3.7.7 \(\int \frac {(-1+x^2) \sqrt {1+x^4}}{(1+x^2)^3} \, dx\)

Optimal. Leaf size=48 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}-\frac {\sqrt {x^4+1} x}{2 \left (x^2+1\right )^2} \]

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Rubi [A]  time = 0.36, antiderivative size = 72, normalized size of antiderivative = 1.50, number of steps used = 20, number of rules used = 10, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {1721, 220, 1224, 1697, 1713, 1196, 1701, 1699, 203, 1211} \begin {gather*} -\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}-\frac {\sqrt {x^4+1} x}{2 \left (x^2+1\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((-1 + x^2)*Sqrt[1 + x^4])/(1 + x^2)^3,x]

[Out]

-1/2*(x*Sqrt[1 + x^4])/(1 + x^2)^2 + (3*ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]])/(2*Sqrt[2]) - 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 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 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 1224

Int[((d_) + (e_.)*(x_)^2)^(q_)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> -Simp[(e^2*x*(d + e*x^2)^(q + 1)*Sqrt[a
+ c*x^4])/(2*d*(q + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*d*(q + 1)*(c*d^2 + a*e^2)), Int[((d + e*x^2)^(q + 1)*
Simp[a*e^2*(2*q + 3) + 2*c*d^2*(q + 1) - 2*e*c*d*(q + 1)*x^2 + c*e^2*(2*q + 5)*x^4, x])/Sqrt[a + c*x^4], x], x
] /; FreeQ[{a, c, d, e}, x] && ILtQ[q, -1]

Rule 1697

Int[((P4x_)*((d_) + (e_.)*(x_)^2)^(q_))/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{A = Coeff[P4x, x, 0], B
= Coeff[P4x, x, 2], C = Coeff[P4x, x, 4]}, -Simp[((C*d^2 - B*d*e + A*e^2)*x*(d + e*x^2)^(q + 1)*Sqrt[a + c*x^4
])/(2*d*(q + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*d*(q + 1)*(c*d^2 + a*e^2)), Int[((d + e*x^2)^(q + 1)*Simp[a*d
*(C*d - B*e) + A*(a*e^2*(2*q + 3) + 2*c*d^2*(q + 1)) + 2*d*(B*c*d - A*c*e + a*C*e)*(q + 1)*x^2 + c*(C*d^2 - B*
d*e + A*e^2)*(2*q + 5)*x^4, x])/Sqrt[a + c*x^4], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[P4x, x^2] && LeQ[E
xpon[P4x, x], 4] && NeQ[c*d^2 + a*e^2, 0] && ILtQ[q, -1]

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 1701

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

Rule 1713

Int[(P4x_)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[{A = Coeff[P4x, x, 0], B = Coe
ff[P4x, x, 2], C = Coeff[P4x, x, 4]}, -Dist[C/e^2, Int[(d - e*x^2)/Sqrt[a + c*x^4], x], x] + Dist[1/e^2, Int[(
C*d^2 + A*e^2 + B*e^2*x^2)/((d + e*x^2)*Sqrt[a + c*x^4]), x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[P4x, x^2,
2] && NeQ[c*d^2 + a*e^2, 0] && EqQ[c*d^2 - a*e^2, 0]

Rule 1721

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_), x_Symbol] :> Int[ExpandIntegrand[1/Sqrt[a +
c*x^4], Px*(d + e*x^2)^q*(a + c*x^4)^(p + 1/2), x], x] /; FreeQ[{a, c, d, e}, x] && PolyQ[Px, x^2] && NeQ[c*d^
2 + a*e^2, 0] && IntegerQ[p + 1/2] && IntegerQ[q]

Rubi steps

\begin {align*} \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx &=\int \left (\frac {1}{\sqrt {1+x^4}}-\frac {4}{\left (1+x^2\right )^3 \sqrt {1+x^4}}+\frac {6}{\left (1+x^2\right )^2 \sqrt {1+x^4}}-\frac {4}{\left (1+x^2\right ) \sqrt {1+x^4}}\right ) \, dx\\ &=-\left (4 \int \frac {1}{\left (1+x^2\right )^3 \sqrt {1+x^4}} \, dx\right )-4 \int \frac {1}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx+6 \int \frac {1}{\left (1+x^2\right )^2 \sqrt {1+x^4}} \, dx+\int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}+\frac {3 x \sqrt {1+x^4}}{2 \left (1+x^2\right )}+\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}+\frac {1}{2} \int \frac {-7+4 x^2-x^4}{\left (1+x^2\right )^2 \sqrt {1+x^4}} \, dx-\frac {3}{2} \int \frac {-3+2 x^2+x^4}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-2 \int \frac {1}{\sqrt {1+x^4}} \, dx-2 \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx\\ &=-\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {1}{8} \int \frac {16-20 x^2-12 x^4}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx+\frac {3}{2} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx-\frac {3}{2} \int \frac {-2+2 x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-2 \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )\\ &=-\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}-\frac {3 x \sqrt {1+x^4}}{2 \left (1+x^2\right )}-\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )+\frac {3 \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 )}{2 \sqrt {1+x^4}}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {1}{8} \int \frac {4-20 x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx-\frac {3}{2} \int \frac {1-x^2}{\sqrt {1+x^4}} \, dx+3 \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )\\ &=-\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2}}-\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} F\left (2 \tan ^{-1}(x)|\frac {1}{2}\right )}{2 \sqrt {1+x^4}}-\frac {3}{2} \int \frac {1-x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx+\int \frac {1}{\sqrt {1+x^4}} \, dx\\ &=-\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2}}-\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )-\frac {3}{2} \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right )\\ &=-\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}}-\sqrt {2} \tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )\\ \end {align*}

