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

Optimal result

Integrand size = 13, antiderivative size = 64 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {c+d x}{4 d \left (1-(c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 d \left (1-(c+d x)^2\right )}+\frac {3 \text {arctanh}(c+d x)}{8 d} \]

[Out]

1/4*(d*x+c)/d/(1-(d*x+c)^2)^2+3/8*(d*x+c)/d/(1-(d*x+c)^2)+3/8*arctanh(d*x+c)/d

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {253, 205, 212} \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {3 \text {arctanh}(c+d x)}{8 d}+\frac {3 (c+d x)}{8 d \left (1-(c+d x)^2\right )}+\frac {c+d x}{4 d \left (1-(c+d x)^2\right )^2} \]

[In]

Int[(1 - (c + d*x)^2)^(-3),x]

[Out]

(c + d*x)/(4*d*(1 - (c + d*x)^2)^2) + (3*(c + d*x))/(8*d*(1 - (c + d*x)^2)) + (3*ArcTanh[c + d*x])/(8*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 )^3} \, dx,x,c+d x\right )}{d} \\ & = \frac {c+d x}{4 d \left (1-(c+d x)^2\right )^2}+\frac {3 \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^2} \, dx,x,c+d x\right )}{4 d} \\ & = \frac {c+d x}{4 d \left (1-(c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 d \left (1-(c+d x)^2\right )}+\frac {3 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,c+d x\right )}{8 d} \\ & = \frac {c+d x}{4 d \left (1-(c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 d \left (1-(c+d x)^2\right )}+\frac {3 \tanh ^{-1}(c+d x)}{8 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {\frac {4 (c+d x)}{\left (-1+(c+d x)^2\right )^2}-\frac {6 (c+d x)}{-1+(c+d x)^2}-3 \log (1-c-d x)+3 \log (1+c+d x)}{16 d} \]

[In]

Integrate[(1 - (c + d*x)^2)^(-3),x]

[Out]

((4*(c + d*x))/(-1 + (c + d*x)^2)^2 - (6*(c + d*x))/(-1 + (c + d*x)^2) - 3*Log[1 - c - d*x] + 3*Log[1 + c + d*
x])/(16*d)

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.22

method	result	size
default	\(\frac {1}{16 d \left (d x +c -1\right )^{2}}-\frac {3}{16 d \left (d x +c -1\right )}-\frac {3 \ln \left (d x +c -1\right )}{16 d}-\frac {1}{16 d \left (d x +c +1\right )^{2}}-\frac {3}{16 d \left (d x +c +1\right )}+\frac {3 \ln \left (d x +c +1\right )}{16 d}\)	\(78\)
risch	\(\frac {-\frac {3 d^{2} x^{3}}{8}-\frac {9 x^{2} c d}{8}+\left (-\frac {9 c^{2}}{8}+\frac {5}{8}\right ) x -\frac {c \left (3 c^{2}-5\right )}{8 d}}{\left (d^{2} x^{2}+2 c d x +c^{2}-1\right )^{2}}-\frac {3 \ln \left (d x +c -1\right )}{16 d}+\frac {3 \ln \left (-d x -c -1\right )}{16 d}\)	\(87\)
norman	\(\frac {\frac {-3 c^{3} d^{3}+5 d^{3} c}{8 d^{4}}+\frac {\left (-9 c^{2} d^{3}+5 d^{3}\right ) x}{8 d^{3}}-\frac {3 d^{2} x^{3}}{8}-\frac {9 x^{2} c d}{8}}{\left (d^{2} x^{2}+2 c d x +c^{2}-1\right )^{2}}-\frac {3 \ln \left (d x +c -1\right )}{16 d}+\frac {3 \ln \left (d x +c +1\right )}{16 d}\)	\(102\)
parallelrisch	\(-\frac {3 \ln \left (d x +c -1\right ) c^{4} d^{3}+6 d^{6} x^{3}+18 x^{2} c \,d^{5}+18 x \,c^{2} d^{4}-6 \ln \left (d x +c -1\right ) c^{2} d^{3}+6 c^{3} d^{3}-10 d^{3} c +12 \ln \left (d x +c -1\right ) x^{3} c \,d^{6}-12 \ln \left (d x +c +1\right ) x^{3} c \,d^{6}+18 \ln \left (d x +c -1\right ) x^{2} c^{2} d^{5}-18 \ln \left (d x +c +1\right ) x^{2} c^{2} d^{5}+12 \ln \left (d x +c -1\right ) x \,c^{3} d^{4}-12 \ln \left (d x +c +1\right ) x \,c^{3} d^{4}-12 \ln \left (d x +c -1\right ) x c \,d^{4}+12 \ln \left (d x +c +1\right ) x c \,d^{4}+3 \ln \left (d x +c -1\right ) d^{3}-3 \ln \left (d x +c +1\right ) d^{3}-10 x \,d^{4}+6 \ln \left (d x +c +1\right ) c^{2} d^{3}+6 \ln \left (d x +c +1\right ) x^{2} d^{5}-3 \ln \left (d x +c +1\right ) c^{4} d^{3}+3 \ln \left (d x +c -1\right ) x^{4} d^{7}-6 \ln \left (d x +c -1\right ) x^{2} d^{5}-3 \ln \left (d x +c +1\right ) x^{4} d^{7}}{16 d^{4} \left (d^{2} x^{2}+2 c d x +c^{2}-1\right )^{2}}\)	\(344\)

