Integrand size = 38, antiderivative size = 106 \[ \int \frac {-3 b+a x^2}{\left (-b+a x^2+x^3\right ) \sqrt [4]{-b x+a x^3}} \, dx=-\sqrt {2} \arctan \left (\frac {-\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b x+a x^3}}{\sqrt {2}}}{x \sqrt [4]{-b x+a x^3}}\right )+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{-b x+a x^3}}{x^2+\sqrt {-b x+a x^3}}\right ) \]
-2^(1/2)*arctan((-1/2*2^(1/2)*x^2+1/2*(a*x^3-b*x)^(1/2)*2^(1/2))/x/(a*x^3- b*x)^(1/4))+2^(1/2)*arctanh(2^(1/2)*x*(a*x^3-b*x)^(1/4)/(x^2+(a*x^3-b*x)^( 1/2)))
Time = 10.75 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.90 \[ \int \frac {-3 b+a x^2}{\left (-b+a x^2+x^3\right ) \sqrt [4]{-b x+a x^3}} \, dx=\sqrt {2} \left (-\arctan \left (\frac {-x^2+\sqrt {-b x+a x^3}}{\sqrt {2} x \sqrt [4]{-b x+a x^3}}\right )+\text {arctanh}\left (\frac {\sqrt {2} x \sqrt [4]{-b x+a x^3}}{x^2+\sqrt {-b x+a x^3}}\right )\right ) \]
Sqrt[2]*(-ArcTan[(-x^2 + Sqrt[-(b*x) + a*x^3])/(Sqrt[2]*x*(-(b*x) + a*x^3) ^(1/4))] + ArcTanh[(Sqrt[2]*x*(-(b*x) + a*x^3)^(1/4))/(x^2 + Sqrt[-(b*x) + a*x^3])])
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.
\(\displaystyle \int \frac {a x^2-3 b}{\left (a x^2-b+x^3\right ) \sqrt [4]{a x^3-b x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^2-b} \int \frac {3 b-a x^2}{\sqrt [4]{x} \sqrt [4]{a x^2-b} \left (-x^3-a x^2+b\right )}dx}{\sqrt [4]{a x^3-b x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^2-b} \int \frac {\sqrt {x} \left (3 b-a x^2\right )}{\sqrt [4]{a x^2-b} \left (-x^3-a x^2+b\right )}d\sqrt [4]{x}}{\sqrt [4]{a x^3-b x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^2-b} \int \left (\frac {a x^{5/2}}{\sqrt [4]{a x^2-b} \left (x^3+a x^2-b\right )}+\frac {3 b \sqrt {x}}{\sqrt [4]{a x^2-b} \left (-x^3-a x^2+b\right )}\right )d\sqrt [4]{x}}{\sqrt [4]{a x^3-b x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^2-b} \left (3 b \int \frac {\sqrt {x}}{\sqrt [4]{a x^2-b} \left (-x^3-a x^2+b\right )}d\sqrt [4]{x}+a \int \frac {x^{5/2}}{\sqrt [4]{a x^2-b} \left (x^3+a x^2-b\right )}d\sqrt [4]{x}\right )}{\sqrt [4]{a x^3-b x}}\) |
3.16.30.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
\[\int \frac {a \,x^{2}-3 b}{\left (a \,x^{2}+x^{3}-b \right ) \left (a \,x^{3}-b x \right )^{\frac {1}{4}}}d x\]
Timed out. \[ \int \frac {-3 b+a x^2}{\left (-b+a x^2+x^3\right ) \sqrt [4]{-b x+a x^3}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {-3 b+a x^2}{\left (-b+a x^2+x^3\right ) \sqrt [4]{-b x+a x^3}} \, dx=\text {Timed out} \]
\[ \int \frac {-3 b+a x^2}{\left (-b+a x^2+x^3\right ) \sqrt [4]{-b x+a x^3}} \, dx=\int { \frac {a x^{2} - 3 \, b}{{\left (a x^{3} - b x\right )}^{\frac {1}{4}} {\left (a x^{2} + x^{3} - b\right )}} \,d x } \]
\[ \int \frac {-3 b+a x^2}{\left (-b+a x^2+x^3\right ) \sqrt [4]{-b x+a x^3}} \, dx=\int { \frac {a x^{2} - 3 \, b}{{\left (a x^{3} - b x\right )}^{\frac {1}{4}} {\left (a x^{2} + x^{3} - b\right )}} \,d x } \]
Timed out. \[ \int \frac {-3 b+a x^2}{\left (-b+a x^2+x^3\right ) \sqrt [4]{-b x+a x^3}} \, dx=\int -\frac {3\,b-a\,x^2}{{\left (a\,x^3-b\,x\right )}^{1/4}\,\left (x^3+a\,x^2-b\right )} \,d x \]