\(\int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} (-2 b+a x^4+x^8)} \, dx\) [2129]

Optimal result

Integrand size = 41, antiderivative size = 154 \[ \int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}-\frac {3}{4} \text {RootSum}\left [3 a^2-b-5 a \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {3 a \log (x)-3 a \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )-2 \log (x) \text {$\#$1}^4+2 \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{5 a \text {$\#$1}-4 \text {$\#$1}^5}\&\right ] \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(549\) vs. \(2(154)=308\).

Time = 0.57 (sec) , antiderivative size = 549, normalized size of antiderivative = 3.56, number of steps used = 16, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {6860, 246, 218, 212, 209, 385, 214, 211} \[ \int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx=-\frac {3 \left (a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right ) \arctan \left (\frac {x \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2+8 b}} \sqrt [4]{a x^4+b}}\right )}{2 \left (a-\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-2 b}}-\frac {3 \left (\frac {a^2+4 b}{\sqrt {a^2+8 b}}+a\right ) \arctan \left (\frac {x \sqrt [4]{a \sqrt {a^2+8 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2+8 b}+a} \sqrt [4]{a x^4+b}}\right )}{2 \left (\sqrt {a^2+8 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2+8 b}+a^2-2 b}}-\frac {3 \left (a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right ) \text {arctanh}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-2 b}}{\sqrt [4]{a-\sqrt {a^2+8 b}} \sqrt [4]{a x^4+b}}\right )}{2 \left (a-\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+8 b}+a^2-2 b}}-\frac {3 \left (\frac {a^2+4 b}{\sqrt {a^2+8 b}}+a\right ) \text {arctanh}\left (\frac {x \sqrt [4]{a \sqrt {a^2+8 b}+a^2-2 b}}{\sqrt [4]{\sqrt {a^2+8 b}+a} \sqrt [4]{a x^4+b}}\right )}{2 \left (\sqrt {a^2+8 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2+8 b}+a^2-2 b}}+\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{a x^4+b}}\right )}{\sqrt [4]{a}} \]

[In]

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

[Out]

ArcTan[(a^(1/4)*x)/(b + a*x^4)^(1/4)]/a^(1/4) - (3*(a - (a^2 + 4*b)/Sqrt[a^2 + 8*b])*ArcTan[((a^2 - 2*b - a*Sq
rt[a^2 + 8*b])^(1/4)*x)/((a - Sqrt[a^2 + 8*b])^(1/4)*(b + a*x^4)^(1/4))])/(2*(a - Sqrt[a^2 + 8*b])^(3/4)*(a^2
- 2*b - a*Sqrt[a^2 + 8*b])^(1/4)) - (3*(a + (a^2 + 4*b)/Sqrt[a^2 + 8*b])*ArcTan[((a^2 - 2*b + a*Sqrt[a^2 + 8*b
])^(1/4)*x)/((a + Sqrt[a^2 + 8*b])^(1/4)*(b + a*x^4)^(1/4))])/(2*(a + Sqrt[a^2 + 8*b])^(3/4)*(a^2 - 2*b + a*Sq
rt[a^2 + 8*b])^(1/4)) + ArcTanh[(a^(1/4)*x)/(b + a*x^4)^(1/4)]/a^(1/4) - (3*(a - (a^2 + 4*b)/Sqrt[a^2 + 8*b])*
ArcTanh[((a^2 - 2*b - a*Sqrt[a^2 + 8*b])^(1/4)*x)/((a - Sqrt[a^2 + 8*b])^(1/4)*(b + a*x^4)^(1/4))])/(2*(a - Sq
rt[a^2 + 8*b])^(3/4)*(a^2 - 2*b - a*Sqrt[a^2 + 8*b])^(1/4)) - (3*(a + (a^2 + 4*b)/Sqrt[a^2 + 8*b])*ArcTanh[((a
^2 - 2*b + a*Sqrt[a^2 + 8*b])^(1/4)*x)/((a + Sqrt[a^2 + 8*b])^(1/4)*(b + a*x^4)^(1/4))])/(2*(a + Sqrt[a^2 + 8*
b])^(3/4)*(a^2 - 2*b + a*Sqrt[a^2 + 8*b])^(1/4))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 246

