3.3.84 \(\int \frac {x^2}{1+2 x^4+x^8} \, dx\) [284]

Optimal. Leaf size=99 \[ \frac {x^3}{4 \left (1+x^4\right )}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}} \]

[Out]

1/4*x^3/(x^4+1)+1/16*arctan(-1+x*2^(1/2))*2^(1/2)+1/16*arctan(1+x*2^(1/2))*2^(1/2)+1/32*ln(1+x^2-x*2^(1/2))*2^
(1/2)-1/32*ln(1+x^2+x*2^(1/2))*2^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {28, 296, 303, 1176, 631, 210, 1179, 642} \begin {gather*} -\frac {\text {ArcTan}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\text {ArcTan}\left (\sqrt {2} x+1\right )}{8 \sqrt {2}}+\frac {\log \left (x^2-\sqrt {2} x+1\right )}{16 \sqrt {2}}-\frac {\log \left (x^2+\sqrt {2} x+1\right )}{16 \sqrt {2}}+\frac {x^3}{4 \left (x^4+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(1 + 2*x^4 + x^8),x]

[Out]

x^3/(4*(1 + x^4)) - ArcTan[1 - Sqrt[2]*x]/(8*Sqrt[2]) + ArcTan[1 + Sqrt[2]*x]/(8*Sqrt[2]) + Log[1 - Sqrt[2]*x
+ x^2]/(16*Sqrt[2]) - Log[1 + Sqrt[2]*x + x^2]/(16*Sqrt[2])

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 210

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

Rule 296

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-(c*x)^(m + 1))*((a + b*x^n)^(p + 1)/
(a*c*n*(p + 1))), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; Free
Q[{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 303

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

Rule 631

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 642

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 1176

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 1179

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 {x^2}{1+2 x^4+x^8} \, dx &=\int \frac {x^2}{\left (1+x^4\right )^2} \, dx\\ &=\frac {x^3}{4 \left (1+x^4\right )}+\frac {1}{4} \int \frac {x^2}{1+x^4} \, dx\\ &=\frac {x^3}{4 \left (1+x^4\right )}-\frac {1}{8} \int \frac {1-x^2}{1+x^4} \, dx+\frac {1}{8} \int \frac {1+x^2}{1+x^4} \, dx\\ &=\frac {x^3}{4 \left (1+x^4\right )}+\frac {1}{16} \int \frac {1}{1-\sqrt {2} x+x^2} \, dx+\frac {1}{16} \int \frac {1}{1+\sqrt {2} x+x^2} \, dx+\frac {\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}+\frac {\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx}{16 \sqrt {2}}\\ &=\frac {x^3}{4 \left (1+x^4\right )}+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} x\right )}{8 \sqrt {2}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} x\right )}{8 \sqrt {2}}\\ &=\frac {x^3}{4 \left (1+x^4\right )}-\frac {\tan ^{-1}\left (1-\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\tan ^{-1}\left (1+\sqrt {2} x\right )}{8 \sqrt {2}}+\frac {\log \left (1-\sqrt {2} x+x^2\right )}{16 \sqrt {2}}-\frac {\log \left (1+\sqrt {2} x+x^2\right )}{16 \sqrt {2}}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 92, normalized size = 0.93 \begin {gather*} \frac {1}{32} \left (\frac {8 x^3}{1+x^4}-2 \sqrt {2} \tan ^{-1}\left (1-\sqrt {2} x\right )+2 \sqrt {2} \tan ^{-1}\left (1+\sqrt {2} x\right )+\sqrt {2} \log \left (1-\sqrt {2} x+x^2\right )-\sqrt {2} \log \left (1+\sqrt {2} x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(1 + 2*x^4 + x^8),x]

[Out]

((8*x^3)/(1 + x^4) - 2*Sqrt[2]*ArcTan[1 - Sqrt[2]*x] + 2*Sqrt[2]*ArcTan[1 + Sqrt[2]*x] + Sqrt[2]*Log[1 - Sqrt[
2]*x + x^2] - Sqrt[2]*Log[1 + Sqrt[2]*x + x^2])/32

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Maple [A]
time = 0.01, size = 65, normalized size = 0.66

method	result	size
risch	\(\frac {x^{3}}{4 x^{4}+4}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{4}+1\right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{16}\)	\(35\)
default	\(\frac {x^{3}}{4 x^{4}+4}+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+x^{2}-\sqrt {2}\, x}{1+x^{2}+\sqrt {2}\, x}\right )+2 \arctan \left (\sqrt {2}\, x +1\right )+2 \arctan \left (\sqrt {2}\, x -1\right )\right )}{32}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(x^8+2*x^4+1),x,method=_RETURNVERBOSE)

[Out]

