\(\int \frac {(1+x^2)^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} (-1+x^2+x^4)} \, dx\) [2902]

Optimal result

Integrand size = 44, antiderivative size = 321 \[ \int \frac {\left (1+x^2\right )^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx=\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )-\frac {\text {RootSum}\left [1+2 \text {$\#$1}^2-6 \text {$\#$1}^4-2 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {\log \left (-1+x^2+\sqrt {1+x^4}\right )-\log \left (\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}-\text {$\#$1}+x^2 \text {$\#$1}+\sqrt {1+x^4} \text {$\#$1}\right )-2 \log \left (-1+x^2+\sqrt {1+x^4}\right ) \text {$\#$1}^4+2 \log \left (\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}-\text {$\#$1}+x^2 \text {$\#$1}+\sqrt {1+x^4} \text {$\#$1}\right ) \text {$\#$1}^4-\log \left (-1+x^2+\sqrt {1+x^4}\right ) \text {$\#$1}^6+\log \left (\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}-\text {$\#$1}+x^2 \text {$\#$1}+\sqrt {1+x^4} \text {$\#$1}\right ) \text {$\#$1}^6}{\text {$\#$1}-6 \text {$\#$1}^3-3 \text {$\#$1}^5+2 \text {$\#$1}^7}\&\right ]}{\sqrt {2}} \]

[Out]

Unintegrable

Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.65 (sec) , antiderivative size = 710, normalized size of antiderivative = 2.21, number of steps used = 30, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {6860, 2157, 212, 6857, 2158, 739} \[ \int \frac {\left (1+x^2\right )^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx=-\frac {i \left (3+\sqrt {5}\right ) \arctan \left (\frac {(1+i) \left (\sqrt {2}-i \sqrt {\sqrt {5}-1} x\right )}{\sqrt {2 \left (\sqrt {5}+(-1-2 i)\right )} \sqrt {1+i x^2}}\right )}{4 \sqrt {(-5-5 i) \left (\sqrt {5}+(-2+i)\right )}}+\frac {i \left (3+\sqrt {5}\right ) \arctan \left (\frac {(1+i) \left (\sqrt {2}+i \sqrt {\sqrt {5}-1} x\right )}{\sqrt {2 \left (\sqrt {5}+(-1-2 i)\right )} \sqrt {1+i x^2}}\right )}{4 \sqrt {(-5-5 i) \left (\sqrt {5}+(-2+i)\right )}}-\frac {\left (3-\sqrt {5}\right ) \arctan \left (\frac {(1+i) \left (\sqrt {2}-\sqrt {1+\sqrt {5}} x\right )}{\sqrt {-2 \sqrt {5}+(-2-4 i)} \sqrt {1+i x^2}}\right )}{4 \sqrt {(-5-5 i) \left (\sqrt {5}+(2-i)\right )}}+\frac {\left (3-\sqrt {5}\right ) \arctan \left (\frac {(1+i) \left (\sqrt {1+\sqrt {5}} x+\sqrt {2}\right )}{\sqrt {-2 \sqrt {5}+(-2-4 i)} \sqrt {1+i x^2}}\right )}{4 \sqrt {(-5-5 i) \left (\sqrt {5}+(2-i)\right )}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (3+\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {2}-i \sqrt {\sqrt {5}-1} x}{\sqrt {(2+i)-i \sqrt {5}} \sqrt {1-i x^2}}\right )}{\sqrt {(10+10 i) \left (\sqrt {5}+(-2-i)\right )}}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (3+\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {2}+i \sqrt {\sqrt {5}-1} x}{\sqrt {(2+i)-i \sqrt {5}} \sqrt {1-i x^2}}\right )}{\sqrt {(10+10 i) \left (\sqrt {5}+(-2-i)\right )}}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (3-\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {2}-\sqrt {1+\sqrt {5}} x}{\sqrt {(2+i)+i \sqrt {5}} \sqrt {1-i x^2}}\right )}{\sqrt {(10+10 i) \left (\sqrt {5}+(2+i)\right )}}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (3-\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {1+\sqrt {5}} x+\sqrt {2}}{\sqrt {(2+i)+i \sqrt {5}} \sqrt {1-i x^2}}\right )}{\sqrt {(10+10 i) \left (\sqrt {5}+(2+i)\right )}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {\sqrt {x^4+1}+x^2}}\right )}{\sqrt {2}} \]

