3.11.18 \(\int \frac {x^2}{\sqrt {-1+x^4} (1+x^4)} \, dx\) [1018]

Optimal. Leaf size=49 \[ -\frac {1}{4} \tan ^{-1}\left (\frac {1+x^2}{x \sqrt {-1+x^4}}\right )-\frac {1}{4} \tanh ^{-1}\left (\frac {1-x^2}{x \sqrt {-1+x^4}}\right ) \]

[Out]

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

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Rubi [C] Result contains complex when optimal does not.
time = 0.08, antiderivative size = 47, normalized size of antiderivative = 0.96, number of steps used = 9, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {504, 1225, 228, 1713, 212, 209} \begin {gather*} \left (\frac {1}{8}+\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) x}{\sqrt {x^4-1}}\right )-\left (\frac {1}{8}+\frac {i}{8}\right ) \text {ArcTan}\left (\frac {(1+i) x}{\sqrt {x^4-1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(Sqrt[-1 + x^4]*(1 + x^4)),x]

[Out]

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

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 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 228

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[(-a)*b, 2]}, Simp[Sqrt[-a + q*x^2]*(Sqrt[(a + q*x^2
)/q]/(Sqrt[2]*Sqrt[-a]*Sqrt[a + b*x^4]))*EllipticF[ArcSin[x/Sqrt[(a + q*x^2)/(2*q)]], 1/2], x] /; IntegerQ[q]]
/; FreeQ[{a, b}, x] && LtQ[a, 0] && GtQ[b, 0]

Rule 504

Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s
= Denominator[Rt[-a/b, 2]]}, Dist[s/(2*b), Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Dist[s/(2*b), Int[1/
((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1225

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

Rule 1713

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.15, size = 53, normalized size = 1.08 \begin {gather*} \left (-\frac {1}{8}-\frac {i}{8}\right ) \tan ^{-1}\left (\frac {(1+i) x}{\sqrt {-1+x^4}}\right )+\left (\frac {1}{8}-\frac {i}{8}\right ) \tan ^{-1}\left (\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {-1+x^4}}{x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(100\) vs. \(2(41)=82\).
time = 0.50, size = 101, normalized size = 2.06

method	result	size
default	\(\frac {\left (\frac {\arctan \left (\frac {\sqrt {x^{4}-1}}{x}+1\right ) \sqrt {2}}{8}+\frac {\arctan \left (\frac {\sqrt {x^{4}-1}}{x}-1\right ) \sqrt {2}}{8}+\frac {\sqrt {2}\, \ln \left (\frac {1+\frac {x^{4}-1}{2 x^{2}}+\frac {\sqrt {x^{4}-1}}{x}}{1+\frac {x^{4}-1}{2 x^{2}}-\frac {\sqrt {x^{4}-1}}{x}}\right )}{16}\right ) \sqrt {2}}{2}\)	\(101\)
elliptic	\(\frac {\left (\frac {\arctan \left (\frac {\sqrt {x^{4}-1}}{x}+1\right ) \sqrt {2}}{8}+\frac {\arctan \left (\frac {\sqrt {x^{4}-1}}{x}-1\right ) \sqrt {2}}{8}+\frac {\sqrt {2}\, \ln \left (\frac {1+\frac {x^{4}-1}{2 x^{2}}+\frac {\sqrt {x^{4}-1}}{x}}{1+\frac {x^{4}-1}{2 x^{2}}-\frac {\sqrt {x^{4}-1}}{x}}\right )}{16}\right ) \sqrt {2}}{2}\)	\(101\)
trager	\(-\frac {\ln \left (\frac {8 \RootOf \left (32 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) x -\sqrt {x^{4}-1}+2 x}{8 x^{2} \RootOf \left (32 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )+x^{2}+1}\right )}{4}-\ln \left (\frac {8 \RootOf \left (32 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) x -\sqrt {x^{4}-1}+2 x}{8 x^{2} \RootOf \left (32 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )+x^{2}+1}\right ) \RootOf \left (32 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )+\RootOf \left (32 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) \ln \left (-\frac {8 \RootOf \left (32 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right ) x +\sqrt {x^{4}-1}}{8 x^{2} \RootOf \left (32 \textit {\_Z}^{2}+8 \textit {\_Z} +1\right )+x^{2}-1}\right )\)	\(181\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(1/8*arctan((x^4-1)^(1/2)/x+1)*2^(1/2)+1/8*arctan((x^4-1)^(1/2)/x-1)*2^(1/2)+1/16*2^(1/2)*ln((1+1/2*(x^4-1
)/x^2+(x^4-1)^(1/2)/x)/(1+1/2*(x^4-1)/x^2-(x^4-1)^(1/2)/x)))*2^(1/2)

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

[Out]

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

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Fricas [A]
time = 0.40, size = 51, normalized size = 1.04 \begin {gather*} \frac {1}{4} \, \arctan \left (\frac {\sqrt {x^{4} - 1} x}{x^{2} + 1}\right ) + \frac {1}{8} \, \log \left (\frac {x^{4} + 2 \, x^{2} + 2 \, \sqrt {x^{4} - 1} x - 1}{x^{4} + 1}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*arctan(sqrt(x^4 - 1)*x/(x^2 + 1)) + 1/8*log((x^4 + 2*x^2 + 2*sqrt(x^4 - 1)*x - 1)/(x^4 + 1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^2}{\sqrt {x^4-1}\,\left (x^4+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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