\(\int \frac {1}{x^4 (2-3 x^2)^{3/4} (4-3 x^2)} \, dx\) [1072]

Optimal result

Integrand size = 24, antiderivative size = 184 \[ \int \frac {1}{x^4 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {\sqrt [4]{2-3 x^2}}{24 x^3}-\frac {\sqrt [4]{2-3 x^2}}{4 x}+\frac {3 \sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}-\frac {3 \sqrt {3} \text {arctanh}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}+\frac {11 \sqrt {3} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{32 \sqrt [4]{2}} \]

[Out]

-1/24*(-3*x^2+2)^(1/4)/x^3-1/4*(-3*x^2+2)^(1/4)/x+3/128*2^(3/4)*arctan(1/3*(2^(3/4)-2^(1/4)*(-3*x^2+2)^(1/2))/
x/(-3*x^2+2)^(1/4)*3^(1/2))*3^(1/2)-3/128*2^(3/4)*arctanh(1/3*(2^(3/4)+2^(1/4)*(-3*x^2+2)^(1/2))/x/(-3*x^2+2)^
(1/4)*3^(1/2))*3^(1/2)+11/64*2^(3/4)*(cos(1/2*arcsin(1/2*x*6^(1/2)))^2)^(1/2)/cos(1/2*arcsin(1/2*x*6^(1/2)))*E
llipticF(sin(1/2*arcsin(1/2*x*6^(1/2))),2^(1/2))*3^(1/2)

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {454, 331, 238, 409, 452} \[ \int \frac {1}{x^4 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\frac {11 \sqrt {3} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\sqrt {\frac {3}{2}} x\right ),2\right )}{32 \sqrt [4]{2}}+\frac {3 \sqrt {3} \arctan \left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}-\frac {3 \sqrt {3} \text {arctanh}\left (\frac {\sqrt [4]{2} \sqrt {2-3 x^2}+2^{3/4}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}-\frac {\sqrt [4]{2-3 x^2}}{4 x}-\frac {\sqrt [4]{2-3 x^2}}{24 x^3} \]

[In]

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

[Out]

-1/24*(2 - 3*x^2)^(1/4)/x^3 - (2 - 3*x^2)^(1/4)/(4*x) + (3*Sqrt[3]*ArcTan[(2^(3/4) - 2^(1/4)*Sqrt[2 - 3*x^2])/
(Sqrt[3]*x*(2 - 3*x^2)^(1/4))])/(64*2^(1/4)) - (3*Sqrt[3]*ArcTanh[(2^(3/4) + 2^(1/4)*Sqrt[2 - 3*x^2])/(Sqrt[3]
*x*(2 - 3*x^2)^(1/4))])/(64*2^(1/4)) + (11*Sqrt[3]*EllipticF[ArcSin[Sqrt[3/2]*x]/2, 2])/(32*2^(1/4))

Rule 238

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2]))*EllipticF[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/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 452

Int[(x_)^2/(((a_) + (b_.)*(x_)^2)^(3/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[(-b/(a*d*Rt[b^2/a, 4]^3))*Ar
cTan[(b + Rt[b^2/a, 4]^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))], x] + Simp[(b/(a*d*Rt[b^2/a, 4
]^3))*ArcTanh[(b - Rt[b^2/a, 4]^2*Sqrt[a + b*x^2])/(Rt[b^2/a, 4]^3*x*(a + b*x^2)^(1/4))], x] /; FreeQ[{a, b, c
, d}, x] && EqQ[b*c - 2*a*d, 0] && PosQ[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}{4 x^4 \left (2-3 x^2\right )^{3/4}}+\frac {3}{16 x^2 \left (2-3 x^2\right )^{3/4}}-\frac {9}{16 \left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )}\right ) \, dx \\ & = \frac {3}{16} \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4}} \, dx+\frac {1}{4} \int \frac {1}{x^4 \left (2-3 x^2\right )^{3/4}} \, dx-\frac {9}{16} \int \frac {1}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx \\ & = -\frac {\sqrt [4]{2-3 x^2}}{24 x^3}-\frac {3 \sqrt [4]{2-3 x^2}}{32 x}+2 \left (\frac {9}{64} \int \frac {1}{\left (2-3 x^2\right )^{3/4}} \, dx\right )+\frac {5}{16} \int \frac {1}{x^2 \left (2-3 x^2\right )^{3/4}} \, dx-\frac {27}{64} \int \frac {x^2}{\left (2-3 x^2\right )^{3/4} \left (-4+3 x^2\right )} \, dx \\ & = -\frac {\sqrt [4]{2-3 x^2}}{24 x^3}-\frac {\sqrt [4]{2-3 x^2}}{4 x}+\frac {3 \sqrt {3} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}-\frac {3 \sqrt {3} \tanh ^{-1}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}+\frac {3 \sqrt {3} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{16 \sqrt [4]{2}}+\frac {15}{64} \int \frac {1}{\left (2-3 x^2\right )^{3/4}} \, dx \\ & = -\frac {\sqrt [4]{2-3 x^2}}{24 x^3}-\frac {\sqrt [4]{2-3 x^2}}{4 x}+\frac {3 \sqrt {3} \tan ^{-1}\left (\frac {2^{3/4}-\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}-\frac {3 \sqrt {3} \tanh ^{-1}\left (\frac {2^{3/4}+\sqrt [4]{2} \sqrt {2-3 x^2}}{\sqrt {3} x \sqrt [4]{2-3 x^2}}\right )}{64 \sqrt [4]{2}}+\frac {11 \sqrt {3} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right |2\right )}{32 \sqrt [4]{2}} \\ \end{align*}

Mathematica [C] (verified)

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

Time = 11.05 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.20 \[ \int \frac {1}{x^4 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {\operatorname {AppellF1}\left (-\frac {3}{2},\frac {3}{4},1,-\frac {1}{2},\frac {3 x^2}{2},\frac {3 x^2}{4}\right )}{12\ 2^{3/4} x^3} \]

[In]

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

[Out]

-1/12*AppellF1[-3/2, 3/4, 1, -1/2, (3*x^2)/2, (3*x^2)/4]/(2^(3/4)*x^3)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

integral((-3*x^2 + 2)^(1/4)/(9*x^8 - 18*x^6 + 8*x^4), x)

Sympy [F]

\[ \int \frac {1}{x^4 \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=- \int \frac {1}{3 x^{6} \left (2 - 3 x^{2}\right )^{\frac {3}{4}} - 4 x^{4} \left (2 - 3 x^{2}\right )^{\frac {3}{4}}}\, dx \]

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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