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Mathematica [C]  time = 0.25, size = 68, normalized size = 1.42 \begin {gather*} -\frac {\sqrt {x^4+1} x}{2 \left (x^2+1\right )^2}-\frac {1}{2} \sqrt [4]{-1} F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} x\right )\right |-1\right )+\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])/(1 + x^2)^3,x]

[Out]

-1/2*(x*Sqrt[1 + x^4])/(1 + x^2)^2 - ((-1)^(1/4)*EllipticF[I*ArcSinh[(-1)^(1/4)*x], -1])/2 + (-1)^(1/4)*Ellipt
icPi[-I, I*ArcSinh[(-1)^(1/4)*x], -1]

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IntegrateAlgebraic [A]  time = 0.37, size = 48, normalized size = 1.00 \begin {gather*} -\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}-\frac {\tan ^{-1}\left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((-1 + x^2)*Sqrt[1 + x^4])/(1 + x^2)^3,x]

[Out]

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

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fricas [A]  time = 0.49, size = 52, normalized size = 1.08 \begin {gather*} -\frac {\sqrt {2} {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + 2 \, \sqrt {x^{4} + 1} x}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)*(x^4+1)^(1/2)/(x^2+1)^3,x, algorithm="fricas")

[Out]

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

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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 )}^{3}}\,{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+1)^3,x, algorithm="giac")

[Out]

integrate(sqrt(x^4 + 1)*(x^2 - 1)/(x^2 + 1)^3, x)

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maple [C]  time = 0.27, size = 53, normalized size = 1.10

method	result	size
trager	\(-\frac {x \sqrt {x^{4}+1}}{2 \left (x^{2}+1\right )^{2}}-\frac {\RootOf \left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {-\RootOf \left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{4}\)	\(53\)
elliptic	\(\frac {\left (-\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{4 \left (\frac {x^{4}+1}{2 x^{2}}+1\right ) x}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{2}\right ) \sqrt {2}}{2}\)	\(54\)
risch	\(-\frac {x \sqrt {x^{4}+1}}{2 \left (x^{2}+1\right )^{2}}+\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{\sqrt {x^{4}+1}}\)	\(128\)
default	\(\frac {\sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \left (\EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )-\EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {x \sqrt {x^{4}+1}}{2 \left (x^{2}+1\right )^{2}}+\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticF \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}-\frac {i \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticE \left (x \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ), i\right )}{2 \left (\frac {\sqrt {2}}{2}+\frac {i \sqrt {2}}{2}\right ) \sqrt {x^{4}+1}}+\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {-i x^{2}+1}\, \sqrt {i x^{2}+1}\, \EllipticPi \left (\left (-1\right )^{\frac {1}{4}} x , i, -\sqrt {-i}\, \left (-1\right )^{\frac {3}{4}}\right )}{\sqrt {x^{4}+1}}\)	\(331\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)*(x^4+1)^(1/2)/(x^2+1)^3,x,method=_RETURNVERBOSE)

[Out]

-1/2*x*(x^4+1)^(1/2)/(x^2+1)^2-1/4*RootOf(_Z^2+2)*ln((-RootOf(_Z^2+2)*x+(x^4+1)^(1/2))/(x^2+1))

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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 )}^{3}}\,{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+1)^3,x, algorithm="maxima")

[Out]

integrate(sqrt(x^4 + 1)*(x^2 - 1)/(x^2 + 1)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {\left (x^2-1\right )\,\sqrt {x^4+1}}{{\left (x^2+1\right )}^3} \,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 + 1)^3,x)

[Out]

int(((x^2 - 1)*(x^4 + 1)^(1/2))/(x^2 + 1)^3, 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}}{\left (x^{2} + 1\right )^{3}}\, 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+1)**3,x)

[Out]

Integral((x - 1)*(x + 1)*sqrt(x**4 + 1)/(x**2 + 1)**3, x)

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