Integrand size = 53, antiderivative size = 177 \[ \int \frac {c x^6 \left (-4 b+a x^5\right )}{\left (b+a x^5\right )^{3/4} \left (b^2+2 a b x^5-c^2 x^8+a^2 x^{10}\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b+a x^5}}\right )}{c^{3/4}}-\frac {\arctan \left (\frac {-\frac {\sqrt [4]{c} x^2}{\sqrt {2}}+\frac {\sqrt {b+a x^5}}{\sqrt {2} \sqrt [4]{c}}}{x \sqrt [4]{b+a x^5}}\right )}{\sqrt {2} c^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b+a x^5}}\right )}{c^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{c} x \sqrt [4]{b+a x^5}}{\sqrt {c} x^2+\sqrt {b+a x^5}}\right )}{\sqrt {2} c^{3/4}} \]
arctan(c^(1/4)*x/(a*x^5+b)^(1/4))/c^(3/4)-1/2*arctan((-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))*2^(1/2)/c^(3/ 4)-arctanh(c^(1/4)*x/(a*x^5+b)^(1/4))/c^(3/4)-1/2*arctanh(2^(1/2)*c^(1/4)* x*(a*x^5+b)^(1/4)/(c^(1/2)*x^2+(a*x^5+b)^(1/2)))*2^(1/2)/c^(3/4)
Time = 7.62 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.90 \[ \int \frac {c x^6 \left (-4 b+a x^5\right )}{\left (b+a x^5\right )^{3/4} \left (b^2+2 a b x^5-c^2 x^8+a^2 x^{10}\right )} \, dx=\frac {2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b+a x^5}}\right )+\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} x \sqrt [4]{b+a x^5}}{-\sqrt {c} x^2+\sqrt {b+a x^5}}\right )-2 \text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b+a x^5}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {c} x^2+\sqrt {b+a x^5}}{\sqrt {2} \sqrt [4]{c} x \sqrt [4]{b+a x^5}}\right )}{2 c^{3/4}} \]
Integrate[(c*x^6*(-4*b + a*x^5))/((b + a*x^5)^(3/4)*(b^2 + 2*a*b*x^5 - c^2 *x^8 + a^2*x^10)),x]
(2*ArcTan[(c^(1/4)*x)/(b + a*x^5)^(1/4)] + Sqrt[2]*ArcTan[(Sqrt[2]*c^(1/4) *x*(b + a*x^5)^(1/4))/(-(Sqrt[c]*x^2) + Sqrt[b + a*x^5])] - 2*ArcTanh[(c^( 1/4)*x)/(b + a*x^5)^(1/4)] - Sqrt[2]*ArcTanh[(Sqrt[c]*x^2 + Sqrt[b + a*x^5 ])/(Sqrt[2]*c^(1/4)*x*(b + a*x^5)^(1/4))])/(2*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.
| \(\displaystyle \int \frac {c x^6 \left (a x^5-4 b\right )}{\left (a x^5+b\right )^{3/4} \left (a^2 x^{10}+2 a b x^5+b^2-c^2 x^8\right )} \, dx\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle c \int -\frac {x^6 \left (4 b-a x^5\right )}{\left (a x^5+b\right )^{3/4} \left (a^2 x^{10}-c^2 x^8+2 a b x^5+b^2\right )}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -c \int \frac {x^6 \left (4 b-a x^5\right )}{\left (a x^5+b\right )^{3/4} \left (a^2 x^{10}-c^2 x^8+2 a b x^5+b^2\right )}dx\) |
| \(\Big \downarrow \) 2463 |
