Integrand size = 41, antiderivative size = 477 \[ \int \frac {x^3 (-5 b+6 a x)}{\sqrt [4]{-b x+a x^2} \left (c-b x^5+a x^6\right )} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {-2^{2/3} x \sqrt [4]{-b x+a x^2}+\sqrt [3]{c} x^2 \sqrt [4]{-b x+a x^2}}{2 \sqrt [6]{2} \sqrt [4]{c}-\sqrt {2} c^{7/12} x+2^{2/3} x \sqrt [4]{-b x+a x^2}-\sqrt [3]{c} x^2 \sqrt [4]{-b x+a x^2}}\right )}{\sqrt [4]{c}}+\frac {\sqrt {2} \arctan \left (\frac {-2^{2/3} x \sqrt [4]{-b x+a x^2}+\sqrt [3]{c} x^2 \sqrt [4]{-b x+a x^2}}{-2 \sqrt [6]{2} \sqrt [4]{c}+\sqrt {2} c^{7/12} x+2^{2/3} x \sqrt [4]{-b x+a x^2}-\sqrt [3]{c} x^2 \sqrt [4]{-b x+a x^2}}\right )}{\sqrt [4]{c}}-\frac {\sqrt {2} \text {arctanh}\left (\frac {2\ 2^{5/6} \sqrt [4]{c} x \sqrt [4]{-b x+a x^2}-4 \sqrt [6]{2} c^{7/12} x^2 \sqrt [4]{-b x+a x^2}+\sqrt {2} c^{11/12} x^3 \sqrt [4]{-b x+a x^2}}{2 \sqrt [3]{2} \sqrt {c}-2\ 2^{2/3} c^{5/6} x+c^{7/6} x^2+2 \sqrt [3]{2} x^2 \sqrt {-b x+a x^2}-2\ 2^{2/3} \sqrt [3]{c} x^3 \sqrt {-b x+a x^2}+c^{2/3} x^4 \sqrt {-b x+a x^2}}\right )}{\sqrt [4]{c}} \]
-2^(1/2)*arctan((-2^(2/3)*x*(a*x^2-b*x)^(1/4)+c^(1/3)*x^2*(a*x^2-b*x)^(1/4 ))/(2*2^(1/6)*c^(1/4)-2^(1/2)*c^(7/12)*x+2^(2/3)*x*(a*x^2-b*x)^(1/4)-c^(1/ 3)*x^2*(a*x^2-b*x)^(1/4)))/c^(1/4)+2^(1/2)*arctan((-2^(2/3)*x*(a*x^2-b*x)^ (1/4)+c^(1/3)*x^2*(a*x^2-b*x)^(1/4))/(-2*2^(1/6)*c^(1/4)+2^(1/2)*c^(7/12)* x+2^(2/3)*x*(a*x^2-b*x)^(1/4)-c^(1/3)*x^2*(a*x^2-b*x)^(1/4)))/c^(1/4)-2^(1 /2)*arctanh((2*2^(5/6)*c^(1/4)*x*(a*x^2-b*x)^(1/4)-4*2^(1/6)*c^(7/12)*x^2* (a*x^2-b*x)^(1/4)+2^(1/2)*c^(11/12)*x^3*(a*x^2-b*x)^(1/4))/(2*2^(1/3)*c^(1 /2)-2*2^(2/3)*c^(5/6)*x+c^(7/6)*x^2+2*2^(1/3)*x^2*(a*x^2-b*x)^(1/2)-2*2^(2 /3)*c^(1/3)*x^3*(a*x^2-b*x)^(1/2)+c^(2/3)*x^4*(a*x^2-b*x)^(1/2)))/c^(1/4)
\[ \int \frac {x^3 (-5 b+6 a x)}{\sqrt [4]{-b x+a x^2} \left (c-b x^5+a x^6\right )} \, dx=\int \frac {x^3 (-5 b+6 a x)}{\sqrt [4]{-b x+a x^2} \left (c-b x^5+a x^6\right )} \, dx \]
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.
\(\displaystyle \int \frac {x^3 (6 a x-5 b)}{\sqrt [4]{a x^2-b x} \left (a x^6-b x^5+c\right )} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x-b} \int -\frac {x^{11/4} (5 b-6 a x)}{\sqrt [4]{a x-b} \left (a x^6-b x^5+c\right )}dx}{\sqrt [4]{a x^2-b x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{a x-b} \int \frac {x^{11/4} (5 b-6 a x)}{\sqrt [4]{a x-b} \left (a x^6-b x^5+c\right )}dx}{\sqrt [4]{a x^2-b x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x-b} \int \frac {x^{7/2} (5 b-6 a x)}{\sqrt [4]{a x-b} \left (a x^6-b x^5+c\right )}d\sqrt [4]{x}}{\sqrt [4]{a x^2-b x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x-b} \int \left (-\frac {6 a x^{9/2}}{\sqrt [4]{a x-b} \left (a x^6-b x^5+c\right )}-\frac {5 b x^{7/2}}{\sqrt [4]{a x-b} \left (-a x^6+b x^5-c\right )}\right )d\sqrt [4]{x}}{\sqrt [4]{a x^2-b x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x-b} \left (-5 b \int \frac {x^{7/2}}{\sqrt [4]{a x-b} \left (-a x^6+b x^5-c\right )}d\sqrt [4]{x}-6 a \int \frac {x^{9/2}}{\sqrt [4]{a x-b} \left (a x^6-b x^5+c\right )}d\sqrt [4]{x}\right )}{\sqrt [4]{a x^2-b x}}\) |
3.31.66.3.1 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]
\[\int \frac {x^{3} \left (6 a x -5 b \right )}{\left (a \,x^{2}-b x \right )^{\frac {1}{4}} \left (a \,x^{6}-b \,x^{5}+c \right )}d x\]
Timed out. \[ \int \frac {x^3 (-5 b+6 a x)}{\sqrt [4]{-b x+a x^2} \left (c-b x^5+a x^6\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^3 (-5 b+6 a x)}{\sqrt [4]{-b x+a x^2} \left (c-b x^5+a x^6\right )} \, dx=\int \frac {x^{3} \cdot \left (6 a x - 5 b\right )}{\sqrt [4]{x \left (a x - b\right )} \left (a x^{6} - b x^{5} + c\right )}\, dx \]
\[ \int \frac {x^3 (-5 b+6 a x)}{\sqrt [4]{-b x+a x^2} \left (c-b x^5+a x^6\right )} \, dx=\int { \frac {{\left (6 \, a x - 5 \, b\right )} x^{3}}{{\left (a x^{6} - b x^{5} + c\right )} {\left (a x^{2} - b x\right )}^{\frac {1}{4}}} \,d x } \]
\[ \int \frac {x^3 (-5 b+6 a x)}{\sqrt [4]{-b x+a x^2} \left (c-b x^5+a x^6\right )} \, dx=\int { \frac {{\left (6 \, a x - 5 \, b\right )} x^{3}}{{\left (a x^{6} - b x^{5} + c\right )} {\left (a x^{2} - b x\right )}^{\frac {1}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^3 (-5 b+6 a x)}{\sqrt [4]{-b x+a x^2} \left (c-b x^5+a x^6\right )} \, dx=\int -\frac {x^3\,\left (5\,b-6\,a\,x\right )}{{\left (a\,x^2-b\,x\right )}^{1/4}\,\left (a\,x^6-b\,x^5+c\right )} \,d x \]