\(\int \frac {x^2}{2+(1+x^2)^4} \, dx\) [389]

Optimal result

Integrand size = 15, antiderivative size = 188 \[ \int \frac {x^2}{2+\left (1+x^2\right )^4} \, dx=\frac {\sqrt [4]{-1} \sqrt {1-\sqrt [4]{-2}} \arctan \left (\frac {x}{\sqrt {1-\sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}-\frac {(-1)^{3/4} \sqrt {1+i \sqrt [4]{-2}} \arctan \left (\frac {x}{\sqrt {1+i \sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}-\frac {\sqrt [4]{-1} \sqrt {1+\sqrt [4]{-2}} \arctan \left (\frac {x}{\sqrt {1+\sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}+\frac {1}{8} i \left (\sqrt [4]{-2}+\sqrt {2}\right ) \sqrt {\frac {1+i}{(1+i)+2^{3/4}}} \arctan \left (\sqrt {\frac {1+i}{(1+i)+2^{3/4}}} x\right ) \]

[Out]

1/8*(-1)^(1/4)*arctan(x/(1-(-2)^(1/4))^(1/2))*(1-(-2)^(1/4))^(1/2)*2^(1/4)-1/8*(-1)^(3/4)*2^(1/4)*arctan(x/(1+
I*(-2)^(1/4))^(1/2))*(1+I*(-2)^(1/4))^(1/2)-1/8*(-1)^(1/4)*arctan(x/(1+(-2)^(1/4))^(1/2))*(1+(-2)^(1/4))^(1/2)
*2^(1/4)+1/8*I*arctan(x*((1+I)/(1+I+2^(3/4)))^(1/2))*((-2)^(1/4)+2^(1/2))*((1+I)/(1+I+2^(3/4)))^(1/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6872, 210, 209, 1997, 211} \[ \int \frac {x^2}{2+\left (1+x^2\right )^4} \, dx=\frac {\sqrt [4]{-1} \sqrt {1-\sqrt [4]{-2}} \arctan \left (\frac {x}{\sqrt {1-\sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}-\frac {(-1)^{3/4} \sqrt {1+i \sqrt [4]{-2}} \arctan \left (\frac {x}{\sqrt {1+i \sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}-\frac {\sqrt [4]{-1} \sqrt {1+\sqrt [4]{-2}} \arctan \left (\frac {x}{\sqrt {1+\sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}+\frac {1}{8} i \left (\sqrt [4]{-2}+\sqrt {2}\right ) \sqrt {\frac {1+i}{2^{3/4}+(1+i)}} \arctan \left (\sqrt {\frac {1+i}{2^{3/4}+(1+i)}} x\right ) \]

[In]

Int[x^2/(2 + (1 + x^2)^4),x]

[Out]

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

Rule 209

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

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 211

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

Rule 1997

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && BinomialQ[u, x] &&  !BinomialMatchQ[
u, x]

Rule 6872

Int[(v_)/((a_) + (b_.)*(u_)^(n_.)), x_Symbol] :> Int[ExpandIntegrand[PolynomialInSubst[v, u, x]/(a + b*x^n), x
] /. x -> u, x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && PolynomialInQ[v, u, x]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {-\sqrt [4]{-2}+i \sqrt {2}}{8 \left (-1+\sqrt [4]{-2}-x^2\right )}+\frac {-\sqrt [4]{-2}-i \sqrt {2}}{8 \left (1+\sqrt [4]{-2}+x^2\right )}+\frac {-\sqrt [4]{-2}+\sqrt {2}}{8 \left (\sqrt [4]{-2}-i \left (1+x^2\right )\right )}+\frac {-\sqrt [4]{-2}-\sqrt {2}}{8 \left (\sqrt [4]{-2}+i \left (1+x^2\right )\right )}\right ) \, dx \\ & = \frac {1}{8} \left (-\sqrt [4]{-2}-\sqrt {2}\right ) \int \frac {1}{\sqrt [4]{-2}+i \left (1+x^2\right )} \, dx+\frac {1}{8} \left (-\sqrt [4]{-2}-i \sqrt {2}\right ) \int \frac {1}{1+\sqrt [4]{-2}+x^2} \, dx+\frac {1}{8} \left (-\sqrt [4]{-2}+i \sqrt {2}\right ) \int \frac {1}{-1+\sqrt [4]{-2}-x^2} \, dx+\frac {1}{8} \left (-\sqrt [4]{-2}+\sqrt {2}\right ) \int \frac {1}{\sqrt [4]{-2}-i \left (1+x^2\right )} \, dx \\ & = \frac {\sqrt [4]{-1} \sqrt {1-\sqrt [4]{-2}} \tan ^{-1}\left (\frac {x}{\sqrt {1-\sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}-\frac {\sqrt [4]{-1} \sqrt {1+\sqrt [4]{-2}} \tan ^{-1}\left (\frac {x}{\sqrt {1+\sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}+\frac {1}{8} \left (-\sqrt [4]{-2}-\sqrt {2}\right ) \int \frac {1}{i+\sqrt [4]{-2}+i x^2} \, dx+\frac {1}{8} \left (-\sqrt [4]{-2}+\sqrt {2}\right ) \int \frac {1}{-i+\sqrt [4]{-2}-i x^2} \, dx \\ & = \frac {\sqrt [4]{-1} \sqrt {1-\sqrt [4]{-2}} \tan ^{-1}\left (\frac {x}{\sqrt {1-\sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}-\frac {(-1)^{3/4} \sqrt {1+i \sqrt [4]{-2}} \tan ^{-1}\left (\frac {x}{\sqrt {1+i \sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}-\frac {\sqrt [4]{-1} \sqrt {1+\sqrt [4]{-2}} \tan ^{-1}\left (\frac {x}{\sqrt {1+\sqrt [4]{-2}}}\right )}{4\ 2^{3/4}}+\frac {1}{8} i \left (\sqrt [4]{-2}+\sqrt {2}\right ) \sqrt {\frac {1+i}{(1+i)+2^{3/4}}} \tan ^{-1}\left (\sqrt {\frac {1+i}{(1+i)+2^{3/4}}} x\right ) \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.

