Integrand size = 17, antiderivative size = 31 \[ \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx=-\frac {\text {csch}^3(a+b x)}{3 b}-\frac {\text {csch}^5(a+b x)}{5 b} \] Output:
-1/3*csch(b*x+a)^3/b-1/5*csch(b*x+a)^5/b
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx=-\frac {\text {csch}^3(a+b x)}{3 b}-\frac {\text {csch}^5(a+b x)}{5 b} \] Input:
Integrate[Coth[a + b*x]^3*Csch[a + b*x]^3,x]
Output:
-1/3*Csch[a + b*x]^3/b - Csch[a + b*x]^5/(5*b)
Result contains complex when optimal does not.
Time = 0.23 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.16, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 3086, 25, 244, 2009}
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.
\(\displaystyle \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan \left (i a+i b x-\frac {\pi }{2}\right )^3 \sec \left (i a+i b x-\frac {\pi }{2}\right )^3dx\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle -\frac {i \int \text {csch}^2(a+b x) \left (\text {csch}^2(a+b x)+1\right )d(-i \text {csch}(a+b x))}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {i \int -\text {csch}^2(a+b x) \left (\text {csch}^2(a+b x)+1\right )d(-i \text {csch}(a+b x))}{b}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {i \int \left (-\text {csch}^4(a+b x)-\text {csch}^2(a+b x)\right )d(-i \text {csch}(a+b x))}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {1}{5} i \text {csch}^5(a+b x)-\frac {1}{3} i \text {csch}^3(a+b x)\right )}{b}\) |
Input:
Int[Coth[a + b*x]^3*Csch[a + b*x]^3,x]
Output:
((-I)*((-1/3*I)*Csch[a + b*x]^3 - (I/5)*Csch[a + b*x]^5))/b
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
Time = 0.18 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.87
method | result | size |
derivativedivides | \(-\frac {\frac {\operatorname {csch}\left (b x +a \right )^{5}}{5}+\frac {\operatorname {csch}\left (b x +a \right )^{3}}{3}}{b}\) | \(27\) |
default | \(-\frac {\frac {\operatorname {csch}\left (b x +a \right )^{5}}{5}+\frac {\operatorname {csch}\left (b x +a \right )^{3}}{3}}{b}\) | \(27\) |
risch | \(-\frac {8 \,{\mathrm e}^{3 b x +3 a} \left (5 \,{\mathrm e}^{4 b x +4 a}+2 \,{\mathrm e}^{2 b x +2 a}+5\right )}{15 b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{5}}\) | \(52\) |
Input:
int(coth(b*x+a)^3*csch(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
-1/b*(1/5*csch(b*x+a)^5+1/3*csch(b*x+a)^3)
Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (27) = 54\).
Time = 0.09 (sec) , antiderivative size = 343, normalized size of antiderivative = 11.06 \[ \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx=-\frac {8 \, {\left (5 \, \cosh \left (b x + a\right )^{4} + 20 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + 5 \, \sinh \left (b x + a\right )^{4} + 2 \, {\left (15 \, \cosh \left (b x + a\right )^{2} + 1\right )} \sinh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (5 \, \cosh \left (b x + a\right )^{3} + \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 5\right )}}{15 \, {\left (b \cosh \left (b x + a\right )^{7} + 7 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{6} + b \sinh \left (b x + a\right )^{7} - 5 \, b \cosh \left (b x + a\right )^{5} + {\left (21 \, b \cosh \left (b x + a\right )^{2} - 5 \, b\right )} \sinh \left (b x + a\right )^{5} + 5 \, {\left (7 \, b \cosh \left (b x + a\right )^{3} - 5 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{4} + 9 \, b \cosh \left (b x + a\right )^{3} + {\left (35 \, b \cosh \left (b x + a\right )^{4} - 50 \, b \cosh \left (b x + a\right )^{2} + 11 \, b\right )} \sinh \left (b x + a\right )^{3} + {\left (21 \, b \cosh \left (b x + a\right )^{5} - 50 \, b \cosh \left (b x + a\right )^{3} + 27 \, b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right )^{2} - 5 \, b \cosh \left (b x + a\right ) + {\left (7 \, b \cosh \left (b x + a\right )^{6} - 25 \, b \cosh \left (b x + a\right )^{4} + 33 \, b \cosh \left (b x + a\right )^{2} - 15 \, b\right )} \sinh \left (b x + a\right )\right )}} \] Input:
integrate(coth(b*x+a)^3*csch(b*x+a)^3,x, algorithm="fricas")
Output:
-8/15*(5*cosh(b*x + a)^4 + 20*cosh(b*x + a)*sinh(b*x + a)^3 + 5*sinh(b*x + a)^4 + 2*(15*cosh(b*x + a)^2 + 1)*sinh(b*x + a)^2 + 2*cosh(b*x + a)^2 + 4 *(5*cosh(b*x + a)^3 + cosh(b*x + a))*sinh(b*x + a) + 5)/(b*cosh(b*x + a)^7 + 7*b*cosh(b*x + a)*sinh(b*x + a)^6 + b*sinh(b*x + a)^7 - 5*b*cosh(b*x + a)^5 + (21*b*cosh(b*x + a)^2 - 5*b)*sinh(b*x + a)^5 + 5*(7*b*cosh(b*x + a) ^3 - 5*b*cosh(b*x + a))*sinh(b*x + a)^4 + 9*b*cosh(b*x + a)^3 + (35*b*cosh (b*x + a)^4 - 50*b*cosh(b*x + a)^2 + 11*b)*sinh(b*x + a)^3 + (21*b*cosh(b* x + a)^5 - 50*b*cosh(b*x + a)^3 + 27*b*cosh(b*x + a))*sinh(b*x + a)^2 - 5* b*cosh(b*x + a) + (7*b*cosh(b*x + a)^6 - 25*b*cosh(b*x + a)^4 + 33*b*cosh( b*x + a)^2 - 15*b)*sinh(b*x + a))
\[ \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx=\int \coth ^{3}{\left (a + b x \right )} \operatorname {csch}^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(coth(b*x+a)**3*csch(b*x+a)**3,x)
Output:
Integral(coth(a + b*x)**3*csch(a + b*x)**3, x)
Leaf count of result is larger than twice the leaf count of optimal. 214 vs. \(2 (27) = 54\).
Time = 0.05 (sec) , antiderivative size = 214, normalized size of antiderivative = 6.90 \[ \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx=\frac {8 \, e^{\left (-3 \, b x - 3 \, a\right )}}{3 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} - 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} - 1\right )}} + \frac {16 \, e^{\left (-5 \, b x - 5 \, a\right )}}{15 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} - 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} - 1\right )}} + \frac {8 \, e^{\left (-7 \, b x - 7 \, a\right )}}{3 \, b {\left (5 \, e^{\left (-2 \, b x - 2 \, a\right )} - 10 \, e^{\left (-4 \, b x - 4 \, a\right )} + 10 \, e^{\left (-6 \, b x - 6 \, a\right )} - 5 \, e^{\left (-8 \, b x - 8 \, a\right )} + e^{\left (-10 \, b x - 10 \, a\right )} - 1\right )}} \] Input:
integrate(coth(b*x+a)^3*csch(b*x+a)^3,x, algorithm="maxima")
Output:
8/3*e^(-3*b*x - 3*a)/(b*(5*e^(-2*b*x - 2*a) - 10*e^(-4*b*x - 4*a) + 10*e^( -6*b*x - 6*a) - 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) - 1)) + 16/15*e^(- 5*b*x - 5*a)/(b*(5*e^(-2*b*x - 2*a) - 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x - 6*a) - 5*e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) - 1)) + 8/3*e^(-7*b*x - 7* a)/(b*(5*e^(-2*b*x - 2*a) - 10*e^(-4*b*x - 4*a) + 10*e^(-6*b*x - 6*a) - 5* e^(-8*b*x - 8*a) + e^(-10*b*x - 10*a) - 1))
Time = 0.15 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.58 \[ \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx=-\frac {8 \, {\left (5 \, {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{2} + 12\right )}}{15 \, b {\left (e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}\right )}^{5}} \] Input:
integrate(coth(b*x+a)^3*csch(b*x+a)^3,x, algorithm="giac")
Output:
-8/15*(5*(e^(b*x + a) - e^(-b*x - a))^2 + 12)/(b*(e^(b*x + a) - e^(-b*x - a))^5)
Time = 1.05 (sec) , antiderivative size = 252, normalized size of antiderivative = 8.13 \[ \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx=-\frac {\frac {4\,{\mathrm {e}}^{a+b\,x}}{5\,b}+\frac {12\,{\mathrm {e}}^{3\,a+3\,b\,x}}{5\,b}+\frac {12\,{\mathrm {e}}^{5\,a+5\,b\,x}}{5\,b}+\frac {4\,{\mathrm {e}}^{7\,a+7\,b\,x}}{5\,b}}{5\,{\mathrm {e}}^{2\,a+2\,b\,x}-10\,{\mathrm {e}}^{4\,a+4\,b\,x}+10\,{\mathrm {e}}^{6\,a+6\,b\,x}-5\,{\mathrm {e}}^{8\,a+8\,b\,x}+{\mathrm {e}}^{10\,a+10\,b\,x}-1}-\frac {28\,{\mathrm {e}}^{a+b\,x}}{15\,b\,\left ({\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1\right )}-\frac {64\,{\mathrm {e}}^{a+b\,x}}{15\,b\,\left (3\,{\mathrm {e}}^{2\,a+2\,b\,x}-3\,{\mathrm {e}}^{4\,a+4\,b\,x}+{\mathrm {e}}^{6\,a+6\,b\,x}-1\right )}-\frac {16\,{\mathrm {e}}^{a+b\,x}}{5\,b\,\left (6\,{\mathrm {e}}^{4\,a+4\,b\,x}-4\,{\mathrm {e}}^{2\,a+2\,b\,x}-4\,{\mathrm {e}}^{6\,a+6\,b\,x}+{\mathrm {e}}^{8\,a+8\,b\,x}+1\right )} \] Input:
int(coth(a + b*x)^3/sinh(a + b*x)^3,x)
Output:
- ((4*exp(a + b*x))/(5*b) + (12*exp(3*a + 3*b*x))/(5*b) + (12*exp(5*a + 5* b*x))/(5*b) + (4*exp(7*a + 7*b*x))/(5*b))/(5*exp(2*a + 2*b*x) - 10*exp(4*a + 4*b*x) + 10*exp(6*a + 6*b*x) - 5*exp(8*a + 8*b*x) + exp(10*a + 10*b*x) - 1) - (28*exp(a + b*x))/(15*b*(exp(4*a + 4*b*x) - 2*exp(2*a + 2*b*x) + 1) ) - (64*exp(a + b*x))/(15*b*(3*exp(2*a + 2*b*x) - 3*exp(4*a + 4*b*x) + exp (6*a + 6*b*x) - 1)) - (16*exp(a + b*x))/(5*b*(6*exp(4*a + 4*b*x) - 4*exp(2 *a + 2*b*x) - 4*exp(6*a + 6*b*x) + exp(8*a + 8*b*x) + 1))
Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \coth ^3(a+b x) \text {csch}^3(a+b x) \, dx=\frac {\mathrm {csch}\left (b x +a \right )^{3} \left (-3 \coth \left (b x +a \right )^{2}-2\right )}{15 b} \] Input:
int(coth(b*x+a)^3*csch(b*x+a)^3,x)
Output:
(csch(a + b*x)**3*( - 3*coth(a + b*x)**2 - 2))/(15*b)