\(\int \frac {(-1+x^2) \sqrt {1+x^4}}{x^2 (1+x^2)} \, dx\) [457]

Optimal result

Integrand size = 25, antiderivative size = 37 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^2 \left (1+x^2\right )} \, dx=\frac {\sqrt {1+x^4}}{x}+\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {6857, 283, 311, 226, 1210, 1223, 1225, 1713, 209} \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^2 \left (1+x^2\right )} \, dx=\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )+\frac {\sqrt {x^4+1}}{x} \]

[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 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 283

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 311

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 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 1223

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 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]

Rule 6857

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^2 \left (1+x^2\right )} \, dx=\frac {\sqrt {1+x^4}}{x}+\sqrt {2} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right ) \]

[In]

Integrate[((-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]]

Maple [A] (verified)

Time = 3.09 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81

method	result	size
risch	\(\frac {\sqrt {x^{4}+1}}{x}+\arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) \sqrt {2}\)	\(30\)
default	\(\frac {\arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) \sqrt {2}\, x +\sqrt {x^{4}+1}}{x}\)	\(31\)
pseudoelliptic	\(\frac {\arctan \left (\frac {\sqrt {2}\, x}{\sqrt {x^{4}+1}}\right ) \sqrt {2}\, x +\sqrt {x^{4}+1}}{x}\)	\(31\)
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	\(\frac {\sqrt {x^{4}+1}}{x}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) \ln \left (-\frac {\operatorname {RootOf}\left (\textit {\_Z}^{2}+2\right ) x -\sqrt {x^{4}+1}}{x^{2}+1}\right )\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \frac {\left (-1+x^2\right ) \sqrt {1+x^4}}{x^2 \left (1+x^2\right )} \, dx=\frac {\sqrt {2} x \arctan \left (\frac {\sqrt {2} x}{\sqrt {x^{4} + 1}}\right ) + \sqrt {x^{4} + 1}}{x} \]

[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

Sympy [F]

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

[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)

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[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)