Time = 0.01 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.32 \[ \int \frac {x^2}{2+\left (1+x^2\right )^4} \, dx=\frac {1}{8} \text {RootSum}\left [3+4 \text {$\#$1}^2+6 \text {$\#$1}^4+4 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {\log (x-\text {$\#$1}) \text {$\#$1}}{1+3 \text {$\#$1}^2+3 \text {$\#$1}^4+\text {$\#$1}^6}\&\right ] \]

[In]

Integrate[x^2/(2 + (1 + x^2)^4),x]

[Out]

RootSum[3 + 4*#1^2 + 6*#1^4 + 4*#1^6 + #1^8 & , (Log[x - #1]*#1)/(1 + 3*#1^2 + 3*#1^4 + #1^6) & ]/8

Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 0.06 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.29

method	result	size
default	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+4 \textit {\_Z}^{6}+6 \textit {\_Z}^{4}+4 \textit {\_Z}^{2}+3\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{7}+3 \textit {\_R}^{5}+3 \textit {\_R}^{3}+\textit {\_R}}\right )}{8}\)	\(54\)
risch	\(\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{8}+4 \textit {\_Z}^{6}+6 \textit {\_Z}^{4}+4 \textit {\_Z}^{2}+3\right )}{\sum }\frac {\textit {\_R}^{2} \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{7}+3 \textit {\_R}^{5}+3 \textit {\_R}^{3}+\textit {\_R}}\right )}{8}\)	\(54\)

[In]

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

[Out]

1/8*sum(_R^2/(_R^7+3*_R^5+3*_R^3+_R)*ln(x-_R),_R=RootOf(_Z^8+4*_Z^6+6*_Z^4+4*_Z^2+3))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2271 vs. \(2 (118) = 236\).