[In]

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

[Out]

((-1/4*I)*(3 + Sqrt[5])*ArcTan[((1 + I)*(Sqrt[2] - I*Sqrt[-1 + Sqrt[5]]*x))/(Sqrt[2*((-1 - 2*I) + Sqrt[5])]*Sq
rt[1 + I*x^2])])/Sqrt[(-5 - 5*I)*((-2 + I) + Sqrt[5])] + ((I/4)*(3 + Sqrt[5])*ArcTan[((1 + I)*(Sqrt[2] + I*Sqr
t[-1 + Sqrt[5]]*x))/(Sqrt[2*((-1 - 2*I) + Sqrt[5])]*Sqrt[1 + I*x^2])])/Sqrt[(-5 - 5*I)*((-2 + I) + Sqrt[5])] -
((3 - Sqrt[5])*ArcTan[((1 + I)*(Sqrt[2] - Sqrt[1 + Sqrt[5]]*x))/(Sqrt[(-2 - 4*I) - 2*Sqrt[5]]*Sqrt[1 + I*x^2]
)])/(4*Sqrt[(-5 - 5*I)*((2 - I) + Sqrt[5])]) + ((3 - Sqrt[5])*ArcTan[((1 + I)*(Sqrt[2] + Sqrt[1 + Sqrt[5]]*x))
/(Sqrt[(-2 - 4*I) - 2*Sqrt[5]]*Sqrt[1 + I*x^2])])/(4*Sqrt[(-5 - 5*I)*((2 - I) + Sqrt[5])]) - ((1/4 - I/4)*(3 +
Sqrt[5])*ArcTanh[(Sqrt[2] - I*Sqrt[-1 + Sqrt[5]]*x)/(Sqrt[(2 + I) - I*Sqrt[5]]*Sqrt[1 - I*x^2])])/Sqrt[(10 +
10*I)*((-2 - I) + Sqrt[5])] + ((1/4 - I/4)*(3 + Sqrt[5])*ArcTanh[(Sqrt[2] + I*Sqrt[-1 + Sqrt[5]]*x)/(Sqrt[(2 +
I) - I*Sqrt[5]]*Sqrt[1 - I*x^2])])/Sqrt[(10 + 10*I)*((-2 - I) + Sqrt[5])] + ((1/4 + I/4)*(3 - Sqrt[5])*ArcTan
h[(Sqrt[2] - Sqrt[1 + Sqrt[5]]*x)/(Sqrt[(2 + I) + I*Sqrt[5]]*Sqrt[1 - I*x^2])])/Sqrt[(10 + 10*I)*((2 + I) + Sq
rt[5])] - ((1/4 + I/4)*(3 - Sqrt[5])*ArcTanh[(Sqrt[2] + Sqrt[1 + Sqrt[5]]*x)/(Sqrt[(2 + I) + I*Sqrt[5]]*Sqrt[1
- I*x^2])])/Sqrt[(10 + 10*I)*((2 + I) + Sqrt[5])] + ArcTanh[(Sqrt[2]*x)/Sqrt[x^2 + Sqrt[1 + x^4]]]/Sqrt[2]

Rule 212

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

Rule 739

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 2157

Int[Sqrt[(c_.)*(x_)^2 + (d_.)*Sqrt[(a_) + (b_.)*(x_)^4]]/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Dist[d, Subst
[Int[1/(1 - 2*c*x^2), x], x, x/Sqrt[c*x^2 + d*Sqrt[a + b*x^4]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[c^2 - b*d
^2, 0]

Rule 2158

Int[(((c_.) + (d_.)*(x_))^(m_.)*Sqrt[(b_.)*(x_)^2 + Sqrt[(a_) + (e_.)*(x_)^4]])/Sqrt[(a_) + (e_.)*(x_)^4], x_S
ymbol] :> Dist[(1 - I)/2, Int[(c + d*x)^m/Sqrt[Sqrt[a] - I*b*x^2], x], x] + Dist[(1 + I)/2, Int[(c + d*x)^m/Sq
rt[Sqrt[a] + I*b*x^2], x], x] /; FreeQ[{a, b, c, d, m}, x] && EqQ[e, b^2] && GtQ[a, 0]

