Integrand size = 24, antiderivative size = 147 \[ \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}+\frac {\arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )}{8 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\text {arctanh}\left (\frac {2^{3/4} \sqrt [4]{2-3 x^2}}{\sqrt {2}+\sqrt {2-3 x^2}}\right )}{8 \sqrt [4]{2}} \] Output:
-1/8*arctan(1/2*2^(3/4)*(-3*x^2+2)^(1/4))*2^(1/4)+1/16*2^(3/4)*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^(1/4)-1/16*2^(3/4)*arctanh(2^(3/4)*(-3*x^2+2)^(1/4) /(2^(1/2)+(-3*x^2+2)^(1/2)))
Time = 0.17 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\frac {-2 \arctan \left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )+\sqrt {2} \arctan \left (\frac {\sqrt {2}-\sqrt {2-3 x^2}}{2^{3/4} \sqrt [4]{2-3 x^2}}\right )-2 \text {arctanh}\left (\sqrt [4]{1-\frac {3 x^2}{2}}\right )-\sqrt {2} \text {arctanh}\left (\frac {2 \sqrt [4]{4-6 x^2}}{2+\sqrt {4-6 x^2}}\right )}{8\ 2^{3/4}} \] Input:
Integrate[1/(x*(2 - 3*x^2)^(3/4)*(4 - 3*x^2)),x]
Output:
(-2*ArcTan[(1 - (3*x^2)/2)^(1/4)] + Sqrt[2]*ArcTan[(Sqrt[2] - Sqrt[2 - 3*x ^2])/(2^(3/4)*(2 - 3*x^2)^(1/4))] - 2*ArcTanh[(1 - (3*x^2)/2)^(1/4)] - Sqr t[2]*ArcTanh[(2*(4 - 6*x^2)^(1/4))/(2 + Sqrt[4 - 6*x^2])])/(8*2^(3/4))
Time = 0.42 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {352, 2009}
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.
\(\displaystyle \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx\) |
\(\Big \downarrow \) 352 |
\(\displaystyle \int \left (\frac {1}{4 x \left (2-3 x^2\right )^{3/4}}-\frac {3 x}{4 \left (2-3 x^2\right )^{3/4} \left (3 x^2-4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\arctan \left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}-\frac {\arctan \left (\sqrt [4]{4-6 x^2}+1\right )}{8 \sqrt [4]{2}}+\frac {\arctan \left (1-\sqrt [4]{2} \sqrt [4]{2-3 x^2}\right )}{8 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2-3 x^2}}{\sqrt [4]{2}}\right )}{4\ 2^{3/4}}+\frac {\log \left (\sqrt {2-3 x^2}-2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{16 \sqrt [4]{2}}-\frac {\log \left (\sqrt {2-3 x^2}+2^{3/4} \sqrt [4]{2-3 x^2}+\sqrt {2}\right )}{16 \sqrt [4]{2}}\) |
Input:
Int[1/(x*(2 - 3*x^2)^(3/4)*(4 - 3*x^2)),x]
Output:
-1/4*ArcTan[(2 - 3*x^2)^(1/4)/2^(1/4)]/2^(3/4) - ArcTan[1 + (4 - 6*x^2)^(1 /4)]/(8*2^(1/4)) + ArcTan[1 - 2^(1/4)*(2 - 3*x^2)^(1/4)]/(8*2^(1/4)) - Arc Tanh[(2 - 3*x^2)^(1/4)/2^(1/4)]/(4*2^(3/4)) + Log[Sqrt[2] - 2^(3/4)*(2 - 3 *x^2)^(1/4) + Sqrt[2 - 3*x^2]]/(16*2^(1/4)) - Log[Sqrt[2] + 2^(3/4)*(2 - 3 *x^2)^(1/4) + Sqrt[2 - 3*x^2]]/(16*2^(1/4))
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] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c - 2*a*d, 0] && IntegerQ[m] && (PosQ[a] || In tegerQ[m/2])
Time = 7.90 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.10
method | result | size |
pseudoelliptic | \(-\frac {2^{\frac {1}{4}} \left (\frac {\ln \left (\frac {\sqrt {-3 x^{2}+2}+2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}}{\sqrt {-3 x^{2}+2}-2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+\sqrt {2}}\right ) \sqrt {2}}{2}+\arctan \left (2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}-1\right ) \sqrt {2}+\arctan \left (2^{\frac {1}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+1\right ) \sqrt {2}+\ln \left (\frac {-\left (-3 x^{2}+2\right )^{\frac {1}{4}}-2^{\frac {1}{4}}}{-\left (-3 x^{2}+2\right )^{\frac {1}{4}}+2^{\frac {1}{4}}}\right )+2 \arctan \left (\frac {2^{\frac {3}{4}} \left (-3 x^{2}+2\right )^{\frac {1}{4}}}{2}\right )\right )}{16}\) | \(162\) |
