3.26.21 \(\int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} (-2 b-c x^4+2 a x^8)} \, dx\) [2521]

3.26.21.1 Optimal result

Integrand size = 46, antiderivative size = 210 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )}{2 \sqrt [4]{a}}+\frac {1}{4} \text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-3 a \log (x)+c \log (x)+3 a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+3 \log (x) \text {$\#$1}^4-3 \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{4 a \text {$\#$1}+c \text {$\#$1}-4 \text {$\#$1}^5}\&\right ] \]

output

Unintegrable
 

3.26.21.2 Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.96 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\frac {1}{4} \left (\frac {2 \left (\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{-b+a x^4}}\right )\right )}{\sqrt [4]{a}}+\text {RootSum}\left [2 a^2-2 a b+a c-4 a \text {$\#$1}^4-c \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {3 a \log (x)-c \log (x)-3 a \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )+c \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right )-3 \log (x) \text {$\#$1}^4+3 \log \left (\sqrt [4]{-b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-4 a \text {$\#$1}-c \text {$\#$1}+4 \text {$\#$1}^5}\&\right ]\right ) \]

input

Integrate[(b - 2*c*x^4 + 2*a*x^8)/((-b + a*x^4)^(1/4)*(-2*b - c*x^4 + 2*a* 
x^8)),x]
 

output

((2*(ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)] + ArcTanh[(a^(1/4)*x)/(-b + a* 
x^4)^(1/4)]))/a^(1/4) + RootSum[2*a^2 - 2*a*b + a*c - 4*a*#1^4 - c*#1^4 + 
2*#1^8 & , (3*a*Log[x] - c*Log[x] - 3*a*Log[(-b + a*x^4)^(1/4) - x*#1] + c 
*Log[(-b + a*x^4)^(1/4) - x*#1] - 3*Log[x]*#1^4 + 3*Log[(-b + a*x^4)^(1/4) 
 - x*#1]*#1^4)/(-4*a*#1 - c*#1 + 4*#1^5) & ])/4
 

3.26.21.3 Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(611\) vs. \(2(210)=420\).

Time = 1.68 (sec) , antiderivative size = 611, normalized size of antiderivative = 2.91, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, 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.

\(\displaystyle \int \frac {2 a x^8+b-2 c x^4}{\sqrt [4]{a x^4-b} \left (2 a x^8-2 b-c x^4\right )} \, dx\)

\(\Big \downarrow \) 7279

\(\displaystyle \int \left (\frac {3 b-c x^4}{\sqrt [4]{a x^4-b} \left (2 a x^8-2 b-c x^4\right )}+\frac {1}{\sqrt [4]{a x^4-b}}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\left (\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}+c\right ) \arctan \left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \arctan \left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}-\frac {\left (\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}+c\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}{\sqrt [4]{\sqrt {16 a b+c^2}-c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}-c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}+4 b-c}}+\frac {\left (c-\frac {12 a b-c^2}{\sqrt {16 a b+c^2}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a} x \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}{\sqrt [4]{\sqrt {16 a b+c^2}+c} \sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a} \left (\sqrt {16 a b+c^2}+c\right )^{3/4} \sqrt [4]{\sqrt {16 a b+c^2}-4 b+c}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4-b}}\right )}{2 \sqrt [4]{a}}\)

input

Int[(b - 2*c*x^4 + 2*a*x^8)/((-b + a*x^4)^(1/4)*(-2*b - c*x^4 + 2*a*x^8)), 
x]
 

output

ArcTan[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(1/4)) - ((c + (12*a*b - c^2)/ 
Sqrt[16*a*b + c^2])*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)*(-b + a*x^4)^(1/4))])/(2*a^(1/4)*(-c + 
Sqrt[16*a*b + c^2])^(3/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)) + ((c - (1 
2*a*b - c^2)/Sqrt[16*a*b + c^2])*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)*(-b + a*x^4)^(1/4))])/(2*a 
^(1/4)*(c + Sqrt[16*a*b + c^2])^(3/4)*(-4*b + c + Sqrt[16*a*b + c^2])^(1/4 
)) + ArcTanh[(a^(1/4)*x)/(-b + a*x^4)^(1/4)]/(2*a^(1/4)) - ((c + (12*a*b - 
 c^2)/Sqrt[16*a*b + c^2])*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)*(-b + a*x^4)^(1/4))])/(2*a^(1/4) 
*(-c + Sqrt[16*a*b + c^2])^(3/4)*(4*b - c + Sqrt[16*a*b + c^2])^(1/4)) + ( 
(c - (12*a*b - c^2)/Sqrt[16*a*b + c^2])*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)*(-b + a*x^4)^(1/4) 
)])/(2*a^(1/4)*(c + Sqrt[16*a*b + c^2])^(3/4)*(-4*b + c + Sqrt[16*a*b + c^ 
2])^(1/4))
 