| \(\displaystyle -c \int \left (\frac {(a x-c) \left (4 b-a x^5\right ) x^6}{2 b c \left (-a x^5+c x^4-b\right ) \left (a x^5+b\right )^{3/4}}+\frac {(c+a x) \left (4 b-a x^5\right ) x^6}{2 b c \left (a x^5+b\right )^{3/4} \left (a x^5+c x^4+b\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -c \left (-\frac {c^2 \int \frac {x^4}{\left (a x^5+b\right )^{3/4} \left (a x^5-c x^4+b\right )}dx}{2 a^2}-\frac {c^2 \int \frac {x^4}{\left (a x^5+b\right )^{3/4} \left (a x^5+c x^4+b\right )}dx}{2 a^2}+\frac {b c \int \frac {1}{\left (a x^5+b\right )^{3/4} \left (a x^5-c x^4+b\right )}dx}{2 a^2}-\frac {b c \int \frac {1}{\left (a x^5+b\right )^{3/4} \left (a x^5+c x^4+b\right )}dx}{2 a^2}+\frac {b \int \frac {x}{\left (a x^5+b\right )^{3/4} \left (a x^5-c x^4+b\right )}dx}{2 a}+\frac {b \int \frac {x}{\left (a x^5+b\right )^{3/4} \left (a x^5+c x^4+b\right )}dx}{2 a}+\frac {5 b \int \frac {x^2}{\left (a x^5+b\right )^{3/4} \left (a x^5-c x^4+b\right )}dx}{2 c}-\frac {5 b \int \frac {x^2}{\left (a x^5+b\right )^{3/4} \left (a x^5+c x^4+b\right )}dx}{2 c}-\frac {x^2 \left (\frac {a x^5}{b}+1\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {2}{5},\frac {3}{4},\frac {7}{5},-\frac {a x^5}{b}\right )}{2 a \left (a x^5+b\right )^{3/4}}\right )\) |
3.24.12.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px]}, Int[ExpandIntegr and[u, Qx^p, x], x] /; !SumQ[NonfreeFactors[Qx, x]]] /; PolyQ[Px, x] && Gt Q[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0]
Time = 0.68 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.19
| method | result | size |
| pseudoelliptic | \(\frac {-2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{5}+b \right )^{\frac {1}{4}}+c^{\frac {1}{4}} x}{c^{\frac {1}{4}} x}\right ) \sqrt {2}-2 \arctan \left (\frac {\sqrt {2}\, \left (a \,x^{5}+b \right )^{\frac {1}{4}}-c^{\frac {1}{4}} x}{c^{\frac {1}{4}} x}\right ) \sqrt {2}-\ln \left (\frac {\left (a \,x^{5}+b \right )^{\frac {1}{4}} x \,c^{\frac {1}{4}} \sqrt {2}+\sqrt {c}\, x^{2}+\sqrt {a \,x^{5}+b}}{\sqrt {a \,x^{5}+b}-\left (a \,x^{5}+b \right )^{\frac {1}{4}} x \,c^{\frac {1}{4}} \sqrt {2}+\sqrt {c}\, x^{2}}\right ) \sqrt {2}-4 \arctan \left (\frac {\left (a \,x^{5}+b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x}\right )-2 \ln \left (\frac {-c^{\frac {1}{4}} x -\left (a \,x^{5}+b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x -\left (a \,x^{5}+b \right )^{\frac {1}{4}}}\right )}{4 c^{\frac {3}{4}}}\) | \(210\) |
int(c*x^6*(a*x^5-4*b)/(a*x^5+b)^(3/4)/(a^2*x^10-c^2*x^8+2*a*b*x^5+b^2),x,m ethod=_RETURNVERBOSE)
1/4*(-2*arctan((2^(1/2)*(a*x^5+b)^(1/4)+c^(1/4)*x)/c^(1/4)/x)*2^(1/2)-2*ar ctan((2^(1/2)*(a*x^5+b)^(1/4)-c^(1/4)*x)/c^(1/4)/x)*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^(1/2)-4*arctan(1/c^(1/4)/x*( a*x^5+b)^(1/4))-2*ln((-c^(1/4)*x-(a*x^5+b)^(1/4))/(c^(1/4)*x-(a*x^5+b)^(1/ 4))))/c^(3/4)
