\(\int \frac {-1+x^8}{\sqrt [4]{1+x^4} (1+x^8)} \, dx\) [1039]

Optimal result

Integrand size = 22, antiderivative size = 78 \[ \int \frac {-1+x^8}{\sqrt [4]{1+x^4} \left (1+x^8\right )} \, dx=\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\frac {1}{4} \text {RootSum}\left [2-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{1+x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ] \]

[Out]

Unintegrable

Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.21 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.96, number of steps used = 21, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {1600, 6857, 399, 246, 218, 212, 209, 385} \[ \int \frac {-1+x^8}{\sqrt [4]{1+x^4} \left (1+x^8\right )} \, dx=\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {\arctan \left (\frac {\sqrt [4]{1-i} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{1-i}}-\frac {\arctan \left (\frac {\sqrt [4]{1+i} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{1+i}}+\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4+1}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt [4]{1-i} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{1-i}}-\frac {\text {arctanh}\left (\frac {\sqrt [4]{1+i} x}{\sqrt [4]{x^4+1}}\right )}{2 \sqrt [4]{1+i}} \]

[In]

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

[Out]

ArcTan[x/(1 + x^4)^(1/4)]/2 - ArcTan[((1 - I)^(1/4)*x)/(1 + x^4)^(1/4)]/(2*(1 - I)^(1/4)) - ArcTan[((1 + I)^(1
/4)*x)/(1 + x^4)^(1/4)]/(2*(1 + I)^(1/4)) + ArcTanh[x/(1 + x^4)^(1/4)]/2 - ArcTanh[((1 - I)^(1/4)*x)/(1 + x^4)
^(1/4)]/(2*(1 - I)^(1/4)) - ArcTanh[((1 + I)^(1/4)*x)/(1 + x^4)^(1/4)]/(2*(1 + I)^(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 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 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 399

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

Rule 1600

Int[(u_.)*(Px_)^(p_.)*(Qx_)^(q_.), x_Symbol] :> Int[u*PolynomialQuotient[Px, Qx, x]^p*Qx^(p + q), x] /; FreeQ[
q, x] && PolyQ[Px, x] && PolyQ[Qx, x] && EqQ[PolynomialRemainder[Px, Qx, x], 0] && IntegerQ[p] && LtQ[p*q, 0]

Rule 6857

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.94 \[ \int \frac {-1+x^8}{\sqrt [4]{1+x^4} \left (1+x^8\right )} \, dx=\frac {1}{4} \left (2 \left (\arctan \left (\frac {x}{\sqrt [4]{1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{1+x^4}}\right )\right )+\text {RootSum}\left [2-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{1+x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ]\right ) \]

[In]

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

[Out]

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

Maple [N/A] (verified)

Time = 16.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.12

method	result	size
pseudoelliptic	\(-\frac {\ln \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}-x}{x}\right )}{4}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}-2 \textit {\_Z}^{4}+2\right )}{\sum }\frac {\ln \left (\frac {-\textit {\_R} x +\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )}{\textit {\_R}}\right )}{4}-\frac {\arctan \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )}{2}+\frac {\ln \left (\frac {\left (x^{4}+1\right )^{\frac {1}{4}}+x}{x}\right )}{4}\)	\(87\)
trager	\(\text {Expression too large to display}\)	\(1292\)

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

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

Time = 0.30 (sec) , antiderivative size = 410, normalized size of antiderivative = 5.26 \[ \int \frac {-1+x^8}{\sqrt [4]{1+x^4} \left (1+x^8\right )} \, dx=-\frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {2} \sqrt {i + 1}} \log \left (\frac {-\left (i - 1\right ) \, \sqrt {i + 1} \sqrt {\sqrt {2} \sqrt {i + 1}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {2} \sqrt {i + 1}} \log \left (\frac {\left (i - 1\right ) \, \sqrt {i + 1} \sqrt {\sqrt {2} \sqrt {i + 1}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {-\sqrt {2} \sqrt {i + 1}} \log \left (\frac {-\left (i - 1\right ) \, \sqrt {i + 1} \sqrt {-\sqrt {2} \sqrt {i + 1}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{8} \, \sqrt {2} \sqrt {-\sqrt {2} \sqrt {i + 1}} \log \left (\frac {\left (i - 1\right ) \, \sqrt {i + 1} \sqrt {-\sqrt {2} \sqrt {i + 1}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {2} \sqrt {-i + 1}} \log \left (\frac {\left (i + 1\right ) \, \sqrt {-i + 1} \sqrt {\sqrt {2} \sqrt {-i + 1}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {\sqrt {2} \sqrt {-i + 1}} \log \left (\frac {-\left (i + 1\right ) \, \sqrt {-i + 1} \sqrt {\sqrt {2} \sqrt {-i + 1}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{8} \, \sqrt {2} \sqrt {-\sqrt {2} \sqrt {-i + 1}} \log \left (\frac {\left (i + 1\right ) \, \sqrt {-i + 1} \sqrt {-\sqrt {2} \sqrt {-i + 1}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{8} \, \sqrt {2} \sqrt {-\sqrt {2} \sqrt {-i + 1}} \log \left (\frac {-\left (i + 1\right ) \, \sqrt {-i + 1} \sqrt {-\sqrt {2} \sqrt {-i + 1}} x + 2 \, {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {x + {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \log \left (-\frac {x - {\left (x^{4} + 1\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

