Integrand size = 13, antiderivative size = 64 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {c+d x}{4 d \left (1-(c+d x)^2\right )^2}+\frac {3 (c+d x)}{8 d \left (1-(c+d x)^2\right )}+\frac {3 \text {arctanh}(c+d x)}{8 d} \] Output:
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
Time = 0.04 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {\frac {4 (c+d x)}{\left (-1+(c+d x)^2\right )^2}-\frac {6 (c+d x)}{-1+(c+d x)^2}-3 \log (1-c-d x)+3 \log (1+c+d x)}{16 d} \] Input:
Integrate[(1 - (c + d*x)^2)^(-3),x]
Output:
((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)
Time = 0.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {239, 215, 215, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx\) |
\(\Big \downarrow \) 239 |
\(\displaystyle \frac {\int \frac {1}{\left (1-(c+d x)^2\right )^3}d(c+d x)}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3}{4} \int \frac {1}{\left (1-(c+d x)^2\right )^2}d(c+d x)+\frac {c+d x}{4 \left (1-(c+d x)^2\right )^2}}{d}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \int \frac {1}{1-(c+d x)^2}d(c+d x)+\frac {c+d x}{2 \left (1-(c+d x)^2\right )}\right )+\frac {c+d x}{4 \left (1-(c+d x)^2\right )^2}}{d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {3}{4} \left (\frac {1}{2} \text {arctanh}(c+d x)+\frac {c+d x}{2 \left (1-(c+d x)^2\right )}\right )+\frac {c+d x}{4 \left (1-(c+d x)^2\right )^2}}{d}\) |
Input:
Int[(1 - (c + d*x)^2)^(-3),x]
Output:
((c + d*x)/(4*(1 - (c + d*x)^2)^2) + (3*((c + d*x)/(2*(1 - (c + d*x)^2)) + ArcTanh[c + d*x]/2))/4)/d
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Simp[1/Coefficient[v, x, 1 ] Subst[Int[(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, n, p}, x] && Lin earQ[v, x] && NeQ[v, x]
Time = 0.51 (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 | \(\frac {-\frac {3 d^{2} x^{3}}{8}-\frac {9 c d \,x^{2}}{8}+\left (-\frac {9 c^{2}}{8}+\frac {5}{8}\right ) x -\frac {c \left (3 c^{2}-5\right )}{8 d}}{\left (d^{2} x^{2}+2 c d x +c^{2}-1\right )^{2}}+\frac {3 \ln \left (-d x -c -1\right )}{16 d}-\frac {3 \ln \left (d x +c -1\right )}{16 d}\) | \(87\) |
norman | \(\frac {\frac {-3 d^{3} c^{3}+5 c \,d^{3}}{8 d^{4}}+\frac {\left (-9 c^{2} d^{3}+5 d^{3}\right ) x}{8 d^{3}}-\frac {3 d^{2} x^{3}}{8}-\frac {9 c d \,x^{2}}{8}}{\left (d^{2} x^{2}+2 c d x +c^{2}-1\right )^{2}}-\frac {3 \ln \left (d x +c -1\right )}{16 d}+\frac {3 \ln \left (d x +c +1\right )}{16 d}\) | \(102\) |
parallelrisch | \(-\frac {18 c^{2} d^{4} x -10 d^{4} x +12 \ln \left (d x +c -1\right ) x^{3} c \,d^{6}-12 \ln \left (d x +c +1\right ) x^{3} c \,d^{6}+18 \ln \left (d x +c -1\right ) x^{2} c^{2} d^{5}-18 \ln \left (d x +c +1\right ) x^{2} c^{2} d^{5}+12 \ln \left (d x +c -1\right ) x \,c^{3} d^{4}-10 c \,d^{3}+6 d^{3} c^{3}+3 \ln \left (d x +c -1\right ) d^{3}-3 \ln \left (d x +c +1\right ) d^{3}+6 x^{3} d^{6}-12 \ln \left (d x +c +1\right ) x \,c^{3} d^{4}-12 \ln \left (d x +c -1\right ) x c \,d^{4}+12 \ln \left (d x +c +1\right ) x c \,d^{4}-3 \ln \left (d x +c +1\right ) c^{4} d^{3}-6 \ln \left (d x +c -1\right ) c^{2} d^{3}+6 \ln \left (d x +c +1\right ) c^{2} d^{3}+18 x^{2} c \,d^{5}+3 \ln \left (d x +c -1\right ) x^{4} d^{7}-3 \ln \left (d x +c +1\right ) x^{4} d^{7}-6 \ln \left (d x +c -1\right ) x^{2} d^{5}+3 \ln \left (d x +c -1\right ) c^{4} d^{3}+6 \ln \left (d x +c +1\right ) x^{2} d^{5}}{16 d^{4} \left (d^{2} x^{2}+2 c d x +c^{2}-1\right )^{2}}\) | \(344\) |
Input:
int(1/(1-(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
Output:
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)
Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (54) = 108\).
