Integrand size = 12, antiderivative size = 96 \[ \int \frac {1}{(a+a \tanh (c+d x))^4} \, dx=\frac {x}{16 a^4}-\frac {1}{8 d (a+a \tanh (c+d x))^4}-\frac {1}{12 a d (a+a \tanh (c+d x))^3}-\frac {1}{16 d \left (a^2+a^2 \tanh (c+d x)\right )^2}-\frac {1}{16 d \left (a^4+a^4 \tanh (c+d x)\right )} \] Output:
1/16*x/a^4-1/8/d/(a+a*tanh(d*x+c))^4-1/12/a/d/(a+a*tanh(d*x+c))^3-1/16/d/( a^2+a^2*tanh(d*x+c))^2-1/16/d/(a^4+a^4*tanh(d*x+c))
Time = 0.17 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+a \tanh (c+d x))^4} \, dx=\frac {a \left (\frac {\text {arctanh}(\tanh (c+d x))}{16 a^5}-\frac {1}{8 a (a+a \tanh (c+d x))^4}-\frac {1}{12 a^2 (a+a \tanh (c+d x))^3}-\frac {1}{16 a^3 (a+a \tanh (c+d x))^2}-\frac {1}{16 a^4 (a+a \tanh (c+d x))}\right )}{d} \] Input:
Integrate[(a + a*Tanh[c + d*x])^(-4),x]
Output:
(a*(ArcTanh[Tanh[c + d*x]]/(16*a^5) - 1/(8*a*(a + a*Tanh[c + d*x])^4) - 1/ (12*a^2*(a + a*Tanh[c + d*x])^3) - 1/(16*a^3*(a + a*Tanh[c + d*x])^2) - 1/ (16*a^4*(a + a*Tanh[c + d*x]))))/d
Time = 0.48 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {3042, 3960, 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.
\(\displaystyle \int \frac {1}{(a \tanh (c+d x)+a)^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a-i a \tan (i c+i d x))^4}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\int \frac {1}{(\tanh (c+d x) a+a)^3}dx}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{8 d (a \tanh (c+d x)+a)^4}+\frac {\int \frac {1}{(a-i a \tan (i c+i d x))^3}dx}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\int \frac {1}{(\tanh (c+d x) a+a)^2}dx}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{8 d (a \tanh (c+d x)+a)^4}+\frac {-\frac {1}{6 d (a \tanh (c+d x)+a)^3}+\frac {\int \frac {1}{(a-i a \tan (i c+i d x))^2}dx}{2 a}}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{\tanh (c+d x) a+a}dx}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{8 d (a \tanh (c+d x)+a)^4}+\frac {-\frac {1}{6 d (a \tanh (c+d x)+a)^3}+\frac {-\frac {1}{4 d (a \tanh (c+d x)+a)^2}+\frac {\int \frac {1}{a-i a \tan (i c+i d x)}dx}{2 a}}{2 a}}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\frac {\int 1dx}{2 a}-\frac {1}{2 d (a \tanh (c+d x)+a)}}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {\frac {\frac {x}{2 a}-\frac {1}{2 d (a \tanh (c+d x)+a)}}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}\) |
Input:
Int[(a + a*Tanh[c + d*x])^(-4),x]
Output:
-1/8*1/(d*(a + a*Tanh[c + d*x])^4) + (-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))/(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 = 0.26 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.79
method | result | size |
risch | \(\frac {x}{16 a^{4}}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 a^{4} d}-\frac {3 \,{\mathrm e}^{-4 d x -4 c}}{32 a^{4} d}-\frac {{\mathrm e}^{-6 d x -6 c}}{24 a^{4} d}-\frac {{\mathrm e}^{-8 d x -8 c}}{128 a^{4} d}\) | \(76\) |
derivativedivides | \(\frac {-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )^{4}}-\frac {1}{12 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{16 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{16 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{32}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{32}}{d \,a^{4}}\) | \(79\) |
default | \(\frac {-\frac {1}{8 \left (\tanh \left (d x +c \right )+1\right )^{4}}-\frac {1}{12 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{16 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{16 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{32}-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{32}}{d \,a^{4}}\) | \(79\) |
parallelrisch | \(\frac {-16-3 \tanh \left (d x +c \right )^{3}+3 \tanh \left (d x +c \right )^{4} x d -12 \tanh \left (d x +c \right )^{2}-19 \tanh \left (d x +c \right )+18 \tanh \left (d x +c \right )^{2} x d +3 d x +12 \tanh \left (d x +c \right ) x d +12 \tanh \left (d x +c \right )^{3} x d}{48 d \,a^{4} \left (\tanh \left (d x +c \right )+1\right )^{4}}\) | \(99\) |
Input:
int(1/(a+a*tanh(d*x+c))^4,x,method=_RETURNVERBOSE)
Output:
1/16*x/a^4-1/8/a^4/d*exp(-2*d*x-2*c)-3/32/a^4/d*exp(-4*d*x-4*c)-1/24/a^4/d *exp(-6*d*x-6*c)-1/128/a^4/d*exp(-8*d*x-8*c)
Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (86) = 172\).
