Integrand size = 12, antiderivative size = 121 \[ \int \frac {1}{(a+a \tanh (c+d x))^5} \, dx=\frac {x}{32 a^5}-\frac {1}{10 d (a+a \tanh (c+d x))^5}-\frac {1}{16 a d (a+a \tanh (c+d x))^4}-\frac {1}{24 a^2 d (a+a \tanh (c+d x))^3}-\frac {1}{32 a d \left (a^2+a^2 \tanh (c+d x)\right )^2}-\frac {1}{32 d \left (a^5+a^5 \tanh (c+d x)\right )} \]
1/32*x/a^5-1/10/d/(a+a*tanh(d*x+c))^5-1/16/a/d/(a+a*tanh(d*x+c))^4-1/24/a^ 2/d/(a+a*tanh(d*x+c))^3-1/32/a/d/(a^2+a^2*tanh(d*x+c))^2-1/32/d/(a^5+a^5*t anh(d*x+c))
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+a \tanh (c+d x))^5} \, dx=\frac {\text {sech}^5(c+d x) (-500 \cosh (c+d x)-375 \cosh (3 (c+d x))-149 \cosh (5 (c+d x))-100 \sinh (c+d x)-225 \sinh (3 (c+d x))-125 \sinh (5 (c+d x))+120 \text {arctanh}(\tanh (c+d x)) (\cosh (5 (c+d x))+\sinh (5 (c+d x))))}{3840 a^5 d (1+\tanh (c+d x))^5} \]
(Sech[c + d*x]^5*(-500*Cosh[c + d*x] - 375*Cosh[3*(c + d*x)] - 149*Cosh[5* (c + d*x)] - 100*Sinh[c + d*x] - 225*Sinh[3*(c + d*x)] - 125*Sinh[5*(c + d *x)] + 120*ArcTanh[Tanh[c + d*x]]*(Cosh[5*(c + d*x)] + Sinh[5*(c + d*x)])) )/(3840*a^5*d*(1 + Tanh[c + d*x])^5)
Time = 0.58 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.917, Rules used = {3042, 3960, 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.
\(\displaystyle \int \frac {1}{(a \tanh (c+d x)+a)^5} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a-i a \tan (i c+i d x))^5}dx\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\int \frac {1}{(\tanh (c+d x) a+a)^4}dx}{2 a}-\frac {1}{10 d (a \tanh (c+d x)+a)^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{10 d (a \tanh (c+d x)+a)^5}+\frac {\int \frac {1}{(a-i a \tan (i c+i d x))^4}dx}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\int \frac {1}{(\tanh (c+d x) a+a)^3}dx}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}}{2 a}-\frac {1}{10 d (a \tanh (c+d x)+a)^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{10 d (a \tanh (c+d x)+a)^5}+\frac {-\frac {1}{8 d (a \tanh (c+d x)+a)^4}+\frac {\int \frac {1}{(a-i a \tan (i c+i d x))^3}dx}{2 a}}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\int \frac {1}{(\tanh (c+d x) a+a)^2}dx}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}}{2 a}-\frac {1}{10 d (a \tanh (c+d x)+a)^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{10 d (a \tanh (c+d x)+a)^5}+\frac {-\frac {1}{8 d (a \tanh (c+d x)+a)^4}+\frac {-\frac {1}{6 d (a \tanh (c+d x)+a)^3}+\frac {\int \frac {1}{(a-i a \tan (i c+i d x))^2}dx}{2 a}}{2 a}}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\frac {\int \frac {1}{\tanh (c+d x) a+a}dx}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}}{2 a}-\frac {1}{10 d (a \tanh (c+d x)+a)^5}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{10 d (a \tanh (c+d x)+a)^5}+\frac {-\frac {1}{8 d (a \tanh (c+d x)+a)^4}+\frac {-\frac {1}{6 d (a \tanh (c+d x)+a)^3}+\frac {-\frac {1}{4 d (a \tanh (c+d x)+a)^2}+\frac {\int \frac {1}{a-i a \tan (i c+i d x)}dx}{2 a}}{2 a}}{2 a}}{2 a}\) |
