Integrand size = 39, antiderivative size = 215 \[ \int \frac {\sqrt [4]{-b+a x^4} \left (b+c x^4+a x^8\right )}{x^6 \left (b+2 a x^8\right )} \, dx=\frac {\sqrt [4]{-b+a x^4} \left (-b+a x^4-5 c x^4\right )}{5 b x^5}-\frac {a \text {RootSum}\left [a^2+2 a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-a c \log (x)-2 b c \log (x)+a c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+2 b c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+b \log (x) \text {$\#$1}^4+c \log (x) \text {$\#$1}^4-b \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4-c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}^3-\text {$\#$1}^7}\&\right ]}{8 b} \] Output:
Unintegrable
Time = 0.00 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.98 \[ \int \frac {\sqrt [4]{-b+a x^4} \left (b+c x^4+a x^8\right )}{x^6 \left (b+2 a x^8\right )} \, dx=-\frac {8 \sqrt [4]{-b+a x^4} \left (b-(a-5 c) x^4\right )+5 a x^5 \text {RootSum}\left [a^2+2 a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {a c \log (x)+2 b c \log (x)-a c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-2 b c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-b \log (x) \text {$\#$1}^4-c \log (x) \text {$\#$1}^4+b \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4+c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-a \text {$\#$1}^3+\text {$\#$1}^7}\&\right ]}{40 b x^5} \] Input:
Integrate[((-b + a*x^4)^(1/4)*(b + c*x^4 + a*x^8))/(x^6*(b + 2*a*x^8)),x]
Output:
-1/40*(8*(-b + a*x^4)^(1/4)*(b - (a - 5*c)*x^4) + 5*a*x^5*RootSum[a^2 + 2* a*b - 2*a*#1^4 + #1^8 & , (a*c*Log[x] + 2*b*c*Log[x] - a*c*Log[(-b + a*x^4 )^(1/4) - x*#1] - 2*b*c*Log[(-b + a*x^4)^(1/4) - x*#1] - b*Log[x]*#1^4 - c *Log[x]*#1^4 + b*Log[(-b + a*x^4)^(1/4) - x*#1]*#1^4 + c*Log[(-b + a*x^4)^ (1/4) - x*#1]*#1^4)/(-(a*#1^3) + #1^7) & ])/(b*x^5)
Leaf count is larger than twice the leaf count of optimal. \(933\) vs. \(2(215)=430\).
Time = 4.65 (sec) , antiderivative size = 933, normalized size of antiderivative = 4.34, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {7276, 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.
\(\displaystyle \int \frac {\sqrt [4]{a x^4-b} \left (a x^8+b+c x^4\right )}{x^6 \left (2 a x^8+b\right )} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {c \sqrt [4]{a x^4-b}}{b x^2}-\frac {a x^2 \sqrt [4]{a x^4-b} \left (b+2 c x^4\right )}{b \left (2 a x^8+b\right )}+\frac {\sqrt [4]{a x^4-b}}{x^6}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \sqrt [4]{a x^4-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},-\frac {\sqrt {2} \sqrt {-a} x^4}{\sqrt {b}},\frac {a x^4}{b}\right ) x^3}{6 b \sqrt [4]{1-\frac {a x^4}{b}}}-\frac {a \sqrt [4]{a x^4-b} \operatorname {AppellF1}\left (\frac {3}{4},1,-\frac {1}{4},\frac {7}{4},\frac {\sqrt {2} \sqrt {-a} x^4}{\sqrt {b}},\frac {a x^4}{b}\right ) x^3}{6 b \sqrt [4]{1-\frac {a x^4}{b}}}+\frac {\sqrt {-a} c \arctan \left (\frac {\sqrt [4]{\sqrt {2} a-2 \sqrt {-a} \sqrt {b}} x}{\sqrt [8]{2} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [8]{2} \left (\sqrt {2} a-2 \sqrt {-a} \sqrt {b}\right )^{3/4} \sqrt {b}}-\frac {a c \arctan \left (\frac {\sqrt [4]{\sqrt {2} a-2 \sqrt {-a} \sqrt {b}} x}{\sqrt [8]{2} \sqrt [4]{a x^4-b}}\right )}{2\ 2^{5/8} \left (\sqrt {2} a-2 \sqrt {-a} \sqrt {b}\right )^{3/4} b}-\frac {\sqrt {-a} c \arctan \left (\frac {\sqrt [4]{\sqrt {2} a+2 \sqrt {-a} \sqrt {b}} x}{\sqrt [8]{2} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [8]{2} \left (\sqrt {2} a+2 \sqrt {-a} \sqrt {b}\right )^{3/4} \sqrt {b}}-\frac {a c \arctan \left (\frac {\sqrt [4]{\sqrt {2} a+2 \sqrt {-a} \sqrt {b}} x}{\sqrt [8]{2} \sqrt [4]{a x^4-b}}\right )}{2\ 2^{5/8} \left (\sqrt {2} a+2 \sqrt {-a} \sqrt {b}\right )^{3/4} b}-\frac {\sqrt {-a} c \text {arctanh}\left (\frac {\sqrt [4]{\sqrt {2} a-2 \sqrt {-a} \sqrt {b}} x}{\sqrt [8]{2} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [8]{2} \left (\sqrt {2} a-2 \sqrt {-a} \sqrt {b}\right )^{3/4} \sqrt {b}}+\frac {a c \text {arctanh}\left (\frac {\sqrt [4]{\sqrt {2} a-2 \sqrt {-a} \sqrt {b}} x}{\sqrt [8]{2} \sqrt [4]{a x^4-b}}\right )}{2\ 2^{5/8} \left (\sqrt {2} a-2 \sqrt {-a} \sqrt {b}\right )^{3/4} b}+\frac {\sqrt {-a} c \text {arctanh}\left (\frac {\sqrt [4]{\sqrt {2} a+2 \sqrt {-a} \sqrt {b}} x}{\sqrt [8]{2} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [8]{2} \left (\sqrt {2} a+2 \sqrt {-a} \sqrt {b}\right )^{3/4} \sqrt {b}}+\frac {a c \text {arctanh}\left (\frac {\sqrt [4]{\sqrt {2} a+2 \sqrt {-a} \sqrt {b}} x}{\sqrt [8]{2} \sqrt [4]{a x^4-b}}\right )}{2\ 2^{5/8} \left (\sqrt {2} a+2 \sqrt {-a} \sqrt {b}\right )^{3/4} b}-\frac {c \sqrt [4]{a x^4-b}}{b x}+\frac {\left (a x^4-b\right )^{5/4}}{5 b x^5}\) |
Input:
Int[((-b + a*x^4)^(1/4)*(b + c*x^4 + a*x^8))/(x^6*(b + 2*a*x^8)),x]
Output:
-((c*(-b + a*x^4)^(1/4))/(b*x)) + (-b + a*x^4)^(5/4)/(5*b*x^5) - (a*x^3*(- b + a*x^4)^(1/4)*AppellF1[3/4, 1, -1/4, 7/4, -((Sqrt[2]*Sqrt[-a]*x^4)/Sqrt [b]), (a*x^4)/b])/(6*b*(1 - (a*x^4)/b)^(1/4)) - (a*x^3*(-b + a*x^4)^(1/4)* AppellF1[3/4, 1, -1/4, 7/4, (Sqrt[2]*Sqrt[-a]*x^4)/Sqrt[b], (a*x^4)/b])/(6 *b*(1 - (a*x^4)/b)^(1/4)) - (a*c*ArcTan[((Sqrt[2]*a - 2*Sqrt[-a]*Sqrt[b])^ (1/4)*x)/(2^(1/8)*(-b + a*x^4)^(1/4))])/(2*2^(5/8)*(Sqrt[2]*a - 2*Sqrt[-a] *Sqrt[b])^(3/4)*b) + (Sqrt[-a]*c*ArcTan[((Sqrt[2]*a - 2*Sqrt[-a]*Sqrt[b])^ (1/4)*x)/(2^(1/8)*(-b + a*x^4)^(1/4))])/(2*2^(1/8)*(Sqrt[2]*a - 2*Sqrt[-a] *Sqrt[b])^(3/4)*Sqrt[b]) - (a*c*ArcTan[((Sqrt[2]*a + 2*Sqrt[-a]*Sqrt[b])^( 1/4)*x)/(2^(1/8)*(-b + a*x^4)^(1/4))])/(2*2^(5/8)*(Sqrt[2]*a + 2*Sqrt[-a]* Sqrt[b])^(3/4)*b) - (Sqrt[-a]*c*ArcTan[((Sqrt[2]*a + 2*Sqrt[-a]*Sqrt[b])^( 1/4)*x)/(2^(1/8)*(-b + a*x^4)^(1/4))])/(2*2^(1/8)*(Sqrt[2]*a + 2*Sqrt[-a]* Sqrt[b])^(3/4)*Sqrt[b]) + (a*c*ArcTanh[((Sqrt[2]*a - 2*Sqrt[-a]*Sqrt[b])^( 1/4)*x)/(2^(1/8)*(-b + a*x^4)^(1/4))])/(2*2^(5/8)*(Sqrt[2]*a - 2*Sqrt[-a]* Sqrt[b])^(3/4)*b) - (Sqrt[-a]*c*ArcTanh[((Sqrt[2]*a - 2*Sqrt[-a]*Sqrt[b])^ (1/4)*x)/(2^(1/8)*(-b + a*x^4)^(1/4))])/(2*2^(1/8)*(Sqrt[2]*a - 2*Sqrt[-a] *Sqrt[b])^(3/4)*Sqrt[b]) + (a*c*ArcTanh[((Sqrt[2]*a + 2*Sqrt[-a]*Sqrt[b])^ (1/4)*x)/(2^(1/8)*(-b + a*x^4)^(1/4))])/(2*2^(5/8)*(Sqrt[2]*a + 2*Sqrt[-a] *Sqrt[b])^(3/4)*b) + (Sqrt[-a]*c*ArcTanh[((Sqrt[2]*a + 2*Sqrt[-a]*Sqrt[b]) ^(1/4)*x)/(2^(1/8)*(-b + a*x^4)^(1/4))])/(2*2^(1/8)*(Sqrt[2]*a + 2*Sqrt...
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 0.00 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.54
method | result | size |
pseudoelliptic | \(\frac {-5 a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}+2 a b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x}\right ) \left (\left (-b -c \right ) \textit {\_R}^{4}+c \left (a +2 b \right )\right )}{\textit {\_R}^{3} \left (-\textit {\_R}^{4}+a \right )}\right ) x^{5}+8 \left (a \,x^{4}-b \right )^{\frac {1}{4}} \left (\left (a -5 c \right ) x^{4}-b \right )}{40 b \,x^{5}}\) | \(116\) |
Input:
int((a*x^4-b)^(1/4)*(a*x^8+c*x^4+b)/x^6/(2*a*x^8+b),x,method=_RETURNVERBOS E)
Output:
1/40*(-5*a*sum(ln((-_R*x+(a*x^4-b)^(1/4))/x)*((-b-c)*_R^4+c*(a+2*b))/_R^3/ (-_R^4+a),_R=RootOf(_Z^8-2*_Z^4*a+a^2+2*a*b))*x^5+8*(a*x^4-b)^(1/4)*((a-5* c)*x^4-b))/b/x^5
Timed out. \[ \int \frac {\sqrt [4]{-b+a x^4} \left (b+c x^4+a x^8\right )}{x^6 \left (b+2 a x^8\right )} \, dx=\text {Timed out} \] Input:
integrate((a*x^4-b)^(1/4)*(a*x^8+c*x^4+b)/x^6/(2*a*x^8+b),x, algorithm="fr icas")
Output:
Timed out
Not integrable
Time = 88.00 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt [4]{-b+a x^4} \left (b+c x^4+a x^8\right )}{x^6 \left (b+2 a x^8\right )} \, dx=\int \frac {\sqrt [4]{a x^{4} - b} \left (a x^{8} + b + c x^{4}\right )}{x^{6} \cdot \left (2 a x^{8} + b\right )}\, dx \] Input:
integrate((a*x**4-b)**(1/4)*(a*x**8+c*x**4+b)/x**6/(2*a*x**8+b),x)
Output:
Integral((a*x**4 - b)**(1/4)*(a*x**8 + b + c*x**4)/(x**6*(2*a*x**8 + b)), x)
Not integrable
