Integrand size = 41, antiderivative size = 154 \[ \int \frac {\left (-4 b+a x^5\right ) \left (b+a x^5\right )^{3/4}}{x^4 \left (2 b+c x^4+2 a x^5\right )} \, dx=\frac {2 \left (b+a x^5\right )^{3/4}}{3 x^3}+\frac {c^{3/4} \arctan \left (\frac {2^{3/4} \sqrt [4]{c} x \sqrt [4]{b+a x^5}}{-\sqrt {c} x^2+\sqrt {2} \sqrt {b+a x^5}}\right )}{2 \sqrt [4]{2}}+\frac {c^{3/4} \text {arctanh}\left (\frac {\frac {\sqrt [4]{c} x^2}{2^{3/4}}+\frac {\sqrt {b+a x^5}}{\sqrt [4]{2} \sqrt [4]{c}}}{x \sqrt [4]{b+a x^5}}\right )}{2 \sqrt [4]{2}} \]
2/3*(a*x^5+b)^(3/4)/x^3+1/4*c^(3/4)*arctan(2^(3/4)*c^(1/4)*x*(a*x^5+b)^(1/ 4)/(-c^(1/2)*x^2+2^(1/2)*(a*x^5+b)^(1/2)))*2^(3/4)+1/4*c^(3/4)*arctanh((1/ 2*c^(1/4)*x^2*2^(1/4)+1/2*(a*x^5+b)^(1/2)*2^(3/4)/c^(1/4))/x/(a*x^5+b)^(1/ 4))*2^(3/4)
Time = 1.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-4 b+a x^5\right ) \left (b+a x^5\right )^{3/4}}{x^4 \left (2 b+c x^4+2 a x^5\right )} \, dx=\frac {1}{12} \left (\frac {8 \left (b+a x^5\right )^{3/4}}{x^3}+3\ 2^{3/4} c^{3/4} \arctan \left (\frac {\sqrt [4]{c} x}{2^{3/4} \sqrt [4]{b+a x^5}}-\frac {\sqrt [4]{b+a x^5}}{\sqrt [4]{2} \sqrt [4]{c} x}\right )+3\ 2^{3/4} c^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{c} x}{2^{3/4} \sqrt [4]{b+a x^5}}+\frac {\sqrt [4]{b+a x^5}}{\sqrt [4]{2} \sqrt [4]{c} x}\right )\right ) \]
((8*(b + a*x^5)^(3/4))/x^3 + 3*2^(3/4)*c^(3/4)*ArcTan[(c^(1/4)*x)/(2^(3/4) *(b + a*x^5)^(1/4)) - (b + a*x^5)^(1/4)/(2^(1/4)*c^(1/4)*x)] + 3*2^(3/4)*c ^(3/4)*ArcTanh[(c^(1/4)*x)/(2^(3/4)*(b + a*x^5)^(1/4)) + (b + a*x^5)^(1/4) /(2^(1/4)*c^(1/4)*x)])/12
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.