[In]

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

[Out]

1/16/d/(d*x+c-1)^2-3/16/d/(d*x+c-1)-3/16/d*ln(d*x+c-1)-1/16/d/(d*x+c+1)^2-3/16/d/(d*x+c+1)+3/16/d*ln(d*x+c+1)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (54) = 108\).

Time = 0.26 (sec) , antiderivative size = 220, normalized size of antiderivative = 3.44 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=-\frac {6 \, d^{3} x^{3} + 18 \, c d^{2} x^{2} + 6 \, c^{3} + 2 \, {\left (9 \, c^{2} - 5\right )} d x - 3 \, {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} x^{2} + c^{4} + 4 \, {\left (c^{3} - c\right )} d x - 2 \, c^{2} + 1\right )} \log \left (d x + c + 1\right ) + 3 \, {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} x^{2} + c^{4} + 4 \, {\left (c^{3} - c\right )} d x - 2 \, c^{2} + 1\right )} \log \left (d x + c - 1\right ) - 10 \, c}{16 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{3} x^{2} + 4 \, {\left (c^{3} - c\right )} d^{2} x + {\left (c^{4} - 2 \, c^{2} + 1\right )} d\right )}} \]

[In]

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

[Out]

-1/16*(6*d^3*x^3 + 18*c*d^2*x^2 + 6*c^3 + 2*(9*c^2 - 5)*d*x - 3*(d^4*x^4 + 4*c*d^3*x^3 + 2*(3*c^2 - 1)*d^2*x^2
+ c^4 + 4*(c^3 - c)*d*x - 2*c^2 + 1)*log(d*x + c + 1) + 3*(d^4*x^4 + 4*c*d^3*x^3 + 2*(3*c^2 - 1)*d^2*x^2 + c^
4 + 4*(c^3 - c)*d*x - 2*c^2 + 1)*log(d*x + c - 1) - 10*c)/(d^5*x^4 + 4*c*d^4*x^3 + 2*(3*c^2 - 1)*d^3*x^2 + 4*(
c^3 - c)*d^2*x + (c^4 - 2*c^2 + 1)*d)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (53) = 106\).

