3.19.52 \(\int \frac {2+x^4}{\sqrt [4]{x^2+x^4} (-1-x^4+2 x^8)} \, dx\)

Optimal. Leaf size=127 \[ \frac {1}{4} \text {RootSum}\left [\text {$\#$1}^8-2 \text {$\#$1}^4+3\& ,\frac {\log \left (\sqrt [4]{x^4+x^2}-\text {$\#$1} x\right )-\log (x)}{\text {$\#$1}}\& \right ]-\frac {\left (x^4+x^2\right )^{3/4}}{x \left (x^2+1\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+x^2}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^4+x^2}}\right )}{2 \sqrt [4]{2}} \]

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Rubi [C]  time = 0.70, antiderivative size = 441, normalized size of antiderivative = 3.47, number of steps used = 19, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2056, 6715, 6728, 1404, 382, 377, 212, 206, 203, 1429} \begin {gather*} -\frac {x}{\sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {2}-2 i} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{x^2+1}}\right )}{2^{7/8} \sqrt [4]{\sqrt {2}-2 i} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \tan ^{-1}\left (\frac {\sqrt [4]{\sqrt {2}+2 i} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{x^2+1}}\right )}{2^{7/8} \sqrt [4]{\sqrt {2}+2 i} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{x^2+1}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {2}-2 i} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{x^2+1}}\right )}{2^{7/8} \sqrt [4]{\sqrt {2}-2 i} \sqrt [4]{x^4+x^2}}-\frac {\sqrt [4]{x^2+1} \sqrt {x} \tanh ^{-1}\left (\frac {\sqrt [4]{\sqrt {2}+2 i} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{x^2+1}}\right )}{2^{7/8} \sqrt [4]{\sqrt {2}+2 i} \sqrt [4]{x^4+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + x^4)/((x^2 + x^4)^(1/4)*(-1 - x^4 + 2*x^8)),x]

[Out]

-(x/(x^2 + x^4)^(1/4)) - (Sqrt[x]*(1 + x^2)^(1/4)*ArcTan[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)])/(2*2^(1/4)*(x^2 +
x^4)^(1/4)) - (Sqrt[x]*(1 + x^2)^(1/4)*ArcTan[((-2*I + Sqrt[2])^(1/4)*Sqrt[x])/(2^(1/8)*(1 + x^2)^(1/4))])/(2
^(7/8)*(-2*I + Sqrt[2])^(1/4)*(x^2 + x^4)^(1/4)) - (Sqrt[x]*(1 + x^2)^(1/4)*ArcTan[((2*I + Sqrt[2])^(1/4)*Sqrt
[x])/(2^(1/8)*(1 + x^2)^(1/4))])/(2^(7/8)*(2*I + Sqrt[2])^(1/4)*(x^2 + x^4)^(1/4)) - (Sqrt[x]*(1 + x^2)^(1/4)*
ArcTanh[(2^(1/4)*Sqrt[x])/(1 + x^2)^(1/4)])/(2*2^(1/4)*(x^2 + x^4)^(1/4)) - (Sqrt[x]*(1 + x^2)^(1/4)*ArcTanh[(
(-2*I + Sqrt[2])^(1/4)*Sqrt[x])/(2^(1/8)*(1 + x^2)^(1/4))])/(2^(7/8)*(-2*I + Sqrt[2])^(1/4)*(x^2 + x^4)^(1/4))
- (Sqrt[x]*(1 + x^2)^(1/4)*ArcTanh[((2*I + Sqrt[2])^(1/4)*Sqrt[x])/(2^(1/8)*(1 + x^2)^(1/4))])/(2^(7/8)*(2*I
+ Sqrt[2])^(1/4)*(x^2 + x^4)^(1/4))

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]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 1404

Int[((d_) + (e_.)*(x_)^(n_))^(q_.)*((a_) + (c_.)*(x_)^(n2_))^(p_.), x_Symbol] :> Int[(d + e*x^n)^(p + q)*(a/d
+ (c*x^n)/e)^p, x] /; FreeQ[{a, c, d, e, n, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && IntegerQ[p]

Rule 1429

Int[((d_) + (e_.)*(x_)^(n_))^(q_)/((a_) + (c_.)*(x_)^(n2_)), x_Symbol] :> With[{r = Rt[-(a*c), 2]}, -Dist[c/(2
*r), Int[(d + e*x^n)^q/(r - c*x^n), x], x] - Dist[c/(2*r), Int[(d + e*x^n)^q/(r + c*x^n), x], x]] /; FreeQ[{a,
c, d, e, n, q}, x] && EqQ[n2, 2*n] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[q]

Rule 2056

Int[(u_.)*(P_)^(p_.), x_Symbol] :> With[{m = MinimumMonomialExponent[P, x]}, Dist[P^FracPart[p]/(x^(m*FracPart
[p])*Distrib[1/x^m, P]^FracPart[p]), Int[u*x^(m*p)*Distrib[1/x^m, P]^p, x], x]] /; FreeQ[p, x] &&  !IntegerQ[p
] && SumQ[P] && EveryQ[BinomialQ[#1, x] & , P] &&  !PolyQ[P, x, 2]

