\(\int \frac {1}{\sqrt [4]{1+x^4} (2+x^4)} \, dx\) [216]

Optimal result

Integrand size = 17, antiderivative size = 53 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (2+x^4\right )} \, dx=\frac {\arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{2\ 2^{3/4}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{2\ 2^{3/4}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {385, 218, 212, 209} \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (2+x^4\right )} \, dx=\frac {\arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}}+\frac {\text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}} \]

[In]

Int[1/((1 + x^4)^(1/4)*(2 + x^4)),x]

[Out]

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

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 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]

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

Mathematica [A] (verified)

Time = 0.30 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (2+x^4\right )} \, dx=\frac {\arctan \left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )+\text {arctanh}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{1+x^4}}\right )}{2\ 2^{3/4}} \]

[In]

Integrate[1/((1 + x^4)^(1/4)*(2 + x^4)),x]

[Out]

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

Maple [A] (verified)

Time = 4.67 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.15

method	result	size
pseudoelliptic	\(-\frac {2^{\frac {1}{4}} \left (2 \arctan \left (\frac {2^{\frac {1}{4}} \left (x^{4}+1\right )^{\frac {1}{4}}}{x}\right )-\ln \left (\frac {2^{\frac {3}{4}} x +2 \left (x^{4}+1\right )^{\frac {1}{4}}}{-2^{\frac {3}{4}} x +2 \left (x^{4}+1\right )^{\frac {1}{4}}}\right )\right )}{8}\)	\(61\)
trager	\(\frac {\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{3} x^{2}+2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{3}+3 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )}{x^{4}+2}\right )}{8}-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {2 \sqrt {x^{4}+1}\, \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{2}-2 \left (x^{4}+1\right )^{\frac {1}{4}} \operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2} x^{3}-3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right ) x^{4}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 \operatorname {RootOf}\left (\textit {\_Z}^{2}+\operatorname {RootOf}\left (\textit {\_Z}^{4}-2\right )^{2}\right )}{x^{4}+2}\right )}{8}\)	\(211\)

[In]

int(1/(x^4+1)^(1/4)/(x^4+2),x,method=_RETURNVERBOSE)

[Out]

-1/8*2^(1/4)*(2*arctan(1/x*2^(1/4)*(x^4+1)^(1/4))-ln((2^(3/4)*x+2*(x^4+1)^(1/4))/(-2^(3/4)*x+2*(x^4+1)^(1/4)))
)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 6.30 (sec) , antiderivative size = 270, normalized size of antiderivative = 5.09 \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (2+x^4\right )} \, dx=\frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) + \frac {1}{64} i \cdot 8^{\frac {3}{4}} \log \left (-\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 8 i \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (3 i \, x^{4} + 2 i\right )} - 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - \frac {1}{64} i \cdot 8^{\frac {3}{4}} \log \left (-\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 8 i \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (-3 i \, x^{4} - 2 i\right )} - 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) \]

[In]

integrate(1/(x^4+1)^(1/4)/(x^4+2),x, algorithm="fricas")

[Out]

1/64*8^(3/4)*log((8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 8*8^(1/4)*sqrt(x^4 + 1)*x^2 + 8^(3/4)*(3*x^4 + 2) + 16*(x^4
+ 1)^(3/4)*x)/(x^4 + 2)) + 1/64*I*8^(3/4)*log(-(8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 + 8*I*8^(1/4)*sqrt(x^4 + 1)*x^2
- 8^(3/4)*(3*I*x^4 + 2*I) - 16*(x^4 + 1)^(3/4)*x)/(x^4 + 2)) - 1/64*I*8^(3/4)*log(-(8*sqrt(2)*(x^4 + 1)^(1/4)*
x^3 - 8*I*8^(1/4)*sqrt(x^4 + 1)*x^2 - 8^(3/4)*(-3*I*x^4 - 2*I) - 16*(x^4 + 1)^(3/4)*x)/(x^4 + 2)) - 1/64*8^(3/
4)*log((8*sqrt(2)*(x^4 + 1)^(1/4)*x^3 - 8*8^(1/4)*sqrt(x^4 + 1)*x^2 - 8^(3/4)*(3*x^4 + 2) + 16*(x^4 + 1)^(3/4)
*x)/(x^4 + 2))

Sympy [F]

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

[In]

integrate(1/(x**4+1)**(1/4)/(x**4+2),x)

[Out]

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

Maxima [F]

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

[In]

integrate(1/(x^4+1)^(1/4)/(x^4+2),x, algorithm="maxima")

[Out]

integrate(1/((x^4 + 2)*(x^4 + 1)^(1/4)), x)

Giac [F]

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

[In]

integrate(1/(x^4+1)^(1/4)/(x^4+2),x, algorithm="giac")

[Out]

integrate(1/((x^4 + 2)*(x^4 + 1)^(1/4)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt [4]{1+x^4} \left (2+x^4\right )} \, dx=\int \frac {1}{{\left (x^4+1\right )}^{1/4}\,\left (x^4+2\right )} \,d x \]

[In]

int(1/((x^4 + 1)^(1/4)*(x^4 + 2)),x)

[Out]

int(1/((x^4 + 1)^(1/4)*(x^4 + 2)), x)