\(\Big \downarrow \) 3960 |
\(\displaystyle \frac {\frac {\frac {\frac {\frac {\int 1dx}{2 a}-\frac {1}{2 d (a \tanh (c+d x)+a)}}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}}{2 a}-\frac {1}{10 d (a \tanh (c+d x)+a)^5}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {\frac {\frac {\frac {\frac {x}{2 a}-\frac {1}{2 d (a \tanh (c+d x)+a)}}{2 a}-\frac {1}{4 d (a \tanh (c+d x)+a)^2}}{2 a}-\frac {1}{6 d (a \tanh (c+d x)+a)^3}}{2 a}-\frac {1}{8 d (a \tanh (c+d x)+a)^4}}{2 a}-\frac {1}{10 d (a \tanh (c+d x)+a)^5}\) |
-1/10*1/(d*(a + a*Tanh[c + d*x])^5) + (-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))/(2*a)
3.1.49.3.1 Defintions of rubi rules used
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 = 0.09 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{64}-\frac {1}{10 \left (\tanh \left (d x +c \right )+1\right )^{5}}-\frac {1}{16 \left (\tanh \left (d x +c \right )+1\right )^{4}}-\frac {1}{24 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{32 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{32 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{64}}{d \,a^{5}}\) | \(91\) |
default | \(\frac {-\frac {\ln \left (\tanh \left (d x +c \right )-1\right )}{64}-\frac {1}{10 \left (\tanh \left (d x +c \right )+1\right )^{5}}-\frac {1}{16 \left (\tanh \left (d x +c \right )+1\right )^{4}}-\frac {1}{24 \left (\tanh \left (d x +c \right )+1\right )^{3}}-\frac {1}{32 \left (\tanh \left (d x +c \right )+1\right )^{2}}-\frac {1}{32 \left (\tanh \left (d x +c \right )+1\right )}+\frac {\ln \left (\tanh \left (d x +c \right )+1\right )}{64}}{d \,a^{5}}\) | \(91\) |
risch | \(\frac {x}{32 a^{5}}-\frac {5 \,{\mathrm e}^{-2 d x -2 c}}{64 a^{5} d}-\frac {5 \,{\mathrm e}^{-4 d x -4 c}}{64 a^{5} d}-\frac {5 \,{\mathrm e}^{-6 d x -6 c}}{96 a^{5} d}-\frac {5 \,{\mathrm e}^{-8 d x -8 c}}{256 a^{5} d}-\frac {{\mathrm e}^{-10 d x -10 c}}{320 a^{5} d}\) | \(93\) |
parallelrisch | \(\frac {-128-175 \tanh \left (d x +c \right )+75 \tanh \left (d x +c \right )^{4} x d +150 \tanh \left (d x +c \right )^{3} x d +15 d x -15 \tanh \left (d x +c \right )^{4}+150 \tanh \left (d x +c \right )^{2} x d +75 \tanh \left (d x +c \right ) x d -75 \tanh \left (d x +c \right )^{3}+15 \tanh \left (d x +c \right )^{5} x d -155 \tanh \left (d x +c \right )^{2}}{480 d \,a^{5} \left (\tanh \left (d x +c \right )+1\right )^{5}}\) | \(121\) |
1/d/a^5*(-1/64*ln(tanh(d*x+c)-1)-1/10/(tanh(d*x+c)+1)^5-1/16/(tanh(d*x+c)+ 1)^4-1/24/(tanh(d*x+c)+1)^3-1/32/(tanh(d*x+c)+1)^2-1/32/(tanh(d*x+c)+1)+1/ 64*ln(tanh(d*x+c)+1))
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (109) = 218\).
