Integrand size = 38, antiderivative size = 108 \[ \int \frac {3 b+a x^4}{\left (-b+x^3+a x^4\right ) \sqrt [4]{-b x+a x^5}} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} x \sqrt [4]{-b x+a x^5}}{-x^2+\sqrt {-b x+a x^5}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\frac {x^2}{\sqrt {2}}+\frac {\sqrt {-b x+a x^5}}{\sqrt {2}}}{x \sqrt [4]{-b x+a x^5}}\right ) \]
-2^(1/2)*arctan(2^(1/2)*x*(a*x^5-b*x)^(1/4)/(-x^2+(a*x^5-b*x)^(1/2)))-2^(1 /2)*arctanh((1/2*2^(1/2)*x^2+1/2*(a*x^5-b*x)^(1/2)*2^(1/2))/x/(a*x^5-b*x)^ (1/4))
Time = 11.49 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.87 \[ \int \frac {3 b+a x^4}{\left (-b+x^3+a x^4\right ) \sqrt [4]{-b x+a x^5}} \, dx=-\sqrt {2} \left (\arctan \left (\frac {\sqrt {2} x \sqrt [4]{-b x+a x^5}}{-x^2+\sqrt {-b x+a x^5}}\right )+\text {arctanh}\left (\frac {x^2+\sqrt {-b x+a x^5}}{\sqrt {2} x \sqrt [4]{-b x+a x^5}}\right )\right ) \]
-(Sqrt[2]*(ArcTan[(Sqrt[2]*x*(-(b*x) + a*x^5)^(1/4))/(-x^2 + Sqrt[-(b*x) + a*x^5])] + ArcTanh[(x^2 + Sqrt[-(b*x) + a*x^5])/(Sqrt[2]*x*(-(b*x) + a*x^ 5)^(1/4))]))
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.
\(\displaystyle \int \frac {a x^4+3 b}{\left (a x^4-b+x^3\right ) \sqrt [4]{a x^5-b x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt [4]{x} \sqrt [4]{a x^4-b} \int -\frac {a x^4+3 b}{\sqrt [4]{x} \left (-a x^4-x^3+b\right ) \sqrt [4]{a x^4-b}}dx}{\sqrt [4]{a x^5-b x}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\sqrt [4]{x} \sqrt [4]{a x^4-b} \int \frac {a x^4+3 b}{\sqrt [4]{x} \left (-a x^4-x^3+b\right ) \sqrt [4]{a x^4-b}}dx}{\sqrt [4]{a x^5-b x}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^4-b} \int \frac {\sqrt {x} \left (a x^4+3 b\right )}{\left (-a x^4-x^3+b\right ) \sqrt [4]{a x^4-b}}d\sqrt [4]{x}}{\sqrt [4]{a x^5-b x}}\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^4-b} \int \left (\frac {\sqrt {x} \left (4 b-x^3\right )}{\left (-a x^4-x^3+b\right ) \sqrt [4]{a x^4-b}}-\frac {\sqrt {x}}{\sqrt [4]{a x^4-b}}\right )d\sqrt [4]{x}}{\sqrt [4]{a x^5-b x}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {4 \sqrt [4]{x} \sqrt [4]{a x^4-b} \left (4 b \int \frac {\sqrt {x}}{\left (-a x^4-x^3+b\right ) \sqrt [4]{a x^4-b}}d\sqrt [4]{x}+\int \frac {x^{7/2}}{\sqrt [4]{a x^4-b} \left (a x^4+x^3-b\right )}d\sqrt [4]{x}-\frac {x^{3/4} \sqrt [4]{1-\frac {a x^4}{b}} \operatorname {Hypergeometric2F1}\left (\frac {3}{16},\frac {1}{4},\frac {19}{16},\frac {a x^4}{b}\right )}{3 \sqrt [4]{a x^4-b}}\right )}{\sqrt [4]{a x^5-b x}}\) |
3.16.76.3.1 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 = 0.87 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.26
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {-{\left (x \left (a \,x^{4}-b \right )\right )}^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x \left (a \,x^{4}-b \right )}}{{\left (x \left (a \,x^{4}-b \right )\right )}^{\frac {1}{4}} \sqrt {2}\, x +x^{2}+\sqrt {x \left (a \,x^{4}-b \right )}}\right )+2 \arctan \left (\frac {{\left (x \left (a \,x^{4}-b \right )\right )}^{\frac {1}{4}} \sqrt {2}+x}{x}\right )+2 \arctan \left (\frac {{\left (x \left (a \,x^{4}-b \right )\right )}^{\frac {1}{4}} \sqrt {2}-x}{x}\right )\right )}{2}\) | \(136\) |
1/2*2^(1/2)*(ln((-(x*(a*x^4-b))^(1/4)*2^(1/2)*x+x^2+(x*(a*x^4-b))^(1/2))/( (x*(a*x^4-b))^(1/4)*2^(1/2)*x+x^2+(x*(a*x^4-b))^(1/2)))+2*arctan(((x*(a*x^ 4-b))^(1/4)*2^(1/2)+x)/x)+2*arctan(((x*(a*x^4-b))^(1/4)*2^(1/2)-x)/x))
Timed out. \[ \int \frac {3 b+a x^4}{\left (-b+x^3+a x^4\right ) \sqrt [4]{-b x+a x^5}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {3 b+a x^4}{\left (-b+x^3+a x^4\right ) \sqrt [4]{-b x+a x^5}} \, dx=\text {Timed out} \]
\[ \int \frac {3 b+a x^4}{\left (-b+x^3+a x^4\right ) \sqrt [4]{-b x+a x^5}} \, dx=\int { \frac {a x^{4} + 3 \, b}{{\left (a x^{5} - b x\right )}^{\frac {1}{4}} {\left (a x^{4} + x^{3} - b\right )}} \,d x } \]
\[ \int \frac {3 b+a x^4}{\left (-b+x^3+a x^4\right ) \sqrt [4]{-b x+a x^5}} \, dx=\int { \frac {a x^{4} + 3 \, b}{{\left (a x^{5} - b x\right )}^{\frac {1}{4}} {\left (a x^{4} + x^{3} - b\right )}} \,d x } \]
Timed out. \[ \int \frac {3 b+a x^4}{\left (-b+x^3+a x^4\right ) \sqrt [4]{-b x+a x^5}} \, dx=\int \frac {a\,x^4+3\,b}{{\left (a\,x^5-b\,x\right )}^{1/4}\,\left (a\,x^4+x^3-b\right )} \,d x \]