Integrand size = 27, antiderivative size = 176 \[ \int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}-\sqrt [4]{2} x+\sqrt [4]{1-x^2}}\right )}{2 \sqrt [4]{2}}+\frac {\arctan \left (\frac {\sqrt [4]{1-x^2}}{-\sqrt [4]{2}+\sqrt [4]{2} x+\sqrt [4]{1-x^2}}\right )}{2 \sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\left (-2 \sqrt [4]{2}+2 \sqrt [4]{2} x\right ) \sqrt [4]{1-x^2}}{\sqrt {2}-2 \sqrt {2} x+\sqrt {2} x^2+2 \sqrt {1-x^2}}\right )}{2 \sqrt [4]{2}} \]
-1/4*arctan((-x^2+1)^(1/4)/(2^(1/4)-2^(1/4)*x+(-x^2+1)^(1/4)))*2^(3/4)+1/4 *arctan((-x^2+1)^(1/4)/(-2^(1/4)+2^(1/4)*x+(-x^2+1)^(1/4)))*2^(3/4)-1/4*ar ctanh((-2*2^(1/4)+2*2^(1/4)*x)*(-x^2+1)^(1/4)/(2^(1/2)-2*x*2^(1/2)+2^(1/2) *x^2+2*(-x^2+1)^(1/2)))*2^(3/4)
Time = 0.26 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.84 \[ \int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}-\sqrt [4]{2} x-\sqrt [4]{1-x^2}}\right )+\arctan \left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}-\sqrt [4]{2} x+\sqrt [4]{1-x^2}}\right )+\text {arctanh}\left (\frac {2 (-1+x) \sqrt [4]{2-2 x^2}}{\sqrt {2}-2 \sqrt {2} x+\sqrt {2} x^2+2 \sqrt {1-x^2}}\right )}{2 \sqrt [4]{2}} \]
-1/2*(ArcTan[(1 - x^2)^(1/4)/(2^(1/4) - 2^(1/4)*x - (1 - x^2)^(1/4))] + Ar cTan[(1 - x^2)^(1/4)/(2^(1/4) - 2^(1/4)*x + (1 - x^2)^(1/4))] + ArcTanh[(2 *(-1 + x)*(2 - 2*x^2)^(1/4))/(Sqrt[2] - 2*Sqrt[2]*x + Sqrt[2]*x^2 + 2*Sqrt [1 - x^2])])/2^(1/4)
Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
Time = 0.91 (sec) , antiderivative size = 378, normalized size of antiderivative = 2.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {7276, 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.
\(\displaystyle \int \frac {x+2}{(x-3) \sqrt [4]{1-x^2} \left (x^2+1\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {-x-1}{2 \sqrt [4]{1-x^2} \left (x^2+1\right )}+\frac {1}{2 (x-3) \sqrt [4]{1-x^2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sqrt {x^2} \operatorname {EllipticPi}\left (-\frac {1}{\sqrt {2}},\arcsin \left (\sqrt [4]{1-x^2}\right ),-1\right )}{2 \sqrt {2} x}+\frac {3 i \sqrt {x^2} \operatorname {EllipticPi}\left (-\frac {i}{2 \sqrt {2}},\arcsin \left (\sqrt [4]{1-x^2}\right ),-1\right )}{4 \sqrt {2} x}-\frac {3 i \sqrt {x^2} \operatorname {EllipticPi}\left (\frac {i}{2 \sqrt {2}},\arcsin \left (\sqrt [4]{1-x^2}\right ),-1\right )}{4 \sqrt {2} x}+\frac {\sqrt {x^2} \operatorname {EllipticPi}\left (\frac {1}{\sqrt {2}},\arcsin \left (\sqrt [4]{1-x^2}\right ),-1\right )}{2 \sqrt {2} x}-\frac {\arctan \left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}-\frac {\arctan \left (1-\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{4 \sqrt [4]{2}}+\frac {\arctan \left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}+1\right )}{4 \sqrt [4]{2}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{1-x^2}}{\sqrt [4]{2}}\right )}{2 \sqrt [4]{2}}+\frac {\log \left (\sqrt {1-x^2}-2 \sqrt [4]{2} \sqrt [4]{1-x^2}+2 \sqrt {2}\right )}{8 \sqrt [4]{2}}-\frac {\log \left (\sqrt {1-x^2}+2 \sqrt [4]{2} \sqrt [4]{1-x^2}+2 \sqrt {2}\right )}{8 \sqrt [4]{2}}\) |
