\(\int \frac {1}{\sqrt [4]{b+a x^4} (-b+a x^8)} \, dx\) [741]

Optimal result

Integrand size = 23, antiderivative size = 57 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx=\frac {\text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]}{8 b} \]

[Out]

Unintegrable

Rubi [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(253\) vs. \(2(57)=114\).

Time = 0.17 (sec) , antiderivative size = 253, normalized size of antiderivative = 4.44, number of steps used = 9, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {1443, 385, 218, 212, 209} \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt [8]{a} x \sqrt [4]{\sqrt {a}-\sqrt {b}}}{\sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{a} b \sqrt [4]{\sqrt {a}-\sqrt {b}}}-\frac {\arctan \left (\frac {\sqrt [8]{a} x \sqrt [4]{\sqrt {a}+\sqrt {b}}}{\sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{a} b \sqrt [4]{\sqrt {a}+\sqrt {b}}}-\frac {\text {arctanh}\left (\frac {\sqrt [8]{a} x \sqrt [4]{\sqrt {a}-\sqrt {b}}}{\sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{a} b \sqrt [4]{\sqrt {a}-\sqrt {b}}}-\frac {\text {arctanh}\left (\frac {\sqrt [8]{a} x \sqrt [4]{\sqrt {a}+\sqrt {b}}}{\sqrt [4]{a x^4+b}}\right )}{4 \sqrt [8]{a} b \sqrt [4]{\sqrt {a}+\sqrt {b}}} \]

[In]

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

[Out]

-1/4*ArcTan[(a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*x)/(b + a*x^4)^(1/4)]/(a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*b) - A
rcTan[(a^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4)*x)/(b + a*x^4)^(1/4)]/(4*a^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4)*b) - ArcTa
nh[(a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*x)/(b + a*x^4)^(1/4)]/(4*a^(1/8)*(Sqrt[a] - Sqrt[b])^(1/4)*b) - ArcTanh[
(a^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4)*x)/(b + a*x^4)^(1/4)]/(4*a^(1/8)*(Sqrt[a] + Sqrt[b])^(1/4)*b)

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 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 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 1443

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

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

Mathematica [A] (verified)

Time = 0.31 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx=\frac {\text {RootSum}\left [a^2-a b-2 a \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{b+a x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]}{8 b} \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 1.22 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 a \,\textit {\_Z}^{4}+a^{2}-a b \right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (a \,x^{4}+b \right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}}{8 b}\)	\(50\)

[In]

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

[Out]

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

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [N/A]

Not integrable

Time = 5.43 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.33 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx=\int \frac {1}{\sqrt [4]{a x^{4} + b} \left (a x^{8} - b\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx=\int { \frac {1}{{\left (a x^{8} - b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.30 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.40 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx=\int { \frac {1}{{\left (a x^{8} - b\right )} {\left (a x^{4} + b\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 5.50 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.42 \[ \int \frac {1}{\sqrt [4]{b+a x^4} \left (-b+a x^8\right )} \, dx=-\int \frac {1}{{\left (a\,x^4+b\right )}^{1/4}\,\left (b-a\,x^8\right )} \,d x \]

[In]

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

[Out]

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