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

Optimal. Leaf size=273 \[ \frac {1}{2} \sqrt {\frac {1}{2} \left (\sqrt {2}-1\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (\sqrt {2}-1\right )} x \sqrt {\sqrt {x^4+1}+x^2}}{\sqrt {x^4+1}+x^2+1}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (\sqrt {2}-1\right )} \tan ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} x \sqrt {\sqrt {x^4+1}+x^2}}{\sqrt {x^4+1}+x^2+1}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tanh ^{-1}\left (\frac {\sqrt {2 \left (\sqrt {2}-1\right )} x \sqrt {\sqrt {x^4+1}+x^2}}{\sqrt {x^4+1}+x^2+1}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tanh ^{-1}\left (\frac {\sqrt {2 \left (1+\sqrt {2}\right )} x \sqrt {\sqrt {x^4+1}+x^2}}{\sqrt {x^4+1}+x^2+1}\right ) \]

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Rubi [C]  time = 1.69, antiderivative size = 309, normalized size of antiderivative = 1.13, number of steps used = 26, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6725, 2133, 725, 206} \begin {gather*} \frac {1}{8} \sqrt {1+i} \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-i x}{\sqrt {1-i} \sqrt {1-i x^2}}\right )+\frac {1}{8} \sqrt {1-i} \tanh ^{-1}\left (\frac {1+i x}{\sqrt {1-i} \sqrt {1-i x^2}}\right )-\frac {1}{8} \sqrt {1+i} \tanh ^{-1}\left (\frac {x+1}{\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}{8} \sqrt {1+i} \tanh ^{-1}\left (\frac {1-i x}{\sqrt {1+i} \sqrt {1+i x^2}}\right )-\frac {1}{8} \sqrt {1+i} \tanh ^{-1}\left (\frac {1+i x}{\sqrt {1+i} \sqrt {1+i x^2}}\right )-\frac {1}{8} \sqrt {1-i} \tanh ^{-1}\left (\frac {x+1}{\sqrt {1-i} \sqrt {1+i x^2}}\right ) \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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 725

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 2133

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 6725

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]

