Integrand size = 43, antiderivative size = 58 \[ \int \frac {x^3 \left (-5 b+9 a x^4\right )}{\sqrt [4]{-b x+a x^5} \left (-2-b x^5+a x^9\right )} \, dx=2^{3/4} \arctan \left (\frac {x \sqrt [4]{-b x+a x^5}}{\sqrt [4]{2}}\right )-2^{3/4} \text {arctanh}\left (\frac {x \sqrt [4]{-b x+a x^5}}{\sqrt [4]{2}}\right ) \]
2^(3/4)*arctan(1/2*x*(a*x^5-b*x)^(1/4)*2^(3/4))-2^(3/4)*arctanh(1/2*x*(a*x ^5-b*x)^(1/4)*2^(3/4))
\[ \int \frac {x^3 \left (-5 b+9 a x^4\right )}{\sqrt [4]{-b x+a x^5} \left (-2-b x^5+a x^9\right )} \, dx=\int \frac {x^3 \left (-5 b+9 a x^4\right )}{\sqrt [4]{-b x+a x^5} \left (-2-b x^5+a x^9\right )} \, dx \]
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.
\(\displaystyle \int \frac {x^3 \left (9 a x^4-5 b\right )}{\sqrt [4]{a x^5-b x} \left (a x^9-b x^5-2\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^4-b} \int \frac {x^{11/4} \left (5 b-9 a x^4\right )}{\sqrt [4]{a x^4-b} \left (-a x^9+b x^5+2\right )}dx}{\sqrt [4]{a x^5-b x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^4-b} \int \frac {x^{7/2} \left (5 b-9 a x^4\right )}{\sqrt [4]{a x^4-b} \left (-a x^9+b x^5+2\right )}d\sqrt [4]{x}}{\sqrt [4]{a x^5-b x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^4-b} \int \left (\frac {9 a x^{15/2}}{\sqrt [4]{a x^4-b} \left (a x^9-b x^5-2\right )}+\frac {5 b x^{7/2}}{\sqrt [4]{a x^4-b} \left (-a x^9+b x^5+2\right )}\right )d\sqrt [4]{x}}{\sqrt [4]{a x^5-b x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {4 \sqrt [4]{x} \sqrt [4]{a x^4-b} \left (5 b \int \frac {x^{7/2}}{\sqrt [4]{a x^4-b} \left (-a x^9+b x^5+2\right )}d\sqrt [4]{x}+9 a \int \frac {x^{15/2}}{\sqrt [4]{a x^4-b} \left (a x^9-b x^5-2\right )}d\sqrt [4]{x}\right )}{\sqrt [4]{a x^5-b x}}\) |
3.8.51.3.1 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]
\[\int \frac {x^{3} \left (9 a \,x^{4}-5 b \right )}{\left (a \,x^{5}-b x \right )^{\frac {1}{4}} \left (a \,x^{9}-b \,x^{5}-2\right )}d x\]
Timed out. \[ \int \frac {x^3 \left (-5 b+9 a x^4\right )}{\sqrt [4]{-b x+a x^5} \left (-2-b x^5+a x^9\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 \left (-5 b+9 a x^4\right )}{\sqrt [4]{-b x+a x^5} \left (-2-b x^5+a x^9\right )} \, dx=\int \frac {x^{3} \cdot \left (9 a x^{4} - 5 b\right )}{\sqrt [4]{x \left (a x^{4} - b\right )} \left (a x^{9} - b x^{5} - 2\right )}\, dx \]
\[ \int \frac {x^3 \left (-5 b+9 a x^4\right )}{\sqrt [4]{-b x+a x^5} \left (-2-b x^5+a x^9\right )} \, dx=\int { \frac {{\left (9 \, a x^{4} - 5 \, b\right )} x^{3}}{{\left (a x^{9} - b x^{5} - 2\right )} {\left (a x^{5} - b x\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {x^3 \left (-5 b+9 a x^4\right )}{\sqrt [4]{-b x+a x^5} \left (-2-b x^5+a x^9\right )} \, dx=\int { \frac {{\left (9 \, a x^{4} - 5 \, b\right )} x^{3}}{{\left (a x^{9} - b x^{5} - 2\right )} {\left (a x^{5} - b x\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (-5 b+9 a x^4\right )}{\sqrt [4]{-b x+a x^5} \left (-2-b x^5+a x^9\right )} \, dx=\int \frac {x^3\,\left (5\,b-9\,a\,x^4\right )}{{\left (a\,x^5-b\,x\right )}^{1/4}\,\left (-a\,x^9+b\,x^5+2\right )} \,d x \]