\(\int \frac {(-1+x^4)^{3/4}}{1+x^4} \, dx\) [1077]

Optimal result

Integrand size = 17, antiderivative size = 81 \[ \int \frac {\left (-1+x^4\right )^{3/4}}{1+x^4} \, dx=\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt [4]{2}}+\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )}{\sqrt [4]{2}} \]

[Out]

1/2*arctan(x/(x^4-1)^(1/4))-1/2*arctan(2^(1/4)*x/(x^4-1)^(1/4))*2^(3/4)+1/2*arctanh(x/(x^4-1)^(1/4))-1/2*arcta
nh(2^(1/4)*x/(x^4-1)^(1/4))*2^(3/4)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {399, 246, 218, 212, 209, 385} \[ \int \frac {\left (-1+x^4\right )^{3/4}}{1+x^4} \, dx=\frac {1}{2} \arctan \left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{\sqrt [4]{2}}+\frac {1}{2} \text {arctanh}\left (\frac {x}{\sqrt [4]{x^4-1}}\right )-\frac {\text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4-1}}\right )}{\sqrt [4]{2}} \]

[In]

Int[(-1 + x^4)^(3/4)/(1 + x^4),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.95 \[ \int \frac {\left (-1+x^4\right )^{3/4}}{1+x^4} \, dx=\frac {1}{2} \left (\arctan \left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-2^{3/4} \arctan \left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{-1+x^4}}\right )-2^{3/4} \text {arctanh}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{-1+x^4}}\right )\right ) \]

[In]

Integrate[(-1 + x^4)^(3/4)/(1 + x^4),x]

[Out]

(ArcTan[x/(-1 + x^4)^(1/4)] - 2^(3/4)*ArcTan[(2^(1/4)*x)/(-1 + x^4)^(1/4)] + ArcTanh[x/(-1 + x^4)^(1/4)] - 2^(
3/4)*ArcTanh[(2^(1/4)*x)/(-1 + x^4)^(1/4)])/2

Maple [A] (verified)

Time = 2.78 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.32

method	result	size
pseudoelliptic	\(-\frac {\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}-x}{x}\right )}{4}-\frac {\arctan \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}}{x}\right )}{2}+\frac {\ln \left (\frac {\left (x^{4}-1\right )^{\frac {1}{4}}+x}{x}\right )}{4}+\frac {\arctan \left (\frac {2^{\frac {3}{4}} \left (x^{4}-1\right )^{\frac {1}{4}}}{2 x}\right ) 2^{\frac {3}{4}}}{2}-\frac {\ln \left (\frac {2^{\frac {1}{4}} x +\left (x^{4}-1\right )^{\frac {1}{4}}}{-2^{\frac {1}{4}} x +\left (x^{4}-1\right )^{\frac {1}{4}}}\right ) 2^{\frac {3}{4}}}{4}\)	\(107\)

[In]

int((x^4-1)^(3/4)/(x^4+1),x,method=_RETURNVERBOSE)

[Out]

-1/4*ln(((x^4-1)^(1/4)-x)/x)-1/2*arctan((x^4-1)^(1/4)/x)+1/4*ln(((x^4-1)^(1/4)+x)/x)+1/2*arctan(1/2*2^(3/4)/x*
(x^4-1)^(1/4))*2^(3/4)-1/4*ln((2^(1/4)*x+(x^4-1)^(1/4))/(-2^(1/4)*x+(x^4-1)^(1/4)))*2^(3/4)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.81 \[ \int \frac {\left (-1+x^4\right )^{3/4}}{1+x^4} \, dx=-\frac {1}{4} \cdot 2^{\frac {3}{4}} \log \left (\frac {2^{\frac {1}{4}} x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \cdot 2^{\frac {3}{4}} \log \left (-\frac {2^{\frac {1}{4}} x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} i \cdot 2^{\frac {3}{4}} \log \left (\frac {i \cdot 2^{\frac {1}{4}} x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} i \cdot 2^{\frac {3}{4}} \log \left (\frac {-i \cdot 2^{\frac {1}{4}} x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{2} \, \arctan \left (\frac {{\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) + \frac {1}{4} \, \log \left (\frac {x + {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) - \frac {1}{4} \, \log \left (-\frac {x - {\left (x^{4} - 1\right )}^{\frac {1}{4}}}{x}\right ) \]

[In]

integrate((x^4-1)^(3/4)/(x^4+1),x, algorithm="fricas")

[Out]

-1/4*2^(3/4)*log((2^(1/4)*x + (x^4 - 1)^(1/4))/x) + 1/4*2^(3/4)*log(-(2^(1/4)*x - (x^4 - 1)^(1/4))/x) + 1/4*I*
2^(3/4)*log((I*2^(1/4)*x + (x^4 - 1)^(1/4))/x) - 1/4*I*2^(3/4)*log((-I*2^(1/4)*x + (x^4 - 1)^(1/4))/x) - 1/2*a
rctan((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 [F]

\[ \int \frac {\left (-1+x^4\right )^{3/4}}{1+x^4} \, dx=\int \frac {\left (\left (x - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right )\right )^{\frac {3}{4}}}{x^{4} + 1}\, dx \]

[In]

integrate((x**4-1)**(3/4)/(x**4+1),x)

[Out]

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

Maxima [F]

\[ \int \frac {\left (-1+x^4\right )^{3/4}}{1+x^4} \, dx=\int { \frac {{\left (x^{4} - 1\right )}^{\frac {3}{4}}}{x^{4} + 1} \,d x } \]

[In]

integrate((x^4-1)^(3/4)/(x^4+1),x, algorithm="maxima")

[Out]

integrate((x^4 - 1)^(3/4)/(x^4 + 1), x)

Giac [F]

\[ \int \frac {\left (-1+x^4\right )^{3/4}}{1+x^4} \, dx=\int { \frac {{\left (x^{4} - 1\right )}^{\frac {3}{4}}}{x^{4} + 1} \,d x } \]

[In]

integrate((x^4-1)^(3/4)/(x^4+1),x, algorithm="giac")

[Out]

integrate((x^4 - 1)^(3/4)/(x^4 + 1), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-1+x^4\right )^{3/4}}{1+x^4} \, dx=\int \frac {{\left (x^4-1\right )}^{3/4}}{x^4+1} \,d x \]

[In]

int((x^4 - 1)^(3/4)/(x^4 + 1),x)

[Out]

int((x^4 - 1)^(3/4)/(x^4 + 1), x)