Time = 0.98 (sec) , antiderivative size = 2271, normalized size of antiderivative = 12.08 \[ \int \frac {x^2}{2+\left (1+x^2\right )^4} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/16*sqrt(2)*sqrt(sqrt(-12288*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 12288*(-1/256*I*sqrt(2) - 1/
2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(2) + 128*sqrt(1/8192*I*s
qrt(2))) - 1) + 32*sqrt(1/8192*I*sqrt(2)) + 32*sqrt(-1/8192*I*sqrt(2)))*log((16384*sqrt(2)*(-1/256*I*sqrt(2) -
1/2*sqrt(-1/8192*I*sqrt(2)))^2*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) + 16384*(sqrt(2)*(I*sqrt(2) + 128*sq
rt(-1/8192*I*sqrt(2))) - sqrt(2))*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 16384*sqrt(2)*(-1/256*I*s
qrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - sqrt(-12288*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 12288
*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(2
) + 128*sqrt(1/8192*I*sqrt(2))) - 1)*((sqrt(2)*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2))) - sqrt(2))*(-I*sqrt(2
) + 128*sqrt(1/8192*I*sqrt(2))) - sqrt(2)*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))) + sqrt(2))*sqrt(sqrt(-122
88*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 12288*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2
- 1/8*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) - 1) + 32*sqrt(1/81
92*I*sqrt(2)) + 32*sqrt(-1/8192*I*sqrt(2))) + 2*x) + 1/16*sqrt(2)*sqrt(sqrt(-12288*(1/256*I*sqrt(2) - 1/2*sqrt
(1/8192*I*sqrt(2)))^2 - 12288*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8*(I*sqrt(2) + 128*sqrt(-
1/8192*I*sqrt(2)))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) - 1) + 32*sqrt(1/8192*I*sqrt(2)) + 32*sqrt(-1/819
2*I*sqrt(2)))*log(-(16384*sqrt(2)*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2*(-I*sqrt(2) + 128*sqrt(1/
8192*I*sqrt(2))) + 16384*(sqrt(2)*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2))) - sqrt(2))*(1/256*I*sqrt(2) - 1/2*
sqrt(1/8192*I*sqrt(2)))^2 - 16384*sqrt(2)*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - sqrt(-12288*(1/
256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 12288*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8
*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) - 1)*((sqrt(2)*(I*sqrt(2)
+ 128*sqrt(-1/8192*I*sqrt(2))) - sqrt(2))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) - sqrt(2)*(I*sqrt(2) + 12
8*sqrt(-1/8192*I*sqrt(2)))) + sqrt(2))*sqrt(sqrt(-12288*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 122
88*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt
(2) + 128*sqrt(1/8192*I*sqrt(2))) - 1) + 32*sqrt(1/8192*I*sqrt(2)) + 32*sqrt(-1/8192*I*sqrt(2))) + 2*x) - 1/16
*sqrt(2)*sqrt(-sqrt(-12288*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 12288*(-1/256*I*sqrt(2) - 1/2*sq
rt(-1/8192*I*sqrt(2)))^2 - 1/8*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(
2))) - 1) + 32*sqrt(1/8192*I*sqrt(2)) + 32*sqrt(-1/8192*I*sqrt(2)))*log((16384*sqrt(2)*(-1/256*I*sqrt(2) - 1/2
*sqrt(-1/8192*I*sqrt(2)))^2*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) + 16384*(sqrt(2)*(I*sqrt(2) + 128*sqrt(-
1/8192*I*sqrt(2))) - sqrt(2))*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 16384*sqrt(2)*(-1/256*I*sqrt(
2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 + sqrt(-12288*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 12288*(-1
/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(2) +
128*sqrt(1/8192*I*sqrt(2))) - 1)*((sqrt(2)*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2))) - sqrt(2))*(-I*sqrt(2) +
128*sqrt(1/8192*I*sqrt(2))) - sqrt(2)*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))) + sqrt(2))*sqrt(-sqrt(-12288*
(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 12288*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 -
1/8*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) - 1) + 32*sqrt(1/8192*
I*sqrt(2)) + 32*sqrt(-1/8192*I*sqrt(2))) + 2*x) + 1/16*sqrt(2)*sqrt(-sqrt(-12288*(1/256*I*sqrt(2) - 1/2*sqrt(1
/8192*I*sqrt(2)))^2 - 12288*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8*(I*sqrt(2) + 128*sqrt(-1/
8192*I*sqrt(2)))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) - 1) + 32*sqrt(1/8192*I*sqrt(2)) + 32*sqrt(-1/8192*
I*sqrt(2)))*log(-(16384*sqrt(2)*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2*(-I*sqrt(2) + 128*sqrt(1/81
92*I*sqrt(2))) + 16384*(sqrt(2)*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2))) - sqrt(2))*(1/256*I*sqrt(2) - 1/2*sq
rt(1/8192*I*sqrt(2)))^2 - 16384*sqrt(2)*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 + sqrt(-12288*(1/25
6*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 12288*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8*(
I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) - 1)*((sqrt(2)*(I*sqrt(2) +
128*sqrt(-1/8192*I*sqrt(2))) - sqrt(2))*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) - sqrt(2)*(I*sqrt(2) + 128*
sqrt(-1/8192*I*sqrt(2)))) + sqrt(2))*sqrt(-sqrt(-12288*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2 - 1228
8*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 1/8*(I*sqrt(2) + 128*sqrt(-1/8192*I*sqrt(2)))*(-I*sqrt(
2) + 128*sqrt(1/8192*I*sqrt(2))) - 1) + 32*sqrt(1/8192*I*sqrt(2)) + 32*sqrt(-1/8192*I*sqrt(2))) + 2*x) - sqrt(
1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))*log(8*(8388608*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^
3 - 32768*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) + 32768
*(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2*(-I*sqrt(2) - 128*sqrt(-1/8192*I*sqrt(2)) + 1) - 2*I*sqrt(2)
- 256*sqrt(-1/8192*I*sqrt(2)) + 3)*sqrt(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2))) + x) + sqrt(1/256*I*sqr
t(2) - 1/2*sqrt(1/8192*I*sqrt(2)))*log(-8*(8388608*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^3 - 32768*
(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2*(-I*sqrt(2) + 128*sqrt(1/8192*I*sqrt(2))) + 32768*(1/256*I*
sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2)))^2*(-I*sqrt(2) - 128*sqrt(-1/8192*I*sqrt(2)) + 1) - 2*I*sqrt(2) - 256*sqr
t(-1/8192*I*sqrt(2)) + 3)*sqrt(1/256*I*sqrt(2) - 1/2*sqrt(1/8192*I*sqrt(2))) + x) + sqrt(-1/256*I*sqrt(2) - 1/
2*sqrt(-1/8192*I*sqrt(2)))*log(8*(8388608*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^3 - 32768*(-1/256*I
*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2 - 2*I*sqrt(2) - 256*sqrt(-1/8192*I*sqrt(2)) + 5)*sqrt(-1/256*I*sqrt(
2) - 1/2*sqrt(-1/8192*I*sqrt(2))) + x) - sqrt(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))*log(-8*(8388608*
(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^3 - 32768*(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2)))^2
- 2*I*sqrt(2) - 256*sqrt(-1/8192*I*sqrt(2)) + 5)*sqrt(-1/256*I*sqrt(2) - 1/2*sqrt(-1/8192*I*sqrt(2))) + x)