trager | \(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{2}+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}-4 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \sqrt {-3 x^{2}+2}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (-3 x^{2}+2\right )^{\frac {1}{4}}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )}{x^{2}}\right )}{16}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (-\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}-4 \sqrt {-3 x^{2}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (-3 x^{2}+2\right )^{\frac {1}{4}}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3}}{x^{2}}\right )}{16}-\frac {\ln \left (\frac {4 \sqrt {-3 x^{2}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3}+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (-3 x^{2}+2\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )}{3 x^{2}-4}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3}}{32}-\frac {\ln \left (\frac {4 \sqrt {-3 x^{2}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3}+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \left (-3 x^{2}+2\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) x^{2}+4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )}{3 x^{2}-4}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )}{32}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (\frac {-2 \sqrt {-3 x^{2}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )+2 \sqrt {-3 x^{2}+2}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3}+4 \left (-3 x^{2}+2\right )^{\frac {3}{4}}-4 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \left (-3 x^{2}+2\right )^{\frac {1}{4}}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) x^{2}}{3 x^{2}-4}\right )}{16}\) | \(561\) |
Input:
int(1/x/(-3*x^2+2)^(3/4)/(-3*x^2+4),x,method=_RETURNVERBOSE)
Output:
-1/16*2^(1/4)*(1/2*ln(((-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^(3/4)*(-3*x^2+2)^(1/4)+2^(1/2)))*2^(1/2)+arctan(2^(1/4 )*(-3*x^2+2)^(1/4)-1)*2^(1/2)+arctan(2^(1/4)*(-3*x^2+2)^(1/4)+1)*2^(1/2)+l n((-(-3*x^2+2)^(1/4)-2^(1/4))/(-(-3*x^2+2)^(1/4)+2^(1/4)))+2*arctan(1/2*2^ (3/4)*(-3*x^2+2)^(1/4)))
Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.17 \[ \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {1}{32} \cdot 8^{\frac {3}{4}} \arctan \left (\frac {1}{2} \cdot 8^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{16} \cdot 2^{\frac {3}{4}} \arctan \left (2^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 1\right ) - \frac {1}{16} \cdot 2^{\frac {3}{4}} \arctan \left (2^{\frac {1}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} - 1\right ) - \frac {1}{32} \cdot 2^{\frac {3}{4}} \log \left (2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) + \frac {1}{32} \cdot 2^{\frac {3}{4}} \log \left (-2^{\frac {3}{4}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {2} + \sqrt {-3 \, x^{2} + 2}\right ) - \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (8^{\frac {3}{4}} + 4 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) + \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (-8^{\frac {3}{4}} + 4 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) \] Input:
integrate(1/x/(-3*x^2+2)^(3/4)/(-3*x^2+4),x, algorithm="fricas")
Output:
-1/32*8^(3/4)*arctan(1/2*8^(1/4)*(-3*x^2 + 2)^(1/4)) - 1/16*2^(3/4)*arctan (2^(1/4)*(-3*x^2 + 2)^(1/4) + 1) - 1/16*2^(3/4)*arctan(2^(1/4)*(-3*x^2 + 2 )^(1/4) - 1) - 1/32*2^(3/4)*log(2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqr t(-3*x^2 + 2)) + 1/32*2^(3/4)*log(-2^(3/4)*(-3*x^2 + 2)^(1/4) + sqrt(2) + sqrt(-3*x^2 + 2)) - 1/64*8^(3/4)*log(8^(3/4) + 4*(-3*x^2 + 2)^(1/4)) + 1/6 4*8^(3/4)*log(-8^(3/4) + 4*(-3*x^2 + 2)^(1/4))