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {2+x^4}{\sqrt [4]{x^2+x^4} \left (-1-x^4+2 x^8\right )} \, dx &=\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \int \frac {2+x^4}{\sqrt {x} \sqrt [4]{1+x^2} \left (-1-x^4+2 x^8\right )} \, dx}{\sqrt [4]{x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {2+x^8}{\sqrt [4]{1+x^4} \left (-1-x^8+2 x^{16}\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}\\ &=\frac {\left (2 \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \left (\frac {4}{\sqrt [4]{1+x^4} \left (-4+4 x^8\right )}-\frac {2}{\sqrt [4]{1+x^4} \left (2+4 x^8\right )}\right ) \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}\\ &=-\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4} \left (2+4 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}+\frac {\left (8 \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4} \left (-4+4 x^8\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}\\ &=\frac {\left (8 \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+x^4\right )^{5/4} \left (-4+4 x^4\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}-\frac {\left (2 i \sqrt {2} \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (2 i \sqrt {2}-4 x^4\right ) \sqrt [4]{1+x^4}} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}-\frac {\left (2 i \sqrt {2} \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4} \left (2 i \sqrt {2}+4 x^4\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}\\ &=-\frac {x}{\sqrt [4]{x^2+x^4}}+\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{1+x^4} \left (-4+4 x^4\right )} \, dx,x,\sqrt {x}\right )}{\sqrt [4]{x^2+x^4}}-\frac {\left (2 i \sqrt {2} \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{2 i \sqrt {2}-\left (-4+2 i \sqrt {2}\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}-\frac {\left (2 i \sqrt {2} \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{2 i \sqrt {2}-\left (4+2 i \sqrt {2}\right ) x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}\\ &=-\frac {x}{\sqrt [4]{x^2+x^4}}+\frac {\left (4 \sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-4+8 x^4} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{\sqrt [4]{x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2}-\sqrt {-2 i+\sqrt {2}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt [4]{x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2}+\sqrt {-2 i+\sqrt {2}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt [4]{x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2}-\sqrt {2 i+\sqrt {2}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt [4]{x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt [4]{2}+\sqrt {2 i+\sqrt {2}} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2^{3/4} \sqrt [4]{x^2+x^4}}\\ &=-\frac {x}{\sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{-2 i+\sqrt {2}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{1+x^2}}\right )}{2^{7/8} \sqrt [4]{-2 i+\sqrt {2}} \sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{2 i+\sqrt {2}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{1+x^2}}\right )}{2^{7/8} \sqrt [4]{2 i+\sqrt {2}} \sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{-2 i+\sqrt {2}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{1+x^2}}\right )}{2^{7/8} \sqrt [4]{-2 i+\sqrt {2}} \sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{2 i+\sqrt {2}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{1+x^2}}\right )}{2^{7/8} \sqrt [4]{2 i+\sqrt {2}} \sqrt [4]{x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2 \sqrt [4]{x^2+x^4}}-\frac {\left (\sqrt {x} \sqrt [4]{1+x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2 \sqrt [4]{x^2+x^4}}\\ &=-\frac {x}{\sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{-2 i+\sqrt {2}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{1+x^2}}\right )}{2^{7/8} \sqrt [4]{-2 i+\sqrt {2}} \sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tan ^{-1}\left (\frac {\sqrt [4]{2 i+\sqrt {2}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{1+x^2}}\right )}{2^{7/8} \sqrt [4]{2 i+\sqrt {2}} \sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{2} \sqrt {x}}{\sqrt [4]{1+x^2}}\right )}{2 \sqrt [4]{2} \sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{-2 i+\sqrt {2}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{1+x^2}}\right )}{2^{7/8} \sqrt [4]{-2 i+\sqrt {2}} \sqrt [4]{x^2+x^4}}-\frac {\sqrt {x} \sqrt [4]{1+x^2} \tanh ^{-1}\left (\frac {\sqrt [4]{2 i+\sqrt {2}} \sqrt {x}}{\sqrt [8]{2} \sqrt [4]{1+x^2}}\right )}{2^{7/8} \sqrt [4]{2 i+\sqrt {2}} \sqrt [4]{x^2+x^4}}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 130, normalized size = 1.02 \begin {gather*} \frac {x \left (\sqrt [4]{\frac {1}{x^2}+1} \left (2 \text {RootSum}\left [\text {$\#$1}^8-2 \text {$\#$1}^4+3\&,\frac {\log \left (\sqrt [4]{\frac {1}{x^2}+1}-\text {$\#$1}\right )}{\text {$\#$1}}\&\right ]+2^{3/4} \left (\log \left (2-2^{3/4} \sqrt [4]{\frac {1}{x^2}+1}\right )-\log \left (2^{3/4} \sqrt [4]{\frac {1}{x^2}+1}+2\right )+2 \tan ^{-1}\left (\frac {\sqrt [4]{\frac {1}{x^2}+1}}{\sqrt [4]{2}}\right )\right )\right )-8\right )}{8 \sqrt [4]{x^4+x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + x^4)/((x^2 + x^4)^(1/4)*(-1 - x^4 + 2*x^8)),x]

