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

Optimal. Leaf size=64 \[ \frac{3 (c+d x)}{8 d \left (1-(c+d x)^2\right )}+\frac{c+d x}{4 d \left (1-(c+d x)^2\right )^2}+\frac{3 \tanh ^{-1}(c+d x)}{8 d} \]

[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)

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Rubi [A]  time = 0.0223646, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {247, 199, 206} \[ \frac{3 (c+d x)}{8 d \left (1-(c+d x)^2\right )}+\frac{c+d x}{4 d \left (1-(c+d x)^2\right )^2}+\frac{3 \tanh ^{-1}(c+d x)}{8 d} \]

Antiderivative was successfully verified.

[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 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 )^3} \, dx &=\frac{\operatorname{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 \operatorname{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 \operatorname{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]  time = 0.0266841, size = 65, normalized size = 1.02 \[ \frac{-\frac{6 (c+d x)}{(c+d x)^2-1}+\frac{4 (c+d x)}{\left ((c+d x)^2-1\right )^2}-3 \log (-c-d x+1)+3 \log (c+d x+1)}{16 d} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.008, size = 78, normalized size = 1.2 \begin{align*} -{\frac{1}{16\,d \left ( dx+c+1 \right ) ^{2}}}-{\frac{3}{16\,d \left ( dx+c+1 \right ) }}+{\frac{3\,\ln \left ( dx+c+1 \right ) }{16\,d}}+{\frac{1}{16\,d \left ( dx+c-1 \right ) ^{2}}}-{\frac{3}{16\,d \left ( dx+c-1 \right ) }}-{\frac{3\,\ln \left ( dx+c-1 \right ) }{16\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [B]  time = 1.01374, size = 165, normalized size = 2.58 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [B]  time = 1.69552, size = 498, normalized size = 7.78 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Sympy [B]  time = 1.30668, size = 141, normalized size = 2.2 \begin{align*} - \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} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Giac [A]  time = 1.13959, size = 119, normalized size = 1.86 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)