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + 1/n), Subst[Int[1/(1 - b*x^n)^(p + 1/n + 1), x], x
, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[p, -2^(-1)] && IntegerQ[p
+ 1/n]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 6860

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] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2}{\sqrt [4]{b+a x^4}}+\frac {3 \left (2 b-a x^4\right )}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )}\right ) \, dx \\ & = 2 \int \frac {1}{\sqrt [4]{b+a x^4}} \, dx+3 \int \frac {2 b-a x^4}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx \\ & = 2 \text {Subst}\left (\int \frac {1}{1-a x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )+3 \int \left (\frac {-a+\frac {a^2+4 b}{\sqrt {a^2+8 b}}}{\left (a-\sqrt {a^2+8 b}+2 x^4\right ) \sqrt [4]{b+a x^4}}+\frac {-a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}}{\left (a+\sqrt {a^2+8 b}+2 x^4\right ) \sqrt [4]{b+a x^4}}\right ) \, dx \\ & = -\left (\left (3 \left (a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right )\right ) \int \frac {1}{\left (a-\sqrt {a^2+8 b}+2 x^4\right ) \sqrt [4]{b+a x^4}} \, dx\right )-\left (3 \left (a+\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right )\right ) \int \frac {1}{\left (a+\sqrt {a^2+8 b}+2 x^4\right ) \sqrt [4]{b+a x^4}} \, dx+\text {Subst}\left (\int \frac {1}{1-\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )+\text {Subst}\left (\int \frac {1}{1+\sqrt {a} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right ) \\ & = \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}-\left (3 \left (a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right )\right ) \text {Subst}\left (\int \frac {1}{a-\sqrt {a^2+8 b}-\left (-2 b+a \left (a-\sqrt {a^2+8 b}\right )\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )-\left (3 \left (a+\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right )\right ) \text {Subst}\left (\int \frac {1}{a+\sqrt {a^2+8 b}-\left (-2 b+a \left (a+\sqrt {a^2+8 b}\right )\right ) x^4} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right ) \\ & = \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}-\frac {\left (3 \left (a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-\sqrt {a^2+8 b}}-\sqrt {a^2-2 b-a \sqrt {a^2+8 b}} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{2 \sqrt {a-\sqrt {a^2+8 b}}}-\frac {\left (3 \left (a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a-\sqrt {a^2+8 b}}+\sqrt {a^2-2 b-a \sqrt {a^2+8 b}} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{2 \sqrt {a-\sqrt {a^2+8 b}}}-\frac {\left (3 \left (a+\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\sqrt {a^2+8 b}}-\sqrt {a^2-2 b+a \sqrt {a^2+8 b}} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{2 \sqrt {a+\sqrt {a^2+8 b}}}-\frac {\left (3 \left (a+\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+\sqrt {a^2+8 b}}+\sqrt {a^2-2 b+a \sqrt {a^2+8 b}} x^2} \, dx,x,\frac {x}{\sqrt [4]{b+a x^4}}\right )}{2 \sqrt {a+\sqrt {a^2+8 b}}} \\ & = \frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}-\frac {3 \left (a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right ) \arctan \left (\frac {\sqrt [4]{a^2-2 b-a \sqrt {a^2+8 b}} x}{\sqrt [4]{a-\sqrt {a^2+8 b}} \sqrt [4]{b+a x^4}}\right )}{2 \left (a-\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{a^2-2 b-a \sqrt {a^2+8 b}}}-\frac {3 \left (a+\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right ) \arctan \left (\frac {\sqrt [4]{a^2-2 b+a \sqrt {a^2+8 b}} x}{\sqrt [4]{a+\sqrt {a^2+8 b}} \sqrt [4]{b+a x^4}}\right )}{2 \left (a+\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{a^2-2 b+a \sqrt {a^2+8 b}}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}-\frac {3 \left (a-\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a^2-2 b-a \sqrt {a^2+8 b}} x}{\sqrt [4]{a-\sqrt {a^2+8 b}} \sqrt [4]{b+a x^4}}\right )}{2 \left (a-\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{a^2-2 b-a \sqrt {a^2+8 b}}}-\frac {3 \left (a+\frac {a^2+4 b}{\sqrt {a^2+8 b}}\right ) \text {arctanh}\left (\frac {\sqrt [4]{a^2-2 b+a \sqrt {a^2+8 b}} x}{\sqrt [4]{a+\sqrt {a^2+8 b}} \sqrt [4]{b+a x^4}}\right )}{2 \left (a+\sqrt {a^2+8 b}\right )^{3/4} \sqrt [4]{a^2-2 b+a \sqrt {a^2+8 b}}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.97 \[ \int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{a} x}{\sqrt [4]{b+a x^4}}\right )}{\sqrt [4]{a}}-\frac {3}{4} \text {RootSum}\left [3 a^2-b-5 a \text {$\#$1}^4+2 \text {$\#$1}^8\&,\frac {-3 a \log (x)+3 a \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )+2 \log (x) \text {$\#$1}^4-2 \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-5 a \text {$\#$1}+4 \text {$\#$1}^5}\&\right ] \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 0.39 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.90

