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

Optimal. Leaf size=53 \[ \frac {\tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}} \]

[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)

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {377, 212, 206, 203} \[ \frac {\tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}}+\frac {\tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}} \]

Antiderivative was successfully verified.

[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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 212

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] &&
!GtQ[a/b, 0]

Rule 377

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

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Mathematica [A]  time = 0.02, size = 44, normalized size = 0.83 \[ \frac {\tan ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )+\tanh ^{-1}\left (\frac {x}{\sqrt [4]{2} \sqrt [4]{x^4+1}}\right )}{2\ 2^{3/4}} \]

Antiderivative was successfully verified.

[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))

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fricas [B]  time = 16.99, size = 208, normalized size = 3.92 \[ -\frac {1}{16} \cdot 8^{\frac {3}{4}} \arctan \left (-\frac {8^{\frac {3}{4}} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 4 \cdot 8^{\frac {1}{4}} {\left (x^{4} + 1\right )}^{\frac {3}{4}} x - 2^{\frac {1}{4}} {\left (8^{\frac {3}{4}} \sqrt {x^{4} + 1} x^{2} + 8^{\frac {1}{4}} {\left (3 \, x^{4} + 2\right )}\right )}}{2 \, {\left (x^{4} + 2\right )}}\right ) + \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} + 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} + 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) - \frac {1}{64} \cdot 8^{\frac {3}{4}} \log \left (\frac {8 \, \sqrt {2} {\left (x^{4} + 1\right )}^{\frac {1}{4}} x^{3} - 8 \cdot 8^{\frac {1}{4}} \sqrt {x^{4} + 1} x^{2} - 8^{\frac {3}{4}} {\left (3 \, x^{4} + 2\right )} + 16 \, {\left (x^{4} + 1\right )}^{\frac {3}{4}} x}{x^{4} + 2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} + 2\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [C]  time = 3.24, size = 211, normalized size = 3.98 \[ -\frac {\RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \ln \left (-\frac {-3 x^{4} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )-2 \left (x^{4}+1\right )^{\frac {1}{4}} x^{3} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}+2 \sqrt {x^{4}+1}\, x^{2} \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right ) \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x -2 \RootOf \left (\textit {\_Z}^{2}+\RootOf \left (\textit {\_Z}^{4}-2\right )^{2}\right )}{x^{4}+2}\right )}{8}+\frac {\RootOf \left (\textit {\_Z}^{4}-2\right ) \ln \left (\frac {3 x^{4} \RootOf \left (\textit {\_Z}^{4}-2\right )+2 \left (x^{4}+1\right )^{\frac {1}{4}} x^{3} \RootOf \left (\textit {\_Z}^{4}-2\right )^{2}+2 \sqrt {x^{4}+1}\, x^{2} \RootOf \left (\textit {\_Z}^{4}-2\right )^{3}+4 \left (x^{4}+1\right )^{\frac {3}{4}} x +2 \RootOf \left (\textit {\_Z}^{4}-2\right )}{x^{4}+2}\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4+1)^(1/4)/(x^4+2),x)

[Out]

-1/8*RootOf(_Z^2+RootOf(_Z^4-2)^2)*ln(-(2*(x^4+1)^(1/2)*RootOf(_Z^4-2)^2*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^2-2*(
x^4+1)^(1/4)*RootOf(_Z^4-2)^2*x^3-3*RootOf(_Z^2+RootOf(_Z^4-2)^2)*x^4+4*(x^4+1)^(3/4)*x-2*RootOf(_Z^2+RootOf(_
Z^4-2)^2))/(x^4+2))+1/8*RootOf(_Z^4-2)*ln((2*(x^4+1)^(1/2)*RootOf(_Z^4-2)^3*x^2+2*(x^4+1)^(1/4)*RootOf(_Z^4-2)
^2*x^3+3*RootOf(_Z^4-2)*x^4+4*(x^4+1)^(3/4)*x+2*RootOf(_Z^4-2))/(x^4+2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (x^{4} + 2\right )} {\left (x^{4} + 1\right )}^{\frac {1}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{{\left (x^4+1\right )}^{1/4}\,\left (x^4+2\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^4 + 1)^(1/4)*(x^4 + 2)),x)

[Out]

int(1/((x^4 + 1)^(1/4)*(x^4 + 2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt [4]{x^{4} + 1} \left (x^{4} + 2\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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