\(\displaystyle \int \frac {\left (a x^5-4 b\right ) \left (a x^5+b\right )^{3/4}}{x^4 \left (2 a x^5+2 b+c x^4\right )} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {\left (a x^5+b\right )^{3/4} (5 a x+2 c)}{2 a x^5+2 b+c x^4}-\frac {2 \left (a x^5+b\right )^{3/4}}{x^4}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle 2 c \int \frac {\left (a x^5+b\right )^{3/4}}{2 a x^5+c x^4+2 b}dx+5 a \int \frac {x \left (a x^5+b\right )^{3/4}}{2 a x^5+c x^4+2 b}dx+\frac {2 \left (a x^5+b\right )^{3/4} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},-\frac {3}{5},\frac {2}{5},-\frac {a x^5}{b}\right )}{3 x^3 \left (\frac {a x^5}{b}+1\right )^{3/4}}\) |
3.22.23.3.1 Defintions of rubi rules used
Time = 1.54 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.21
method | result | size |
pseudoelliptic | \(\frac {\left (8 \left (a \,x^{5}+b \right )^{\frac {3}{4}} 2^{\frac {1}{4}} c^{\frac {1}{4}}-3 \ln \left (\frac {\sqrt {2}\, \sqrt {c}\, x^{2}-2 \,2^{\frac {1}{4}} c^{\frac {1}{4}} \left (a \,x^{5}+b \right )^{\frac {1}{4}} x +2 \sqrt {a \,x^{5}+b}}{\sqrt {2}\, \sqrt {c}\, x^{2}+2 \,2^{\frac {1}{4}} c^{\frac {1}{4}} \left (a \,x^{5}+b \right )^{\frac {1}{4}} x +2 \sqrt {a \,x^{5}+b}}\right ) c \,x^{3}-6 \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{5}+b \right )^{\frac {1}{4}}+c^{\frac {1}{4}} x}{c^{\frac {1}{4}} x}\right ) c \,x^{3}-6 \arctan \left (\frac {2^{\frac {3}{4}} \left (a \,x^{5}+b \right )^{\frac {1}{4}}-c^{\frac {1}{4}} x}{c^{\frac {1}{4}} x}\right ) c \,x^{3}\right ) 2^{\frac {3}{4}}}{24 x^{3} c^{\frac {1}{4}}}\) | \(187\) |
1/24*(8*(a*x^5+b)^(3/4)*2^(1/4)*c^(1/4)-3*ln((2^(1/2)*c^(1/2)*x^2-2*2^(1/4 )*c^(1/4)*(a*x^5+b)^(1/4)*x+2*(a*x^5+b)^(1/2))/(2^(1/2)*c^(1/2)*x^2+2*2^(1 /4)*c^(1/4)*(a*x^5+b)^(1/4)*x+2*(a*x^5+b)^(1/2)))*c*x^3-6*arctan((2^(3/4)* (a*x^5+b)^(1/4)+c^(1/4)*x)/c^(1/4)/x)*c*x^3-6*arctan((2^(3/4)*(a*x^5+b)^(1 /4)-c^(1/4)*x)/c^(1/4)/x)*c*x^3)*2^(3/4)/x^3/c^(1/4)
Timed out. \[ \int \frac {\left (-4 b+a x^5\right ) \left (b+a x^5\right )^{3/4}}{x^4 \left (2 b+c x^4+2 a x^5\right )} \, dx=\text {Timed out} \]
\[ \int \frac {\left (-4 b+a x^5\right ) \left (b+a x^5\right )^{3/4}}{x^4 \left (2 b+c x^4+2 a x^5\right )} \, dx=\int \frac {\left (a x^{5} - 4 b\right ) \left (a x^{5} + b\right )^{\frac {3}{4}}}{x^{4} \cdot \left (2 a x^{5} + 2 b + c x^{4}\right )}\, dx \]
\[ \int \frac {\left (-4 b+a x^5\right ) \left (b+a x^5\right )^{3/4}}{x^4 \left (2 b+c x^4+2 a x^5\right )} \, dx=\int { \frac {{\left (a x^{5} + b\right )}^{\frac {3}{4}} {\left (a x^{5} - 4 \, b\right )}}{{\left (2 \, a x^{5} + c x^{4} + 2 \, b\right )} x^{4}} \,d x } \]
\[ \int \frac {\left (-4 b+a x^5\right ) \left (b+a x^5\right )^{3/4}}{x^4 \left (2 b+c x^4+2 a x^5\right )} \, dx=\int { \frac {{\left (a x^{5} + b\right )}^{\frac {3}{4}} {\left (a x^{5} - 4 \, b\right )}}{{\left (2 \, a x^{5} + c x^{4} + 2 \, b\right )} x^{4}} \,d x } \]
Timed out. \[ \int \frac {\left (-4 b+a x^5\right ) \left (b+a x^5\right )^{3/4}}{x^4 \left (2 b+c x^4+2 a x^5\right )} \, dx=\int -\frac {{\left (a\,x^5+b\right )}^{3/4}\,\left (4\,b-a\,x^5\right )}{x^4\,\left (2\,a\,x^5+c\,x^4+2\,b\right )} \,d x \]