\[ \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=- \int \frac {1}{3 x^{3} \left (2 - 3 x^{2}\right )^{\frac {3}{4}} - 4 x \left (2 - 3 x^{2}\right )^{\frac {3}{4}}}\, dx \] Input:
integrate(1/x/(-3*x**2+2)**(3/4)/(-3*x**2+4),x)
Output:
-Integral(1/(3*x**3*(2 - 3*x**2)**(3/4) - 4*x*(2 - 3*x**2)**(3/4)), x)
\[ \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=\int { -\frac {1}{{\left (3 \, x^{2} - 4\right )} {\left (-3 \, x^{2} + 2\right )}^{\frac {3}{4}} x} \,d x } \] Input:
integrate(1/x/(-3*x^2+2)^(3/4)/(-3*x^2+4),x, algorithm="maxima")
Output:
-integrate(1/((3*x^2 - 4)*(-3*x^2 + 2)^(3/4)*x), x)
Time = 0.16 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.43 \[ \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {1}{16} \cdot 4^{\frac {1}{8}} \sqrt {2} \arctan \left (\frac {1}{8} \cdot 4^{\frac {7}{8}} \sqrt {2} {\left (4^{\frac {1}{8}} \sqrt {2} + 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{16} \cdot 4^{\frac {1}{8}} \sqrt {2} \arctan \left (-\frac {1}{8} \cdot 4^{\frac {7}{8}} \sqrt {2} {\left (4^{\frac {1}{8}} \sqrt {2} - 2 \, {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right )}\right ) - \frac {1}{32} \cdot 4^{\frac {1}{8}} \sqrt {2} \log \left (4^{\frac {1}{8}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {-3 \, x^{2} + 2} + 4^{\frac {1}{4}}\right ) + \frac {1}{32} \cdot 4^{\frac {1}{8}} \sqrt {2} \log \left (-4^{\frac {1}{8}} \sqrt {2} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + \sqrt {-3 \, x^{2} + 2} + 4^{\frac {1}{4}}\right ) - \frac {1}{8} \cdot 4^{\frac {1}{8}} \arctan \left (\frac {1}{4} \cdot 4^{\frac {7}{8}} {\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}}\right ) - \frac {1}{16} \cdot 4^{\frac {1}{8}} \log \left ({\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4^{\frac {1}{8}}\right ) + \frac {1}{16} \cdot 4^{\frac {1}{8}} \log \left (-{\left (-3 \, x^{2} + 2\right )}^{\frac {1}{4}} + 4^{\frac {1}{8}}\right ) \] Input:
integrate(1/x/(-3*x^2+2)^(3/4)/(-3*x^2+4),x, algorithm="giac")
Output:
-1/16*4^(1/8)*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^(1/8)*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^(1/8)*sqrt(2)*log(4^(1/8)*sqr t(2)*(-3*x^2 + 2)^(1/4) + sqrt(-3*x^2 + 2) + 4^(1/4)) + 1/32*4^(1/8)*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)*arctan(1/4*4^(7/8)*(-3*x^2 + 2)^(1/4)) - 1/16*4^(1/8)*log((-3 *x^2 + 2)^(1/4) + 4^(1/8)) + 1/16*4^(1/8)*log(-(-3*x^2 + 2)^(1/4) + 4^(1/8 ))
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\frac {2^{1/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}}{2}\right )}{8}+\frac {2^{1/4}\,\mathrm {atan}\left (\frac {2^{3/4}\,{\left (2-3\,x^2\right )}^{1/4}\,1{}\mathrm {i}}{2}\right )\,1{}\mathrm {i}}{8}+2^{3/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )+2^{3/4}\,\mathrm {atan}\left (2^{1/4}\,{\left (2-3\,x^2\right )}^{1/4}\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (-\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right ) \] Input:
int(-1/(x*(2 - 3*x^2)^(3/4)*(3*x^2 - 4)),x)
Output:
(2^(1/4)*atan((2^(3/4)*(2 - 3*x^2)^(1/4)*1i)/2)*1i)/8 - (2^(1/4)*atan((2^( 3/4)*(2 - 3*x^2)^(1/4))/2))/8 - 2^(3/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/ 2 - 1i/2))*(1/16 + 1i/16) - 2^(3/4)*atan(2^(1/4)*(2 - 3*x^2)^(1/4)*(1/2 + 1i/2))*(1/16 - 1i/16)
\[ \int \frac {1}{x \left (2-3 x^2\right )^{3/4} \left (4-3 x^2\right )} \, dx=-\left (\int \frac {1}{3 \left (-3 x^{2}+2\right )^{\frac {3}{4}} x^{3}-4 \left (-3 x^{2}+2\right )^{\frac {3}{4}} x}d x \right ) \] Input:
int(1/x/(-3*x^2+2)^(3/4)/(-3*x^2+4),x)
Output:
- int(1/(3*( - 3*x**2 + 2)**(3/4)*x**3 - 4*( - 3*x**2 + 2)**(3/4)*x),x)