Time = 0.26 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.37 \[ \int \frac {1}{(a+a \tanh (c+d x))^5} \, dx=\frac {12 \, {\left (10 \, d x - 1\right )} \cosh \left (d x + c\right )^{5} + 60 \, {\left (10 \, d x - 1\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 12 \, {\left (10 \, d x + 1\right )} \sinh \left (d x + c\right )^{5} + 15 \, {\left (8 \, {\left (10 \, d x + 1\right )} \cosh \left (d x + c\right )^{2} - 15\right )} \sinh \left (d x + c\right )^{3} - 375 \, \cosh \left (d x + c\right )^{3} + 15 \, {\left (8 \, {\left (10 \, d x - 1\right )} \cosh \left (d x + c\right )^{3} - 75 \, \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 5 \, {\left (12 \, {\left (10 \, d x + 1\right )} \cosh \left (d x + c\right )^{4} - 135 \, \cosh \left (d x + c\right )^{2} - 20\right )} \sinh \left (d x + c\right ) - 500 \, \cosh \left (d x + c\right )}{3840 \, {\left (a^{5} d \cosh \left (d x + c\right )^{5} + 5 \, a^{5} d \cosh \left (d x + c\right )^{4} \sinh \left (d x + c\right ) + 10 \, a^{5} d \cosh \left (d x + c\right )^{3} \sinh \left (d x + c\right )^{2} + 10 \, a^{5} d \cosh \left (d x + c\right )^{2} \sinh \left (d x + c\right )^{3} + 5 \, a^{5} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + a^{5} d \sinh \left (d x + c\right )^{5}\right )}} \]
1/3840*(12*(10*d*x - 1)*cosh(d*x + c)^5 + 60*(10*d*x - 1)*cosh(d*x + c)*si nh(d*x + c)^4 + 12*(10*d*x + 1)*sinh(d*x + c)^5 + 15*(8*(10*d*x + 1)*cosh( d*x + c)^2 - 15)*sinh(d*x + c)^3 - 375*cosh(d*x + c)^3 + 15*(8*(10*d*x - 1 )*cosh(d*x + c)^3 - 75*cosh(d*x + c))*sinh(d*x + c)^2 + 5*(12*(10*d*x + 1) *cosh(d*x + c)^4 - 135*cosh(d*x + c)^2 - 20)*sinh(d*x + c) - 500*cosh(d*x + c))/(a^5*d*cosh(d*x + c)^5 + 5*a^5*d*cosh(d*x + c)^4*sinh(d*x + c) + 10* a^5*d*cosh(d*x + c)^3*sinh(d*x + c)^2 + 10*a^5*d*cosh(d*x + c)^2*sinh(d*x + c)^3 + 5*a^5*d*cosh(d*x + c)*sinh(d*x + c)^4 + a^5*d*sinh(d*x + c)^5)
Leaf count of result is larger than twice the leaf count of optimal. 1018 vs. \(2 (102) = 204\).
Time = 1.31 (sec) , antiderivative size = 1018, normalized size of antiderivative = 8.41 \[ \int \frac {1}{(a+a \tanh (c+d x))^5} \, dx=\text {Too large to display} \]
Piecewise((15*d*x*tanh(c + d*x)**5/(480*a**5*d*tanh(c + d*x)**5 + 2400*a** 5*d*tanh(c + d*x)**4 + 4800*a**5*d*tanh(c + d*x)**3 + 4800*a**5*d*tanh(c + d*x)**2 + 2400*a**5*d*tanh(c + d*x) + 480*a**5*d) + 75*d*x*tanh(c + d*x)* *4/(480*a**5*d*tanh(c + d*x)**5 + 2400*a**5*d*tanh(c + d*x)**4 + 4800*a**5 *d*tanh(c + d*x)**3 + 4800*a**5*d*tanh(c + d*x)**2 + 2400*a**5*d*tanh(c + d*x) + 480*a**5*d) + 150*d*x*tanh(c + d*x)**3/(480*a**5*d*tanh(c + d*x)**5 + 