Integrand size = 26, antiderivative size = 149 \[ \int \frac {a+b x}{\sqrt [4]{1-x^2} \left (2-x^2\right )} \, dx=\frac {b \arctan \left (\frac {1-\sqrt {1-x^2}}{\sqrt {2} \sqrt [4]{1-x^2}}\right )}{\sqrt {2}}+\frac {1}{2} a \arctan \left (\frac {1-\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right )+\frac {b \text {arctanh}\left (\frac {1+\sqrt {1-x^2}}{\sqrt {2} \sqrt [4]{1-x^2}}\right )}{\sqrt {2}}+\frac {1}{2} a \text {arctanh}\left (\frac {1+\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right ) \]
1/2*a*arctan((1-(-x^2+1)^(1/2))/x/(-x^2+1)^(1/4))+1/2*a*arctanh((1+(-x^2+1 )^(1/2))/x/(-x^2+1)^(1/4))+1/2*b*arctan(1/2*(1-(-x^2+1)^(1/2))/(-x^2+1)^(1 /4)*2^(1/2))*2^(1/2)+1/2*b*arctanh(1/2*(1+(-x^2+1)^(1/2))/(-x^2+1)^(1/4)*2 ^(1/2))*2^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 10.18 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.97 \[ \int \frac {a+b x}{\sqrt [4]{1-x^2} \left (2-x^2\right )} \, dx=\frac {1}{4} b x^2 \operatorname {AppellF1}\left (1,\frac {1}{4},1,2,x^2,\frac {x^2}{2}\right )-\frac {6 a x \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},x^2,\frac {x^2}{2}\right )}{\sqrt [4]{1-x^2} \left (-2+x^2\right ) \left (6 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},1,\frac {3}{2},x^2,\frac {x^2}{2}\right )+x^2 \left (2 \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},2,\frac {5}{2},x^2,\frac {x^2}{2}\right )+\operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},1,\frac {5}{2},x^2,\frac {x^2}{2}\right )\right )\right )} \]
(b*x^2*AppellF1[1, 1/4, 1, 2, x^2, x^2/2])/4 - (6*a*x*AppellF1[1/2, 1/4, 1 , 3/2, x^2, x^2/2])/((1 - x^2)^(1/4)*(-2 + x^2)*(6*AppellF1[1/2, 1/4, 1, 3 /2, x^2, x^2/2] + x^2*(2*AppellF1[3/2, 1/4, 2, 5/2, x^2, x^2/2] + AppellF1 [3/2, 5/4, 1, 5/2, x^2, x^2/2])))
Time = 0.22 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1343, 308, 348}
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.
\(\displaystyle \int \frac {a+b x}{\sqrt [4]{1-x^2} \left (2-x^2\right )} \, dx\) |
\(\Big \downarrow \) 1343 |
\(\displaystyle a \int \frac {1}{\sqrt [4]{1-x^2} \left (2-x^2\right )}dx+b \int \frac {x}{\sqrt [4]{1-x^2} \left (2-x^2\right )}dx\) |
\(\Big \downarrow \) 308 |
\(\displaystyle b \int \frac {x}{\sqrt [4]{1-x^2} \left (2-x^2\right )}dx+a \left (\frac {1}{2} \arctan \left (\frac {1-\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {1-x^2}+1}{x \sqrt [4]{1-x^2}}\right )\right )\) |
\(\Big \downarrow \) 348 |
\(\displaystyle a \left (\frac {1}{2} \arctan \left (\frac {1-\sqrt {1-x^2}}{x \sqrt [4]{1-x^2}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {\sqrt {1-x^2}+1}{x \sqrt [4]{1-x^2}}\right )\right )+b \left (\frac {\arctan \left (\frac {1-\sqrt {1-x^2}}{\sqrt {2} \sqrt [4]{1-x^2}}\right )}{\sqrt {2}}+\frac {\text {arctanh}\left (\frac {\sqrt {1-x^2}+1}{\sqrt {2} \sqrt [4]{1-x^2}}\right )}{\sqrt {2}}\right )\) |
b*(ArcTan[(1 - Sqrt[1 - x^2])/(Sqrt[2]*(1 - x^2)^(1/4))]/Sqrt[2] + ArcTanh [(1 + Sqrt[1 - x^2])/(Sqrt[2]*(1 - x^2)^(1/4))]/Sqrt[2]) + a*(ArcTan[(1 - Sqrt[1 - x^2])/(x*(1 - x^2)^(1/4))]/2 + ArcTanh[(1 + Sqrt[1 - x^2])/(x*(1 - x^2)^(1/4))]/2)
3.1.71.3.1 Defintions of rubi rules used
Int[1/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Wit h[{q = Rt[b^2/a, 4]}, Simp[(-b/(2*a*d*q))*ArcTan[(b + q^2*Sqrt[a + b*x^2])/ (q^3*x*(a + b*x^2)^(1/4))], x] - Simp[(b/(2*a*d*q))*ArcTanh[(b - q^2*Sqrt[a + b*x^2])/(q^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]
Int[(x_)/(((a_) + (b_.)*(x_)^2)^(1/4)*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Simp[(-(Sqrt[2]*Rt[a, 4]*d)^(-1))*ArcTan[(Rt[a, 4]^2 - Sqrt[a + b*x^2])/(Sq rt[2]*Rt[a, 4]*(a + b*x^2)^(1/4))], x] - Simp[(1/(Sqrt[2]*Rt[a, 4]*d))*ArcT anh[(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]
Int[((g_) + (h_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_)*((d_) + (f_.)*(x_)^2)^(q _), x_Symbol] :> Simp[g Int[(a + c*x^2)^p*(d + f*x^2)^q, x], x] + Simp[h Int[x*(a + c*x^2)^p*(d + f*x^2)^q, x], x] /; FreeQ[{a, c, d, f, g, h, p, q}, x]
\[\int \frac {b x +a}{\left (-x^{2}+1\right )^{\frac {1}{4}} \left (-x^{2}+2\right )}d x\]
Timed out. \[ \int \frac {a+b x}{\sqrt [4]{1-x^2} \left (2-x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {a+b x}{\sqrt [4]{1-x^2} \left (2-x^2\right )} \, dx=- \int \frac {a}{x^{2} \sqrt [4]{1 - x^{2}} - 2 \sqrt [4]{1 - x^{2}}}\, dx - \int \frac {b x}{x^{2} \sqrt [4]{1 - x^{2}} - 2 \sqrt [4]{1 - x^{2}}}\, dx \]
-Integral(a/(x**2*(1 - x**2)**(1/4) - 2*(1 - x**2)**(1/4)), x) - Integral( b*x/(x**2*(1 - x**2)**(1/4) - 2*(1 - x**2)**(1/4)), x)
\[ \int \frac {a+b x}{\sqrt [4]{1-x^2} \left (2-x^2\right )} \, dx=\int { -\frac {b x + a}{{\left (x^{2} - 2\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {a+b x}{\sqrt [4]{1-x^2} \left (2-x^2\right )} \, dx=\int { -\frac {b x + a}{{\left (x^{2} - 2\right )} {\left (-x^{2} + 1\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {a+b x}{\sqrt [4]{1-x^2} \left (2-x^2\right )} \, dx=\int -\frac {a+b\,x}{{\left (1-x^2\right )}^{1/4}\,\left (x^2-2\right )} \,d x \]