3.26.21.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7279

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]
 

3.26.21.4 Maple [N/A] (verified)

Time = 0.14 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.75

method	result	size
pseudoelliptic	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}+\left (-4 a -c \right ) \textit {\_Z}^{4}+2 a^{2}-2 a b +a c \right )}{\sum }\frac {\left (3 \textit {\_R}^{4}-3 a +c \right ) \ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (4 \textit {\_R}^{4}-4 a -c \right )}\right ) a^{\frac {1}{4}}-2 \arctan \left (\frac {\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+\ln \left (\frac {-x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}{x \,a^{\frac {1}{4}}-\left (a \,x^{4}-b \right )^{\frac {1}{4}}}\right )}{4 a^{\frac {1}{4}}}\)	\(157\)

input

int((2*a*x^8-2*c*x^4+b)/(a*x^4-b)^(1/4)/(2*a*x^8-c*x^4-2*b),x,method=_RETU 
RNVERBOSE)
 

output

1/4*(sum(1/_R*(3*_R^4-3*a+c)*ln((-_R*x+(a*x^4-b)^(1/4))/x)/(4*_R^4-4*a-c), 
_R=RootOf(2*_Z^8+(-4*a-c)*_Z^4+2*a^2-2*a*b+a*c))*a^(1/4)-2*arctan(1/a^(1/4 
)/x*(a*x^4-b)^(1/4))+ln((-x*a^(1/4)-(a*x^4-b)^(1/4))/(x*a^(1/4)-(a*x^4-b)^ 
(1/4))))/a^(1/4)
 

3.26.21.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 6.91 (sec) , antiderivative size = 16424, normalized size of antiderivative = 78.21 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\text {Too large to display} \]

input

integrate((2*a*x^8-2*c*x^4+b)/(a*x^4-b)^(1/4)/(2*a*x^8-c*x^4-2*b),x, algor 
ithm="fricas")
 

output

Too large to include
 

3.26.21.6 Sympy [F(-1)]

Timed out. \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\text {Timed out} \]

input

integrate((2*a*x**8-2*c*x**4+b)/(a*x**4-b)**(1/4)/(2*a*x**8-c*x**4-2*b),x)
 

output

Timed out
 

3.26.21.7 Maxima [N/A]

Not integrable

Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int { \frac {2 \, a x^{8} - 2 \, c x^{4} + b}{{\left (2 \, a x^{8} - c x^{4} - 2 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \]

input

integrate((2*a*x^8-2*c*x^4+b)/(a*x^4-b)^(1/4)/(2*a*x^8-c*x^4-2*b),x, algor 
ithm="maxima")
 

output

integrate((2*a*x^8 - 2*c*x^4 + b)/((2*a*x^8 - c*x^4 - 2*b)*(a*x^4 - b)^(1/ 
4)), x)
 

3.26.21.8 Giac [N/A]

Not integrable

Time = 1.75 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int { \frac {2 \, a x^{8} - 2 \, c x^{4} + b}{{\left (2 \, a x^{8} - c x^{4} - 2 \, b\right )} {\left (a x^{4} - b\right )}^{\frac {1}{4}}} \,d x } \]

input

integrate((2*a*x^8-2*c*x^4+b)/(a*x^4-b)^(1/4)/(2*a*x^8-c*x^4-2*b),x, algor 
ithm="giac")
 

output

integrate((2*a*x^8 - 2*c*x^4 + b)/((2*a*x^8 - c*x^4 - 2*b)*(a*x^4 - b)^(1/ 
4)), x)
 

3.26.21.9 Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.22 \[ \int \frac {b-2 c x^4+2 a x^8}{\sqrt [4]{-b+a x^4} \left (-2 b-c x^4+2 a x^8\right )} \, dx=\int -\frac {2\,a\,x^8-2\,c\,x^4+b}{{\left (a\,x^4-b\right )}^{1/4}\,\left (-2\,a\,x^8+c\,x^4+2\,b\right )} \,d x \]

input

int(-(b + 2*a*x^8 - 2*c*x^4)/((a*x^4 - b)^(1/4)*(2*b - 2*a*x^8 + c*x^4)),x 
)
 

output

int(-(b + 2*a*x^8 - 2*c*x^4)/((a*x^4 - b)^(1/4)*(2*b - 2*a*x^8 + c*x^4)), 
x)