Integrand size = 24, antiderivative size = 63 \[ \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx=\arctan \left (\frac {1-\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right )-\text {arctanh}\left (\frac {x \sqrt [4]{1-x^2}}{1+\sqrt {1-x^2}}\right ) \] Output:
arctan((1-(-x^2+1)^(1/2))/x/(-x^2+1)^(1/4))-arctanh(x*(-x^2+1)^(1/4)/(1+(- x^2+1)^(1/2)))
Time = 1.55 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.38 \[ \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx=-\frac {1}{2} \arctan \left (\frac {x}{x-2 \sqrt [4]{1-x^2}}\right )+\frac {1}{2} \arctan \left (\frac {x}{x+2 \sqrt [4]{1-x^2}}\right )-\frac {1}{2} \text {arctanh}\left (\frac {2 x \sqrt [4]{1-x^2}}{x^2+2 \sqrt {1-x^2}}\right ) \] Input:
Integrate[x^2/((1 - x^2)^(3/4)*(2 - x^2)),x]
Output:
-1/2*ArcTan[x/(x - 2*(1 - x^2)^(1/4))] + ArcTan[x/(x + 2*(1 - x^2)^(1/4))] /2 - ArcTanh[(2*x*(1 - x^2)^(1/4))/(x^2 + 2*Sqrt[1 - x^2])]/2
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {350}
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.
\(\displaystyle \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx\) |
\(\Big \downarrow \) 350 |
\(\displaystyle \arctan \left (\frac {1-\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right )-\text {arctanh}\left (\frac {\sqrt {1-x^2}+1}{x \sqrt [4]{1-x^2}}\right )\) |
Input:
Int[x^2/((1 - x^2)^(3/4)*(2 - x^2)),x]
Output:
ArcTan[(1 - Sqrt[1 - x^2])/(x*(1 - x^2)^(1/4))] - ArcTanh[(1 + Sqrt[1 - x^ 2])/(x*(1 - x^2)^(1/4))]
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))*ArcTan[(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 ]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 1.42 (sec) , antiderivative size = 310, normalized size of antiderivative = 4.92
method | result | size |
trager | \(\operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \ln \left (\frac {2 \left (-x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-2 \left (-x^{2}+1\right )^{\frac {3}{4}}+x \sqrt {-x^{2}+1}-2 \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (-x^{2}+1\right )^{\frac {1}{4}}+2 x \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-x}{x^{2}-2}\right )-\ln \left (-\frac {2 \left (-x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-x \sqrt {-x^{2}+1}-2 \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (-x^{2}+1\right )^{\frac {1}{4}}+2 x \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )+2 \left (-x^{2}+1\right )^{\frac {1}{4}}-x}{x^{2}-2}\right ) \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )+\ln \left (-\frac {2 \left (-x^{2}+1\right )^{\frac {3}{4}} \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )-x \sqrt {-x^{2}+1}-2 \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right ) \left (-x^{2}+1\right )^{\frac {1}{4}}+2 x \operatorname {RootOf}\left (2 \textit {\_Z}^{2}-2 \textit {\_Z} +1\right )+2 \left (-x^{2}+1\right )^{\frac {1}{4}}-x}{x^{2}-2}\right )\) | \(310\) |
Input:
int(x^2/(-x^2+1)^(3/4)/(-x^2+2),x,method=_RETURNVERBOSE)
Output:
RootOf(2*_Z^2-2*_Z+1)*ln((2*(-x^2+1)^(3/4)*RootOf(2*_Z^2-2*_Z+1)-2*(-x^2+1 )^(3/4)+x*(-x^2+1)^(1/2)-2*RootOf(2*_Z^2-2*_Z+1)*(-x^2+1)^(1/4)+2*x*RootOf (2*_Z^2-2*_Z+1)-x)/(x^2-2))-ln(-(2*(-x^2+1)^(3/4)*RootOf(2*_Z^2-2*_Z+1)-x* (-x^2+1)^(1/2)-2*RootOf(2*_Z^2-2*_Z+1)*(-x^2+1)^(1/4)+2*x*RootOf(2*_Z^2-2* _Z+1)+2*(-x^2+1)^(1/4)-x)/(x^2-2))*RootOf(2*_Z^2-2*_Z+1)+ln(-(2*(-x^2+1)^( 3/4)*RootOf(2*_Z^2-2*_Z+1)-x*(-x^2+1)^(1/2)-2*RootOf(2*_Z^2-2*_Z+1)*(-x^2+ 1)^(1/4)+2*x*RootOf(2*_Z^2-2*_Z+1)+2*(-x^2+1)^(1/4)-x)/(x^2-2))
Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (54) = 108\).
