\(\int \frac {-1+2 x^4}{(1+2 x^2+2 x^4) \sqrt {1+3 x^2+2 x^4}} \, dx\) [200]

Optimal result

Integrand size = 38, antiderivative size = 21 \[ \int \frac {-1+2 x^4}{\left (1+2 x^2+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}} \, dx=-\text {arctanh}\left (\frac {x}{\sqrt {1+3 x^2+2 x^4}}\right ) \]

[Out]

-arctanh(x/(2*x^4+3*x^2+1)^(1/2))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2137, 212} \[ \int \frac {-1+2 x^4}{\left (1+2 x^2+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}} \, dx=-\text {arctanh}\left (\frac {x}{\sqrt {2 x^4+3 x^2+1}}\right ) \]

[In]

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

[Out]

-ArcTanh[x/Sqrt[1 + 3*x^2 + 2*x^4]]

Rule 212

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

Rule 2137

Int[((u_)*((A_) + (B_.)*(x_)^4))/Sqrt[v_], x_Symbol] :> With[{a = Coeff[v, x, 0], b = Coeff[v, x, 2], c = Coef
f[v, x, 4], d = Coeff[1/u, x, 0], e = Coeff[1/u, x, 2], f = Coeff[1/u, x, 4]}, Dist[A, Subst[Int[1/(d - (b*d -
a*e)*x^2), x], x, x/Sqrt[v]], x] /; EqQ[a*B + A*c, 0] && EqQ[c*d - a*f, 0]] /; FreeQ[{A, B}, x] && PolyQ[v, x
^2, 2] && PolyQ[1/u, x^2, 2]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,\frac {x}{\sqrt {1+3 x^2+2 x^4}}\right ) \\ & = -\text {arctanh}\left (\frac {x}{\sqrt {1+3 x^2+2 x^4}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int \frac {-1+2 x^4}{\left (1+2 x^2+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}} \, dx=-\text {arctanh}\left (\frac {x}{\sqrt {1+3 x^2+2 x^4}}\right ) \]

[In]

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

[Out]

-ArcTanh[x/Sqrt[1 + 3*x^2 + 2*x^4]]

Maple [A] (verified)

Time = 3.69 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05

method	result	size
elliptic	\(-\operatorname {arctanh}\left (\frac {\sqrt {2 x^{4}+3 x^{2}+1}}{x}\right )\)	\(22\)
trager	\(\frac {\ln \left (-\frac {-2 x^{4}+2 x \sqrt {2 x^{4}+3 x^{2}+1}-4 x^{2}-1}{2 x^{4}+2 x^{2}+1}\right )}{2}\)	\(49\)
default	\(\frac {\operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}+2 x \sqrt {2}-3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}-2 x \sqrt {2}+3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )}{2}\)	\(92\)
pseudoelliptic	\(\frac {\operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}+2 x \sqrt {2}-3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )}{2}-\frac {\operatorname {arctanh}\left (\frac {\left (2 x^{2}+\sqrt {2}\right ) \sqrt {\sqrt {2}-1}-2 x \sqrt {2}+3 x}{\sqrt {2 x^{4}+3 x^{2}+1}}\right )}{2}\)	\(92\)

[In]

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

[Out]

-arctanh((2*x^4+3*x^2+1)^(1/2)/x)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 47 vs. \(2 (19) = 38\).

Time = 0.26 (sec) , antiderivative size = 47, normalized size of antiderivative = 2.24 \[ \int \frac {-1+2 x^4}{\left (1+2 x^2+2 x^4\right ) \sqrt {1+3 x^2+2 x^4}} \, dx=\frac {1}{2} \, \log \left (\frac {2 \, x^{4} + 4 \, x^{2} - 2 \, \sqrt {2 \, x^{4} + 3 \, x^{2} + 1} x + 1}{2 \, x^{4} + 2 \, x^{2} + 1}\right ) \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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