\(\int \frac {1}{(1-(c+d x)^2)^2} \, dx\) [93]

Optimal result

Integrand size = 13, antiderivative size = 39 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^2} \, dx=\frac {c+d x}{2 d \left (1-(c+d x)^2\right )}+\frac {\text {arctanh}(c+d x)}{2 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {253, 205, 212} \[ \int \frac {1}{\left (1-(c+d x)^2\right )^2} \, dx=\frac {\text {arctanh}(c+d x)}{2 d}+\frac {c+d x}{2 d \left (1-(c+d x)^2\right )} \]

[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 205

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] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 253

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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 {\text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^2} \, dx=\frac {-\frac {2 (c+d x)}{-1+(c+d x)^2}-\log (1-c-d x)+\log (1+c+d x)}{4 d} \]

[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)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.33

method	result	size
default	\(-\frac {1}{4 d \left (d x +c -1\right )}-\frac {\ln \left (d x +c -1\right )}{4 d}-\frac {1}{4 d \left (d x +c +1\right )}+\frac {\ln \left (d x +c +1\right )}{4 d}\)	\(52\)
norman	\(\frac {-\frac {c}{2 d}-\frac {x}{2}}{d^{2} x^{2}+2 c d x +c^{2}-1}-\frac {\ln \left (d x +c -1\right )}{4 d}+\frac {\ln \left (d x +c +1\right )}{4 d}\)	\(56\)
risch	\(\frac {-\frac {c}{2 d}-\frac {x}{2}}{d^{2} x^{2}+2 c d x +c^{2}-1}-\frac {\ln \left (d x +c -1\right )}{4 d}+\frac {\ln \left (-d x -c -1\right )}{4 d}\)	\(59\)
parallelrisch	\(-\frac {\ln \left (d x +c -1\right ) x^{2} d^{3}-\ln \left (d x +c +1\right ) x^{2} d^{3}+2 \ln \left (d x +c -1\right ) x c \,d^{2}-2 \ln \left (d x +c +1\right ) x c \,d^{2}+\ln \left (d x +c -1\right ) c^{2} d -\ln \left (d x +c +1\right ) c^{2} d +2 d^{2} x -\ln \left (d x +c -1\right ) d +\ln \left (d x +c +1\right ) d +2 c d}{4 d^{2} \left (d^{2} x^{2}+2 c d x +c^{2}-1\right )}\)	\(137\)

[In]

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

[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)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 85 vs. \(2 (33) = 66\).

Time = 0.29 (sec) , antiderivative size = 85, normalized size of antiderivative = 2.18 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^2} \, dx=-\frac {2 \, d x - {\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )} \log \left (d x + c + 1\right ) + {\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )} \log \left (d x + c - 1\right ) + 2 \, c}{4 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + {\left (c^{2} - 1\right )} d\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^2} \, dx=\frac {- c - d x}{2 c^{2} d + 4 c d^{2} x + 2 d^{3} x^{2} - 2 d} + \frac {- \frac {\log {\left (x + \frac {c - 1}{d} \right )}}{4} + \frac {\log {\left (x + \frac {c + 1}{d} \right )}}{4}}{d} \]

[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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^2} \, dx=-\frac {d x + c}{2 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + {\left (c^{2} - 1\right )} d\right )}} + \frac {\log \left (d x + c + 1\right )}{4 \, d} - \frac {\log \left (d x + c - 1\right )}{4 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.44 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^2} \, dx=\frac {\log \left ({\left | d x + c + 1 \right |}\right )}{4 \, d} - \frac {\log \left ({\left | d x + c - 1 \right |}\right )}{4 \, d} - \frac {d x + c}{2 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )} d} \]

[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)

Mupad [B] (verification not implemented)

Time = 10.36 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^2} \, dx=\frac {\mathrm {atanh}\left (c+d\,x\right )}{2\,d}-\frac {\frac {x}{2}+\frac {c}{2\,d}}{c^2+2\,c\,d\,x+d^2\,x^2-1} \]

[In]

int(1/((c + d*x)^2 - 1)^2,x)

[Out]

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