1/4*x^3/(x^4+1)+1/32*2^(1/2)*(ln((1+x^2-2^(1/2)*x)/(1+x^2+2^(1/2)*x))+2*arctan(2^(1/2)*x+1)+2*arctan(2^(1/2)*x
-1))

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Maxima [A]
time = 0.51, size = 84, normalized size = 0.85 \begin {gather*} \frac {x^{3}}{4 \, {\left (x^{4} + 1\right )}} + \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^8+2*x^4+1),x, algorithm="maxima")

[Out]

1/4*x^3/(x^4 + 1) + 1/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x -
sqrt(2))) - 1/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 1/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1)

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Fricas [A]
time = 0.39, size = 133, normalized size = 1.34 \begin {gather*} \frac {8 \, x^{3} - 4 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + \sqrt {2} x + 1} - 1\right ) - 4 \, \sqrt {2} {\left (x^{4} + 1\right )} \arctan \left (-\sqrt {2} x + \sqrt {2} \sqrt {x^{2} - \sqrt {2} x + 1} + 1\right ) - \sqrt {2} {\left (x^{4} + 1\right )} \log \left (4 \, x^{2} + 4 \, \sqrt {2} x + 4\right ) + \sqrt {2} {\left (x^{4} + 1\right )} \log \left (4 \, x^{2} - 4 \, \sqrt {2} x + 4\right )}{32 \, {\left (x^{4} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^8+2*x^4+1),x, algorithm="fricas")

[Out]

1/32*(8*x^3 - 4*sqrt(2)*(x^4 + 1)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 + sqrt(2)*x + 1) - 1) - 4*sqrt(2)*(x^4
+ 1)*arctan(-sqrt(2)*x + sqrt(2)*sqrt(x^2 - sqrt(2)*x + 1) + 1) - sqrt(2)*(x^4 + 1)*log(4*x^2 + 4*sqrt(2)*x +
4) + sqrt(2)*(x^4 + 1)*log(4*x^2 - 4*sqrt(2)*x + 4))/(x^4 + 1)

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Sympy [A]
time = 0.06, size = 83, normalized size = 0.84 \begin {gather*} \frac {x^{3}}{4 x^{4} + 4} + \frac {\sqrt {2} \log {\left (x^{2} - \sqrt {2} x + 1 \right )}}{32} - \frac {\sqrt {2} \log {\left (x^{2} + \sqrt {2} x + 1 \right )}}{32} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x - 1 \right )}}{16} + \frac {\sqrt {2} \operatorname {atan}{\left (\sqrt {2} x + 1 \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(x**8+2*x**4+1),x)

[Out]

x**3/(4*x**4 + 4) + sqrt(2)*log(x**2 - sqrt(2)*x + 1)/32 - sqrt(2)*log(x**2 + sqrt(2)*x + 1)/32 + sqrt(2)*atan
(sqrt(2)*x - 1)/16 + sqrt(2)*atan(sqrt(2)*x + 1)/16

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Giac [A]
time = 4.34, size = 84, normalized size = 0.85 \begin {gather*} \frac {x^{3}}{4 \, {\left (x^{4} + 1\right )}} + \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x + \sqrt {2}\right )}\right ) + \frac {1}{16} \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (2 \, x - \sqrt {2}\right )}\right ) - \frac {1}{32} \, \sqrt {2} \log \left (x^{2} + \sqrt {2} x + 1\right ) + \frac {1}{32} \, \sqrt {2} \log \left (x^{2} - \sqrt {2} x + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(x^8+2*x^4+1),x, algorithm="giac")

[Out]

1/4*x^3/(x^4 + 1) + 1/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x + sqrt(2))) + 1/16*sqrt(2)*arctan(1/2*sqrt(2)*(2*x -
sqrt(2))) - 1/32*sqrt(2)*log(x^2 + sqrt(2)*x + 1) + 1/32*sqrt(2)*log(x^2 - sqrt(2)*x + 1)

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Mupad [B]
time = 0.05, size = 46, normalized size = 0.46 \begin {gather*} \frac {x^3}{4\,\left (x^4+1\right )}+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}-\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{16}-\frac {1}{16}{}\mathrm {i}\right )+\sqrt {2}\,\mathrm {atan}\left (\sqrt {2}\,x\,\left (\frac {1}{2}+\frac {1}{2}{}\mathrm {i}\right )\right )\,\left (\frac {1}{16}+\frac {1}{16}{}\mathrm {i}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(2*x^4 + x^8 + 1),x)

[Out]

2^(1/2)*atan(2^(1/2)*x*(1/2 - 1i/2))*(1/16 - 1i/16) + 2^(1/2)*atan(2^(1/2)*x*(1/2 + 1i/2))*(1/16 + 1i/16) + x^
3/(4*(x^4 + 1))

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