\(\int \frac {1}{\sqrt [4]{b+a x^4} (-2 b-2 a x^4+x^8)} \, dx\) [1338]

Optimal result

Integrand size = 27, antiderivative size = 96 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-2 b-2 a x^4+x^8\right )} \, dx=-\frac {\text {RootSum}\left [b+2 a \text {$\#$1}^4-2 \text {$\#$1}^8\&,\frac {a \log (x)-a \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )-\log (x) \text {$\#$1}^4+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{a \text {$\#$1}-2 \text {$\#$1}^5}\&\right ]}{8 b} \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(441\) vs. \(2(96)=192\).

Time = 0.16 (sec) , antiderivative size = 441, normalized size of antiderivative = 4.59, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.185, Rules used = {1442, 385, 218, 214, 211} \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-2 b-2 a x^4+x^8\right )} \, dx=\frac {\arctan \left (\frac {x \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{a-\sqrt {a^2+2 b}} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2+2 b} \left (a-\sqrt {a^2+2 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}-\frac {\arctan \left (\frac {x \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{\sqrt {a^2+2 b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2+2 b} \left (\sqrt {a^2+2 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}+\frac {\text {arctanh}\left (\frac {x \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{a-\sqrt {a^2+2 b}} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2+2 b} \left (a-\sqrt {a^2+2 b}\right )^{3/4} \sqrt [4]{-a \sqrt {a^2+2 b}+a^2+b}}-\frac {\text {arctanh}\left (\frac {x \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}}{\sqrt [4]{\sqrt {a^2+2 b}+a} \sqrt [4]{a x^4+b}}\right )}{4 \sqrt {a^2+2 b} \left (\sqrt {a^2+2 b}+a\right )^{3/4} \sqrt [4]{a \sqrt {a^2+2 b}+a^2+b}} \]

[In]

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

[Out]

ArcTan[((a^2 + b - a*Sqrt[a^2 + 2*b])^(1/4)*x)/((a - Sqrt[a^2 + 2*b])^(1/4)*(b + a*x^4)^(1/4))]/(4*Sqrt[a^2 +
2*b]*(a - Sqrt[a^2 + 2*b])^(3/4)*(a^2 + b - a*Sqrt[a^2 + 2*b])^(1/4)) - ArcTan[((a^2 + b + a*Sqrt[a^2 + 2*b])^
(1/4)*x)/((a + Sqrt[a^2 + 2*b])^(1/4)*(b + a*x^4)^(1/4))]/(4*Sqrt[a^2 + 2*b]*(a + Sqrt[a^2 + 2*b])^(3/4)*(a^2
+ b + a*Sqrt[a^2 + 2*b])^(1/4)) + ArcTanh[((a^2 + b - a*Sqrt[a^2 + 2*b])^(1/4)*x)/((a - Sqrt[a^2 + 2*b])^(1/4)
*(b + a*x^4)^(1/4))]/(4*Sqrt[a^2 + 2*b]*(a - Sqrt[a^2 + 2*b])^(3/4)*(a^2 + b - a*Sqrt[a^2 + 2*b])^(1/4)) - Arc
Tanh[((a^2 + b + a*Sqrt[a^2 + 2*b])^(1/4)*x)/((a + Sqrt[a^2 + 2*b])^(1/4)*(b + a*x^4)^(1/4))]/(4*Sqrt[a^2 + 2*
b]*(a + Sqrt[a^2 + 2*b])^(3/4)*(a^2 + b + a*Sqrt[a^2 + 2*b])^(1/4))

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 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 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 1442

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (b_.)*(x_)^(n_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[b^2 -
4*a*c, 2]}, Dist[2*(c/r), Int[(d + e*x^n)^q/(b - r + 2*c*x^n), x], x] - Dist[2*(c/r), Int[(d + e*x^n)^q/(b +
r + 2*c*x^n), x], x]] /; FreeQ[{a, b, c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] &&  !IntegerQ[q]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-2 b-2 a x^4+x^8\right )} \, dx=-\frac {\text {RootSum}\left [b+2 a \text {$\#$1}^4-2 \text {$\#$1}^8\&,\frac {-a \log (x)+a \log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )+\log (x) \text {$\#$1}^4-\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right ) \text {$\#$1}^4}{-a \text {$\#$1}+2 \text {$\#$1}^5}\&\right ]}{8 b} \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 0.00 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69

method	result	size
pseudoelliptic	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (2 \textit {\_Z}^{8}-2 \textit {\_Z}^{4} a -b \right )}{\sum }\frac {\left (\textit {\_R}^{4}-a \right ) \ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R} \left (2 \textit {\_R}^{4}-a \right )}}{8 b}\)	\(66\)

[In]

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

[Out]

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

Fricas [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-2 b-2 a x^4+x^8\right )} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 6.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.27 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-2 b-2 a x^4+x^8\right )} \, dx=\int \frac {1}{\sqrt [4]{a x^{4} + b} \left (- 2 a x^{4} - 2 b + x^{8}\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.23 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.28 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-2 b-2 a x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} - 2 \, a x^{4} - 2 \, b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.92 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.28 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-2 b-2 a x^4+x^8\right )} \, dx=\int { \frac {1}{{\left (x^{8} - 2 \, a x^{4} - 2 \, b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.32 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-2 b-2 a x^4+x^8\right )} \, dx=-\int \frac {1}{{\left (a\,x^4+b\right )}^{1/4}\,\left (-x^8+2\,a\,x^4+2\,b\right )} \,d x \]

[In]

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

[Out]

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