Integrand size = 12, antiderivative size = 73 \[ \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx=\frac {x}{8 a^3}-\frac {1}{6 d (a+a \tanh (c+d x))^3}-\frac {1}{8 a d (a+a \tanh (c+d x))^2}-\frac {1}{8 d \left (a^3+a^3 \tanh (c+d x)\right )} \] Output:
1/8*x/a^3-1/6/d/(a+a*tanh(d*x+c))^3-1/8/a/d/(a+a*tanh(d*x+c))^2-1/8/d/(a^3 +a^3*tanh(d*x+c))
Time = 0.17 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx=-\frac {10+9 \tanh (c+d x)+3 \tanh ^2(c+d x)-3 \text {arctanh}(\tanh (c+d x)) (1+\tanh (c+d x))^3}{24 a^3 d (1+\tanh (c+d x))^3} \] Input:
Integrate[(a + a*Tanh[c + d*x])^(-3),x]
Output:
-1/24*(10 + 9*Tanh[c + d*x] + 3*Tanh[c + d*x]^2 - 3*ArcTanh[Tanh[c + d*x]] *(1 + Tanh[c + d*x])^3)/(a^3*d*(1 + Tanh[c + d*x])^3)
Time = 0.38 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.583, Rules used = {3042, 3960, 3042, 3960, 3042, 3960, 24}
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.
\(\displaystyle \int \frac {1}{(a \tanh (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a-i a \tan (i c+i d x))^3}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\int \frac {1}{(\tanh (c+d x) a+a)^2}dx}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{6 d (a \tanh (c+d x)+a)^3}+\frac {\int \frac {1}{(a-i a \tan (i c+i d x))^2}dx}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\int \frac {1}{\tanh (c+d x) a+a}dx}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{6 d (a \tanh (c+d x)+a)^3}+\frac {-\frac {1}{4 d (a \tanh (c+d x)+a)^2}+\frac {\int \frac {1}{a-i a \tan (i c+i d x)}dx}{2 a}}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\int 1dx}{2 a}-\frac {1}{2 d (a \tanh (c+d x)+a)}}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {\frac {x}{2 a}-\frac {1}{2 d (a \tanh (c+d x)+a)}}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}\) |
Input:
Int[(a + a*Tanh[c + d*x])^(-3),x]
Output:
-1/6*1/(d*(a + a*Tanh[c + d*x])^3) + (-1/4*1/(d*(a + a*Tanh[c + d*x])^2) + (x/(2*a) - 1/(2*d*(a + a*Tanh[c + d*x])))/(2*a))/(2*a)
Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a) Int[(a + b*Tan[c + d*x])^ (n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.81
method | result | size |
risch | \(\frac {x}{8 a^{3}}-\frac {3 \,{\mathrm e}^{-2 d x -2 c}}{16 a^{3} d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c}}{32 a^{3} d}-\frac {{\mathrm e}^{-6 d x -6 c}}{48 a^{3} d}\) | \(59\) |
derivativedivides | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{16}-\frac {1}{6 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{16}}{d \,a^{3}}\) | \(67\) |
default | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{16}-\frac {1}{6 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{16}}{d \,a^{3}}\) | \(67\) |
parallelrisch | \(\frac {-10-3 \tanh \left (d x +c \right )^{2}-9 \tanh \left (d x +c \right )+9 \tanh \left (d x +c \right )^{2} x d +3 d x +9 \tanh \left (d x +c \right ) x d +3 \tanh \left (d x +c \right )^{3} x d}{24 d \,a^{3} \left (\tanh \left (d x +c \right )+1\right )^{3}}\) | \(77\) |
Input:
int(1/(a+a*tanh(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/8*x/a^3-3/16/a^3/d*exp(-2*d*x-2*c)-3/32/a^3/d*exp(-4*d*x-4*c)-1/48/a^3/d *exp(-6*d*x-6*c)
Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (65) = 130\).
Time = 0.10 (sec) , antiderivative size = 160, normalized size of antiderivative = 2.19 \[ \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx=\frac {2 \, {\left (6 \, d x - 1\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (6 \, d x - 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 2 \, {\left (6 \, d x + 1\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (2 \, {\left (6 \, d x + 1\right )} \cosh \left (d x + c\right )^{2} - 3\right )} \sinh \left (d x + c\right ) - 27 \, \cosh \left (d x + c\right )}{96 \, {\left (a^{3} d \cosh \left (d x + c\right )^{3} + 3 \, a^{3} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right ) + 3 \, a^{3} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + a^{3} d \sinh \left (d x + c\right )^{3}\right )}} \] Input:
integrate(1/(a+a*tanh(d*x+c))^3,x, algorithm="fricas")
Output:
1/96*(2*(6*d*x - 1)*cosh(d*x + c)^3 + 6*(6*d*x - 1)*cosh(d*x + c)*sinh(d*x + c)^2 + 2*(6*d*x + 1)*sinh(d*x + c)^3 + 3*(2*(6*d*x + 1)*cosh(d*x + c)^2 - 3)*sinh(d*x + c) - 27*cosh(d*x + c))/(a^3*d*cosh(d*x + c)^3 + 3*a^3*d*c osh(d*x + c)^2*sinh(d*x + c) + 3*a^3*d*cosh(d*x + c)*sinh(d*x + c)^2 + a^3 *d*sinh(d*x + c)^3)
Leaf count of result is larger than twice the leaf count of optimal. 430 vs. \(2 (60) = 120\).
