3.24.75 \(\int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{(1+x^2)^2 \sqrt {1+x^4}} \, dx\) [2375]

Optimal. Leaf size=189 \[ \frac {-x^2 \left (-1+x^2\right )-x^2 \sqrt {1+x^4}}{4 x \left (1+x^2\right ) \sqrt {x^2+\sqrt {1+x^4}}}+\frac {1}{4} \sqrt {-1+5 \sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right )+\frac {1}{4} \sqrt {1+5 \sqrt {2}} \tanh ^{-1}\left (\frac {\sqrt {2 \left (-1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}{1+x^2+\sqrt {1+x^4}}\right ) \]

[Out]

1/4*(-x^2*(x^2-1)-(x^4+1)^(1/2)*x^2)/x/(x^2+1)/(x^2+(x^4+1)^(1/2))^(1/2)+1/4*(-1+5*2^(1/2))^(1/2)*arctan((2+2*
2^(1/2))^(1/2)*x*(x^2+(x^4+1)^(1/2))^(1/2)/(1+x^2+(x^4+1)^(1/2)))+1/4*(1+5*2^(1/2))^(1/2)*arctanh((-2+2*2^(1/2
))^(1/2)*x*(x^2+(x^4+1)^(1/2))^(1/2)/(1+x^2+(x^4+1)^(1/2)))

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Rubi [C] Result contains complex when optimal does not.
time = 0.97, antiderivative size = 405, normalized size of antiderivative = 2.14, number of steps used = 28, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {6874, 2158, 745, 739, 212, 6857} \begin {gather*} \frac {i \sqrt {1-i x^2}}{8 (-x+i)}-\frac {i \sqrt {1-i x^2}}{8 (x+i)}-\frac {i \sqrt {1+i x^2}}{8 (-x+i)}+\frac {i \sqrt {1+i x^2}}{8 (x+i)}-\frac {1}{8} \sqrt {1+i} \tanh ^{-1}\left (\frac {1-x}{\sqrt {1+i} \sqrt {1-i x^2}}\right )+\frac {\tanh ^{-1}\left (\frac {1-x}{\sqrt {1+i} \sqrt {1-i x^2}}\right )}{4 (1+i)^{5/2}}+\frac {1}{8} \sqrt {1+i} \tanh ^{-1}\left (\frac {x+1}{\sqrt {1+i} \sqrt {1-i x^2}}\right )-\frac {\tanh ^{-1}\left (\frac {x+1}{\sqrt {1+i} \sqrt {1-i x^2}}\right )}{4 (1+i)^{5/2}}-\frac {1}{8} \sqrt {1-i} \tanh ^{-1}\left (\frac {1-x}{\sqrt {1-i} \sqrt {1+i x^2}}\right )+\frac {\tanh ^{-1}\left (\frac {1-x}{\sqrt {1-i} \sqrt {1+i x^2}}\right )}{4 (1-i)^{5/2}}+\frac {1}{8} \sqrt {1-i} \tanh ^{-1}\left (\frac {x+1}{\sqrt {1-i} \sqrt {1+i x^2}}\right )-\frac {\tanh ^{-1}\left (\frac {x+1}{\sqrt {1-i} \sqrt {1+i x^2}}\right )}{4 (1-i)^{5/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((I/8)*Sqrt[1 - I*x^2])/(I - x) - ((I/8)*Sqrt[1 - I*x^2])/(I + x) - ((I/8)*Sqrt[1 + I*x^2])/(I - x) + ((I/8)*S
qrt[1 + I*x^2])/(I + x) + ArcTanh[(1 - x)/(Sqrt[1 + I]*Sqrt[1 - I*x^2])]/(4*(1 + I)^(5/2)) - (Sqrt[1 + I]*ArcT
anh[(1 - x)/(Sqrt[1 + I]*Sqrt[1 - I*x^2])])/8 - ArcTanh[(1 + x)/(Sqrt[1 + I]*Sqrt[1 - I*x^2])]/(4*(1 + I)^(5/2
)) + (Sqrt[1 + I]*ArcTanh[(1 + x)/(Sqrt[1 + I]*Sqrt[1 - I*x^2])])/8 + ArcTanh[(1 - x)/(Sqrt[1 - I]*Sqrt[1 + I*
x^2])]/(4*(1 - I)^(5/2)) - (Sqrt[1 - I]*ArcTanh[(1 - x)/(Sqrt[1 - I]*Sqrt[1 + I*x^2])])/8 - ArcTanh[(1 + x)/(S
qrt[1 - I]*Sqrt[1 + I*x^2])]/(4*(1 - I)^(5/2)) + (Sqrt[1 - I]*ArcTanh[(1 + x)/(Sqrt[1 - I]*Sqrt[1 + I*x^2])])/
8

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 745

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m + 1)*((a + c*x^2)^(p
+ 1)/((m + 1)*(c*d^2 + a*e^2))), x] + Dist[c*(d/(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x]
/; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 3, 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 6874