Time = 0.10 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.29 \[ \int \frac {1}{(a+a \tanh (c+d x))^4} \, dx=\frac {3 \, {\left (8 \, d x - 1\right )} \cosh \left (d x + c\right )^{4} + 12 \, {\left (8 \, d x + 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + 3 \, {\left (8 \, d x - 1\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (9 \, {\left (8 \, d x - 1\right )} \cosh \left (d x + c\right )^{2} - 32\right )} \sinh \left (d x + c\right )^{2} - 64 \, \cosh \left (d x + c\right )^{2} + 4 \, {\left (3 \, {\left (8 \, d x + 1\right )} \cosh \left (d x + c\right )^{3} - 16 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - 36}{384 \, {\left (a^{4} d \cosh \left (d x + c\right )^{4} + 4 \, a^{4} d \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right ) + 6 \, a^{4} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{2} + 4 \, a^{4} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{4} d \sinh \left (d x + c\right )^{4}\right )}} \] Input:
integrate(1/(a+a*tanh(d*x+c))^4,x, algorithm="fricas")
Output:
1/384*(3*(8*d*x - 1)*cosh(d*x + c)^4 + 12*(8*d*x + 1)*cosh(d*x + c)*sinh(d *x + c)^3 + 3*(8*d*x - 1)*sinh(d*x + c)^4 + 2*(9*(8*d*x - 1)*cosh(d*x + c) ^2 - 32)*sinh(d*x + c)^2 - 64*cosh(d*x + c)^2 + 4*(3*(8*d*x + 1)*cosh(d*x + c)^3 - 16*cosh(d*x + c))*sinh(d*x + c) - 36)/(a^4*d*cosh(d*x + c)^4 + 4* a^4*d*cosh(d*x + c)^3*sinh(d*x + c) + 6*a^4*d*cosh(d*x + c)^2*sinh(d*x + c )^2 + 4*a^4*d*cosh(d*x + c)*sinh(d*x + c)^3 + a^4*d*sinh(d*x + c)^4)
Leaf count of result is larger than twice the leaf count of optimal. 694 vs. \(2 (80) = 160\).