Time = 0.08 (sec) , antiderivative size = 220, normalized size of antiderivative = 3.44 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=-\frac {6 \, d^{3} x^{3} + 18 \, c d^{2} x^{2} + 6 \, c^{3} + 2 \, {\left (9 \, c^{2} - 5\right )} d x - 3 \, {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} x^{2} + c^{4} + 4 \, {\left (c^{3} - c\right )} d x - 2 \, c^{2} + 1\right )} \log \left (d x + c + 1\right ) + 3 \, {\left (d^{4} x^{4} + 4 \, c d^{3} x^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{2} x^{2} + c^{4} + 4 \, {\left (c^{3} - c\right )} d x - 2 \, c^{2} + 1\right )} \log \left (d x + c - 1\right ) - 10 \, c}{16 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{3} x^{2} + 4 \, {\left (c^{3} - c\right )} d^{2} x + {\left (c^{4} - 2 \, c^{2} + 1\right )} d\right )}} \] Input:
integrate(1/(1-(d*x+c)^2)^3,x, algorithm="fricas")
Output:
-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)
Leaf count of result is larger than twice the leaf count of optimal. 141 vs. \(2 (53) = 106\).
Time = 0.47 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.20 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=- \frac {3 c^{3} + 9 c d^{2} x^{2} - 5 c + 3 d^{3} x^{3} + x \left (9 c^{2} d - 5 d\right )}{8 c^{4} d - 16 c^{2} d + 32 c d^{4} x^{3} + 8 d^{5} x^{4} + 8 d + x^{2} \cdot \left (48 c^{2} d^{3} - 16 d^{3}\right ) + x \left (32 c^{3} d^{2} - 32 c d^{2}\right )} - \frac {\frac {3 \log {\left (x + \frac {3 c - 3}{3 d} \right )}}{16} - \frac {3 \log {\left (x + \frac {3 c + 3}{3 d} \right )}}{16}}{d} \] Input:
integrate(1/(1-(d*x+c)**2)**3,x)
Output:
-(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
Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (54) = 108\).
Time = 0.03 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.91 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=-\frac {3 \, d^{3} x^{3} + 9 \, c d^{2} x^{2} + 3 \, c^{3} + {\left (9 \, c^{2} - 5\right )} d x - 5 \, c}{8 \, {\left (d^{5} x^{4} + 4 \, c d^{4} x^{3} + 2 \, {\left (3 \, c^{2} - 1\right )} d^{3} x^{2} + 4 \, {\left (c^{3} - c\right )} d^{2} x + {\left (c^{4} - 2 \, c^{2} + 1\right )} d\right )}} + \frac {3 \, \log \left (d x + c + 1\right )}{16 \, d} - \frac {3 \, \log \left (d x + c - 1\right )}{16 \, d} \] Input:
integrate(1/(1-(d*x+c)^2)^3,x, algorithm="maxima")
Output:
-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
Time = 0.13 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.38 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {3 \, \log \left ({\left | d x + c + 1 \right |}\right )}{16 \, d} - \frac {3 \, \log \left ({\left | d x + c - 1 \right |}\right )}{16 \, d} - \frac {3 \, d^{3} x^{3} + 9 \, c d^{2} x^{2} + 9 \, c^{2} d x + 3 \, c^{3} - 5 \, d x - 5 \, c}{8 \, {\left (d^{2} x^{2} + 2 \, c d x + c^{2} - 1\right )}^{2} d} \] Input:
integrate(1/(1-(d*x+c)^2)^3,x, algorithm="giac")
Output:
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)
Time = 0.13 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.78 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {3\,\mathrm {atanh}\left (c+d\,x\right )}{8\,d}-\frac {x\,\left (\frac {9\,c^2}{8}-\frac {5}{8}\right )-\frac {5\,c-3\,c^3}{8\,d}+\frac {3\,d^2\,x^3}{8}+\frac {9\,c\,d\,x^2}{8}}{c^4-2\,c^2-x^2\,\left (2\,d^2-6\,c^2\,d^2\right )-x\,\left (4\,c\,d-4\,c^3\,d\right )+d^4\,x^4+4\,c\,d^3\,x^3+1} \] Input:
int(-1/((c + d*x)^2 - 1)^3,x)
Output:
(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)
Time = 0.21 (sec) , antiderivative size = 378, normalized size of antiderivative = 5.91 \[ \int \frac {1}{\left (1-(c+d x)^2\right )^3} \, dx=\frac {3+12 \,\mathrm {log}\left (d x +c -1\right ) c \,d^{2} x^{2}-12 \,\mathrm {log}\left (d x +c +1\right ) c \,d^{2} x^{2}+24 \,\mathrm {log}\left (d x +c -1\right ) c^{2} d x -24 \,\mathrm {log}\left (d x +c +1\right ) c^{2} d x +8 c d x -24 c^{3} d x -24 \,\mathrm {log}\left (d x +c -1\right ) c^{4} d x -36 \,\mathrm {log}\left (d x +c -1\right ) c^{3} d^{2} x^{2}-24 \,\mathrm {log}\left (d x +c -1\right ) c^{2} d^{3} x^{3}-6 \,\mathrm {log}\left (d x +c -1\right ) c \,d^{4} x^{4}+24 \,\mathrm {log}\left (d x +c +1\right ) c^{4} d x +36 \,\mathrm {log}\left (d x +c +1\right ) c^{3} d^{2} x^{2}+24 \,\mathrm {log}\left (d x +c +1\right ) c^{2} d^{3} x^{3}+6 \,\mathrm {log}\left (d x +c +1\right ) c \,d^{4} x^{4}+3 d^{4} x^{4}-6 d^{2} x^{2}+12 \,\mathrm {log}\left (d x +c -1\right ) c^{3}-6 \,\mathrm {log}\left (d x +c -1\right ) c -12 \,\mathrm {log}\left (d x +c +1\right ) c^{3}+6 \,\mathrm {log}\left (d x +c +1\right ) c -9 c^{4}-18 c^{2} d^{2} x^{2}+14 c^{2}-6 \,\mathrm {log}\left (d x +c -1\right ) c^{5}+6 \,\mathrm {log}\left (d x +c +1\right ) c^{5}}{32 c d \left (d^{4} x^{4}+4 c \,d^{3} x^{3}+6 c^{2} d^{2} x^{2}+4 c^{3} d x +c^{4}-2 d^{2} x^{2}-4 c d x -2 c^{2}+1\right )} \] Input:
int(1/(1-(d*x+c)^2)^3,x)
Output:
( - 6*log(c + d*x - 1)*c**5 - 24*log(c + d*x - 1)*c**4*d*x - 36*log(c + d* x - 1)*c**3*d**2*x**2 + 12*log(c + d*x - 1)*c**3 - 24*log(c + d*x - 1)*c** 2*d**3*x**3 + 24*log(c + d*x - 1)*c**2*d*x - 6*log(c + d*x - 1)*c*d**4*x** 4 + 12*log(c + d*x - 1)*c*d**2*x**2 - 6*log(c + d*x - 1)*c + 6*log(c + d*x + 1)*c**5 + 24*log(c + d*x + 1)*c**4*d*x + 36*log(c + d*x + 1)*c**3*d**2* x**2 - 12*log(c + d*x + 1)*c**3 + 24*log(c + d*x + 1)*c**2*d**3*x**3 - 24* log(c + d*x + 1)*c**2*d*x + 6*log(c + d*x + 1)*c*d**4*x**4 - 12*log(c + d* x + 1)*c*d**2*x**2 + 6*log(c + d*x + 1)*c - 9*c**4 - 24*c**3*d*x - 18*c**2 *d**2*x**2 + 14*c**2 + 8*c*d*x + 3*d**4*x**4 - 6*d**2*x**2 + 3)/(32*c*d*(c **4 + 4*c**3*d*x + 6*c**2*d**2*x**2 - 2*c**2 + 4*c*d**3*x**3 - 4*c*d*x + d **4*x**4 - 2*d**2*x**2 + 1))