\(\int \frac {-1+x^2}{(1+x^2) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx\) [574]

Optimal result

Integrand size = 42, antiderivative size = 44 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx=-\frac {\left (\left (1+x^4\right )^5\right )^{9/10} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2} \left (1+x^4\right )^{9/2}} \]

[Out]

-1/2*((x^4+1)^5)^(9/10)*arctan(2^(1/2)*x/(x^4+1)^(1/2))*2^(1/2)/(x^4+1)^(9/2)

Rubi [A] (warning: unable to verify)

Time = 0.17 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.55, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {6820, 1972, 1713, 209} \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx=-\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {x^4+1}}\right )}{\sqrt {2}} \]

[In]

Int[(-1 + x^2)/((1 + x^2)*(1 + 5*x^4 + 10*x^8 + 10*x^12 + 5*x^16 + x^20)^(1/10)),x]

[Out]

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

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 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 1972

Int[(u_.)*((c_.)*((a_.) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(a + b*x^n)
^(p*q)], Int[u*(a + b*x^n)^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] && GeQ[a, 0]

Rule 6820

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rubi steps \begin{align*} \text {integral}& = \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{\left (1+x^4\right )^5}} \, dx \\ & = \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx \\ & = -\text {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\frac {x}{\sqrt {1+x^4}}\right ) \\ & = -\frac {\arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx=-\frac {\sqrt {1+x^4} \arctan \left (\frac {\sqrt {2} x}{\sqrt {1+x^4}}\right )}{\sqrt {2} \sqrt [10]{\left (1+x^4\right )^5}} \]

[In]

Integrate[(-1 + x^2)/((1 + x^2)*(1 + 5*x^4 + 10*x^8 + 10*x^12 + 5*x^16 + x^20)^(1/10)),x]

[Out]

-((Sqrt[1 + x^4]*ArcTan[(Sqrt[2]*x)/Sqrt[1 + x^4]])/(Sqrt[2]*((1 + x^4)^5)^(1/10)))

Maple [F]

\[\int \frac {x^{2}-1}{\left (x^{2}+1\right ) \left (x^{20}+5 x^{16}+10 x^{12}+10 x^{8}+5 x^{4}+1\right )^{\frac {1}{10}}}d x\]

[In]

int((x^2-1)/(x^2+1)/(x^20+5*x^16+10*x^12+10*x^8+5*x^4+1)^(1/10),x)

[Out]

int((x^2-1)/(x^2+1)/(x^20+5*x^16+10*x^12+10*x^8+5*x^4+1)^(1/10),x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.02 \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx=-\frac {1}{2} \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (x^{20} + 5 \, x^{16} + 10 \, x^{12} + 10 \, x^{8} + 5 \, x^{4} + 1\right )}^{\frac {1}{10}} x}{x^{4} + 1}\right ) \]

[In]

integrate((x^2-1)/(x^2+1)/(x^20+5*x^16+10*x^12+10*x^8+5*x^4+1)^(1/10),x, algorithm="fricas")

[Out]

-1/2*sqrt(2)*arctan(sqrt(2)*(x^20 + 5*x^16 + 10*x^12 + 10*x^8 + 5*x^4 + 1)^(1/10)*x/(x^4 + 1))

Sympy [F]

\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx=\int \frac {\left (x - 1\right ) \left (x + 1\right )}{\left (x^{2} + 1\right ) \sqrt [10]{\left (x^{4} + 1\right )^{5}}}\, dx \]

[In]

integrate((x**2-1)/(x**2+1)/(x**20+5*x**16+10*x**12+10*x**8+5*x**4+1)**(1/10),x)

[Out]

Integral((x - 1)*(x + 1)/((x**2 + 1)*((x**4 + 1)**5)**(1/10)), x)

Maxima [F]

\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{20} + 5 \, x^{16} + 10 \, x^{12} + 10 \, x^{8} + 5 \, x^{4} + 1\right )}^{\frac {1}{10}} {\left (x^{2} + 1\right )}} \,d x } \]

[In]

integrate((x^2-1)/(x^2+1)/(x^20+5*x^16+10*x^12+10*x^8+5*x^4+1)^(1/10),x, algorithm="maxima")

[Out]

integrate((x^2 - 1)/((x^20 + 5*x^16 + 10*x^12 + 10*x^8 + 5*x^4 + 1)^(1/10)*(x^2 + 1)), x)

Giac [F]

\[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx=\int { \frac {x^{2} - 1}{{\left (x^{20} + 5 \, x^{16} + 10 \, x^{12} + 10 \, x^{8} + 5 \, x^{4} + 1\right )}^{\frac {1}{10}} {\left (x^{2} + 1\right )}} \,d x } \]

[In]

integrate((x^2-1)/(x^2+1)/(x^20+5*x^16+10*x^12+10*x^8+5*x^4+1)^(1/10),x, algorithm="giac")

[Out]

integrate((x^2 - 1)/((x^20 + 5*x^16 + 10*x^12 + 10*x^8 + 5*x^4 + 1)^(1/10)*(x^2 + 1)), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {-1+x^2}{\left (1+x^2\right ) \sqrt [10]{1+5 x^4+10 x^8+10 x^{12}+5 x^{16}+x^{20}}} \, dx=\int \frac {x^2-1}{\left (x^2+1\right )\,{\left (x^{20}+5\,x^{16}+10\,x^{12}+10\,x^8+5\,x^4+1\right )}^{1/10}} \,d x \]

[In]

int((x^2 - 1)/((x^2 + 1)*(5*x^4 + 10*x^8 + 10*x^12 + 5*x^16 + x^20 + 1)^(1/10)),x)

[Out]

int((x^2 - 1)/((x^2 + 1)*(5*x^4 + 10*x^8 + 10*x^12 + 5*x^16 + x^20 + 1)^(1/10)), x)