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]

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))

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Rubi [A]  time = 0.018443, antiderivative size = 53, normalized size of antiderivative = 1., 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 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]

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 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 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])

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*}

Mathematica [A]  time = 0.0147199, 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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Maple [F]  time = 0.388, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}+2}{\frac{1}{\sqrt [4]{{x}^{4}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{4} + 2\right )}{\left (x^{4} + 1\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

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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Fricas [B]  time = 44.1397, size = 583, normalized size = 11. \begin{align*} -\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 ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt [4]{x^{4} + 1} \left (x^{4} + 2\right )}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (x^{4} + 2\right )}{\left (x^{4} + 1\right )}^{\frac{1}{4}}}\,{d x} \end{align*}

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)