Time = 0.74 (sec) , antiderivative size = 430, normalized size of antiderivative = 5.89 \[ \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx=\begin {cases} \frac {3 d x \tanh ^{3}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {9 d x \tanh ^{2}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {9 d x \tanh {\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} + \frac {3 d x}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {3 \tanh ^{2}{\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {9 \tanh {\left (c + d x \right )}}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} - \frac {10}{24 a^{3} d \tanh ^{3}{\left (c + d x \right )} + 72 a^{3} d \tanh ^{2}{\left (c + d x \right )} + 72 a^{3} d \tanh {\left (c + d x \right )} + 24 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \tanh {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(a+a*tanh(d*x+c))**3,x)
Output:
Piecewise((3*d*x*tanh(c + d*x)**3/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d* tanh(c + d*x)**2 + 72*a**3*d*tanh(c + d*x) + 24*a**3*d) + 9*d*x*tanh(c + d *x)**2/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2 + 72*a**3* d*tanh(c + d*x) + 24*a**3*d) + 9*d*x*tanh(c + d*x)/(24*a**3*d*tanh(c + d*x )**3 + 72*a**3*d*tanh(c + d*x)**2 + 72*a**3*d*tanh(c + d*x) + 24*a**3*d) + 3*d*x/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2 + 72*a**3* d*tanh(c + d*x) + 24*a**3*d) - 3*tanh(c + d*x)**2/(24*a**3*d*tanh(c + d*x) **3 + 72*a**3*d*tanh(c + d*x)**2 + 72*a**3*d*tanh(c + d*x) + 24*a**3*d) - 9*tanh(c + d*x)/(24*a**3*d*tanh(c + d*x)**3 + 72*a**3*d*tanh(c + d*x)**2 + 72*a**3*d*tanh(c + d*x) + 24*a**3*d) - 10/(24*a**3*d*tanh(c + d*x)**3 + 7 2*a**3*d*tanh(c + d*x)**2 + 72*a**3*d*tanh(c + d*x) + 24*a**3*d), Ne(d, 0) ), (x/(a*tanh(c) + a)**3, True))
Time = 0.04 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx=\frac {d x + c}{8 \, a^{3} d} - \frac {18 \, e^{\left (-2 \, d x - 2 \, c\right )} + 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2 \, e^{\left (-6 \, d x - 6 \, c\right )}}{96 \, a^{3} d} \] Input:
integrate(1/(a+a*tanh(d*x+c))^3,x, algorithm="maxima")
Output:
1/8*(d*x + c)/(a^3*d) - 1/96*(18*e^(-2*d*x - 2*c) + 9*e^(-4*d*x - 4*c) + 2 *e^(-6*d*x - 6*c))/(a^3*d)
Time = 0.13 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx=-\frac {\frac {{\left (18 \, e^{\left (4 \, d x + 4 \, c\right )} + 9 \, e^{\left (2 \, d x + 2 \, c\right )} + 2\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{a^{3}} - \frac {12 \, {\left (d x + c\right )}}{a^{3}}}{96 \, d} \] Input:
integrate(1/(a+a*tanh(d*x+c))^3,x, algorithm="giac")
Output:
-1/96*((18*e^(4*d*x + 4*c) + 9*e^(2*d*x + 2*c) + 2)*e^(-6*d*x - 6*c)/a^3 - 12*(d*x + c)/a^3)/d
Time = 1.98 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx=\frac {x}{8\,a^3}-\frac {3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{16\,a^3\,d}-\frac {3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{32\,a^3\,d}-\frac {{\mathrm {e}}^{-6\,c-6\,d\,x}}{48\,a^3\,d} \] Input:
int(1/(a + a*tanh(c + d*x))^3,x)
Output:
x/(8*a^3) - (3*exp(- 2*c - 2*d*x))/(16*a^3*d) - (3*exp(- 4*c - 4*d*x))/(32 *a^3*d) - exp(- 6*c - 6*d*x)/(48*a^3*d)
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {1}{(a+a \tanh (c+d x))^3} \, dx=\frac {12 e^{6 d x +6 c} d x -18 e^{4 d x +4 c}-9 e^{2 d x +2 c}-2}{96 e^{6 d x +6 c} a^{3} d} \] Input:
int(1/(a+a*tanh(d*x+c))^3,x)
Output:
(12*e**(6*c + 6*d*x)*d*x - 18*e**(4*c + 4*d*x) - 9*e**(2*c + 2*d*x) - 2)/( 96*e**(6*c + 6*d*x)*a**3*d)