Rule 6857

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

Rule 6860

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*} \text {integral}& = \int \left (\frac {\sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4}}+\frac {\left (2+x^2\right ) \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )}\right ) \, dx \\ & = \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4}} \, dx+\int \frac {\left (2+x^2\right ) \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx \\ & = \int \left (\frac {\left (1+\frac {3}{\sqrt {5}}\right ) \sqrt {x^2+\sqrt {1+x^4}}}{\left (1-\sqrt {5}+2 x^2\right ) \sqrt {1+x^4}}+\frac {\left (1-\frac {3}{\sqrt {5}}\right ) \sqrt {x^2+\sqrt {1+x^4}}}{\left (1+\sqrt {5}+2 x^2\right ) \sqrt {1+x^4}}\right ) \, dx+\text {Subst}\left (\int \frac {1}{1-2 x^2} \, dx,x,\frac {x}{\sqrt {x^2+\sqrt {1+x^4}}}\right ) \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}}+\frac {1}{5} \left (5-3 \sqrt {5}\right ) \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\left (1+\sqrt {5}+2 x^2\right ) \sqrt {1+x^4}} \, dx+\frac {1}{5} \left (5+3 \sqrt {5}\right ) \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\left (1-\sqrt {5}+2 x^2\right ) \sqrt {1+x^4}} \, dx \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}}+\frac {1}{5} \left (5-3 \sqrt {5}\right ) \int \left (\frac {i \sqrt {x^2+\sqrt {1+x^4}}}{2 \sqrt {1+\sqrt {5}} \left (i \sqrt {1+\sqrt {5}}-\sqrt {2} x\right ) \sqrt {1+x^4}}+\frac {i \sqrt {x^2+\sqrt {1+x^4}}}{2 \sqrt {1+\sqrt {5}} \left (i \sqrt {1+\sqrt {5}}+\sqrt {2} x\right ) \sqrt {1+x^4}}\right ) \, dx+\frac {1}{5} \left (5+3 \sqrt {5}\right ) \int \left (\frac {\sqrt {-1+\sqrt {5}} \sqrt {x^2+\sqrt {1+x^4}}}{2 \left (1-\sqrt {5}\right ) \left (\sqrt {-1+\sqrt {5}}-\sqrt {2} x\right ) \sqrt {1+x^4}}+\frac {\sqrt {-1+\sqrt {5}} \sqrt {x^2+\sqrt {1+x^4}}}{2 \left (1-\sqrt {5}\right ) \left (\sqrt {-1+\sqrt {5}}+\sqrt {2} x\right ) \sqrt {1+x^4}}\right ) \, dx \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}}--\frac {\left (-5-3 \sqrt {5}\right ) \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\left (\sqrt {-1+\sqrt {5}}-\sqrt {2} x\right ) \sqrt {1+x^4}} \, dx}{10 \sqrt {-1+\sqrt {5}}}--\frac {\left (-5-3 \sqrt {5}\right ) \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\left (\sqrt {-1+\sqrt {5}}+\sqrt {2} x\right ) \sqrt {1+x^4}} \, dx}{10 \sqrt {-1+\sqrt {5}}}+\frac {\left (i \left (5-3 \sqrt {5}\right )\right ) \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\left (i \sqrt {1+\sqrt {5}}-\sqrt {2} x\right ) \sqrt {1+x^4}} \, dx}{10 \sqrt {1+\sqrt {5}}}+\frac {\left (i \left (5-3 \sqrt {5}\right )\right ) \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\left (i \sqrt {1+\sqrt {5}}+\sqrt {2} x\right ) \sqrt {1+x^4}} \, dx}{10 \sqrt {1+\sqrt {5}}} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}}--\frac {\left (\left (\frac {1}{20}+\frac {i}{20}\right ) \left (-5-3 \sqrt {5}\right )\right ) \int \frac {1}{\left (\sqrt {-1+\sqrt {5}}-\sqrt {2} x\right ) \sqrt {1+i x^2}} \, dx}{\sqrt {-1+\sqrt {5}}}--\frac {\left (\left (\frac {1}{20}+\frac {i}{20}\right ) \left (-5-3 \sqrt {5}\right )\right ) \int \frac {1}{\left (\sqrt {-1+\sqrt {5}}+\sqrt {2} x\right ) \sqrt {1+i x^2}} \, dx}{\sqrt {-1+\sqrt {5}}}--\frac {\left (\left (\frac {1}{20}-\frac {i}{20}\right ) \left (-5-3 \sqrt {5}\right )\right ) \int \frac {1}{\left (\sqrt {-1+\sqrt {5}}-\sqrt {2} x\right ) \sqrt {1-i x^2}} \, dx}{\sqrt {-1+\sqrt {5}}}--\frac {\left (\left (\frac {1}{20}-\frac {i}{20}\right ) \left (-5-3 \sqrt {5}\right )\right ) \int \frac {1}{\left (\sqrt {-1+\sqrt {5}}+\sqrt {2} x\right ) \sqrt {1-i x^2}} \, dx}{\sqrt {-1+\sqrt {5}}}+-\frac {\left (\left (\frac {1}{20}-\frac {i}{20}\right ) \left (5-3 \sqrt {5}\right )\right ) \int \frac {1}{\left (i \sqrt {1+\sqrt {5}}-\sqrt {2} x\right ) \sqrt {1+i x^2}} \, dx}{\sqrt {1+\sqrt {5}}}+-\frac {\left (\left (\frac {1}{20}-\frac {i}{20}\right ) \left (5-3 \sqrt {5}\right )\right ) \int \frac {1}{\left (i \sqrt {1+\sqrt {5}}+\sqrt {2} x\right ) \sqrt {1+i x^2}} \, dx}{\sqrt {1+\sqrt {5}}}+\frac {\left (\left (\frac {1}{20}+\frac {i}{20}\right ) \left (5-3 \sqrt {5}\right )\right ) \int \frac {1}{\left (i \sqrt {1+\sqrt {5}}-\sqrt {2} x\right ) \sqrt {1-i x^2}} \, dx}{\sqrt {1+\sqrt {5}}}+\frac {\left (\left (\frac {1}{20}+\frac {i}{20}\right ) \left (5-3 \sqrt {5}\right )\right ) \int \frac {1}{\left (i \sqrt {1+\sqrt {5}}+\sqrt {2} x\right ) \sqrt {1-i x^2}} \, dx}{\sqrt {1+\sqrt {5}}} \\ & = \frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}}-\frac {\left (\left (\frac {1}{20}-\frac {i}{20}\right ) \left (-5-3 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{2-i \left (-1+\sqrt {5}\right )-x^2} \, dx,x,\frac {-\sqrt {2}+i \sqrt {-1+\sqrt {5}} x}{\sqrt {1-i x^2}}\right )}{\sqrt {-1+\sqrt {5}}}-\frac {\left (\left (\frac {1}{20}-\frac {i}{20}\right ) \left (-5-3 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{2-i \left (-1+\sqrt {5}\right )-x^2} \, dx,x,\frac {\sqrt {2}+i \sqrt {-1+\sqrt {5}} x}{\sqrt {1-i x^2}}\right )}{\sqrt {-1+\sqrt {5}}}-\frac {\left (\left (\frac {1}{20}+\frac {i}{20}\right ) \left (-5-3 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{2+i \left (-1+\sqrt {5}\right )-x^2} \, dx,x,\frac {-\sqrt {2}-i \sqrt {-1+\sqrt {5}} x}{\sqrt {1+i x^2}}\right )}{\sqrt {-1+\sqrt {5}}}-\frac {\left (\left (\frac {1}{20}+\frac {i}{20}\right ) \left (-5-3 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{2+i \left (-1+\sqrt {5}\right )-x^2} \, dx,x,\frac {\sqrt {2}-i \sqrt {-1+\sqrt {5}} x}{\sqrt {1+i x^2}}\right )}{\sqrt {-1+\sqrt {5}}}+-\frac {\left (\left (\frac {1}{20}+\frac {i}{20}\right ) \left (5-3 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{2-i \left (-1-\sqrt {5}\right )-x^2} \, dx,x,\frac {-\sqrt {2}-\sqrt {1+\sqrt {5}} x}{\sqrt {1-i x^2}}\right )}{\sqrt {1+\sqrt {5}}}+-\frac {\left (\left (\frac {1}{20}+\frac {i}{20}\right ) \left (5-3 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{2-i \left (-1-\sqrt {5}\right )-x^2} \, dx,x,\frac {\sqrt {2}-\sqrt {1+\sqrt {5}} x}{\sqrt {1-i x^2}}\right )}{\sqrt {1+\sqrt {5}}}+\frac {\left (\left (\frac {1}{20}-\frac {i}{20}\right ) \left (5-3 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{2+i \left (-1-\sqrt {5}\right )-x^2} \, dx,x,\frac {-\sqrt {2}+\sqrt {1+\sqrt {5}} x}{\sqrt {1+i x^2}}\right )}{\sqrt {1+\sqrt {5}}}+\frac {\left (\left (\frac {1}{20}-\frac {i}{20}\right ) \left (5-3 \sqrt {5}\right )\right ) \text {Subst}\left (\int \frac {1}{2+i \left (-1-\sqrt {5}\right )-x^2} \, dx,x,\frac {\sqrt {2}+\sqrt {1+\sqrt {5}} x}{\sqrt {1+i x^2}}\right )}{\sqrt {1+\sqrt {5}}} \\ & = -\frac {i \left (3+\sqrt {5}\right ) \arctan \left (\frac {(1+i) \left (\sqrt {2}-i \sqrt {-1+\sqrt {5}} x\right )}{\sqrt {2 \left ((-1-2 i)+\sqrt {5}\right )} \sqrt {1+i x^2}}\right )}{4 \sqrt {(-5-5 i) \left ((-2+i)+\sqrt {5}\right )}}+\frac {i \left (3+\sqrt {5}\right ) \arctan \left (\frac {(1+i) \left (\sqrt {2}+i \sqrt {-1+\sqrt {5}} x\right )}{\sqrt {2 \left ((-1-2 i)+\sqrt {5}\right )} \sqrt {1+i x^2}}\right )}{4 \sqrt {(-5-5 i) \left ((-2+i)+\sqrt {5}\right )}}-\frac {\left (3-\sqrt {5}\right ) \arctan \left (\frac {(1+i) \left (\sqrt {2}-\sqrt {1+\sqrt {5}} x\right )}{\sqrt {(-2-4 i)-2 \sqrt {5}} \sqrt {1+i x^2}}\right )}{4 \sqrt {(-5-5 i) \left ((2-i)+\sqrt {5}\right )}}+\frac {\left (3-\sqrt {5}\right ) \arctan \left (\frac {(1+i) \left (\sqrt {2}+\sqrt {1+\sqrt {5}} x\right )}{\sqrt {(-2-4 i)-2 \sqrt {5}} \sqrt {1+i x^2}}\right )}{4 \sqrt {(-5-5 i) \left ((2-i)+\sqrt {5}\right )}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (3+\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {2}-i \sqrt {-1+\sqrt {5}} x}{\sqrt {(2+i)-i \sqrt {5}} \sqrt {1-i x^2}}\right )}{\sqrt {(10+10 i) \left ((-2-i)+\sqrt {5}\right )}}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (3+\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {2}+i \sqrt {-1+\sqrt {5}} x}{\sqrt {(2+i)-i \sqrt {5}} \sqrt {1-i x^2}}\right )}{\sqrt {(10+10 i) \left ((-2-i)+\sqrt {5}\right )}}+\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (3-\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {2}-\sqrt {1+\sqrt {5}} x}{\sqrt {(2+i)+i \sqrt {5}} \sqrt {1-i x^2}}\right )}{\sqrt {(10+10 i) \left ((2+i)+\sqrt {5}\right )}}-\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \left (3-\sqrt {5}\right ) \text {arctanh}\left (\frac {\sqrt {2}+\sqrt {1+\sqrt {5}} x}{\sqrt {(2+i)+i \sqrt {5}} \sqrt {1-i x^2}}\right )}{\sqrt {(10+10 i) \left ((2+i)+\sqrt {5}\right )}}+\frac {\text {arctanh}\left (\frac {\sqrt {2} x}{\sqrt {x^2+\sqrt {1+x^4}}}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.01 \[ \int \frac {\left (1+x^2\right )^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx=\frac {2 \text {arctanh}\left (\frac {-1+x^2+\sqrt {1+x^4}}{\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}}\right )+\text {RootSum}\left [1-2 \text {$\#$1}^2-6 \text {$\#$1}^4+2 \text {$\#$1}^6+\text {$\#$1}^8\&,\frac {\log \left (1+x^2+\sqrt {1+x^4}\right )-\log \left (\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}-\text {$\#$1}-x^2 \text {$\#$1}-\sqrt {1+x^4} \text {$\#$1}\right )+2 \log \left (1+x^2+\sqrt {1+x^4}\right ) \text {$\#$1}^2-2 \log \left (\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}-\text {$\#$1}-x^2 \text {$\#$1}-\sqrt {1+x^4} \text {$\#$1}\right ) \text {$\#$1}^2-\log \left (1+x^2+\sqrt {1+x^4}\right ) \text {$\#$1}^6+\log \left (\sqrt {2} x \sqrt {x^2+\sqrt {1+x^4}}-\text {$\#$1}-x^2 \text {$\#$1}-\sqrt {1+x^4} \text {$\#$1}\right ) \text {$\#$1}^6}{-\text {$\#$1}-6 \text {$\#$1}^3+3 \text {$\#$1}^5+2 \text {$\#$1}^7}\&\right ]}{\sqrt {2}} \]

