3.93 \(\int \frac{1}{(1-(c+d x)^2)^2} \, dx\)

Optimal. Leaf size=39 \[ \frac{c+d x}{2 d \left (1-(c+d x)^2\right )}+\frac{\tanh ^{-1}(c+d x)}{2 d} \]

[Out]

(c + d*x)/(2*d*(1 - (c + d*x)^2)) + ArcTanh[c + d*x]/(2*d)

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Rubi [A]  time = 0.0126135, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {247, 199, 206} \[ \frac{c+d x}{2 d \left (1-(c+d x)^2\right )}+\frac{\tanh ^{-1}(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(1 - (c + d*x)^2)^(-2),x]

[Out]

(c + d*x)/(2*d*(1 - (c + d*x)^2)) + ArcTanh[c + d*x]/(2*d)

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 206

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

Rubi steps

\begin{align*} \int \frac{1}{\left (1-(c+d x)^2\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (1-x^2\right )^2} \, dx,x,c+d x\right )}{d}\\ &=\frac{c+d x}{2 d \left (1-(c+d x)^2\right )}+\frac{\operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,c+d x\right )}{2 d}\\ &=\frac{c+d x}{2 d \left (1-(c+d x)^2\right )}+\frac{\tanh ^{-1}(c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0197483, size = 45, normalized size = 1.15 \[ \frac{-\frac{2 (c+d x)}{(c+d x)^2-1}-\log (-c-d x+1)+\log (c+d x+1)}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(1 - (c + d*x)^2)^(-2),x]

[Out]

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

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Maple [A]  time = 0.008, size = 52, normalized size = 1.3 \begin{align*} -{\frac{1}{4\,d \left ( dx+c+1 \right ) }}+{\frac{\ln \left ( dx+c+1 \right ) }{4\,d}}-{\frac{1}{4\,d \left ( dx+c-1 \right ) }}-{\frac{\ln \left ( dx+c-1 \right ) }{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(1-(d*x+c)^2)^2,x)

[Out]

-1/4/d/(d*x+c+1)+1/4/d*ln(d*x+c+1)-1/4/d/(d*x+c-1)-1/4/d*ln(d*x+c-1)

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Maxima [A]  time = 1.05549, size = 76, normalized size = 1.95 \begin{align*} -\frac{d x + c}{2 \,{\left (d^{3} x^{2} + 2 \, c d^{2} x +{\left (c^{2} - 1\right )} d\right )}} + \frac{\log \left (d x + c + 1\right )}{4 \, d} - \frac{\log \left (d x + c - 1\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.71631, size = 208, normalized size = 5.33 \begin{align*} -\frac{2 \, d x -{\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )} \log \left (d x + c + 1\right ) +{\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )} \log \left (d x + c - 1\right ) + 2 \, c}{4 \,{\left (d^{3} x^{2} + 2 \, c d^{2} x +{\left (c^{2} - 1\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.595988, size = 53, normalized size = 1.36 \begin{align*} - \frac{c + d x}{2 c^{2} d + 4 c d^{2} x + 2 d^{3} x^{2} - 2 d} + \frac{- \frac{\log{\left (x + \frac{c - 1}{d} \right )}}{4} + \frac{\log{\left (x + \frac{c + 1}{d} \right )}}{4}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(1-(d*x+c)**2)**2,x)

[Out]

-(c + d*x)/(2*c**2*d + 4*c*d**2*x + 2*d**3*x**2 - 2*d) + (-log(x + (c - 1)/d)/4 + log(x + (c + 1)/d)/4)/d

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Giac [A]  time = 1.15003, size = 76, normalized size = 1.95 \begin{align*} \frac{\log \left ({\left | d x + c + 1 \right |}\right )}{4 \, d} - \frac{\log \left ({\left | d x + c - 1 \right |}\right )}{4 \, d} - \frac{d x + c}{2 \,{\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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