\(\int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx\) [94]

Optimal result

Integrand size = 20, antiderivative size = 21 \[ \int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx=\frac {5 x}{2 \left (1-x^2\right )}+\frac {\text {arctanh}(x)}{2} \]

[Out]

5/2*x/(-x^2+1)+1/2*arctanh(x)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {28, 393, 213} \[ \int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx=\frac {\text {arctanh}(x)}{2}+\frac {5 x}{2 \left (1-x^2\right )} \]

[In]

Int[(3 + 2*x^2)/(1 - 2*x^2 + x^4),x]

[Out]

(5*x)/(2*(1 - x^2)) + ArcTanh[x]/2

Rule 28

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/c^p, Int[u*(b/2 + c*x^n)^(2*
p), x], x] /; FreeQ[{a, b, c, n}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 213

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

Rule 393

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d))*x*((a + b*x^n)^(p
+ 1)/(a*b*n*(p + 1))), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x]
/; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

Rubi steps \begin{align*} \text {integral}& = \int \frac {3+2 x^2}{\left (-1+x^2\right )^2} \, dx \\ & = \frac {5 x}{2 \left (1-x^2\right )}-\frac {1}{2} \int \frac {1}{-1+x^2} \, dx \\ & = \frac {5 x}{2 \left (1-x^2\right )}+\frac {1}{2} \tanh ^{-1}(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.29 \[ \int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx=\frac {1}{4} \left (-\frac {10 x}{-1+x^2}-\log (1-x)+\log (1+x)\right ) \]

[In]

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

[Out]

((-10*x)/(-1 + x^2) - Log[1 - x] + Log[1 + x])/4

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.14

method	result	size
norman	\(-\frac {5 x}{2 \left (x^{2}-1\right )}-\frac {\ln \left (x -1\right )}{4}+\frac {\ln \left (x +1\right )}{4}\)	\(24\)
risch	\(-\frac {5 x}{2 \left (x^{2}-1\right )}-\frac {\ln \left (x -1\right )}{4}+\frac {\ln \left (x +1\right )}{4}\)	\(24\)
default	\(-\frac {5}{4 \left (x +1\right )}+\frac {\ln \left (x +1\right )}{4}-\frac {5}{4 \left (x -1\right )}-\frac {\ln \left (x -1\right )}{4}\)	\(28\)
parallelrisch	\(-\frac {\ln \left (x -1\right ) x^{2}-\ln \left (x +1\right ) x^{2}-\ln \left (x -1\right )+\ln \left (x +1\right )+10 x}{4 \left (x^{2}-1\right )}\)	\(41\)

[In]

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

[Out]

-5/2*x/(x^2-1)-1/4*ln(x-1)+1/4*ln(x+1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 34 vs. \(2 (15) = 30\).

Time = 0.25 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.62 \[ \int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx=\frac {{\left (x^{2} - 1\right )} \log \left (x + 1\right ) - {\left (x^{2} - 1\right )} \log \left (x - 1\right ) - 10 \, x}{4 \, {\left (x^{2} - 1\right )}} \]

[In]

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

[Out]

1/4*((x^2 - 1)*log(x + 1) - (x^2 - 1)*log(x - 1) - 10*x)/(x^2 - 1)

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.05 \[ \int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx=- \frac {5 x}{2 x^{2} - 2} - \frac {\log {\left (x - 1 \right )}}{4} + \frac {\log {\left (x + 1 \right )}}{4} \]

[In]

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

[Out]

-5*x/(2*x**2 - 2) - log(x - 1)/4 + log(x + 1)/4

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.10 \[ \int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx=-\frac {5 \, x}{2 \, {\left (x^{2} - 1\right )}} + \frac {1}{4} \, \log \left (x + 1\right ) - \frac {1}{4} \, \log \left (x - 1\right ) \]

[In]

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

[Out]

-5/2*x/(x^2 - 1) + 1/4*log(x + 1) - 1/4*log(x - 1)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.19 \[ \int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx=-\frac {5 \, x}{2 \, {\left (x^{2} - 1\right )}} + \frac {1}{4} \, \log \left ({\left | x + 1 \right |}\right ) - \frac {1}{4} \, \log \left ({\left | x - 1 \right |}\right ) \]

[In]

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

[Out]

-5/2*x/(x^2 - 1) + 1/4*log(abs(x + 1)) - 1/4*log(abs(x - 1))

Mupad [B] (verification not implemented)

Time = 13.45 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.81 \[ \int \frac {3+2 x^2}{1-2 x^2+x^4} \, dx=\frac {\mathrm {atanh}\left (x\right )}{2}-\frac {5\,x}{2\,\left (x^2-1\right )} \]

[In]

int((2*x^2 + 3)/(x^4 - 2*x^2 + 1),x)

[Out]

atanh(x)/2 - (5*x)/(2*(x^2 - 1))