{8} \int \frac {x^2}{\left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac {2 \sqrt [4]{-1+3 x^2}}{x}+\frac {3}{8} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {15}{8} \int \frac {1}{\left (-1+3 x^2\right )^{3/4}} \, dx-2 \frac {\left (3 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x} \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac {2 \sqrt [4]{-1+3 x^2}}{x}+\frac {3}{8} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3 \sqrt {3} \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 )}{4 x}-\frac {\left (5 \sqrt {3} \sqrt {x^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+x^4}} \, dx,x,\sqrt [4]{-1+3 x^2}\right )}{4 x} \\ & = -\frac {\sqrt [4]{-1+3 x^2}}{6 x^3}-\frac {2 \sqrt [4]{-1+3 x^2}}{x}+\frac {3}{8} \sqrt {\frac {3}{2}} \tan ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {3}{8} \sqrt {\frac {3}{2}} \tanh ^{-1}\left (\frac {\sqrt {\frac {3}{2}} x}{\sqrt [4]{-1+3 x^2}}\right )-\frac {11 \sqrt {3} \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 )}{8 x} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 10.04 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.32 \[ \int \frac {1}{x^4 \left (-2+3 x^2\right ) \left (-1+3 x^2\right )^{3/4}} \, dx=\frac {\left (1-3 x^2\right )^{3/4} \operatorname {AppellF1}\left (-\frac {3}{2},\frac {3}{4},1,-\frac {1}{2},3 x^2,\frac {3 x^2}{2}\right )}{6 x^3 \left (-1+3 x^2\right )^{3/4}} \]

[In]

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

[Out]

((1 - 3*x^2)^(3/4)*AppellF1[-3/2, 3/4, 1, -1/2, 3*x^2, (3*x^2)/2])/(6*x^3*(-1 + 3*x^2)^(3/4))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((3*x^2 - 1)^(1/4)/(9*x^8 - 9*x^6 + 2*x^4), x)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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