-1/2*ArcTan[(1 - x^2)^(1/4)/2^(1/4)]/2^(1/4) - ArcTan[1 - (1 - x^2)^(1/4)/ 2^(1/4)]/(4*2^(1/4)) + ArcTan[1 + (1 - x^2)^(1/4)/2^(1/4)]/(4*2^(1/4)) + A rcTanh[(1 - x^2)^(1/4)/2^(1/4)]/(2*2^(1/4)) - (Sqrt[x^2]*EllipticPi[-(1/Sq rt[2]), ArcSin[(1 - x^2)^(1/4)], -1])/(2*Sqrt[2]*x) + (((3*I)/4)*Sqrt[x^2] *EllipticPi[(-1/2*I)/Sqrt[2], ArcSin[(1 - x^2)^(1/4)], -1])/(Sqrt[2]*x) - (((3*I)/4)*Sqrt[x^2]*EllipticPi[(I/2)/Sqrt[2], ArcSin[(1 - x^2)^(1/4)], -1 ])/(Sqrt[2]*x) + (Sqrt[x^2]*EllipticPi[1/Sqrt[2], ArcSin[(1 - x^2)^(1/4)], -1])/(2*Sqrt[2]*x) + Log[2*Sqrt[2] - 2*2^(1/4)*(1 - x^2)^(1/4) + Sqrt[1 - x^2]]/(8*2^(1/4)) - Log[2*Sqrt[2] + 2*2^(1/4)*(1 - x^2)^(1/4) + Sqrt[1 - x^2]]/(8*2^(1/4))
3.23.99.3.1 Defintions of rubi rules used
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.68 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.32
method | result | size |
trager | \(-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right ) \ln \left (-\frac {2 \sqrt {-x^{2}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{3} x -2 \sqrt {-x^{2}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{3}+2 \left (-x^{2}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x^{2}-4 \left (-x^{2}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right ) x^{3}+2 \left (-x^{2}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}-3 \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right ) x^{2}-4 \left (-x^{2}+1\right )^{\frac {3}{4}}+5 \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right ) x +\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )}{\left (-3+x \right ) \left (x^{2}+1\right )}\right )}{4}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) \ln \left (-\frac {2 \sqrt {-x^{2}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x -2 \sqrt {-x^{2}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}-2 \left (-x^{2}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x^{2}+4 \left (-x^{2}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2} x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) x^{3}-2 \left (-x^{2}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}-4 \left (-x^{2}+1\right )^{\frac {3}{4}}+3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) x^{2}-5 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right ) x -\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}+2\right )^{2}\right )}{\left (-3+x \right ) \left (x^{2}+1\right )}\right )}{4}\) | \(408\) |
-1/4*RootOf(_Z^4+2)*ln(-(2*(-x^2+1)^(1/2)*RootOf(_Z^4+2)^3*x-2*(-x^2+1)^(1 /2)*RootOf(_Z^4+2)^3+2*(-x^2+1)^(1/4)*RootOf(_Z^4+2)^2*x^2-4*(-x^2+1)^(1/4 )*RootOf(_Z^4+2)^2*x+RootOf(_Z^4+2)*x^3+2*(-x^2+1)^(1/4)*RootOf(_Z^4+2)^2- 3*RootOf(_Z^4+2)*x^2-4*(-x^2+1)^(3/4)+5*RootOf(_Z^4+2)*x+RootOf(_Z^4+2))/( -3+x)/(x^2+1))+1/4*RootOf(_Z^2+RootOf(_Z^4+2)^2)*ln(-(2*(-x^2+1)^(1/2)*Roo tOf(_Z^2+RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2*x-2*(-x^2+1)^(1/2)*RootOf(_Z^2 +RootOf(_Z^4+2)^2)*RootOf(_Z^4+2)^2-2*(-x^2+1)^(1/4)*RootOf(_Z^4+2)^2*x^2+ 4*(-x^2+1)^(1/4)*RootOf(_Z^4+2)^2*x-RootOf(_Z^2+RootOf(_Z^4+2)^2)*x^3-2*(- x^2+1)^(1/4)*RootOf(_Z^4+2)^2-4*(-x^2+1)^(3/4)+3*RootOf(_Z^2+RootOf(_Z^4+2 )^2)*x^2-5*RootOf(_Z^2+RootOf(_Z^4+2)^2)*x-RootOf(_Z^2+RootOf(_Z^4+2)^2))/ (-3+x)/(x^2+1))
Result contains complex when optimal does not.