Time = 0.08 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.75 \[ \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx=-\frac {1}{2} \, \arctan \left (\frac {x + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \arctan \left (-\frac {x - 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \log \left (\frac {x^{2} + 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} x + 2 \, \sqrt {-x^{2} + 1}}{x^{2}}\right ) + \frac {1}{4} \, \log \left (\frac {x^{2} - 2 \, {\left (-x^{2} + 1\right )}^{\frac {1}{4}} x + 2 \, \sqrt {-x^{2} + 1}}{x^{2}}\right ) \] Input:
integrate(x^2/(-x^2+1)^(3/4)/(-x^2+2),x, algorithm="fricas")
Output:
-1/2*arctan((x + 2*(-x^2 + 1)^(1/4))/x) - 1/2*arctan(-(x - 2*(-x^2 + 1)^(1 /4))/x) - 1/4*log((x^2 + 2*(-x^2 + 1)^(1/4)*x + 2*sqrt(-x^2 + 1))/x^2) + 1 /4*log((x^2 - 2*(-x^2 + 1)^(1/4)*x + 2*sqrt(-x^2 + 1))/x^2)
\[ \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx=- \int \frac {x^{2}}{x^{2} \left (1 - x^{2}\right )^{\frac {3}{4}} - 2 \left (1 - x^{2}\right )^{\frac {3}{4}}}\, dx \] Input:
integrate(x**2/(-x**2+1)**(3/4)/(-x**2+2),x)
Output:
-Integral(x**2/(x**2*(1 - x**2)**(3/4) - 2*(1 - x**2)**(3/4)), x)
\[ \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx=\int { -\frac {x^{2}}{{\left (x^{2} - 2\right )} {\left (-x^{2} + 1\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(x^2/(-x^2+1)^(3/4)/(-x^2+2),x, algorithm="maxima")
Output:
-integrate(x^2/((x^2 - 2)*(-x^2 + 1)^(3/4)), x)
\[ \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx=\int { -\frac {x^{2}}{{\left (x^{2} - 2\right )} {\left (-x^{2} + 1\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate(x^2/(-x^2+1)^(3/4)/(-x^2+2),x, algorithm="giac")
Output:
integrate(-x^2/((x^2 - 2)*(-x^2 + 1)^(3/4)), x)
Timed out. \[ \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx=-\int \frac {x^2}{{\left (1-x^2\right )}^{3/4}\,\left (x^2-2\right )} \,d x \] Input:
int(-x^2/((1 - x^2)^(3/4)*(x^2 - 2)),x)
Output:
-int(x^2/((1 - x^2)^(3/4)*(x^2 - 2)), x)
\[ \int \frac {x^2}{\left (1-x^2\right )^{3/4} \left (2-x^2\right )} \, dx=-\left (\int \frac {x^{2}}{\left (-x^{2}+1\right )^{\frac {3}{4}} x^{2}-2 \left (-x^{2}+1\right )^{\frac {3}{4}}}d x \right ) \] Input:
int(x^2/(-x^2+1)^(3/4)/(-x^2+2),x)
Output:
- int(x**2/(( - x**2 + 1)**(3/4)*x**2 - 2*( - x**2 + 1)**(3/4)),x)