Time = 0.98 (sec) , antiderivative size = 694, normalized size of antiderivative = 7.23 \[ \int \frac {1}{(a+a \tanh (c+d x))^4} \, dx =\text {Too large to display} \] Input:
integrate(1/(a+a*tanh(d*x+c))**4,x)
Output:
Piecewise((3*d*x*tanh(c + d*x)**4/(48*a**4*d*tanh(c + d*x)**4 + 192*a**4*d *tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) + 12*d*x*tanh(c + d*x)**3/(48*a**4*d*tanh(c + d*x)**4 + 192* a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) + 18*d*x*tanh(c + d*x)**2/(48*a**4*d*tanh(c + d*x)**4 + 192*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*t anh(c + d*x) + 48*a**4*d) + 12*d*x*tanh(c + d*x)/(48*a**4*d*tanh(c + d*x)* *4 + 192*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4* d*tanh(c + d*x) + 48*a**4*d) + 3*d*x/(48*a**4*d*tanh(c + d*x)**4 + 192*a** 4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d *x) + 48*a**4*d) - 3*tanh(c + d*x)**3/(48*a**4*d*tanh(c + d*x)**4 + 192*a* *4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) - 12*tanh(c + d*x)**2/(48*a**4*d*tanh(c + d*x)**4 + 192* a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) - 19*tanh(c + d*x)/(48*a**4*d*tanh(c + d*x)**4 + 192*a **4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4*d) - 16/(48*a**4*d*tanh(c + d*x)**4 + 192*a**4*d*tanh(c + d*x)**3 + 288*a**4*d*tanh(c + d*x)**2 + 192*a**4*d*tanh(c + d*x) + 48*a**4 *d), Ne(d, 0)), (x/(a*tanh(c) + a)**4, True))
Time = 0.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(a+a \tanh (c+d x))^4} \, dx=\frac {d x + c}{16 \, a^{4} d} - \frac {48 \, e^{\left (-2 \, d x - 2 \, c\right )} + 36 \, e^{\left (-4 \, d x - 4 \, c\right )} + 16 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{384 \, a^{4} d} \] Input:
integrate(1/(a+a*tanh(d*x+c))^4,x, algorithm="maxima")
Output:
1/16*(d*x + c)/(a^4*d) - 1/384*(48*e^(-2*d*x - 2*c) + 36*e^(-4*d*x - 4*c) + 16*e^(-6*d*x - 6*c) + 3*e^(-8*d*x - 8*c))/(a^4*d)
Time = 0.13 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.67 \[ \int \frac {1}{(a+a \tanh (c+d x))^4} \, dx=-\frac {\frac {{\left (48 \, e^{\left (6 \, d x + 6 \, c\right )} + 36 \, e^{\left (4 \, d x + 4 \, c\right )} + 16 \, e^{\left (2 \, d x + 2 \, c\right )} + 3\right )} e^{\left (-8 \, d x - 8 \, c\right )}}{a^{4}} - \frac {24 \, {\left (d x + c\right )}}{a^{4}}}{384 \, d} \] Input:
integrate(1/(a+a*tanh(d*x+c))^4,x, algorithm="giac")
Output:
-1/384*((48*e^(6*d*x + 6*c) + 36*e^(4*d*x + 4*c) + 16*e^(2*d*x + 2*c) + 3) *e^(-8*d*x - 8*c)/a^4 - 24*(d*x + c)/a^4)/d
Time = 1.97 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+a \tanh (c+d x))^4} \, dx=\frac {x}{16\,a^4}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,a^4\,d}-\frac {3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{32\,a^4\,d}-\frac {{\mathrm {e}}^{-6\,c-6\,d\,x}}{24\,a^4\,d}-\frac {{\mathrm {e}}^{-8\,c-8\,d\,x}}{128\,a^4\,d} \] Input:
int(1/(a + a*tanh(c + d*x))^4,x)
Output:
x/(16*a^4) - exp(- 2*c - 2*d*x)/(8*a^4*d) - (3*exp(- 4*c - 4*d*x))/(32*a^4 *d) - exp(- 6*c - 6*d*x)/(24*a^4*d) - exp(- 8*c - 8*d*x)/(128*a^4*d)
Time = 0.22 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.75 \[ \int \frac {1}{(a+a \tanh (c+d x))^4} \, dx=\frac {24 e^{8 d x +8 c} d x -48 e^{6 d x +6 c}-36 e^{4 d x +4 c}-16 e^{2 d x +2 c}-3}{384 e^{8 d x +8 c} a^{4} d} \] Input:
int(1/(a+a*tanh(d*x+c))^4,x)
Output:
(24*e**(8*c + 8*d*x)*d*x - 48*e**(6*c + 6*d*x) - 36*e**(4*c + 4*d*x) - 16* e**(2*c + 2*d*x) - 3)/(384*e**(8*c + 8*d*x)*a**4*d)