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

Optimal. Leaf size=141 \[ -\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{2 \sqrt {2}}-\frac {\log \left (-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+\frac {x^2}{\sqrt {x^4+2}}+1\right )}{4 \sqrt {2}}+\frac {\log \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+\frac {x^2}{\sqrt {x^4+2}}+1\right )}{4 \sqrt {2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 141, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {377, 211, 1165, 628, 1162, 617, 204} \[ -\frac {\log \left (\frac {x^2}{\sqrt {x^4+2}}-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{4 \sqrt {2}}+\frac {\log \left (\frac {x^2}{\sqrt {x^4+2}}+\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{4 \sqrt {2}}-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )}{2 \sqrt {2}}+\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((1 + x^4)*(2 + x^4)^(1/4)),x]

[Out]

-ArcTan[1 - (Sqrt[2]*x)/(2 + x^4)^(1/4)]/(2*Sqrt[2]) + ArcTan[1 + (Sqrt[2]*x)/(2 + x^4)^(1/4)]/(2*Sqrt[2]) - L
og[1 + x^2/Sqrt[2 + x^4] - (Sqrt[2]*x)/(2 + x^4)^(1/4)]/(4*Sqrt[2]) + Log[1 + x^2/Sqrt[2 + x^4] + (Sqrt[2]*x)/
(2 + x^4)^(1/4)]/(4*Sqrt[2])

Rule 204

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

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 120, normalized size = 0.85 \[ \frac {-2 \tan ^{-1}\left (1-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}\right )+2 \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+1\right )-\log \left (-\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+\frac {x^2}{\sqrt {x^4+2}}+1\right )+\log \left (\frac {\sqrt {2} x}{\sqrt [4]{x^4+2}}+\frac {x^2}{\sqrt {x^4+2}}+1\right )}{4 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.21, size = 85, normalized size = 0.60 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{x^4+2}}{\sqrt {x^4+2}-x^2}\right )}{2 \sqrt {2}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {2} x \sqrt [4]{x^4+2}}{\sqrt {x^4+2}+x^2}\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 6.77, size = 388, normalized size = 2.75 \[ \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {5}{4}} - {\left (2 \, x^{5} - \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {5}{4}} + 4 \, x\right )} \sqrt {\frac {x^{4} + \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {x^{4} + 2} x^{2} + \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x + 1}{x^{4} + 1}}}{2 \, {\left (x^{5} + 2 \, x\right )}}\right ) + \frac {1}{4} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x^{2} - \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {5}{4}} + {\left (2 \, x^{5} + \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x^{2} + \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {5}{4}} + 4 \, x\right )} \sqrt {\frac {x^{4} - \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {x^{4} + 2} x^{2} - \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x + 1}{x^{4} + 1}}}{2 \, {\left (x^{5} + 2 \, x\right )}}\right ) + \frac {1}{16} \, \sqrt {2} \log \left (\frac {4 \, {\left (x^{4} + \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {x^{4} + 2} x^{2} + \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x + 1\right )}}{x^{4} + 1}\right ) - \frac {1}{16} \, \sqrt {2} \log \left (\frac {4 \, {\left (x^{4} - \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {1}{4}} x^{3} + 2 \, \sqrt {x^{4} + 2} x^{2} - \sqrt {2} {\left (x^{4} + 2\right )}^{\frac {3}{4}} x + 1\right )}}{x^{4} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(2)*arctan(1/2*(sqrt(2)*(x^4 + 2)^(3/4)*x^2 - sqrt(2)*(x^4 + 2)^(5/4) - (2*x^5 - sqrt(2)*(x^4 + 2)^(3/
4)*x^2 - sqrt(2)*(x^4 + 2)^(5/4) + 4*x)*sqrt((x^4 + sqrt(2)*(x^4 + 2)^(1/4)*x^3 + 2*sqrt(x^4 + 2)*x^2 + sqrt(2
)*(x^4 + 2)^(3/4)*x + 1)/(x^4 + 1)))/(x^5 + 2*x)) + 1/4*sqrt(2)*arctan(1/2*(sqrt(2)*(x^4 + 2)^(3/4)*x^2 - sqrt
(2)*(x^4 + 2)^(5/4) + (2*x^5 + sqrt(2)*(x^4 + 2)^(3/4)*x^2 + sqrt(2)*(x^4 + 2)^(5/4) + 4*x)*sqrt((x^4 - sqrt(2
)*(x^4 + 2)^(1/4)*x^3 + 2*sqrt(x^4 + 2)*x^2 - sqrt(2)*(x^4 + 2)^(3/4)*x + 1)/(x^4 + 1)))/(x^5 + 2*x)) + 1/16*s
qrt(2)*log(4*(x^4 + sqrt(2)*(x^4 + 2)^(1/4)*x^3 + 2*sqrt(x^4 + 2)*x^2 + sqrt(2)*(x^4 + 2)^(3/4)*x + 1)/(x^4 +
1)) - 1/16*sqrt(2)*log(4*(x^4 - sqrt(2)*(x^4 + 2)^(1/4)*x^3 + 2*sqrt(x^4 + 2)*x^2 - sqrt(2)*(x^4 + 2)^(3/4)*x
+ 1)/(x^4 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.34, size = 150, normalized size = 1.06

method	result	size
trager	\(-\frac {\RootOf \left (\textit {\_Z}^{4}+1\right ) \ln \left (\frac {\sqrt {x^{4}+2}\, \RootOf \left (\textit {\_Z}^{4}+1\right )^{3} x^{2}-\left (x^{4}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{3}+\left (x^{4}+2\right )^{\frac {3}{4}} x -\RootOf \left (\textit {\_Z}^{4}+1\right )}{x^{4}+1}\right )}{4}+\frac {\RootOf \left (\textit {\_Z}^{4}+1\right )^{3} \ln \left (\frac {\left (x^{4}+2\right )^{\frac {1}{4}} \RootOf \left (\textit {\_Z}^{4}+1\right )^{2} x^{3}-\sqrt {x^{4}+2}\, \RootOf \left (\textit {\_Z}^{4}+1\right ) x^{2}+\left (x^{4}+2\right )^{\frac {3}{4}} x +\RootOf \left (\textit {\_Z}^{4}+1\right )^{3}}{x^{4}+1}\right )}{4}\)	\(150\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\left (x^4+1\right )\,{\left (x^4+2\right )}^{1/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((x^4 + 1)*(x^4 + 2)^(1/4)),x)

[Out]

int(1/((x^4 + 1)*(x^4 + 2)^(1/4)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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