Rubi steps

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

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Mathematica [F]  time = 0.24, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {x^2+\sqrt {1+x^4}}}{\left (-1+x^4\right ) \sqrt {1+x^4}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.60, size = 389, normalized size = 1.42 \begin {gather*} \frac {1}{2} \sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {-\sqrt {-\frac {1}{2}+\frac {1}{\sqrt {2}}}+\sqrt {-\frac {1}{2}+\frac {1}{\sqrt {2}}} x^2+\sqrt {-\frac {1}{2}+\frac {1}{\sqrt {2}}} \sqrt {1+x^4}}{x \sqrt {x^2+\sqrt {1+x^4}}}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (-1+\sqrt {2}\right )} \tan ^{-1}\left (\frac {-\sqrt {\frac {1}{2}+\frac {1}{\sqrt {2}}}+\sqrt {\frac {1}{2}+\frac {1}{\sqrt {2}}} x^2+\sqrt {\frac {1}{2}+\frac {1}{\sqrt {2}}} \sqrt {1+x^4}}{x \sqrt {x^2+\sqrt {1+x^4}}}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tanh ^{-1}\left (\frac {-\sqrt {-\frac {1}{2}+\frac {1}{\sqrt {2}}}+\sqrt {-\frac {1}{2}+\frac {1}{\sqrt {2}}} x^2+\sqrt {-\frac {1}{2}+\frac {1}{\sqrt {2}}} \sqrt {1+x^4}}{x \sqrt {x^2+\sqrt {1+x^4}}}\right )-\frac {1}{2} \sqrt {\frac {1}{2} \left (1+\sqrt {2}\right )} \tanh ^{-1}\left (\frac {-\sqrt {\frac {1}{2}+\frac {1}{\sqrt {2}}}+\sqrt {\frac {1}{2}+\frac {1}{\sqrt {2}}} x^2+\sqrt {\frac {1}{2}+\frac {1}{\sqrt {2}}} \sqrt {1+x^4}}{x \sqrt {x^2+\sqrt {1+x^4}}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 12.46, size = 373, normalized size = 1.37 \begin {gather*} -\frac {1}{4} \, \sqrt {2} \sqrt {\sqrt {2} - 1} \arctan \left (\frac {x^{8} - 2 \, x^{4} - 2 \, {\left (2 \, x^{7} - 2 \, x^{3} + \sqrt {2} {\left (3 \, x^{7} + x^{3}\right )} - {\left (4 \, \sqrt {2} x^{5} + 5 \, x^{5} - x\right )} \sqrt {x^{4} + 1}\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} \sqrt {\sqrt {2} - 1} - 2 \, \sqrt {2} {\left (x^{8} + 3 \, x^{4}\right )} - 2 \, {\left (3 \, x^{6} + x^{2} + \sqrt {2} {\left (x^{6} - x^{2}\right )}\right )} \sqrt {x^{4} + 1} + 1}{7 \, x^{8} + 10 \, x^{4} - 1}\right ) - \frac {1}{16} \, \sqrt {2} \sqrt {\sqrt {2} + 1} \log \left (\frac {2 \, {\left (\sqrt {2} x^{3} + 2 \, x^{3} + \sqrt {x^{4} + 1} {\left (\sqrt {2} x + x\right )}\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} + {\left (2 \, \sqrt {2} x^{4} + 3 \, x^{4} + 2 \, \sqrt {x^{4} + 1} {\left (\sqrt {2} x^{2} + x^{2}\right )} + 1\right )} \sqrt {\sqrt {2} + 1}}{x^{4} - 1}\right ) + \frac {1}{16} \, \sqrt {2} \sqrt {\sqrt {2} + 1} \log \left (\frac {2 \, {\left (\sqrt {2} x^{3} + 2 \, x^{3} + \sqrt {x^{4} + 1} {\left (\sqrt {2} x + x\right )}\right )} \sqrt {x^{2} + \sqrt {x^{4} + 1}} - {\left (2 \, \sqrt {2} x^{4} + 3 \, x^{4} + 2 \, \sqrt {x^{4} + 1} {\left (\sqrt {2} x^{2} + x^{2}\right )} + 1\right )} \sqrt {\sqrt {2} + 1}}{x^{4} - 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^4-1)/(x^4+1)^(1/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(2)*sqrt(sqrt(2) - 1)*arctan((x^8 - 2*x^4 - 2*(2*x^7 - 2*x^3 + sqrt(2)*(3*x^7 + x^3) - (4*sqrt(2)*x^5
+ 5*x^5 - x)*sqrt(x^4 + 1))*sqrt(x^2 + sqrt(x^4 + 1))*sqrt(sqrt(2) - 1) - 2*sqrt(2)*(x^8 + 3*x^4) - 2*(3*x^6
+ x^2 + sqrt(2)*(x^6 - x^2))*sqrt(x^4 + 1) + 1)/(7*x^8 + 10*x^4 - 1)) - 1/16*sqrt(2)*sqrt(sqrt(2) + 1)*log((2*
(sqrt(2)*x^3 + 2*x^3 + sqrt(x^4 + 1)*(sqrt(2)*x + x))*sqrt(x^2 + sqrt(x^4 + 1)) + (2*sqrt(2)*x^4 + 3*x^4 + 2*s
qrt(x^4 + 1)*(sqrt(2)*x^2 + x^2) + 1)*sqrt(sqrt(2) + 1))/(x^4 - 1)) + 1/16*sqrt(2)*sqrt(sqrt(2) + 1)*log((2*(s
qrt(2)*x^3 + 2*x^3 + sqrt(x^4 + 1)*(sqrt(2)*x + x))*sqrt(x^2 + sqrt(x^4 + 1)) - (2*sqrt(2)*x^4 + 3*x^4 + 2*sqr
t(x^4 + 1)*(sqrt(2)*x^2 + x^2) + 1)*sqrt(sqrt(2) + 1))/(x^4 - 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\sqrt {x^{2}+\sqrt {x^{4}+1}}}{\left (x^{4}-1\right ) \sqrt {x^{4}+1}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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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 - 1\right ) \left (x + 1\right ) \left (x^{2} + 1\right ) \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**4-1)/(x**4+1)**(1/2),x)

[Out]

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

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