Timed out. \[ \int \frac {c x^6 \left (-4 b+a x^5\right )}{\left (b+a x^5\right )^{3/4} \left (b^2+2 a b x^5-c^2 x^8+a^2 x^{10}\right )} \, dx=\text {Timed out} \]
integrate(c*x^6*(a*x^5-4*b)/(a*x^5+b)^(3/4)/(a^2*x^10-c^2*x^8+2*a*b*x^5+b^ 2),x, algorithm="fricas")
\[ \int \frac {c x^6 \left (-4 b+a x^5\right )}{\left (b+a x^5\right )^{3/4} \left (b^2+2 a b x^5-c^2 x^8+a^2 x^{10}\right )} \, dx=c \left (\int \frac {a x^{11}}{a^{2} x^{10} \left (a x^{5} + b\right )^{\frac {3}{4}} + 2 a b x^{5} \left (a x^{5} + b\right )^{\frac {3}{4}} + b^{2} \left (a x^{5} + b\right )^{\frac {3}{4}} - c^{2} x^{8} \left (a x^{5} + b\right )^{\frac {3}{4}}}\, dx + \int \left (- \frac {4 b x^{6}}{a^{2} x^{10} \left (a x^{5} + b\right )^{\frac {3}{4}} + 2 a b x^{5} \left (a x^{5} + b\right )^{\frac {3}{4}} + b^{2} \left (a x^{5} + b\right )^{\frac {3}{4}} - c^{2} x^{8} \left (a x^{5} + b\right )^{\frac {3}{4}}}\right )\, dx\right ) \]
c*(Integral(a*x**11/(a**2*x**10*(a*x**5 + b)**(3/4) + 2*a*b*x**5*(a*x**5 + b)**(3/4) + b**2*(a*x**5 + b)**(3/4) - c**2*x**8*(a*x**5 + b)**(3/4)), x) + Integral(-4*b*x**6/(a**2*x**10*(a*x**5 + b)**(3/4) + 2*a*b*x**5*(a*x**5 + b)**(3/4) + b**2*(a*x**5 + b)**(3/4) - c**2*x**8*(a*x**5 + b)**(3/4)), x))
\[ \int \frac {c x^6 \left (-4 b+a x^5\right )}{\left (b+a x^5\right )^{3/4} \left (b^2+2 a b x^5-c^2 x^8+a^2 x^{10}\right )} \, dx=\int { \frac {{\left (a x^{5} - 4 \, b\right )} c x^{6}}{{\left (a^{2} x^{10} - c^{2} x^{8} + 2 \, a b x^{5} + b^{2}\right )} {\left (a x^{5} + b\right )}^{\frac {3}{4}}} \,d x } \]
integrate(c*x^6*(a*x^5-4*b)/(a*x^5+b)^(3/4)/(a^2*x^10-c^2*x^8+2*a*b*x^5+b^ 2),x, algorithm="maxima")
c*integrate((a*x^5 - 4*b)*x^6/((a^2*x^10 - c^2*x^8 + 2*a*b*x^5 + b^2)*(a*x ^5 + b)^(3/4)), x)
\[ \int \frac {c x^6 \left (-4 b+a x^5\right )}{\left (b+a x^5\right )^{3/4} \left (b^2+2 a b x^5-c^2 x^8+a^2 x^{10}\right )} \, dx=\int { \frac {{\left (a x^{5} - 4 \, b\right )} c x^{6}}{{\left (a^{2} x^{10} - c^{2} x^{8} + 2 \, a b x^{5} + b^{2}\right )} {\left (a x^{5} + b\right )}^{\frac {3}{4}}} \,d x } \]
integrate(c*x^6*(a*x^5-4*b)/(a*x^5+b)^(3/4)/(a^2*x^10-c^2*x^8+2*a*b*x^5+b^ 2),x, algorithm="giac")
integrate((a*x^5 - 4*b)*c*x^6/((a^2*x^10 - c^2*x^8 + 2*a*b*x^5 + b^2)*(a*x ^5 + b)^(3/4)), x)
Timed out. \[ \int \frac {c x^6 \left (-4 b+a x^5\right )}{\left (b+a x^5\right )^{3/4} \left (b^2+2 a b x^5-c^2 x^8+a^2 x^{10}\right )} \, dx=-\int \frac {c\,x^6\,\left (4\,b-a\,x^5\right )}{{\left (a\,x^5+b\right )}^{3/4}\,\left (a^2\,x^{10}+2\,a\,b\,x^5+b^2-c^2\,x^8\right )} \,d x \]