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

Optimal. Leaf size=37 \[ \frac {\sqrt {x^4+1}}{x}+\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right ) \]

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Rubi [A]  time = 0.33, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {6725, 277, 305, 220, 1196, 1209, 1211, 1699, 203} \begin {gather*} \frac {\sqrt {x^4+1}}{x}+\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 220

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

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rule 1209

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

Rule 1211

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 1699

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]

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 {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^2 \left (1+x^2\right )} \, dx &=\int \left (-\frac {\sqrt {1+x^4}}{x^2}+\frac {2 \sqrt {1+x^4}}{1+x^2}\right ) \, dx\\ &=2 \int \frac {\sqrt {1+x^4}}{1+x^2} \, dx-\int \frac {\sqrt {1+x^4}}{x^2} \, dx\\ &=\frac {\sqrt {1+x^4}}{x}-2 \int \frac {x^2}{\sqrt {1+x^4}} \, dx-2 \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx+4 \int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx\\ &=\frac {\sqrt {1+x^4}}{x}+\frac {2 x \sqrt {1+x^4}}{1+x^2}-\frac {2 \left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} E\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{\sqrt {1+x^4}}+2 \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx+2 \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx\\ &=\frac {\sqrt {1+x^4}}{x}+2 \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )\\ &=\frac {\sqrt {1+x^4}}{x}+\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.12, size = 72, normalized size = 1.95 \begin {gather*} \frac {1}{\sqrt {x^4+1} x}+\frac {x^3}{\sqrt {x^4+1}}+2 \sqrt [4]{-1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )-4 \sqrt [4]{-1} \Pi \left (-i;\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1/(x*Sqrt[1 + x^4]) + x^3/Sqrt[1 + x^4] + 2*(-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1] - 4*(-1)^(1/4)*E
llipticPi[-I, I*ArcSinh[(-1)^(1/4)*x], -1]

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IntegrateAlgebraic [A]  time = 0.26, size = 37, normalized size = 1.00 \begin {gather*} \frac {\sqrt {1+x^4}}{x}+\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.55, size = 30, normalized size = 0.81 \begin {gather*} \frac {\sqrt {2} x \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + \sqrt {x^{4} + 1}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.33, size = 39, normalized size = 1.05

method	result	size
elliptic	\(\frac {\left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{x}-2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )\right ) \sqrt {2}}{2}\)	\(39\)
trager	\(\frac {\sqrt {x^{4}+1}}{x}-\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )\)	\(46\)
risch	\(\frac {\sqrt {x^{4}+1}}{x}-\frac {2 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {4 \left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{\sqrt {x^{4}+1}}\)	\(123\)
default	\(\frac {\sqrt {x^{4}+1}}{x}-\frac {2 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {2 \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {2 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {2 i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{\left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {4 \left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{\sqrt {x^{4}+1}}\)	\(326\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\left (x^2-1\right )\,\sqrt {x^4+1}}{x^2\,\left (x^2+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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