Time = 0.09 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt [4]{-b+a x^4} \left (b+c x^4+a x^8\right )}{x^6 \left (b+2 a x^8\right )} \, dx=\int { \frac {{\left (a x^{8} + c x^{4} + b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}}{{\left (2 \, a x^{8} + b\right )} x^{6}} \,d x } \] Input:
integrate((a*x^4-b)^(1/4)*(a*x^8+c*x^4+b)/x^6/(2*a*x^8+b),x, algorithm="ma xima")
Output:
integrate((a*x^8 + c*x^4 + b)*(a*x^4 - b)^(1/4)/((2*a*x^8 + b)*x^6), x)
Exception generated. \[ \int \frac {\sqrt [4]{-b+a x^4} \left (b+c x^4+a x^8\right )}{x^6 \left (b+2 a x^8\right )} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a*x^4-b)^(1/4)*(a*x^8+c*x^4+b)/x^6/(2*a*x^8+b),x, algorithm="gi ac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{1,[0,1,4,1,0]%%%}+%%%{1,[0,1,0,0,1]%%%} / %%%{1,[0,0,0,1,0 ]%%%} Err
Not integrable
Time = 0.00 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.18 \[ \int \frac {\sqrt [4]{-b+a x^4} \left (b+c x^4+a x^8\right )}{x^6 \left (b+2 a x^8\right )} \, dx=\int \frac {{\left (a\,x^4-b\right )}^{1/4}\,\left (a\,x^8+c\,x^4+b\right )}{x^6\,\left (2\,a\,x^8+b\right )} \,d x \] Input:
int(((a*x^4 - b)^(1/4)*(b + a*x^8 + c*x^4))/(x^6*(b + 2*a*x^8)),x)
Output:
int(((a*x^4 - b)^(1/4)*(b + a*x^8 + c*x^4))/(x^6*(b + 2*a*x^8)), x)
Not integrable
Time = 0.19 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt [4]{-b+a x^4} \left (b+c x^4+a x^8\right )}{x^6 \left (b+2 a x^8\right )} \, dx=\frac {-3 \left (a \,x^{4}-b \right )^{\frac {1}{4}} a \,x^{4}-2 \left (a \,x^{4}-b \right )^{\frac {1}{4}} b +5 \left (\int \frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{2} x^{14}-2 a b \,x^{10}+a b \,x^{6}-b^{2} x^{2}}d x \right ) a \,b^{2} x^{5}-10 \left (\int \frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{2 a^{2} x^{14}-2 a b \,x^{10}+a b \,x^{6}-b^{2} x^{2}}d x \right ) b^{2} c \,x^{5}+10 \left (\int \frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}} x^{2}}{2 a^{2} x^{12}-2 a b \,x^{8}+a b \,x^{4}-b^{2}}d x \right ) a \,b^{2} x^{5}+10 \left (\int \frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}} x^{2}}{2 a^{2} x^{12}-2 a b \,x^{8}+a b \,x^{4}-b^{2}}d x \right ) a b c \,x^{5}}{10 b \,x^{5}} \] Input:
int((a*x^4-b)^(1/4)*(a*x^8+c*x^4+b)/x^6/(2*a*x^8+b),x)
Output:
( - 3*(a*x**4 - b)**(1/4)*a*x**4 - 2*(a*x**4 - b)**(1/4)*b + 5*int((a*x**4 - b)**(1/4)/(2*a**2*x**14 - 2*a*b*x**10 + a*b*x**6 - b**2*x**2),x)*a*b**2 *x**5 - 10*int((a*x**4 - b)**(1/4)/(2*a**2*x**14 - 2*a*b*x**10 + a*b*x**6 - b**2*x**2),x)*b**2*c*x**5 + 10*int(((a*x**4 - b)**(1/4)*x**2)/(2*a**2*x* *12 - 2*a*b*x**8 + a*b*x**4 - b**2),x)*a*b**2*x**5 + 10*int(((a*x**4 - b)* *(1/4)*x**2)/(2*a**2*x**12 - 2*a*b*x**8 + a*b*x**4 - b**2),x)*a*b*c*x**5)/ (10*b*x**5)