Integrand size = 42, antiderivative size = 270 \[ \int \frac {\sqrt [4]{2 b+a x^4} \left (-4 b+a x^8\right )}{x^6 \left (-4 b+c x^4+a x^8\right )} \, dx=\frac {\sqrt [4]{2 b+a x^4} \left (-4 b-2 a x^4-5 c x^4\right )}{20 b x^5}-\frac {c \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-2 a^2 \log (x)+2 a b \log (x)-a c \log (x)+2 a^2 \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )-2 a b \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )+a c \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )+2 a \log (x) \text {$\#$1}^4+c \log (x) \text {$\#$1}^4-2 a \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4-c \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{4 a \text {$\#$1}^3+c \text {$\#$1}^3-4 \text {$\#$1}^7}\&\right ]}{16 b} \]
Time = 0.70 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt [4]{2 b+a x^4} \left (-4 b+a x^8\right )}{x^6 \left (-4 b+c x^4+a x^8\right )} \, dx=\frac {\sqrt [4]{2 b+a x^4} \left (-4 b-2 a x^4-5 c x^4\right )}{20 b x^5}-\frac {c \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {2 a^2 \log (x)-2 a b \log (x)+a c \log (x)-2 a^2 \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )+2 a b \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )-a c \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right )-2 a \log (x) \text {$\#$1}^4-c \log (x) \text {$\#$1}^4+2 a \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+c \log \left (\sqrt [4]{2 b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-4 a \text {$\#$1}^3-c \text {$\#$1}^3+4 \text {$\#$1}^7}\&\right ]}{16 b} \]
((2*b + a*x^4)^(1/4)*(-4*b - 2*a*x^4 - 5*c*x^4))/(20*b*x^5) - (c*RootSum[2 *a^2 - 2*a*b + a*c - 4*a*#1^4 - c*#1^4 + 2*#1^8 & , (2*a^2*Log[x] - 2*a*b* Log[x] + a*c*Log[x] - 2*a^2*Log[(2*b + a*x^4)^(1/4) - x*#1] + 2*a*b*Log[(2 *b + a*x^4)^(1/4) - x*#1] - a*c*Log[(2*b + a*x^4)^(1/4) - x*#1] - 2*a*Log[ x]*#1^4 - c*Log[x]*#1^4 + 2*a*Log[(2*b + a*x^4)^(1/4) - x*#1]*#1^4 + c*Log [(2*b + a*x^4)^(1/4) - x*#1]*#1^4)/(-4*a*#1^3 - c*#1^3 + 4*#1^7) & ])/(16* b)
Leaf count is larger than twice the leaf count of optimal. \(1308\) vs. \(2(270)=540\).
Time = 5.71 (sec) , antiderivative size = 1308, normalized size of antiderivative = 4.84, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, 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.
| \(\displaystyle \int \frac {\sqrt [4]{a x^4+2 b} \left (a x^8-4 b\right )}{x^6 \left (a x^8-4 b+c x^4\right )} \, dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {c \sqrt [4]{a x^4+2 b}}{4 b x^2}+\frac {c x^2 \sqrt [4]{a x^4+2 b} \left (a x^4+c\right )}{4 b \left (-a x^8+4 b-c x^4\right )}+\frac {\sqrt [4]{a x^4+2 b}}{x^6}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a c^2 \sqrt [4]{a x^4+2 b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {2 a x^4}{c-\sqrt {c^2+16 a b}},-\frac {a x^4}{2 b}\right ) x^3}{3\ 2^{3/4} b \left (16 a b+c \left (c-\sqrt {c^2+16 a b}\right )\right ) \sqrt [4]{\frac {a x^4}{b}+2}}+\frac {a c^2 \sqrt [4]{a x^4+2 b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {2 a x^4}{c+\sqrt {c^2+16 a b}},-\frac {a x^4}{2 b}\right ) x^3}{3\ 2^{3/4} b \left (16 a b+c \left (c+\sqrt {c^2+16 a b}\right )\right ) \sqrt [4]{\frac {a x^4}{b}+2}}-\frac {\sqrt [4]{a} (2 b-c) c \left (\sqrt {c^2+16 a b}-c\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{\sqrt {c^2+16 a b}-c} \sqrt [4]{a x^4+2 b}}\right )}{8 b \sqrt {c^2+16 a b} \left (4 b-c+\sqrt {c^2+16 a b}\right )^{3/4}}-\frac {a^{5/4} c \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{\sqrt {c^2+16 a b}-c} \sqrt [4]{a x^4+2 b}}\right )}{\sqrt {c^2+16 a b} \sqrt [4]{\sqrt {c^2+16 a b}-c} \left (4 b-c+\sqrt {c^2+16 a b}\right )^{3/4}}+\frac {\sqrt [4]{a} (2 b-c) c \left (c+\sqrt {c^2+16 a b}\right )^{3/4} \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{c+\sqrt {c^2+16 a b}} \sqrt [4]{a x^4+2 b}}\right )}{8 b \sqrt {c^2+16 a b} \left (-4 b+c+\sqrt {c^2+16 a b}\right )^{3/4}}-\frac {a^{5/4} c \arctan \left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{c+\sqrt {c^2+16 a b}} \sqrt [4]{a x^4+2 b}}\right )}{\sqrt {c^2+16 a b} \sqrt [4]{c+\sqrt {c^2+16 a b}} \left (-4 b+c+\sqrt {c^2+16 a b}\right )^{3/4}}+\frac {\sqrt [4]{a} (2 b-c) c \left (\sqrt {c^2+16 a b}-c\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{\sqrt {c^2+16 a b}-c} \sqrt [4]{a x^4+2 b}}\right )}{8 b \sqrt {c^2+16 a b} \left (4 b-c+\sqrt {c^2+16 a b}\right )^{3/4}}+\frac {a^{5/4} c \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{4 b-c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{\sqrt {c^2+16 a b}-c} \sqrt [4]{a x^4+2 b}}\right )}{\sqrt {c^2+16 a b} \sqrt [4]{\sqrt {c^2+16 a b}-c} \left (4 b-c+\sqrt {c^2+16 a b}\right )^{3/4}}-\frac {\sqrt [4]{a} (2 b-c) c \left (c+\sqrt {c^2+16 a b}\right )^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{c+\sqrt {c^2+16 a b}} \sqrt [4]{a x^4+2 b}}\right )}{8 b \sqrt {c^2+16 a b} \left (-4 b+c+\sqrt {c^2+16 a b}\right )^{3/4}}+\frac {a^{5/4} c \text {arctanh}\left (\frac {\sqrt [4]{a} \sqrt [4]{-4 b+c+\sqrt {c^2+16 a b}} x}{\sqrt [4]{c+\sqrt {c^2+16 a b}} \sqrt [4]{a x^4+2 b}}\right )}{\sqrt {c^2+16 a b} \sqrt [4]{c+\sqrt {c^2+16 a b}} \left (-4 b+c+\sqrt {c^2+16 a b}\right )^{3/4}}-\frac {c \sqrt [4]{a x^4+2 b}}{4 b x}-\frac {\left (a x^4+2 b\right )^{5/4}}{10 b x^5}\) |
-1/4*(c*(2*b + a*x^4)^(1/4))/(b*x) - (2*b + a*x^4)^(5/4)/(10*b*x^5) + (a*c ^2*x^3*(2*b + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, (-2*a*x^4)/(c - Sqr t[16*a*b + c^2]), -1/2*(a*x^4)/b])/(3*2^(3/4)*b*(16*a*b + c*(c - Sqrt[16*a *b + c^2]))*(2 + (a*x^4)/b)^(1/4)) + (a*c^2*x^3*(2*b + a*x^4)^(1/4)*Appell F1[3/4, 1, -1/4, 7/4, (-2*a*x^4)/(c + Sqrt[16*a*b + c^2]), -1/2*(a*x^4)/b] )/(3*2^(3/4)*b*(16*a*b + c*(c + Sqrt[16*a*b + c^2]))*(2 + (a*x^4)/b)^(1/4) ) - (a^(5/4)*c*ArcTan[(a^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)*x)/((- c + Sqrt[16*a*b + c^2])^(1/4)*(2*b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^2]*( -c + Sqrt[16*a*b + c^2])^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(3/4)) - (a^ (1/4)*(2*b - c)*c*(-c + Sqrt[16*a*b + c^2])^(3/4)*ArcTan[(a^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)*x)/((-c + Sqrt[16*a*b + c^2])^(1/4)*(2*b + a* x^4)^(1/4))])/(8*b*Sqrt[16*a*b + c^2]*(4*b - c + Sqrt[16*a*b + c^2])^(3/4) ) - (a^(5/4)*c*ArcTan[(a^(1/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(1/4)*x)/(( c + Sqrt[16*a*b + c^2])^(1/4)*(2*b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^2]*( c + Sqrt[16*a*b + c^2])^(1/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(3/4)) + (a^ (1/4)*(2*b - c)*c*(c + Sqrt[16*a*b + c^2])^(3/4)*ArcTan[(a^(1/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(1/4)*x)/((c + Sqrt[16*a*b + c^2])^(1/4)*(2*b + a*x ^4)^(1/4))])/(8*b*Sqrt[16*a*b + c^2]*(-4*b + c + Sqrt[16*a*b + c^2])^(3/4) ) + (a^(5/4)*c*ArcTanh[(a^(1/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)*x)/(( -c + Sqrt[16*a*b + c^2])^(1/4)*(2*b + a*x^4)^(1/4))])/(Sqrt[16*a*b + c^...
