Integrand size = 42, antiderivative size = 133 \[ \int \frac {x^2 \left (4 b+a x^5\right )}{\left (-b+a x^5\right )^{3/4} \left (-b+c x^4+a x^5\right )} \, dx=-\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x \sqrt [4]{-b+a x^5}}{-\sqrt {c} x^2+\sqrt {-b+a x^5}}\right )}{c^{3/4}}+\frac {\sqrt {2} \text {arctanh}\left (\frac {\frac {\sqrt [4]{c} x^2}{\sqrt {2}}+\frac {\sqrt {-b+a x^5}}{\sqrt {2} \sqrt [4]{c}}}{x \sqrt [4]{-b+a x^5}}\right )}{c^{3/4}} \]
-2^(1/2)*arctan(2^(1/2)*c^(1/4)*x*(a*x^5-b)^(1/4)/(-c^(1/2)*x^2+(a*x^5-b)^ (1/2)))/c^(3/4)+2^(1/2)*arctanh((1/2*c^(1/4)*x^2*2^(1/2)+1/2*(a*x^5-b)^(1/ 2)*2^(1/2)/c^(1/4))/x/(a*x^5-b)^(1/4))/c^(3/4)
Time = 7.16 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.87 \[ \int \frac {x^2 \left (4 b+a x^5\right )}{\left (-b+a x^5\right )^{3/4} \left (-b+c x^4+a x^5\right )} \, dx=\frac {\sqrt {2} \left (\arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x \sqrt [4]{-b+a x^5}}{\sqrt {c} x^2-\sqrt {-b+a x^5}}\right )+\text {arctanh}\left (\frac {\sqrt {c} x^2+\sqrt {-b+a x^5}}{\sqrt {2} \sqrt [4]{c} x \sqrt [4]{-b+a x^5}}\right )\right )}{c^{3/4}} \]
(Sqrt[2]*(ArcTan[(Sqrt[2]*c^(1/4)*x*(-b + a*x^5)^(1/4))/(Sqrt[c]*x^2 - Sqr t[-b + a*x^5])] + ArcTanh[(Sqrt[c]*x^2 + Sqrt[-b + a*x^5])/(Sqrt[2]*c^(1/4 )*x*(-b + a*x^5)^(1/4))]))/c^(3/4)
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.
\(\displaystyle \int \frac {x^2 \left (a x^5+4 b\right )}{\left (a x^5-b\right )^{3/4} \left (a x^5-b+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {c^2}{a^2 \left (a x^5-b\right )^{3/4}}+\frac {5 a^2 b x^2-a b c x+b c^2-c^3 x^4}{a^2 \left (a x^5-b\right )^{3/4} \left (a x^5-b+c x^4\right )}-\frac {c x}{a \left (a x^5-b\right )^{3/4}}+\frac {x^2}{\left (a x^5-b\right )^{3/4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {c^3 \int \frac {x^4}{\left (a x^5-b\right )^{3/4} \left (a x^5+c x^4-b\right )}dx}{a^2}-\frac {b c^2 \int \frac {1}{\left (-a x^5-c x^4+b\right ) \left (a x^5-b\right )^{3/4}}dx}{a^2}-\frac {b c \int \frac {x}{\left (a x^5-b\right )^{3/4} \left (a x^5+c x^4-b\right )}dx}{a}+5 b \int \frac {x^2}{\left (a x^5-b\right )^{3/4} \left (a x^5+c x^4-b\right )}dx+\frac {c^2 x \left (1-\frac {a x^5}{b}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{5},\frac {3}{4},\frac {6}{5},\frac {a x^5}{b}\right )}{a^2 \left (a x^5-b\right )^{3/4}}-\frac {c x^2 \left (1-\frac {a x^5}{b}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {2}{5},\frac {3}{4},\frac {7}{5},\frac {a x^5}{b}\right )}{2 a \left (a x^5-b\right )^{3/4}}+\frac {x^3 \left (1-\frac {a x^5}{b}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},\frac {3}{4},\frac {8}{5},\frac {a x^5}{b}\right )}{3 \left (a x^5-b\right )^{3/4}}\) |
3.20.18.3.1 Defintions of rubi rules used
Time = 1.15 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.16
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {\left (a \,x^{5}-b \right )^{\frac {1}{4}} x \,c^{\frac {1}{4}} \sqrt {2}+\sqrt {c}\, x^{2}+\sqrt {a \,x^{5}-b}}{\sqrt {a \,x^{5}-b}-\left (a \,x^{5}-b \right )^{\frac {1}{4}} x \,c^{\frac {1}{4}} \sqrt {2}+\sqrt {c}\, x^{2}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{5}-b \right )^{\frac {1}{4}}+c^{\frac {1}{4}} x}{c^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{5}-b \right )^{\frac {1}{4}}-c^{\frac {1}{4}} x}{c^{\frac {1}{4}} x}\right )\right )}{2 c^{\frac {3}{4}}}\) | \(154\) |
1/2/c^(3/4)*2^(1/2)*(ln(((a*x^5-b)^(1/4)*x*c^(1/4)*2^(1/2)+c^(1/2)*x^2+(a* x^5-b)^(1/2))/((a*x^5-b)^(1/2)-(a*x^5-b)^(1/4)*x*c^(1/4)*2^(1/2)+c^(1/2)*x ^2))+2*arctan((2^(1/2)*(a*x^5-b)^(1/4)+c^(1/4)*x)/c^(1/4)/x)+2*arctan((2^( 1/2)*(a*x^5-b)^(1/4)-c^(1/4)*x)/c^(1/4)/x))
Timed out. \[ \int \frac {x^2 \left (4 b+a x^5\right )}{\left (-b+a x^5\right )^{3/4} \left (-b+c x^4+a x^5\right )} \, dx=\text {Timed out} \]
\[ \int \frac {x^2 \left (4 b+a x^5\right )}{\left (-b+a x^5\right )^{3/4} \left (-b+c x^4+a x^5\right )} \, dx=\int \frac {x^{2} \left (a x^{5} + 4 b\right )}{\left (a x^{5} - b\right )^{\frac {3}{4}} \left (a x^{5} - b + c x^{4}\right )}\, dx \]
\[ \int \frac {x^2 \left (4 b+a x^5\right )}{\left (-b+a x^5\right )^{3/4} \left (-b+c x^4+a x^5\right )} \, dx=\int { \frac {{\left (a x^{5} + 4 \, b\right )} x^{2}}{{\left (a x^{5} + c x^{4} - b\right )} {\left (a x^{5} - b\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {x^2 \left (4 b+a x^5\right )}{\left (-b+a x^5\right )^{3/4} \left (-b+c x^4+a x^5\right )} \, dx=\int { \frac {{\left (a x^{5} + 4 \, b\right )} x^{2}}{{\left (a x^{5} + c x^{4} - b\right )} {\left (a x^{5} - b\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {x^2 \left (4 b+a x^5\right )}{\left (-b+a x^5\right )^{3/4} \left (-b+c x^4+a x^5\right )} \, dx=\int \frac {x^2\,\left (a\,x^5+4\,b\right )}{{\left (a\,x^5-b\right )}^{3/4}\,\left (a\,x^5+c\,x^4-b\right )} \,d x \]