[In]

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

[Out]

(2*ArcTanh[(-1 + x^2 + Sqrt[1 + x^4])/(Sqrt[2]*x*Sqrt[x^2 + Sqrt[1 + x^4]])] + RootSum[1 - 2*#1^2 - 6*#1^4 + 2
*#1^6 + #1^8 & , (Log[1 + x^2 + Sqrt[1 + x^4]] - Log[Sqrt[2]*x*Sqrt[x^2 + Sqrt[1 + x^4]] - #1 - x^2*#1 - Sqrt[
1 + x^4]*#1] + 2*Log[1 + x^2 + Sqrt[1 + x^4]]*#1^2 - 2*Log[Sqrt[2]*x*Sqrt[x^2 + Sqrt[1 + x^4]] - #1 - x^2*#1 -
Sqrt[1 + x^4]*#1]*#1^2 - Log[1 + x^2 + Sqrt[1 + x^4]]*#1^6 + Log[Sqrt[2]*x*Sqrt[x^2 + Sqrt[1 + x^4]] - #1 - x
^2*#1 - Sqrt[1 + x^4]*#1]*#1^6)/(-#1 - 6*#1^3 + 3*#1^5 + 2*#1^7) & ])/Sqrt[2]

Maple [N/A] (verified)

Not integrable

Time = 0.00 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.12

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

[In]

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

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 3 vs. order 1.

Time = 20.91 (sec) , antiderivative size = 10253, normalized size of antiderivative = 31.94 \[ \int \frac {\left (1+x^2\right )^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

Too large to include

Sympy [N/A]

Not integrable

Time = 29.59 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.12 \[ \int \frac {\left (1+x^2\right )^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx=\int \frac {\left (x^{2} + 1\right )^{2} \sqrt {x^{2} + \sqrt {x^{4} + 1}}}{\sqrt {x^{4} + 1} \left (x^{4} + x^{2} - 1\right )}\, dx \]

[In]

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

[Out]

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

Maxima [N/A]

Not integrable

Time = 0.34 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.12 \[ \int \frac {\left (1+x^2\right )^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx=\int { \frac {\sqrt {x^{2} + \sqrt {x^{4} + 1}} {\left (x^{2} + 1\right )}^{2}}{{\left (x^{4} + x^{2} - 1\right )} \sqrt {x^{4} + 1}} \,d x } \]

[In]

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

[Out]

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

Giac [N/A]

Not integrable

Time = 0.44 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.12 \[ \int \frac {\left (1+x^2\right )^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx=\int { \frac {\sqrt {x^{2} + \sqrt {x^{4} + 1}} {\left (x^{2} + 1\right )}^{2}}{{\left (x^{4} + x^{2} - 1\right )} \sqrt {x^{4} + 1}} \,d x } \]

[In]

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

[Out]

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

Mupad [N/A]

Not integrable

Time = 0.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.12 \[ \int \frac {\left (1+x^2\right )^2 \sqrt {x^2+\sqrt {1+x^4}}}{\sqrt {1+x^4} \left (-1+x^2+x^4\right )} \, dx=\int \frac {{\left (x^2+1\right )}^2\,\sqrt {\sqrt {x^4+1}+x^2}}{\sqrt {x^4+1}\,\left (x^4+x^2-1\right )} \,d x \]

[In]

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

[Out]

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