3.28.95.3.1 Defintions of rubi rules used
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 = 0.42 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.51
| method | result | size |
| pseudoelliptic | \(\frac {-5 c \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}+\left (-4 a -c \right ) \textit {\_Z}^{4}+2 a^{2}-2 a b +a c \right )}{\sum }\frac {2 \left (a^{2}+\left (-\textit {\_R}^{4}-b +\frac {c}{2}\right ) a -\frac {\textit {\_R}^{4} c}{2}\right ) \ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}+2 b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}^{3} \left (-4 \textit {\_R}^{4}+4 a +c \right )}\right ) x^{5}-8 \left (\left (a +\frac {5 c}{2}\right ) x^{4}+2 b \right ) \left (a \,x^{4}+2 b \right )^{\frac {1}{4}}}{80 b \,x^{5}}\) | \(137\) |
1/80*(-5*c*sum(2*(a^2+(-_R^4-b+1/2*c)*a-1/2*_R^4*c)*ln((-_R*x+(a*x^4+2*b)^ (1/4))/x)/_R^3/(-4*_R^4+4*a+c),_R=RootOf(2*_Z^8+(-4*a-c)*_Z^4+2*a^2-2*a*b+ a*c))*x^5-8*((a+5/2*c)*x^4+2*b)*(a*x^4+2*b)^(1/4))/b/x^5
Timed out. \[ \int \frac {\sqrt [4]{2 b+a x^4} \left (-4 b+a x^8\right )}{x^6 \left (-4 b+c x^4+a x^8\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt [4]{2 b+a x^4} \left (-4 b+a x^8\right )}{x^6 \left (-4 b+c x^4+a x^8\right )} \, dx=\text {Timed out} \]
Not integrable
Time = 0.23 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt [4]{2 b+a x^4} \left (-4 b+a x^8\right )}{x^6 \left (-4 b+c x^4+a x^8\right )} \, dx=\int { \frac {{\left (a x^{8} - 4 \, b\right )} {\left (a x^{4} + 2 \, b\right )}^{\frac {1}{4}}}{{\left (a x^{8} + c x^{4} - 4 \, b\right )} x^{6}} \,d x } \]
Not integrable
Time = 3.65 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt [4]{2 b+a x^4} \left (-4 b+a x^8\right )}{x^6 \left (-4 b+c x^4+a x^8\right )} \, dx=\int { \frac {{\left (a x^{8} - 4 \, b\right )} {\left (a x^{4} + 2 \, b\right )}^{\frac {1}{4}}}{{\left (a x^{8} + c x^{4} - 4 \, b\right )} x^{6}} \,d x } \]
Not integrable
Time = 8.67 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt [4]{2 b+a x^4} \left (-4 b+a x^8\right )}{x^6 \left (-4 b+c x^4+a x^8\right )} \, dx=\int -\frac {{\left (a\,x^4+2\,b\right )}^{1/4}\,\left (4\,b-a\,x^8\right )}{x^6\,\left (a\,x^8+c\,x^4-4\,b\right )} \,d x \]