3.44 \(\int \frac{(-1+x^2)^4}{(1+x^2)^5} \, dx\)

Optimal. Leaf size=47 \[ \frac{x \left (1-x^2\right )^3}{4 \left (x^2+1\right )^4}+\frac{3 x \left (1-x^2\right )}{8 \left (x^2+1\right )^2}+\frac{3}{8} \tan ^{-1}(x) \]

[Out]

(x*(1 - x^2)^3)/(4*(1 + x^2)^4) + (3*x*(1 - x^2))/(8*(1 + x^2)^2) + (3*ArcTan[x])/8

________________________________________________________________________________________

Rubi [A]  time = 0.0163951, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {413, 21, 203} \[ \frac{x \left (1-x^2\right )^3}{4 \left (x^2+1\right )^4}+\frac{3 x \left (1-x^2\right )}{8 \left (x^2+1\right )^2}+\frac{3}{8} \tan ^{-1}(x) \]

Antiderivative was successfully verified.

[In]

Int[(-1 + x^2)^4/(1 + x^2)^5,x]

[Out]

(x*(1 - x^2)^3)/(4*(1 + x^2)^4) + (3*x*(1 - x^2))/(8*(1 + x^2)^2) + (3*ArcTan[x])/8

Rule 413

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((a*d - c*b)*x*(a + b*x^n)^
(p + 1)*(c + d*x^n)^(q - 1))/(a*b*n*(p + 1)), x] - Dist[1/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)
^(q - 2)*Simp[c*(a*d - c*b*(n*(p + 1) + 1)) + d*(a*d*(n*(q - 1) + 1) - b*c*(n*(p + q) + 1))*x^n, x], x], x] /;
FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, n, p, q
, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 203

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

Rubi steps

\begin{align*} \int \frac{\left (-1+x^2\right )^4}{\left (1+x^2\right )^5} \, dx &=\frac{x \left (1-x^2\right )^3}{4 \left (1+x^2\right )^4}+\frac{1}{8} \int \frac{\left (-1+x^2\right )^2 \left (6+6 x^2\right )}{\left (1+x^2\right )^4} \, dx\\ &=\frac{x \left (1-x^2\right )^3}{4 \left (1+x^2\right )^4}+\frac{3}{4} \int \frac{\left (-1+x^2\right )^2}{\left (1+x^2\right )^3} \, dx\\ &=\frac{x \left (1-x^2\right )^3}{4 \left (1+x^2\right )^4}+\frac{3 x \left (1-x^2\right )}{8 \left (1+x^2\right )^2}+\frac{3}{16} \int \frac{2+2 x^2}{\left (1+x^2\right )^2} \, dx\\ &=\frac{x \left (1-x^2\right )^3}{4 \left (1+x^2\right )^4}+\frac{3 x \left (1-x^2\right )}{8 \left (1+x^2\right )^2}+\frac{3}{8} \int \frac{1}{1+x^2} \, dx\\ &=\frac{x \left (1-x^2\right )^3}{4 \left (1+x^2\right )^4}+\frac{3 x \left (1-x^2\right )}{8 \left (1+x^2\right )^2}+\frac{3}{8} \tan ^{-1}(x)\\ \end{align*}

Mathematica [A]  time = 0.0124781, size = 41, normalized size = 0.87 \[ \frac{-5 x^7+3 x^5-3 x^3+3 \left (x^2+1\right )^4 \tan ^{-1}(x)+5 x}{8 \left (x^2+1\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(-1 + x^2)^4/(1 + x^2)^5,x]

[Out]

(5*x - 3*x^3 + 3*x^5 - 5*x^7 + 3*(1 + x^2)^4*ArcTan[x])/(8*(1 + x^2)^4)

________________________________________________________________________________________

Maple [A]  time = 0.007, size = 33, normalized size = 0.7 \begin{align*}{\frac{1}{ \left ({x}^{2}+1 \right ) ^{4}} \left ( -{\frac{5\,{x}^{7}}{8}}+{\frac{3\,{x}^{5}}{8}}-{\frac{3\,{x}^{3}}{8}}+{\frac{5\,x}{8}} \right ) }+{\frac{3\,\arctan \left ( x \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-1)^4/(x^2+1)^5,x)

[Out]

(-5/8*x^7+3/8*x^5-3/8*x^3+5/8*x)/(x^2+1)^4+3/8*arctan(x)

________________________________________________________________________________________

Maxima [A]  time = 1.46019, size = 65, normalized size = 1.38 \begin{align*} -\frac{5 \, x^{7} - 3 \, x^{5} + 3 \, x^{3} - 5 \, x}{8 \,{\left (x^{8} + 4 \, x^{6} + 6 \, x^{4} + 4 \, x^{2} + 1\right )}} + \frac{3}{8} \, \arctan \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^4/(x^2+1)^5,x, algorithm="maxima")

[Out]

-1/8*(5*x^7 - 3*x^5 + 3*x^3 - 5*x)/(x^8 + 4*x^6 + 6*x^4 + 4*x^2 + 1) + 3/8*arctan(x)

________________________________________________________________________________________

Fricas [A]  time = 1.53216, size = 159, normalized size = 3.38 \begin{align*} -\frac{5 \, x^{7} - 3 \, x^{5} + 3 \, x^{3} - 3 \,{\left (x^{8} + 4 \, x^{6} + 6 \, x^{4} + 4 \, x^{2} + 1\right )} \arctan \left (x\right ) - 5 \, x}{8 \,{\left (x^{8} + 4 \, x^{6} + 6 \, x^{4} + 4 \, x^{2} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^4/(x^2+1)^5,x, algorithm="fricas")

[Out]

-1/8*(5*x^7 - 3*x^5 + 3*x^3 - 3*(x^8 + 4*x^6 + 6*x^4 + 4*x^2 + 1)*arctan(x) - 5*x)/(x^8 + 4*x^6 + 6*x^4 + 4*x^
2 + 1)

________________________________________________________________________________________

Sympy [A]  time = 0.156707, size = 46, normalized size = 0.98 \begin{align*} - \frac{5 x^{7} - 3 x^{5} + 3 x^{3} - 5 x}{8 x^{8} + 32 x^{6} + 48 x^{4} + 32 x^{2} + 8} + \frac{3 \operatorname{atan}{\left (x \right )}}{8} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(5*x**7 - 3*x**5 + 3*x**3 - 5*x)/(8*x**8 + 32*x**6 + 48*x**4 + 32*x**2 + 8) + 3*atan(x)/8

________________________________________________________________________________________

Giac [A]  time = 1.44375, size = 73, normalized size = 1.55 \begin{align*} \frac{3}{32} \, \pi \mathrm{sgn}\left (x\right ) - \frac{5 \,{\left (x - \frac{1}{x}\right )}^{3} + 12 \, x - \frac{12}{x}}{8 \,{\left ({\left (x - \frac{1}{x}\right )}^{2} + 4\right )}^{2}} + \frac{3}{16} \, \arctan \left (\frac{x^{2} - 1}{2 \, x}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-1)^4/(x^2+1)^5,x, algorithm="giac")

[Out]

3/32*pi*sgn(x) - 1/8*(5*(x - 1/x)^3 + 12*x - 12/x)/((x - 1/x)^2 + 4)^2 + 3/16*arctan(1/2*(x^2 - 1)/x)