Integrand size = 42, antiderivative size = 107 \[ \int \frac {x^4 \left (-2 b+a x^2\right )}{\left (-b+a x^2\right )^2 \sqrt [4]{-b+a x^2+c x^4}} \, dx=-\frac {x \left (-b+a x^2+c x^4\right )^{3/4}}{2 c \left (b-a x^2\right )}-\frac {\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{-b+a x^2+c x^4}}\right )}{4 c^{5/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{-b+a x^2+c x^4}}\right )}{4 c^{5/4}} \]
-1/2*x*(c*x^4+a*x^2-b)^(3/4)/c/(-a*x^2+b)-1/4*arctan(c^(1/4)*x/(c*x^4+a*x^ 2-b)^(1/4))/c^(5/4)-1/4*arctanh(c^(1/4)*x/(c*x^4+a*x^2-b)^(1/4))/c^(5/4)
Time = 0.97 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.95 \[ \int \frac {x^4 \left (-2 b+a x^2\right )}{\left (-b+a x^2\right )^2 \sqrt [4]{-b+a x^2+c x^4}} \, dx=\frac {-\frac {2 \sqrt [4]{c} x \left (-b+a x^2+c x^4\right )^{3/4}}{b-a x^2}-\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{-b+a x^2+c x^4}}\right )-\text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{-b+a x^2+c x^4}}\right )}{4 c^{5/4}} \]
((-2*c^(1/4)*x*(-b + a*x^2 + c*x^4)^(3/4))/(b - a*x^2) - ArcTan[(c^(1/4)*x )/(-b + a*x^2 + c*x^4)^(1/4)] - ArcTanh[(c^(1/4)*x)/(-b + a*x^2 + c*x^4)^( 1/4)])/(4*c^(5/4))
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.
\(\displaystyle \int \frac {x^4 \left (a x^2-2 b\right )}{\left (a x^2-b\right )^2 \sqrt [4]{a x^2-b+c x^4}} \, dx\) |
\(\Big \downarrow \) 2250 |
\(\displaystyle \int \frac {x^4 \left (a x^2-2 b\right )}{\left (a x^2-b\right )^2 \sqrt [4]{a x^2-b+c x^4}}dx\) |
3.16.59.3.1 Defintions of rubi rules used
Int[(Px_)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_) ^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Unintegrable[Px*(f*x)^m*(d + e*x^2)^ q*(a + b*x^2 + c*x^4)^p, x] /; FreeQ[{a, b, c, d, e, f, m, p, q}, x] && Pol yQ[Px, x]
Time = 0.70 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.22
method | result | size |
pseudoelliptic | \(-\frac {-4 \left (c \,x^{4}+a \,x^{2}-b \right )^{\frac {3}{4}} x \,c^{\frac {1}{4}}+\left (\ln \left (\frac {-c^{\frac {1}{4}} x -\left (c \,x^{4}+a \,x^{2}-b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x -\left (c \,x^{4}+a \,x^{2}-b \right )^{\frac {1}{4}}}\right )-2 \arctan \left (\frac {\left (c \,x^{4}+a \,x^{2}-b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x}\right )\right ) \left (a \,x^{2}-b \right )}{c^{\frac {5}{4}} \left (8 a \,x^{2}-8 b \right )}\) | \(131\) |
-1/c^(5/4)*(-4*(c*x^4+a*x^2-b)^(3/4)*x*c^(1/4)+(ln((-c^(1/4)*x-(c*x^4+a*x^ 2-b)^(1/4))/(c^(1/4)*x-(c*x^4+a*x^2-b)^(1/4)))-2*arctan(1/c^(1/4)/x*(c*x^4 +a*x^2-b)^(1/4)))*(a*x^2-b))/(8*a*x^2-8*b)