method	result	size
pseudoelliptic	\(\frac {3 \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}-5 a \,\textit {\_Z}^{4}+3 a^{2}-b \right )}{\sum }\frac {\left (2 \textit {\_R}^{4}-3 a \right ) \ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (4 \textit {\_R}^{4}-5 a \right )}\right ) a^{\frac {1}{4}}-4 \arctan \left (\frac {\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x}\right )+2 \ln \left (\frac {-a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{a^{\frac {1}{4}} x -\left (a \,x^{4}+b \right )^{\frac {1}{4}}}\right )}{4 a^{\frac {1}{4}}}\)	\(139\)

[In]

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

[Out]

1/4*(3*sum(1/_R*(2*_R^4-3*a)*ln((-_R*x+(a*x^4+b)^(1/4))/x)/(4*_R^4-5*a),_R=RootOf(2*_Z^8-5*_Z^4*a+3*a^2-b))*a^
(1/4)-4*arctan(1/a^(1/4)/x*(a*x^4+b)^(1/4))+2*ln((-a^(1/4)*x-(a*x^4+b)^(1/4))/(a^(1/4)*x-(a*x^4+b)^(1/4))))/a^
(1/4)

Fricas [C] (verification not implemented)

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

Time = 0.69 (sec) , antiderivative size = 5666, normalized size of antiderivative = 36.79 \[ \int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [N/A]

Not integrable

Time = 111.53 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.23 \[ \int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx=\int \frac {- a x^{4} + 2 b + 2 x^{8}}{\sqrt [4]{a x^{4} + b} \left (a x^{4} - 2 b + x^{8}\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.22 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.27 \[ \int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{8} - a x^{4} + 2 \, b}{{\left (x^{8} + a x^{4} - 2 \, b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.92 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.27 \[ \int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx=\int { \frac {2 \, x^{8} - a x^{4} + 2 \, b}{{\left (x^{8} + a x^{4} - 2 \, b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.27 \[ \int \frac {2 b-a x^4+2 x^8}{\sqrt [4]{b+a x^4} \left (-2 b+a x^4+x^8\right )} \, dx=\int \frac {2\,x^8-a\,x^4+2\,b}{{\left (a\,x^4+b\right )}^{1/4}\,\left (x^8+a\,x^4-2\,b\right )} \,d x \]

[In]

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

[Out]

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