Integrand size = 52, antiderivative size = 146 \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\arctan \left (\frac {\sqrt [4]{-b+a x^3}}{x}\right )-\frac {\arctan \left (\frac {-\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b+a x^3}}{\sqrt {2}}}{x \sqrt [4]{-b+a x^3}}\right )}{\sqrt {2}}+\text {arctanh}\left (\frac {x \left (-b+a x^3\right )^{3/4}}{b-a x^3}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{-b+a x^3}}{x^2+\sqrt {-b+a x^3}}\right )}{\sqrt {2}} \]
arctan((a*x^3-b)^(1/4)/x)-1/2*2^(1/2)*arctan((-1/2*2^(1/2)*x^2+1/2*(a*x^3- b)^(1/2)*2^(1/2))/x/(a*x^3-b)^(1/4))+arctanh(x*(a*x^3-b)^(3/4)/(-a*x^3+b)) +1/2*2^(1/2)*arctanh(2^(1/2)*x*(a*x^3-b)^(1/4)/(x^2+(a*x^3-b)^(1/2)))
Time = 1.57 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.90 \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\arctan \left (\frac {\sqrt [4]{-b+a x^3}}{x}\right )-\frac {\arctan \left (\frac {-x^2+\sqrt {-b+a x^3}}{\sqrt {2} x \sqrt [4]{-b+a x^3}}\right )}{\sqrt {2}}-\text {arctanh}\left (\frac {x}{\sqrt [4]{-b+a x^3}}\right )+\frac {\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{-b+a x^3}}{x^2+\sqrt {-b+a x^3}}\right )}{\sqrt {2}} \]
ArcTan[(-b + a*x^3)^(1/4)/x] - ArcTan[(-x^2 + Sqrt[-b + a*x^3])/(Sqrt[2]*x *(-b + a*x^3)^(1/4))]/Sqrt[2] - ArcTanh[x/(-b + a*x^3)^(1/4)] + ArcTanh[(S qrt[2]*x*(-b + a*x^3)^(1/4))/(x^2 + Sqrt[-b + a*x^3])]/Sqrt[2]
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.
\(\displaystyle \int \frac {x^4 \left (a x^3-4 b\right )}{\sqrt [4]{a x^3-b} \left (-a^2 x^6+2 a b x^3-b^2+x^8\right )} \, dx\) |
\(\Big \downarrow \) 2463 |
\(\displaystyle \int \left (\frac {x^4 \left (-a^2 x^2-a x^3-b\right ) \left (a x^3-4 b\right )}{2 b^2 \sqrt [4]{a x^3-b} \left (-a x^3+b-x^4\right )}+\frac {x^4 \left (a^2 x^2-a x^3-b\right ) \left (a x^3-4 b\right )}{2 b^2 \sqrt [4]{a x^3-b} \left (-a x^3+b+x^4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 b \int \frac {1}{\sqrt [4]{a x^3-b} \left (-x^4-a x^3+b\right )}dx-\frac {1}{2} a \int \frac {x^3}{\sqrt [4]{a x^3-b} \left (-x^4+a x^3-b\right )}dx-2 b \int \frac {1}{\sqrt [4]{a x^3-b} \left (x^4-a x^3+b\right )}dx+\frac {1}{2} a \int \frac {x^3}{\sqrt [4]{a x^3-b} \left (x^4+a x^3-b\right )}dx\) |
3.21.44.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
\[\int \frac {x^{4} \left (a \,x^{3}-4 b \right )}{\left (a \,x^{3}-b \right )^{\frac {1}{4}} \left (-a^{2} x^{6}+x^{8}+2 a b \,x^{3}-b^{2}\right )}d x\]
Timed out. \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\int { -\frac {{\left (a x^{3} - 4 \, b\right )} x^{4}}{{\left (a^{2} x^{6} - x^{8} - 2 \, a b x^{3} + b^{2}\right )} {\left (a x^{3} - b\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=\int { -\frac {{\left (a x^{3} - 4 \, b\right )} x^{4}}{{\left (a^{2} x^{6} - x^{8} - 2 \, a b x^{3} + b^{2}\right )} {\left (a x^{3} - b\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^4 \left (-4 b+a x^3\right )}{\sqrt [4]{-b+a x^3} \left (-b^2+2 a b x^3-a^2 x^6+x^8\right )} \, dx=-\int -\frac {x^4\,\left (4\,b-a\,x^3\right )}{{\left (a\,x^3-b\right )}^{1/4}\,\left (a^2\,x^6-2\,a\,b\,x^3+b^2-x^8\right )} \,d x \]