Integrand size = 39, antiderivative size = 110 \[ \int \frac {-3 b+2 a x^5}{\left (2 b+x^3+2 a x^5\right ) \sqrt [4]{b x+a x^6}} \, dx=-\frac {\arctan \left (\frac {2^{3/4} x \sqrt [4]{b x+a x^6}}{-x^2+\sqrt {2} \sqrt {b x+a x^6}}\right )}{\sqrt [4]{2}}-\frac {\text {arctanh}\left (\frac {\frac {x^2}{2^{3/4}}+\frac {\sqrt {b x+a x^6}}{\sqrt [4]{2}}}{x \sqrt [4]{b x+a x^6}}\right )}{\sqrt [4]{2}} \]
-1/2*arctan(2^(3/4)*x*(a*x^6+b*x)^(1/4)/(-x^2+2^(1/2)*(a*x^6+b*x)^(1/2)))* 2^(3/4)-1/2*arctanh((1/2*x^2*2^(1/4)+1/2*(a*x^6+b*x)^(1/2)*2^(3/4))/x/(a*x ^6+b*x)^(1/4))*2^(3/4)
\[ \int \frac {-3 b+2 a x^5}{\left (2 b+x^3+2 a x^5\right ) \sqrt [4]{b x+a x^6}} \, dx=\int \frac {-3 b+2 a x^5}{\left (2 b+x^3+2 a x^5\right ) \sqrt [4]{b x+a x^6}} \, dx \]
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.
\(\displaystyle \int \frac {2 a x^5-3 b}{\left (2 a x^5+2 b+x^3\right ) \sqrt [4]{a x^6+b x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^5+b} \int -\frac {3 b-2 a x^5}{\sqrt [4]{x} \sqrt [4]{a x^5+b} \left (2 a x^5+x^3+2 b\right )}dx}{\sqrt [4]{a x^6+b x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{a x^5+b} \int \frac {3 b-2 a x^5}{\sqrt [4]{x} \sqrt [4]{a x^5+b} \left (2 a x^5+x^3+2 b\right )}dx}{\sqrt [4]{a x^6+b x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^5+b} \int \frac {\sqrt {x} \left (3 b-2 a x^5\right )}{\sqrt [4]{a x^5+b} \left (2 a x^5+x^3+2 b\right )}d\sqrt [4]{x}}{\sqrt [4]{a x^6+b x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^5+b} \int \left (\frac {\sqrt {x} \left (x^3+5 b\right )}{\sqrt [4]{a x^5+b} \left (2 a x^5+x^3+2 b\right )}-\frac {\sqrt {x}}{\sqrt [4]{a x^5+b}}\right )d\sqrt [4]{x}}{\sqrt [4]{a x^6+b x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^5+b} \left (5 b \int \frac {\sqrt {x}}{\sqrt [4]{a x^5+b} \left (2 a x^5+x^3+2 b\right )}d\sqrt [4]{x}+\int \frac {x^{7/2}}{\sqrt [4]{a x^5+b} \left (2 a x^5+x^3+2 b\right )}d\sqrt [4]{x}-\frac {x^{3/4} \sqrt [4]{\frac {a x^5}{b}+1} \operatorname {Hypergeometric2F1}\left (\frac {3}{20},\frac {1}{4},\frac {23}{20},-\frac {a x^5}{b}\right )}{3 \sqrt [4]{a x^5+b}}\right )}{\sqrt [4]{a x^6+b x}}\) |
3.17.18.3.1 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]
Time = 1.09 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.25
method | result | size |
pseudoelliptic | \(\frac {2^{\frac {3}{4}} \left (\ln \left (\frac {-2 \,2^{\frac {1}{4}} {\left (x \left (a \,x^{5}+b \right )\right )}^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+2 \sqrt {x \left (a \,x^{5}+b \right )}}{2 \,2^{\frac {1}{4}} {\left (x \left (a \,x^{5}+b \right )\right )}^{\frac {1}{4}} x +\sqrt {2}\, x^{2}+2 \sqrt {x \left (a \,x^{5}+b \right )}}\right )+2 \arctan \left (\frac {{\left (x \left (a \,x^{5}+b \right )\right )}^{\frac {1}{4}} 2^{\frac {3}{4}}+x}{x}\right )+2 \arctan \left (\frac {{\left (x \left (a \,x^{5}+b \right )\right )}^{\frac {1}{4}} 2^{\frac {3}{4}}-x}{x}\right )\right )}{4}\) | \(137\) |
1/4*2^(3/4)*(ln((-2*2^(1/4)*(x*(a*x^5+b))^(1/4)*x+2^(1/2)*x^2+2*(x*(a*x^5+ b))^(1/2))/(2*2^(1/4)*(x*(a*x^5+b))^(1/4)*x+2^(1/2)*x^2+2*(x*(a*x^5+b))^(1 /2)))+2*arctan(((x*(a*x^5+b))^(1/4)*2^(3/4)+x)/x)+2*arctan(((x*(a*x^5+b))^ (1/4)*2^(3/4)-x)/x))
Timed out. \[ \int \frac {-3 b+2 a x^5}{\left (2 b+x^3+2 a x^5\right ) \sqrt [4]{b x+a x^6}} \, dx=\text {Timed out} \]
\[ \int \frac {-3 b+2 a x^5}{\left (2 b+x^3+2 a x^5\right ) \sqrt [4]{b x+a x^6}} \, dx=\int \frac {2 a x^{5} - 3 b}{\sqrt [4]{x \left (a x^{5} + b\right )} \left (2 a x^{5} + 2 b + x^{3}\right )}\, dx \]
\[ \int \frac {-3 b+2 a x^5}{\left (2 b+x^3+2 a x^5\right ) \sqrt [4]{b x+a x^6}} \, dx=\int { \frac {2 \, a x^{5} - 3 \, b}{{\left (a x^{6} + b x\right )}^{\frac {1}{4}} {\left (2 \, a x^{5} + x^{3} + 2 \, b\right )}} \,d x } \]
\[ \int \frac {-3 b+2 a x^5}{\left (2 b+x^3+2 a x^5\right ) \sqrt [4]{b x+a x^6}} \, dx=\int { \frac {2 \, a x^{5} - 3 \, b}{{\left (a x^{6} + b x\right )}^{\frac {1}{4}} {\left (2 \, a x^{5} + x^{3} + 2 \, b\right )}} \,d x } \]
Timed out. \[ \int \frac {-3 b+2 a x^5}{\left (2 b+x^3+2 a x^5\right ) \sqrt [4]{b x+a x^6}} \, dx=\int -\frac {3\,b-2\,a\,x^5}{{\left (a\,x^6+b\,x\right )}^{1/4}\,\left (2\,a\,x^5+x^3+2\,b\right )} \,d x \]