Time = 5.08 (sec) , antiderivative size = 413, normalized size of antiderivative = 2.35 \[ \int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx=\left (\frac {1}{64} i + \frac {1}{64}\right ) \cdot 8^{\frac {3}{4}} \sqrt {2} \log \left (\frac {8^{\frac {3}{4}} \sqrt {2} {\left (\left (i + 1\right ) \, x^{3} - \left (3 i + 3\right ) \, x^{2} + \left (5 i + 5\right ) \, x + i + 1\right )} - 8 \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-x^{2} + 1} {\left (-\left (i - 1\right ) \, x + i - 1\right )} - 16 \, \sqrt {2} {\left (i \, x^{2} - 2 i \, x + i\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} + 32 \, {\left (-x^{2} + 1\right )}^{\frac {3}{4}}}{x^{3} - 3 \, x^{2} + x - 3}\right ) - \left (\frac {1}{64} i - \frac {1}{64}\right ) \cdot 8^{\frac {3}{4}} \sqrt {2} \log \left (\frac {8^{\frac {3}{4}} \sqrt {2} {\left (-\left (i - 1\right ) \, x^{3} + \left (3 i - 3\right ) \, x^{2} - \left (5 i - 5\right ) \, x - i + 1\right )} - 8 \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-x^{2} + 1} {\left (\left (i + 1\right ) \, x - i - 1\right )} - 16 \, \sqrt {2} {\left (-i \, x^{2} + 2 i \, x - i\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} + 32 \, {\left (-x^{2} + 1\right )}^{\frac {3}{4}}}{x^{3} - 3 \, x^{2} + x - 3}\right ) + \left (\frac {1}{64} i - \frac {1}{64}\right ) \cdot 8^{\frac {3}{4}} \sqrt {2} \log \left (\frac {8^{\frac {3}{4}} \sqrt {2} {\left (\left (i - 1\right ) \, x^{3} - \left (3 i - 3\right ) \, x^{2} + \left (5 i - 5\right ) \, x + i - 1\right )} - 8 \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-x^{2} + 1} {\left (-\left (i + 1\right ) \, x + i + 1\right )} - 16 \, \sqrt {2} {\left (-i \, x^{2} + 2 i \, x - i\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} + 32 \, {\left (-x^{2} + 1\right )}^{\frac {3}{4}}}{x^{3} - 3 \, x^{2} + x - 3}\right ) - \left (\frac {1}{64} i + \frac {1}{64}\right ) \cdot 8^{\frac {3}{4}} \sqrt {2} \log \left (\frac {8^{\frac {3}{4}} \sqrt {2} {\left (-\left (i + 1\right ) \, x^{3} + \left (3 i + 3\right ) \, x^{2} - \left (5 i + 5\right ) \, x - i - 1\right )} - 8 \cdot 8^{\frac {1}{4}} \sqrt {2} \sqrt {-x^{2} + 1} {\left (\left (i - 1\right ) \, x - i + 1\right )} - 16 \, \sqrt {2} {\left (i \, x^{2} - 2 i \, x + i\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} + 32 \, {\left (-x^{2} + 1\right )}^{\frac {3}{4}}}{x^{3} - 3 \, x^{2} + x - 3}\right ) \]
(1/64*I + 1/64)*8^(3/4)*sqrt(2)*log((8^(3/4)*sqrt(2)*((I + 1)*x^3 - (3*I + 3)*x^2 + (5*I + 5)*x + I + 1) - 8*8^(1/4)*sqrt(2)*sqrt(-x^2 + 1)*(-(I - 1 )*x + I - 1) - 16*sqrt(2)*(I*x^2 - 2*I*x + I)*(-x^2 + 1)^(1/4) + 32*(-x^2 + 1)^(3/4))/(x^3 - 3*x^2 + x - 3)) - (1/64*I - 1/64)*8^(3/4)*sqrt(2)*log(( 8^(3/4)*sqrt(2)*(-(I - 1)*x^3 + (3*I - 3)*x^2 - (5*I - 5)*x - I + 1) - 8*8 ^(1/4)*sqrt(2)*sqrt(-x^2 + 1)*((I + 1)*x - I - 1) - 16*sqrt(2)*(-I*x^2 + 2 *I*x - I)*(-x^2 + 1)^(1/4) + 32*(-x^2 + 1)^(3/4))/(x^3 - 3*x^2 + x - 3)) + (1/64*I - 1/64)*8^(3/4)*sqrt(2)*log((8^(3/4)*sqrt(2)*((I - 1)*x^3 - (3*I - 3)*x^2 + (5*I - 5)*x + I - 1) - 8*8^(1/4)*sqrt(2)*sqrt(-x^2 + 1)*(-(I + 1)*x + I + 1) - 16*sqrt(2)*(-I*x^2 + 2*I*x - I)*(-x^2 + 1)^(1/4) + 32*(-x^ 2 + 1)^(3/4))/(x^3 - 3*x^2 + x - 3)) - (1/64*I + 1/64)*8^(3/4)*sqrt(2)*log ((8^(3/4)*sqrt(2)*(-(I + 1)*x^3 + (3*I + 3)*x^2 - (5*I + 5)*x - I - 1) - 8 *8^(1/4)*sqrt(2)*sqrt(-x^2 + 1)*((I - 1)*x - I + 1) - 16*sqrt(2)*(I*x^2 - 2*I*x + I)*(-x^2 + 1)^(1/4) + 32*(-x^2 + 1)^(3/4))/(x^3 - 3*x^2 + x - 3))
\[ \int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx=\int \frac {x + 2}{\sqrt [4]{- \left (x - 1\right ) \left (x + 1\right )} \left (x - 3\right ) \left (x^{2} + 1\right )}\, dx \]
\[ \int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx=\int { \frac {x + 2}{{\left (x^{2} + 1\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} {\left (x - 3\right )}} \,d x } \]
\[ \int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx=\int { \frac {x + 2}{{\left (x^{2} + 1\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}} {\left (x - 3\right )}} \,d x } \]
Timed out. \[ \int \frac {2+x}{(-3+x) \sqrt [4]{1-x^2} \left (1+x^2\right )} \, dx=\int \frac {x+2}{{\left (1-x^2\right )}^{1/4}\,\left (x^2+1\right )\,\left (x-3\right )} \,d x \]