\(\int \frac {(-1+x^2) \sqrt {1+x^4}}{(1+x^2)^3} \, dx\) [607]

Optimal result

Integrand size = 22, antiderivative size = 48 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx=-\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.50, number of steps used = 20, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {1735, 226, 1238, 1711, 1727, 1210, 1715, 1713, 209, 1225} \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx=-\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )+\frac {3 \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{2 \sqrt {2}}-\frac {\sqrt {x^4+1} x}{2 \left (x^2+1\right )^2} \]

[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 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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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 1238

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
)/Sqrt[a + c*x^4])*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], x],
x] /; FreeQ[{a, c, d, e}, x] && ILtQ[q, -1]

Rule 1711

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)/Sqrt[a
+ c*x^4])*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], x], x]] /; FreeQ[{a, c, d, e}, x] && PolyQ[P4x, x^2] && LeQ
[Expon[P4x, x], 4] && NeQ[c*d^2 + a*e^2, 0] && ILtQ[q, -1]

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]

Rule 1715

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 1727

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 1735

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*} \text {integral}& = \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}} \operatorname {EllipticF}\left (2 \arctan (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}} \operatorname {EllipticF}\left (2 \arctan (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 \text {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} \arctan \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 \arctan (x)\left |\frac {1}{2}\right .\right )}{2 \sqrt {1+x^4}}-\frac {\left (1+x^2\right ) \sqrt {\frac {1+x^4}{\left (1+x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan (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 \text {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 \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2}}-\sqrt {2} \arctan \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}} \operatorname {EllipticF}\left (2 \arctan (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 \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2}}-\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )-\frac {3}{2} \text {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 \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}}-\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx=-\frac {x \sqrt {1+x^4}}{2 \left (1+x^2\right )^2}-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{2 \sqrt {2}} \]

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

Maple [A] (verified)

Time = 1.83 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {x \sqrt {x^{4}+1}}{2 \left (x^{2}+1\right )^{2}}-\frac {\arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) \sqrt {2}}{4}\)	\(37\)
default	\(\frac {-\sqrt {2}\, \left (x^{2}+1\right )^{2} \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )-2 \sqrt {x^{4}+1}\, x}{4 \left (x^{2}+1\right )^{2}}\)	\(46\)
pseudoelliptic	\(\frac {-\sqrt {2}\, \left (x^{2}+1\right )^{2} \arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right )-2 \sqrt {x^{4}+1}\, x}{4 \left (x^{2}+1\right )^{2}}\)	\(46\)
trager	\(-\frac {x \sqrt {x^{4}+1}}{2 \left (x^{2}+1\right )^{2}}+\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x +\sqrt {x^{4}+1}}{x^{2}+1}\right )}{4}\)	\(52\)
elliptic	\(\frac {\left (-\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{4 \left (\frac {x^{4}+1}{2 x^{2}}+1\right ) x}+\frac {\arctan \left (\frac {\sqrt {2}\, \sqrt {x^{4}+1}}{2 x}\right )}{2}\right ) \sqrt {2}}{2}\)	\(54\)

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx=-\frac {\sqrt {2} {\left (x^{4} + 2 \, x^{2} + 1\right )} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + 2 \, \sqrt {x^{4} + 1} x}{4 \, {\left (x^{4} + 2 \, x^{2} + 1\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right ) \sqrt {x^{4} + 1}}{\left (x^{2} + 1\right )^{3}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx=\int { \frac {\sqrt {x^{4} + 1} {\left (x^{2} - 1\right )}}{{\left (x^{2} + 1\right )}^{3}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx=\int { \frac {\sqrt {x^{4} + 1} {\left (x^{2} - 1\right )}}{{\left (x^{2} + 1\right )}^{3}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{\left (1+x^2\right )^3} \, dx=\int \frac {\left (x^2-1\right )\,\sqrt {x^4+1}}{{\left (x^2+1\right )}^3} \,d x \]

[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)