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 239, normalized size of antiderivative = 2.23 \[ \int \frac {x^4 \left (-2 b+a x^2\right )}{\left (-b+a x^2\right )^2 \sqrt [4]{-b+a x^2+c x^4}} \, dx=-\frac {{\left (a c x^{2} - b c\right )} \frac {1}{c^{5}}^{\frac {1}{4}} \log \left (\frac {c^{4} \frac {1}{c^{5}}^{\frac {3}{4}} x + {\left (c x^{4} + a x^{2} - b\right )}^{\frac {1}{4}}}{x}\right ) - {\left (a c x^{2} - b c\right )} \frac {1}{c^{5}}^{\frac {1}{4}} \log \left (-\frac {c^{4} \frac {1}{c^{5}}^{\frac {3}{4}} x - {\left (c x^{4} + a x^{2} - b\right )}^{\frac {1}{4}}}{x}\right ) + {\left (-i \, a c x^{2} + i \, b c\right )} \frac {1}{c^{5}}^{\frac {1}{4}} \log \left (\frac {i \, c^{4} \frac {1}{c^{5}}^{\frac {3}{4}} x + {\left (c x^{4} + a x^{2} - b\right )}^{\frac {1}{4}}}{x}\right ) + {\left (i \, a c x^{2} - i \, b c\right )} \frac {1}{c^{5}}^{\frac {1}{4}} \log \left (\frac {-i \, c^{4} \frac {1}{c^{5}}^{\frac {3}{4}} x + {\left (c x^{4} + a x^{2} - b\right )}^{\frac {1}{4}}}{x}\right ) - 4 \, {\left (c x^{4} + a x^{2} - b\right )}^{\frac {3}{4}} x}{8 \, {\left (a c x^{2} - b c\right )}} \]
-1/8*((a*c*x^2 - b*c)*(c^(-5))^(1/4)*log((c^4*(c^(-5))^(3/4)*x + (c*x^4 + a*x^2 - b)^(1/4))/x) - (a*c*x^2 - b*c)*(c^(-5))^(1/4)*log(-(c^4*(c^(-5))^( 3/4)*x - (c*x^4 + a*x^2 - b)^(1/4))/x) + (-I*a*c*x^2 + I*b*c)*(c^(-5))^(1/ 4)*log((I*c^4*(c^(-5))^(3/4)*x + (c*x^4 + a*x^2 - b)^(1/4))/x) + (I*a*c*x^ 2 - I*b*c)*(c^(-5))^(1/4)*log((-I*c^4*(c^(-5))^(3/4)*x + (c*x^4 + a*x^2 - b)^(1/4))/x) - 4*(c*x^4 + a*x^2 - b)^(3/4)*x)/(a*c*x^2 - b*c)
\[ \int \frac {x^4 \left (-2 b+a x^2\right )}{\left (-b+a x^2\right )^2 \sqrt [4]{-b+a x^2+c x^4}} \, dx=\int \frac {x^{4} \left (a x^{2} - 2 b\right )}{\left (a x^{2} - b\right )^{2} \sqrt [4]{a x^{2} - b + c x^{4}}}\, dx \]
\[ \int \frac {x^4 \left (-2 b+a x^2\right )}{\left (-b+a x^2\right )^2 \sqrt [4]{-b+a x^2+c x^4}} \, dx=\int { \frac {{\left (a x^{2} - 2 \, b\right )} x^{4}}{{\left (c x^{4} + a x^{2} - b\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}^{2}} \,d x } \]
\[ \int \frac {x^4 \left (-2 b+a x^2\right )}{\left (-b+a x^2\right )^2 \sqrt [4]{-b+a x^2+c x^4}} \, dx=\int { \frac {{\left (a x^{2} - 2 \, b\right )} x^{4}}{{\left (c x^{4} + a x^{2} - b\right )}^{\frac {1}{4}} {\left (a x^{2} - b\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (-2 b+a x^2\right )}{\left (-b+a x^2\right )^2 \sqrt [4]{-b+a x^2+c x^4}} \, dx=-\int \frac {x^4\,\left (2\,b-a\,x^2\right )}{{\left (b-a\,x^2\right )}^2\,{\left (c\,x^4+a\,x^2-b\right )}^{1/4}} \,d x \]