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

[Out]

-1/8*sqrt(2)*sqrt(sqrt(2)*sqrt(I + 1))*log((-(I - 1)*sqrt(I + 1)*sqrt(sqrt(2)*sqrt(I + 1))*x + 2*(x^4 + 1)^(1/
4))/x) + 1/8*sqrt(2)*sqrt(sqrt(2)*sqrt(I + 1))*log(((I - 1)*sqrt(I + 1)*sqrt(sqrt(2)*sqrt(I + 1))*x + 2*(x^4 +
1)^(1/4))/x) + 1/8*sqrt(2)*sqrt(-sqrt(2)*sqrt(I + 1))*log((-(I - 1)*sqrt(I + 1)*sqrt(-sqrt(2)*sqrt(I + 1))*x
+ 2*(x^4 + 1)^(1/4))/x) - 1/8*sqrt(2)*sqrt(-sqrt(2)*sqrt(I + 1))*log(((I - 1)*sqrt(I + 1)*sqrt(-sqrt(2)*sqrt(I
+ 1))*x + 2*(x^4 + 1)^(1/4))/x) - 1/8*sqrt(2)*sqrt(sqrt(2)*sqrt(-I + 1))*log(((I + 1)*sqrt(-I + 1)*sqrt(sqrt(
2)*sqrt(-I + 1))*x + 2*(x^4 + 1)^(1/4))/x) + 1/8*sqrt(2)*sqrt(sqrt(2)*sqrt(-I + 1))*log((-(I + 1)*sqrt(-I + 1)
*sqrt(sqrt(2)*sqrt(-I + 1))*x + 2*(x^4 + 1)^(1/4))/x) + 1/8*sqrt(2)*sqrt(-sqrt(2)*sqrt(-I + 1))*log(((I + 1)*s
qrt(-I + 1)*sqrt(-sqrt(2)*sqrt(-I + 1))*x + 2*(x^4 + 1)^(1/4))/x) - 1/8*sqrt(2)*sqrt(-sqrt(2)*sqrt(-I + 1))*lo
g((-(I + 1)*sqrt(-I + 1)*sqrt(-sqrt(2)*sqrt(-I + 1))*x + 2*(x^4 + 1)^(1/4))/x) - 1/2*arctan((x^4 + 1)^(1/4)/x)
+ 1/4*log((x + (x^4 + 1)^(1/4))/x) - 1/4*log(-(x - (x^4 + 1)^(1/4))/x)

Sympy [N/A]

Not integrable

Time = 66.74 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.33 \[ \int \frac {-1+x^8}{\sqrt [4]{1+x^4} \left (1+x^8\right )} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \left (x^{4} + 1\right )^{\frac {3}{4}}}{x^{8} + 1}\, dx \]

[In]

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

[Out]

Integral((x - 1)*(x + 1)*(x**2 + 1)*(x**4 + 1)**(3/4)/(x**8 + 1), x)

Maxima [N/A]

Not integrable

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.28 \[ \int \frac {-1+x^8}{\sqrt [4]{1+x^4} \left (1+x^8\right )} \, dx=\int { \frac {x^{8} - 1}{{\left (x^{8} + 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.27 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.28 \[ \int \frac {-1+x^8}{\sqrt [4]{1+x^4} \left (1+x^8\right )} \, dx=\int { \frac {x^{8} - 1}{{\left (x^{8} + 1\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.28 \[ \int \frac {-1+x^8}{\sqrt [4]{1+x^4} \left (1+x^8\right )} \, dx=\int \frac {x^8-1}{{\left (x^4+1\right )}^{1/4}\,\left (x^8+1\right )} \,d x \]

[In]

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

[Out]

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