Integrand size = 36, antiderivative size = 57 \[ \int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-c x^4+a x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b+a x^8}}\right )}{2 \sqrt [4]{c}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b+a x^8}}\right )}{2 \sqrt [4]{c}} \] Output:
-1/2*arctan(c^(1/4)*x/(a*x^8+b)^(1/4))/c^(1/4)-1/2*arctanh(c^(1/4)*x/(a*x^ 8+b)^(1/4))/c^(1/4)
Time = 8.70 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84 \[ \int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-c x^4+a x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b+a x^8}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{c} x}{\sqrt [4]{b+a x^8}}\right )}{2 \sqrt [4]{c}} \] Input:
Integrate[(-b + a*x^8)/((b + a*x^8)^(1/4)*(b - c*x^4 + a*x^8)),x]
Output:
-1/2*(ArcTan[(c^(1/4)*x)/(b + a*x^8)^(1/4)] + ArcTanh[(c^(1/4)*x)/(b + a*x ^8)^(1/4)])/c^(1/4)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.42 (sec) , antiderivative size = 461, normalized size of antiderivative = 8.09, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {7279, 2009}
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.
\(\displaystyle \int \frac {a x^8-b}{\sqrt [4]{a x^8+b} \left (a x^8+b-c x^4\right )} \, dx\) |
\(\Big \downarrow \) 7279 |
\(\displaystyle \int \left (\frac {1}{\sqrt [4]{a x^8+b}}-\frac {2 b-c x^4}{\sqrt [4]{a x^8+b} \left (a x^8+b-c x^4\right )}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {x \sqrt [4]{\frac {a x^8}{b}+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},-\frac {2 a^2 x^8}{2 a b-c \left (c-\sqrt {c^2-4 a b}\right )},-\frac {a x^8}{b}\right )}{\sqrt [4]{a x^8+b}}-\frac {x \sqrt [4]{\frac {a x^8}{b}+1} \operatorname {AppellF1}\left (\frac {1}{8},1,\frac {1}{4},\frac {9}{8},-\frac {2 a^2 x^8}{2 a b-c \left (c+\sqrt {c^2-4 a b}\right )},-\frac {a x^8}{b}\right )}{\sqrt [4]{a x^8+b}}+\frac {a x^5 \left (c-\sqrt {c^2-4 a b}\right ) \sqrt [4]{\frac {a x^8}{b}+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},-\frac {2 a^2 x^8}{2 a b-c \left (c-\sqrt {c^2-4 a b}\right )},-\frac {a x^8}{b}\right )}{5 \left (2 a b-c \left (c-\sqrt {c^2-4 a b}\right )\right ) \sqrt [4]{a x^8+b}}+\frac {a x^5 \left (\sqrt {c^2-4 a b}+c\right ) \sqrt [4]{\frac {a x^8}{b}+1} \operatorname {AppellF1}\left (\frac {5}{8},1,\frac {1}{4},\frac {13}{8},-\frac {2 a^2 x^8}{2 a b-c \left (c+\sqrt {c^2-4 a b}\right )},-\frac {a x^8}{b}\right )}{5 \left (2 a b-c \left (\sqrt {c^2-4 a b}+c\right )\right ) \sqrt [4]{a x^8+b}}+\frac {x \sqrt [4]{\frac {a x^8}{b}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{8},\frac {1}{4},\frac {9}{8},-\frac {a x^8}{b}\right )}{\sqrt [4]{a x^8+b}}\) |
Input:
Int[(-b + a*x^8)/((b + a*x^8)^(1/4)*(b - c*x^4 + a*x^8)),x]
Output:
-((x*(1 + (a*x^8)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (-2*a^2*x^8)/(2*a*b - c*(c - Sqrt[-4*a*b + c^2])), -((a*x^8)/b)])/(b + a*x^8)^(1/4)) - (x*(1 + (a*x^8)/b)^(1/4)*AppellF1[1/8, 1, 1/4, 9/8, (-2*a^2*x^8)/(2*a*b - c*(c + Sqrt[-4*a*b + c^2])), -((a*x^8)/b)])/(b + a*x^8)^(1/4) + (a*(c - Sqrt[-4*a *b + c^2])*x^5*(1 + (a*x^8)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (-2*a^2*x ^8)/(2*a*b - c*(c - Sqrt[-4*a*b + c^2])), -((a*x^8)/b)])/(5*(2*a*b - c*(c - Sqrt[-4*a*b + c^2]))*(b + a*x^8)^(1/4)) + (a*(c + Sqrt[-4*a*b + c^2])*x^ 5*(1 + (a*x^8)/b)^(1/4)*AppellF1[5/8, 1, 1/4, 13/8, (-2*a^2*x^8)/(2*a*b - c*(c + Sqrt[-4*a*b + c^2])), -((a*x^8)/b)])/(5*(2*a*b - c*(c + Sqrt[-4*a*b + c^2]))*(b + a*x^8)^(1/4)) + (x*(1 + (a*x^8)/b)^(1/4)*Hypergeometric2F1[ 1/8, 1/4, 9/8, -((a*x^8)/b)])/(b + a*x^8)^(1/4)