Time = 0.51 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.20 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=- \frac {3 c^{3} + 9 c d^{2} x^{2} - 5 c + 3 d^{3} x^{3} + x \left (9 c^{2} d - 5 d\right )}{8 c^{4} d - 16 c^{2} d + 32 c d^{4} x^{3} + 8 d^{5} x^{4} + 8 d + x^{2} \cdot \left (48 c^{2} d^{3} - 16 d^{3}\right ) + x \left (32 c^{3} d^{2} - 32 c d^{2}\right )} - \frac {\frac {3 \log {\left (x + \frac {3 c - 3}{3 d} \right )}}{16} - \frac {3 \log {\left (x + \frac {3 c + 3}{3 d} \right )}}{16}}{d} \]

[In]

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

[Out]

-(3*c**3 + 9*c*d**2*x**2 - 5*c + 3*d**3*x**3 + x*(9*c**2*d - 5*d))/(8*c**4*d - 16*c**2*d + 32*c*d**4*x**3 + 8*
d**5*x**4 + 8*d + x**2*(48*c**2*d**3 - 16*d**3) + x*(32*c**3*d**2 - 32*c*d**2)) - (3*log(x + (3*c - 3)/(3*d))/
16 - 3*log(x + (3*c + 3)/(3*d))/16)/d

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (54) = 108\).

Time = 0.20 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.91 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=-\frac {3 \, d^{3} x^{3} + 9 \, c d^{2} x^{2} + 3 \, c^{3} + {\left (9 \, c^{2} - 5\right )} d x - 5 \, c}{8 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{3} x^{2} + 4 \, {\left (c^{3} - c\right )} d^{2} x + {\left (c^{4} - 2 \, c^{2} + 1\right )} d\right )}} + \frac {3 \, \log \left (d x + c + 1\right )}{16 \, d} - \frac {3 \, \log \left (d x + c - 1\right )}{16 \, d} \]

[In]

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

[Out]

-1/8*(3*d^3*x^3 + 9*c*d^2*x^2 + 3*c^3 + (9*c^2 - 5)*d*x - 5*c)/(d^5*x^4 + 4*c*d^4*x^3 + 2*(3*c^2 - 1)*d^3*x^2
+ 4*(c^3 - c)*d^2*x + (c^4 - 2*c^2 + 1)*d) + 3/16*log(d*x + c + 1)/d - 3/16*log(d*x + c - 1)/d

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {3 \, \log \left ({\left | d x + c + 1 \right |}\right )}{16 \, d} - \frac {3 \, \log \left ({\left | d x + c - 1 \right |}\right )}{16 \, d} - \frac {3 \, d^{3} x^{3} + 9 \, c d^{2} x^{2} + 9 \, c^{2} d x + 3 \, c^{3} - 5 \, d x - 5 \, c}{8 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )}^{2} d} \]

[In]

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

[Out]

3/16*log(abs(d*x + c + 1))/d - 3/16*log(abs(d*x + c - 1))/d - 1/8*(3*d^3*x^3 + 9*c*d^2*x^2 + 9*c^2*d*x + 3*c^3
- 5*d*x - 5*c)/((d^2*x^2 + 2*c*d*x + c^2 - 1)^2*d)

Mupad [B] (verification not implemented)

Time = 10.06 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.78 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {3\,\mathrm {atanh}\left (c+d\,x\right )}{8\,d}-\frac {x\,\left (\frac {9\,c^2}{8}-\frac {5}{8}\right )-\frac {5\,c-3\,c^3}{8\,d}+\frac {3\,d^2\,x^3}{8}+\frac {9\,c\,d\,x^2}{8}}{c^4-2\,c^2-x^2\,\left (2\,d^2-6\,c^2\,d^2\right )-x\,\left (4\,c\,d-4\,c^3\,d\right )+d^4\,x^4+4\,c\,d^3\,x^3+1} \]

[In]

int(-1/((c + d*x)^2 - 1)^3,x)

[Out]

(3*atanh(c + d*x))/(8*d) - (x*((9*c^2)/8 - 5/8) - (5*c - 3*c^3)/(8*d) + (3*d^2*x^3)/8 + (9*c*d*x^2)/8)/(c^4 -
2*c^2 - x^2*(2*d^2 - 6*c^2*d^2) - x*(4*c*d - 4*c^3*d) + d^4*x^4 + 4*c*d^3*x^3 + 1)