\(\int \frac {1}{x \sqrt [4]{2-3 x^2} (4-3 x^2)} \, dx\) [1035]

Optimal result

Integrand size = 24, antiderivative size = 145 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}} \]

[Out]

1/8*arctan(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(3/4)+1/8*2^(1/4)*arctan(1/2*(2^(1/2)-(-3*x^2+2)^(1/2))*2^(1/4)/(-3
*x^2+2)^(1/4))-1/8*arctanh(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(3/4)+1/8*2^(1/4)*arctanh(1/2*(2^(1/2)+(-3*x^2+2)^(
1/2))*2^(1/4)/(-3*x^2+2)^(1/4))

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {451, 272, 65, 304, 209, 212, 450} \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\sqrt {2-3 x^2}+\sqrt {2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}} \]

[In]

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

[Out]

ArcTan[(2 - 3*x^2)^(1/4)/2^(1/4)]/(4*2^(1/4)) + ArcTan[(Sqrt[2] - Sqrt[2 - 3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))
]/(4*2^(3/4)) - ArcTanh[(2 - 3*x^2)^(1/4)/2^(1/4)]/(4*2^(1/4)) + ArcTanh[(Sqrt[2] + Sqrt[2 - 3*x^2])/(2^(3/4)*
(2 - 3*x^2)^(1/4))]/(4*2^(3/4))

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 450

Int[(x_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[(-(Sqrt[2]*Rt[a, 4]*d)^(-1))*A
rcTan[(Rt[a, 4]^2 - Sqrt[a + b*x^2])/(Sqrt[2]*Rt[a, 4]*(a + b*x^2)^(1/4))], x] - Simp[(1/(Sqrt[2]*Rt[a, 4]*d))
*ArcTanh[(Rt[a, 4]^2 + Sqrt[a + b*x^2])/(Sqrt[2]*Rt[a, 4]*(a + b*x^2)^(1/4))], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c - 2*a*d, 0] && PosQ[a]

Rule 451

Int[(x_)^(m_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegrand[x^m/((a +
b*x^2)^(1/4)*(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 \sqrt [4]{2-3 x^2}}-\frac {3 x}{4 \sqrt [4]{2-3 x^2} \left (-4+3 x^2\right )}\right ) \, dx \\ & = \frac {1}{4} \int \frac {1}{x \sqrt [4]{2-3 x^2}} \, dx-\frac {3}{4} \int \frac {x}{\sqrt [4]{2-3 x^2} \left (-4+3 x^2\right )} \, dx \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}+\frac {1}{8} \text {Subst}\left (\int \frac {1}{\sqrt [4]{2-3 x} x} \, dx,x,x^2\right ) \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}-\frac {1}{6} \text {Subst}\left (\int \frac {x^2}{\frac {2}{3}-\frac {x^4}{3}} \, dx,x,\sqrt [4]{2-3 x^2}\right ) \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}-\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {2}-x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right )+\frac {1}{4} \text {Subst}\left (\int \frac {1}{\sqrt {2}+x^2} \, dx,x,\sqrt [4]{2-3 x^2}\right ) \\ & = \frac {\tan ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2}+\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{4\ 2^{3/4}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.81 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {\sqrt {2} \arctan \left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+\arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-\sqrt {2} \text {arctanh}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+\text {arctanh}\left (\frac {2 \sqrt [4]{4-6 x^2}}{2+\sqrt {4-6 x^2}}\right )}{4\ 2^{3/4}} \]

[In]

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

[Out]

(Sqrt[2]*ArcTan[(1 - (3*x^2)/2)^(1/4)] + ArcTan[(Sqrt[2] - Sqrt[2 - 3*x^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))] - Sqr
t[2]*ArcTanh[(1 - (3*x^2)/2)^(1/4)] + ArcTanh[(2*(4 - 6*x^2)^(1/4))/(2 + Sqrt[4 - 6*x^2])])/(4*2^(3/4))

Maple [A] (verified)

Time = 8.58 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.08

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {1}{4}} \left (-2 \arctan \left (\frac {2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}}{2}\right ) \sqrt {2}+\ln \left (\frac {\left (-3 x^{2}+2\right )^{\frac {1}{4}}+2^{\frac {1}{4}}}{\left (-3 x^{2}+2\right )^{\frac {1}{4}}-2^{\frac {1}{4}}}\right ) \sqrt {2}+\ln \left (\frac {-2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}+\sqrt {-3 x^{2}+2}}{2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}+\sqrt {-3 x^{2}+2}}\right )+2 \arctan \left (2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+1\right )+2 \arctan \left (-1+2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}\right )\right )}{16}\)	\(156\)

[In]

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

[Out]

-1/16*2^(1/4)*(-2*arctan(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(1/2)+ln(((-3*x^2+2)^(1/4)+2^(1/4))/((-3*x^2+2)^(1/4)
-2^(1/4)))*2^(1/2)+ln((-2^(3/4)*(-3*x^2+2)^(1/4)+2^(1/2)+(-3*x^2+2)^(1/2))/(2^(3/4)*(-3*x^2+2)^(1/4)+2^(1/2)+(
-3*x^2+2)^(1/2)))+2*arctan(2^(1/4)*(-3*x^2+2)^(1/4)+1)+2*arctan(-1+2^(1/4)*(-3*x^2+2)^(1/4)))