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.55 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.18
method | result | size |
pseudoelliptic | \(\frac {2 \arctan \left (\frac {\left (a \,x^{8}+b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x}\right )-\ln \left (\frac {-c^{\frac {1}{4}} x -\left (a \,x^{8}+b \right )^{\frac {1}{4}}}{c^{\frac {1}{4}} x -\left (a \,x^{8}+b \right )^{\frac {1}{4}}}\right )}{4 c^{\frac {1}{4}}}\) | \(67\) |
Input:
int((a*x^8-b)/(a*x^8+b)^(1/4)/(a*x^8-c*x^4+b),x,method=_RETURNVERBOSE)
Output:
1/4/c^(1/4)*(2*arctan(1/c^(1/4)/x*(a*x^8+b)^(1/4))-ln((-c^(1/4)*x-(a*x^8+b )^(1/4))/(c^(1/4)*x-(a*x^8+b)^(1/4))))
Timed out. \[ \int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-c x^4+a x^8\right )} \, dx=\text {Timed out} \] Input:
integrate((a*x^8-b)/(a*x^8+b)^(1/4)/(a*x^8-c*x^4+b),x, algorithm="fricas")
Output:
Timed out
Timed out. \[ \int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-c x^4+a x^8\right )} \, dx=\text {Timed out} \] Input:
integrate((a*x**8-b)/(a*x**8+b)**(1/4)/(a*x**8-c*x**4+b),x)
Output:
Timed out
\[ \int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-c x^4+a x^8\right )} \, dx=\int { \frac {a x^{8} - b}{{\left (a x^{8} - c x^{4} + b\right )} {\left (a x^{8} + b\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((a*x^8-b)/(a*x^8+b)^(1/4)/(a*x^8-c*x^4+b),x, algorithm="maxima")
Output:
integrate((a*x^8 - b)/((a*x^8 - c*x^4 + b)*(a*x^8 + b)^(1/4)), x)
\[ \int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-c x^4+a x^8\right )} \, dx=\int { \frac {a x^{8} - b}{{\left (a x^{8} - c x^{4} + b\right )} {\left (a x^{8} + b\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((a*x^8-b)/(a*x^8+b)^(1/4)/(a*x^8-c*x^4+b),x, algorithm="giac")
Output:
integrate((a*x^8 - b)/((a*x^8 - c*x^4 + b)*(a*x^8 + b)^(1/4)), x)
Timed out. \[ \int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-c x^4+a x^8\right )} \, dx=\int -\frac {b-a\,x^8}{{\left (a\,x^8+b\right )}^{1/4}\,\left (a\,x^8-c\,x^4+b\right )} \,d x \] Input:
int(-(b - a*x^8)/((b + a*x^8)^(1/4)*(b + a*x^8 - c*x^4)),x)
Output:
int(-(b - a*x^8)/((b + a*x^8)^(1/4)*(b + a*x^8 - c*x^4)), x)
\[ \int \frac {-b+a x^8}{\sqrt [4]{b+a x^8} \left (b-c x^4+a x^8\right )} \, dx=\left (\int \frac {x^{8}}{\left (a \,x^{8}+b \right )^{\frac {1}{4}} a \,x^{8}+\left (a \,x^{8}+b \right )^{\frac {1}{4}} b -\left (a \,x^{8}+b \right )^{\frac {1}{4}} c \,x^{4}}d x \right ) a -\left (\int \frac {1}{\left (a \,x^{8}+b \right )^{\frac {1}{4}} a \,x^{8}+\left (a \,x^{8}+b \right )^{\frac {1}{4}} b -\left (a \,x^{8}+b \right )^{\frac {1}{4}} c \,x^{4}}d x \right ) b \] Input:
int((a*x^8-b)/(a*x^8+b)^(1/4)/(a*x^8-c*x^4+b),x)
Output:
int(x**8/((a*x**8 + b)**(1/4)*a*x**8 + (a*x**8 + b)**(1/4)*b - (a*x**8 + b )**(1/4)*c*x**4),x)*a - int(1/((a*x**8 + b)**(1/4)*a*x**8 + (a*x**8 + b)** (1/4)*b - (a*x**8 + b)**(1/4)*c*x**4),x)*b