Integrand size = 39, antiderivative size = 209 \[ \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{2 a b-c} x \sqrt [4]{b^2+c x^4+a^2 x^8}}{\sqrt {2 a b-c} x^2-\sqrt {b^2+c x^4+a^2 x^8}}\right )}{2 \sqrt {2} \sqrt [4]{2 a b-c}}-\frac {\text {arctanh}\left (\frac {\frac {\sqrt [4]{2 a b-c} x^2}{\sqrt {2}}+\frac {\sqrt {b^2+c x^4+a^2 x^8}}{\sqrt {2} \sqrt [4]{2 a b-c}}}{x \sqrt [4]{b^2+c x^4+a^2 x^8}}\right )}{2 \sqrt {2} \sqrt [4]{2 a b-c}} \]
1/4*arctan(2^(1/2)*(2*a*b-c)^(1/4)*x*(a^2*x^8+c*x^4+b^2)^(1/4)/((2*a*b-c)^ (1/2)*x^2-(a^2*x^8+c*x^4+b^2)^(1/2)))*2^(1/2)/(2*a*b-c)^(1/4)-1/4*arctanh( (1/2*(2*a*b-c)^(1/4)*x^2*2^(1/2)+1/2*(a^2*x^8+c*x^4+b^2)^(1/2)*2^(1/2)/(2* a*b-c)^(1/4))/x/(a^2*x^8+c*x^4+b^2)^(1/4))*2^(1/2)/(2*a*b-c)^(1/4)
Time = 2.06 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.88 \[ \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx=\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{2 a b-c} x \sqrt [4]{b^2+c x^4+a^2 x^8}}{\sqrt {2 a b-c} x^2-\sqrt {b^2+c x^4+a^2 x^8}}\right )-\text {arctanh}\left (\frac {\sqrt {2 a b-c} x^2+\sqrt {b^2+c x^4+a^2 x^8}}{\sqrt {2} \sqrt [4]{2 a b-c} x \sqrt [4]{b^2+c x^4+a^2 x^8}}\right )}{2 \sqrt {2} \sqrt [4]{2 a b-c}} \]
(ArcTan[(Sqrt[2]*(2*a*b - c)^(1/4)*x*(b^2 + c*x^4 + a^2*x^8)^(1/4))/(Sqrt[ 2*a*b - c]*x^2 - Sqrt[b^2 + c*x^4 + a^2*x^8])] - ArcTanh[(Sqrt[2*a*b - c]* x^2 + Sqrt[b^2 + c*x^4 + a^2*x^8])/(Sqrt[2]*(2*a*b - c)^(1/4)*x*(b^2 + c*x ^4 + a^2*x^8)^(1/4))])/(2*Sqrt[2]*(2*a*b - c)^(1/4))
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.
\(\displaystyle \int \frac {a x^4-b}{\left (a x^4+b\right ) \sqrt [4]{a^2 x^8+b^2+c x^4}} \, dx\) |
\(\Big \downarrow \) 7276 |
\(\displaystyle \int \left (\frac {1}{\sqrt [4]{a^2 x^8+b^2+c x^4}}-\frac {2 b}{\left (a x^4+b\right ) \sqrt [4]{a^2 x^8+b^2+c x^4}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x \sqrt [4]{\frac {2 a^2 x^4}{c-\sqrt {c^2-4 a^2 b^2}}+1} \sqrt [4]{\frac {2 a^2 x^4}{\sqrt {c^2-4 a^2 b^2}+c}+1} \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{4},\frac {1}{4},\frac {5}{4},-\frac {2 a^2 x^4}{c-\sqrt {c^2-4 a^2 b^2}},-\frac {2 a^2 x^4}{c+\sqrt {c^2-4 a^2 b^2}}\right )}{\sqrt [4]{a^2 x^8+b^2+c x^4}}-2 b \int \frac {1}{\left (a x^4+b\right ) \sqrt [4]{a^2 x^8+c x^4+b^2}}dx\) |
3.26.10.3.1 Defintions of rubi rules used
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 = 1.38 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.24
method | result | size |
pseudoelliptic | \(\frac {\sqrt {2}\, \left (\ln \left (\frac {-\left (a^{2} x^{8}+c \,x^{4}+b^{2}\right )^{\frac {1}{4}} x \left (2 a b -c \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {2 a b -c}\, x^{2}+\sqrt {a^{2} x^{8}+c \,x^{4}+b^{2}}}{\left (a^{2} x^{8}+c \,x^{4}+b^{2}\right )^{\frac {1}{4}} x \left (2 a b -c \right )^{\frac {1}{4}} \sqrt {2}+\sqrt {2 a b -c}\, x^{2}+\sqrt {a^{2} x^{8}+c \,x^{4}+b^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a^{2} x^{8}+c \,x^{4}+b^{2}\right )^{\frac {1}{4}}+\left (2 a b -c \right )^{\frac {1}{4}} x}{\left (2 a b -c \right )^{\frac {1}{4}} x}\right )+2 \arctan \left (\frac {\sqrt {2}\, \left (a^{2} x^{8}+c \,x^{4}+b^{2}\right )^{\frac {1}{4}}-\left (2 a b -c \right )^{\frac {1}{4}} x}{\left (2 a b -c \right )^{\frac {1}{4}} x}\right )\right )}{8 \left (2 a b -c \right )^{\frac {1}{4}}}\) | \(259\) |
1/8/(2*a*b-c)^(1/4)*2^(1/2)*(ln((-(a^2*x^8+c*x^4+b^2)^(1/4)*x*(2*a*b-c)^(1 /4)*2^(1/2)+(2*a*b-c)^(1/2)*x^2+(a^2*x^8+c*x^4+b^2)^(1/2))/((a^2*x^8+c*x^4 +b^2)^(1/4)*x*(2*a*b-c)^(1/4)*2^(1/2)+(2*a*b-c)^(1/2)*x^2+(a^2*x^8+c*x^4+b ^2)^(1/2)))+2*arctan((2^(1/2)*(a^2*x^8+c*x^4+b^2)^(1/4)+(2*a*b-c)^(1/4)*x) /(2*a*b-c)^(1/4)/x)+2*arctan((2^(1/2)*(a^2*x^8+c*x^4+b^2)^(1/4)-(2*a*b-c)^ (1/4)*x)/(2*a*b-c)^(1/4)/x))
Timed out. \[ \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx=\text {Timed out} \]
\[ \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx=\int \frac {a x^{4} - b}{\left (a x^{4} + b\right ) \sqrt [4]{a^{2} x^{8} + b^{2} + c x^{4}}}\, dx \]
\[ \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx=\int { \frac {a x^{4} - b}{{\left (a^{2} x^{8} + c x^{4} + b^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}} \,d x } \]
\[ \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx=\int { \frac {a x^{4} - b}{{\left (a^{2} x^{8} + c x^{4} + b^{2}\right )}^{\frac {1}{4}} {\left (a x^{4} + b\right )}} \,d x } \]
Timed out. \[ \int \frac {-b+a x^4}{\left (b+a x^4\right ) \sqrt [4]{b^2+c x^4+a^2 x^8}} \, dx=\int -\frac {b-a\,x^4}{\left (a\,x^4+b\right )\,{\left (a^2\,x^8+b^2+c\,x^4\right )}^{1/4}} \,d x \]