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

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

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Mathematica [A]
time = 1.04, size = 172, normalized size = 0.91 \begin {gather*} \frac {1}{4} \left (-\frac {x \left (-1+x^2+\sqrt {1+x^4}\right )}{\left (1+x^2\right ) \sqrt {x^2+\sqrt {1+x^4}}}+\sqrt {-1+5 \sqrt {2}} \text {ArcTan}\left (\frac {\sqrt {\frac {1}{2}+\frac {1}{\sqrt {2}}} \left (-1+x^2+\sqrt {1+x^4}\right )}{x \sqrt {x^2+\sqrt {1+x^4}}}\right )+\sqrt {1+5 \sqrt {2}} \tanh ^{-1}\left (\frac {-1+x^2+\sqrt {1+x^4}}{\sqrt {2 \left (1+\sqrt {2}\right )} x \sqrt {x^2+\sqrt {1+x^4}}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-((x*(-1 + x^2 + Sqrt[1 + x^4]))/((1 + x^2)*Sqrt[x^2 + Sqrt[1 + x^4]])) + Sqrt[-1 + 5*Sqrt[2]]*ArcTan[(Sqrt[1
/2 + 1/Sqrt[2]]*(-1 + x^2 + Sqrt[1 + x^4]))/(x*Sqrt[x^2 + Sqrt[1 + x^4]])] + Sqrt[1 + 5*Sqrt[2]]*ArcTanh[(-1 +
x^2 + Sqrt[1 + x^4])/(Sqrt[2*(1 + Sqrt[2])]*x*Sqrt[x^2 + Sqrt[1 + x^4]])])/4

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {x^{2}+\sqrt {x^{4}+1}}}{\left (x^{2}+1\right )^{2} \sqrt {x^{4}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 384 vs. \(2 (147) = 294\).
time = 4.42, size = 384, normalized size = 2.03 \begin {gather*} \frac {4 \, {\left (x^{2} + 1\right )} \sqrt {5 \, \sqrt {2} - 1} \arctan \left (\frac {{\left (21 \, x^{2} + 7 \, \sqrt {2} {\left (x^{2} + 2\right )} + \sqrt {x^{4} + 1} {\left ({\left (\sqrt {2} + 3\right )} \sqrt {-98 \, \sqrt {2} + 147} - 7 \, \sqrt {2} - 21\right )} - {\left (3 \, x^{2} + \sqrt {2} {\left (x^{2} + 4\right )} + 5\right )} \sqrt {-98 \, \sqrt {2} + 147} - 7\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} \sqrt {5 \, \sqrt {2} - 1}}{98 \, x}\right ) + {\left (x^{2} + 1\right )} \sqrt {5 \, \sqrt {2} + 1} \log \left (\frac {7 \, \sqrt {2} x^{2} + 14 \, x^{2} + {\left (x^{3} + \sqrt {2} {\left (2 \, x^{3} + 3 \, x\right )} - \sqrt {x^{4} + 1} {\left (2 \, \sqrt {2} x + x\right )} + 5 \, x\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} \sqrt {5 \, \sqrt {2} + 1} + 7 \, \sqrt {x^{4} + 1} {\left (\sqrt {2} + 1\right )}}{x^{2} + 1}\right ) - {\left (x^{2} + 1\right )} \sqrt {5 \, \sqrt {2} + 1} \log \left (\frac {7 \, \sqrt {2} x^{2} + 14 \, x^{2} - {\left (x^{3} + \sqrt {2} {\left (2 \, x^{3} + 3 \, x\right )} - \sqrt {x^{4} + 1} {\left (2 \, \sqrt {2} x + x\right )} + 5 \, x\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} \sqrt {5 \, \sqrt {2} + 1} + 7 \, \sqrt {x^{4} + 1} {\left (\sqrt {2} + 1\right )}}{x^{2} + 1}\right ) - 4 \, {\left (x^{3} - \sqrt {x^{4} + 1} x + x\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}}}{16 \, {\left (x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(4*(x^2 + 1)*sqrt(5*sqrt(2) - 1)*arctan(1/98*(21*x^2 + 7*sqrt(2)*(x^2 + 2) + sqrt(x^4 + 1)*((sqrt(2) + 3)
*sqrt(-98*sqrt(2) + 147) - 7*sqrt(2) - 21) - (3*x^2 + sqrt(2)*(x^2 + 4) + 5)*sqrt(-98*sqrt(2) + 147) - 7)*sqrt
(x^2 + sqrt(x^4 + 1))*sqrt(5*sqrt(2) - 1)/x) + (x^2 + 1)*sqrt(5*sqrt(2) + 1)*log((7*sqrt(2)*x^2 + 14*x^2 + (x^
3 + sqrt(2)*(2*x^3 + 3*x) - sqrt(x^4 + 1)*(2*sqrt(2)*x + x) + 5*x)*sqrt(x^2 + sqrt(x^4 + 1))*sqrt(5*sqrt(2) +
1) + 7*sqrt(x^4 + 1)*(sqrt(2) + 1))/(x^2 + 1)) - (x^2 + 1)*sqrt(5*sqrt(2) + 1)*log((7*sqrt(2)*x^2 + 14*x^2 - (
x^3 + sqrt(2)*(2*x^3 + 3*x) - sqrt(x^4 + 1)*(2*sqrt(2)*x + x) + 5*x)*sqrt(x^2 + sqrt(x^4 + 1))*sqrt(5*sqrt(2)
+ 1) + 7*sqrt(x^4 + 1)*(sqrt(2) + 1))/(x^2 + 1)) - 4*(x^3 - sqrt(x^4 + 1)*x + x)*sqrt(x^2 + sqrt(x^4 + 1)))/(x
^2 + 1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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