2400*a**5*d*tanh(c + d*x)**4 + 4800*a**5*d*tanh(c + d*x)**3 + 4800*a**5 *d*tanh(c + d*x)**2 + 2400*a**5*d*tanh(c + d*x) + 480*a**5*d) + 150*d*x*ta nh(c + d*x)**2/(480*a**5*d*tanh(c + d*x)**5 + 2400*a**5*d*tanh(c + d*x)**4 + 4800*a**5*d*tanh(c + d*x)**3 + 4800*a**5*d*tanh(c + d*x)**2 + 2400*a**5 *d*tanh(c + d*x) + 480*a**5*d) + 75*d*x*tanh(c + d*x)/(480*a**5*d*tanh(c + d*x)**5 + 2400*a**5*d*tanh(c + d*x)**4 + 4800*a**5*d*tanh(c + d*x)**3 + 4 800*a**5*d*tanh(c + d*x)**2 + 2400*a**5*d*tanh(c + d*x) + 480*a**5*d) + 15 *d*x/(480*a**5*d*tanh(c + d*x)**5 + 2400*a**5*d*tanh(c + d*x)**4 + 4800*a* *5*d*tanh(c + d*x)**3 + 4800*a**5*d*tanh(c + d*x)**2 + 2400*a**5*d*tanh(c + d*x) + 480*a**5*d) - 15*tanh(c + d*x)**4/(480*a**5*d*tanh(c + d*x)**5 + 2400*a**5*d*tanh(c + d*x)**4 + 4800*a**5*d*tanh(c + d*x)**3 + 4800*a**5*d* tanh(c + d*x)**2 + 2400*a**5*d*tanh(c + d*x) + 480*a**5*d) - 75*tanh(c + d *x)**3/(480*a**5*d*tanh(c + d*x)**5 + 2400*a**5*d*tanh(c + d*x)**4 + 4800* a**5*d*tanh(c + d*x)**3 + 4800*a**5*d*tanh(c + d*x)**2 + 2400*a**5*d*ta...
Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.64 \[ \int \frac {1}{(a+a \tanh (c+d x))^5} \, dx=\frac {d x + c}{32 \, a^{5} d} - \frac {300 \, e^{\left (-2 \, d x - 2 \, c\right )} + 300 \, e^{\left (-4 \, d x - 4 \, c\right )} + 200 \, e^{\left (-6 \, d x - 6 \, c\right )} + 75 \, e^{\left (-8 \, d x - 8 \, c\right )} + 12 \, e^{\left (-10 \, d x - 10 \, c\right )}}{3840 \, a^{5} d} \]
1/32*(d*x + c)/(a^5*d) - 1/3840*(300*e^(-2*d*x - 2*c) + 300*e^(-4*d*x - 4* c) + 200*e^(-6*d*x - 6*c) + 75*e^(-8*d*x - 8*c) + 12*e^(-10*d*x - 10*c))/( a^5*d)
Time = 0.26 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.62 \[ \int \frac {1}{(a+a \tanh (c+d x))^5} \, dx=-\frac {\frac {{\left (300 \, e^{\left (8 \, d x + 8 \, c\right )} + 300 \, e^{\left (6 \, d x + 6 \, c\right )} + 200 \, e^{\left (4 \, d x + 4 \, c\right )} + 75 \, e^{\left (2 \, d x + 2 \, c\right )} + 12\right )} e^{\left (-10 \, d x - 10 \, c\right )}}{a^{5}} - \frac {120 \, {\left (d x + c\right )}}{a^{5}}}{3840 \, d} \]
-1/3840*((300*e^(8*d*x + 8*c) + 300*e^(6*d*x + 6*c) + 200*e^(4*d*x + 4*c) + 75*e^(2*d*x + 2*c) + 12)*e^(-10*d*x - 10*c)/a^5 - 120*(d*x + c)/a^5)/d
Time = 1.75 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(a+a \tanh (c+d x))^5} \, dx=\frac {x}{32\,a^5}-\frac {5\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{64\,a^5\,d}-\frac {5\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,a^5\,d}-\frac {5\,{\mathrm {e}}^{-6\,c-6\,d\,x}}{96\,a^5\,d}-\frac {5\,{\mathrm {e}}^{-8\,c-8\,d\,x}}{256\,a^5\,d}-\frac {{\mathrm {e}}^{-10\,c-10\,d\,x}}{320\,a^5\,d} \]