Sympy [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.21 \[ \int \frac {x^2}{2+\left (1+x^2\right )^4} \, dx=\operatorname {RootSum} {\left (1073741824 t^{8} + 65536 t^{4} + 1024 t^{2} + 3, \left ( t \mapsto t \log {\left (67108864 t^{7} - 262144 t^{5} + 4096 t^{3} + 40 t + x \right )} \right )\right )} \]

[In]

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

[Out]

RootSum(1073741824*_t**8 + 65536*_t**4 + 1024*_t**2 + 3, Lambda(_t, _t*log(67108864*_t**7 - 262144*_t**5 + 409
6*_t**3 + 40*_t + x)))

Maxima [F]

\[ \int \frac {x^2}{2+\left (1+x^2\right )^4} \, dx=\int { \frac {x^{2}}{{\left (x^{2} + 1\right )}^{4} + 2} \,d x } \]

[In]

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

[Out]

integrate(x^2/((x^2 + 1)^4 + 2), x)

Giac [F]

\[ \int \frac {x^2}{2+\left (1+x^2\right )^4} \, dx=\int { \frac {x^{2}}{{\left (x^{2} + 1\right )}^{4} + 2} \,d x } \]

[In]

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

[Out]

integrate(x^2/((x^2 + 1)^4 + 2), x)

Mupad [B] (verification not implemented)

Time = 10.04 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.76 \[ \int \frac {x^2}{2+\left (1+x^2\right )^4} \, dx=\sum _{k=1}^8\ln \left (\mathrm {root}\left (z^8+\frac {z^4}{16384}+\frac {z^2}{1048576}+\frac {3}{1073741824},z,k\right )\,\left (40\,x+\mathrm {root}\left (z^8+\frac {z^4}{16384}+\frac {z^2}{1048576}+\frac {3}{1073741824},z,k\right )\,\left (\mathrm {root}\left (z^8+\frac {z^4}{16384}+\frac {z^2}{1048576}+\frac {3}{1073741824},z,k\right )\,\left (4096\,x-{\mathrm {root}\left (z^8+\frac {z^4}{16384}+\frac {z^2}{1048576}+\frac {3}{1073741824},z,k\right )}^2\,\left (786432\,x-{\mathrm {root}\left (z^8+\frac {z^4}{16384}+\frac {z^2}{1048576}+\frac {3}{1073741824},z,k\right )}^2\,x\,67108864\right )\right )-768\right )\right )-3\right )\,\mathrm {root}\left (z^8+\frac {z^4}{16384}+\frac {z^2}{1048576}+\frac {3}{1073741824},z,k\right ) \]

[In]

int(x^2/((x^2 + 1)^4 + 2),x)

[Out]

symsum(log(root(z^8 + z^4/16384 + z^2/1048576 + 3/1073741824, z, k)*(40*x + root(z^8 + z^4/16384 + z^2/1048576
+ 3/1073741824, z, k)*(root(z^8 + z^4/16384 + z^2/1048576 + 3/1073741824, z, k)*(4096*x - root(z^8 + z^4/1638
4 + z^2/1048576 + 3/1073741824, z, k)^2*(786432*x - 67108864*root(z^8 + z^4/16384 + z^2/1048576 + 3/1073741824
, z, k)^2*x)) - 768)) - 3)*root(z^8 + z^4/16384 + z^2/1048576 + 3/1073741824, z, k), k, 1, 8)