[Out]

(x*(-8 + (1 + x^(-2))^(1/4)*(2^(3/4)*(2*ArcTan[(1 + x^(-2))^(1/4)/2^(1/4)] + Log[2 - 2^(3/4)*(1 + x^(-2))^(1/4
)] - Log[2 + 2^(3/4)*(1 + x^(-2))^(1/4)]) + 2*RootSum[3 - 2*#1^4 + #1^8 & , Log[(1 + x^(-2))^(1/4) - #1]/#1 &
])))/(8*(x^2 + x^4)^(1/4))

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IntegrateAlgebraic [A]  time = 0.00, size = 127, normalized size = 1.00 \begin {gather*} -\frac {\left (x^2+x^4\right )^{3/4}}{x \left (1+x^2\right )}-\frac {\tan ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [4]{2} x}{\sqrt [4]{x^2+x^4}}\right )}{2 \sqrt [4]{2}}+\frac {1}{4} \text {RootSum}\left [3-2 \text {$\#$1}^4+\text {$\#$1}^8\&,\frac {-\log (x)+\log \left (\sqrt [4]{x^2+x^4}-x \text {$\#$1}\right )}{\text {$\#$1}}\&\right ] \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(2 + x^4)/((x^2 + x^4)^(1/4)*(-1 - x^4 + 2*x^8)),x]

[Out]

-((x^2 + x^4)^(3/4)/(x*(1 + x^2))) - ArcTan[(2^(1/4)*x)/(x^2 + x^4)^(1/4)]/(2*2^(1/4)) - ArcTanh[(2^(1/4)*x)/(
x^2 + x^4)^(1/4)]/(2*2^(1/4)) + RootSum[3 - 2*#1^4 + #1^8 & , (-Log[x] + Log[(x^2 + x^4)^(1/4) - x*#1])/#1 & ]
/4

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4} + 2}{{\left (2 \, x^{8} - x^{4} - 1\right )} {\left (x^{4} + x^{2}\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{4}+2}{\left (x^{4}+x^{2}\right )^{\frac {1}{4}} \left (2 x^{8}-x^{4}-1\right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^4+2)/(x^4+x^2)^(1/4)/(2*x^8-x^4-1),x)

[Out]

int((x^4+2)/(x^4+x^2)^(1/4)/(2*x^8-x^4-1),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {2 \, {\left (64 \, x^{9} + 16 \, x^{7} + 11 \, {\left (4 \, x^{5} + x^{3} - 3 \, x\right )} x^{4} - 28 \, x^{5} - 2 \, x^{3} - 22 \, x\right )}}{231 \, {\left (2 \, x^{\frac {17}{2}} - x^{\frac {9}{2}} - \sqrt {x}\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}} + \int \frac {4 \, {\left (256 \, x^{8} + 64 \, x^{6} + {\left (128 \, x^{8} + 32 \, x^{6} + 164 \, x^{4} + 51 \, x^{2} - 209\right )} x^{4} - 68 \, x^{4} + 3 \, x^{2} - 121\right )}}{231 \, {\left (4 \, x^{\frac {33}{2}} - 4 \, x^{\frac {25}{2}} - 3 \, x^{\frac {17}{2}} + 2 \, x^{\frac {9}{2}} + \sqrt {x}\right )} {\left (x^{2} + 1\right )}^{\frac {1}{4}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/231*(64*x^9 + 16*x^7 + 11*(4*x^5 + x^3 - 3*x)*x^4 - 28*x^5 - 2*x^3 - 22*x)/((2*x^(17/2) - x^(9/2) - sqrt(x))
*(x^2 + 1)^(1/4)) + integrate(4/231*(256*x^8 + 64*x^6 + (128*x^8 + 32*x^6 + 164*x^4 + 51*x^2 - 209)*x^4 - 68*x
^4 + 3*x^2 - 121)/((4*x^(33/2) - 4*x^(25/2) - 3*x^(17/2) + 2*x^(9/2) + sqrt(x))*(x^2 + 1)^(1/4)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int -\frac {x^4+2}{{\left (x^4+x^2\right )}^{1/4}\,\left (-2\,x^8+x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(x^4 + 2)/((x^2 + x^4)^(1/4)*(x^4 - 2*x^8 + 1)),x)

[Out]

int(-(x^4 + 2)/((x^2 + x^4)^(1/4)*(x^4 - 2*x^8 + 1)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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