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.21 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{16} i \cdot 2^{\frac {3}{4}} \log \left (i \cdot 2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{16} i \cdot 2^{\frac {3}{4}} \log \left (-i \cdot 2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{16} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {1}{4}} + {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \left (\frac {1}{16} i - \frac {1}{16}\right ) \cdot 2^{\frac {1}{4}} \log \left (\left (i + 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{16} i + \frac {1}{16}\right ) \cdot 2^{\frac {1}{4}} \log \left (-\left (i - 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \left (\frac {1}{16} i + \frac {1}{16}\right ) \cdot 2^{\frac {1}{4}} \log \left (\left (i - 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \left (\frac {1}{16} i - \frac {1}{16}\right ) \cdot 2^{\frac {1}{4}} \log \left (-\left (i + 1\right ) \cdot 2^{\frac {3}{4}} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) \]

[In]

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

[Out]

-1/16*2^(3/4)*log(2^(1/4) + (-3*x^2 + 2)^(1/4)) + 1/16*I*2^(3/4)*log(I*2^(1/4) + (-3*x^2 + 2)^(1/4)) - 1/16*I*
2^(3/4)*log(-I*2^(1/4) + (-3*x^2 + 2)^(1/4)) + 1/16*2^(3/4)*log(-2^(1/4) + (-3*x^2 + 2)^(1/4)) - (1/16*I - 1/1
6)*2^(1/4)*log((I + 1)*2^(3/4) + 2*(-3*x^2 + 2)^(1/4)) + (1/16*I + 1/16)*2^(1/4)*log(-(I - 1)*2^(3/4) + 2*(-3*
x^2 + 2)^(1/4)) - (1/16*I + 1/16)*2^(1/4)*log((I - 1)*2^(3/4) + 2*(-3*x^2 + 2)^(1/4)) + (1/16*I - 1/16)*2^(1/4
)*log(-(I + 1)*2^(3/4) + 2*(-3*x^2 + 2)^(1/4))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.49 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=-\frac {1}{16} \cdot 4^{\frac {3}{8}} \sqrt {2} \arctan \left (\frac {1}{8} \cdot 4^{\frac {7}{8}} \sqrt {2} {\left (4^{\frac {1}{8}} \sqrt {2} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {3}{8}} \sqrt {2} \arctan \left (-\frac {1}{8} \cdot 4^{\frac {7}{8}} \sqrt {2} {\left (4^{\frac {1}{8}} \sqrt {2} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) + \frac {1}{32} \cdot 4^{\frac {3}{8}} \sqrt {2} \log \left (4^{\frac {1}{8}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {-3 \, x^{2} + 2} + 4^{\frac {1}{4}}\right ) - \frac {1}{32} \cdot 4^{\frac {3}{8}} \sqrt {2} \log \left (-4^{\frac {1}{8}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {-3 \, x^{2} + 2} + 4^{\frac {1}{4}}\right ) + \frac {1}{8} \cdot 4^{\frac {1}{8}} \sqrt {2} \arctan \left (\frac {1}{4} \cdot 4^{\frac {7}{8}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{16} \cdot 4^{\frac {1}{8}} \sqrt {2} \log \left (-{\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4^{\frac {1}{8}}\right ) - \frac {1}{16} \cdot 4^{\frac {3}{8}} \log \left ({\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4^{\frac {1}{8}}\right ) \]

[In]

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

[Out]

-1/16*4^(3/8)*sqrt(2)*arctan(1/8*4^(7/8)*sqrt(2)*(4^(1/8)*sqrt(2) + 2*(-3*x^2 + 2)^(1/4))) - 1/16*4^(3/8)*sqrt
(2)*arctan(-1/8*4^(7/8)*sqrt(2)*(4^(1/8)*sqrt(2) - 2*(-3*x^2 + 2)^(1/4))) + 1/32*4^(3/8)*sqrt(2)*log(4^(1/8)*s
qrt(2)*(-3*x^2 + 2)^(1/4) + sqrt(-3*x^2 + 2) + 4^(1/4)) - 1/32*4^(3/8)*sqrt(2)*log(-4^(1/8)*sqrt(2)*(-3*x^2 +
2)^(1/4) + sqrt(-3*x^2 + 2) + 4^(1/4)) + 1/8*4^(1/8)*sqrt(2)*arctan(1/4*4^(7/8)*(-3*x^2 + 2)^(1/4)) + 1/16*4^(
1/8)*sqrt(2)*log(-(-3*x^2 + 2)^(1/4) + 4^(1/8)) - 1/16*4^(3/8)*log((-3*x^2 + 2)^(1/4) + 4^(1/8))

Mupad [B] (verification not implemented)

Time = 5.41 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.63 \[ \int \frac {1}{x \sqrt [4]{2-3 x^2} \left (4-3 x^2\right )} \, dx=\frac {2^{3/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}}{2}\right )}{8}+2^{1/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}+\frac {1}{8}{}\mathrm {i}\right )+2^{1/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{8}-\frac {1}{8}{}\mathrm {i}\right )+\frac {2^{3/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{8} \]

[In]

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

[Out]

(2^(3/4)*atan((2^(3/4)*(2 - 3*x^2)^(1/4))/2))/8 - 2^(1/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2 - 1i/2))*(1/8 -
1i/8) - 2^(1/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2 + 1i/2))*(1/8 + 1i/8) + (2^(3/4)*atan((2^(3/4)*(2 - 3*x^2)
^(1/4)*1i)/2)*1i)/8