Integrand size = 23, antiderivative size = 64 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {a^3 \coth (c+d x)}{d}+\frac {3 a^2 b \tanh (c+d x)}{d}+\frac {a b^2 \tanh ^3(c+d x)}{d}+\frac {b^3 \tanh ^5(c+d x)}{5 d} \] Output:
-a^3*coth(d*x+c)/d+3*a^2*b*tanh(d*x+c)/d+a*b^2*tanh(d*x+c)^3/d+1/5*b^3*tan h(d*x+c)^5/d
Time = 2.19 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.09 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {-5 a^3 \coth (c+d x)+b \left (15 a^2+5 a b+b^2-b (5 a+2 b) \text {sech}^2(c+d x)+b^2 \text {sech}^4(c+d x)\right ) \tanh (c+d x)}{5 d} \] Input:
Integrate[Csch[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^3,x]
Output:
(-5*a^3*Coth[c + d*x] + b*(15*a^2 + 5*a*b + b^2 - b*(5*a + 2*b)*Sech[c + d *x]^2 + b^2*Sech[c + d*x]^4)*Tanh[c + d*x])/(5*d)
Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.88, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 25, 4146, 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.
| \(\displaystyle \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\left (a-b \tan (i c+i d x)^2\right )^3}{\sin (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (a-b \tan (i c+i d x)^2\right )^3}{\sin (i c+i d x)^2}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \coth ^2(c+d x) \left (b \tanh ^2(c+d x)+a\right )^3d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 244 |
| \(\displaystyle \frac {\int \left (b^3 \tanh ^4(c+d x)+3 a b^2 \tanh ^2(c+d x)+a^3 \coth ^2(c+d x)+3 a^2 b\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-a^3 \coth (c+d x)+3 a^2 b \tanh (c+d x)+a b^2 \tanh ^3(c+d x)+\frac {1}{5} b^3 \tanh ^5(c+d x)}{d}\) |
Input:
Int[Csch[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^3,x]
Output:
(-(a^3*Coth[c + d*x]) + 3*a^2*b*Tanh[c + d*x] + a*b^2*Tanh[c + d*x]^3 + (b ^3*Tanh[c + d*x]^5)/5)/d
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[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Leaf count of result is larger than twice the leaf count of optimal. \(140\) vs. \(2(62)=124\).
Time = 12.01 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.20
| method | result | size |
| derivativedivides | \(\frac {-\coth \left (d x +c \right ) a^{3}+3 \tanh \left (d x +c \right ) a^{2} b +3 b^{2} a \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+b^{3} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(141\) |
| default | \(\frac {-\coth \left (d x +c \right ) a^{3}+3 \tanh \left (d x +c \right ) a^{2} b +3 b^{2} a \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+b^{3} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\) | \(141\) |
| risch | \(-\frac {2 \left (5 \,{\mathrm e}^{10 d x +10 c} a^{3}+15 a^{2} b \,{\mathrm e}^{10 d x +10 c}+15 a \,b^{2} {\mathrm e}^{10 d x +10 c}+5 b^{3} {\mathrm e}^{10 d x +10 c}+25 a^{3} {\mathrm e}^{8 d x +8 c}+45 a^{2} b \,{\mathrm e}^{8 d x +8 c}+15 a \,b^{2} {\mathrm e}^{8 d x +8 c}-5 b^{3} {\mathrm e}^{8 d x +8 c}+50 a^{3} {\mathrm e}^{6 d x +6 c}+30 a^{2} b \,{\mathrm e}^{6 d x +6 c}-10 a \,b^{2} {\mathrm e}^{6 d x +6 c}+10 b^{3} {\mathrm e}^{6 d x +6 c}+50 \,{\mathrm e}^{4 d x +4 c} a^{3}-30 a^{2} b \,{\mathrm e}^{4 d x +4 c}-10 a \,b^{2} {\mathrm e}^{4 d x +4 c}-10 b^{3} {\mathrm e}^{4 d x +4 c}+25 \,{\mathrm e}^{2 d x +2 c} a^{3}-45 a^{2} b \,{\mathrm e}^{2 d x +2 c}-5 a \,b^{2} {\mathrm e}^{2 d x +2 c}+b^{3} {\mathrm e}^{2 d x +2 c}+5 a^{3}-15 a^{2} b -5 b^{2} a -b^{3}\right )}{5 d \left ({\mathrm e}^{2 d x +2 c}-1\right ) \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\) | \(344\) |
Input:
int(csch(d*x+c)^2*(a+tanh(d*x+c)^2*b)^3,x,method=_RETURNVERBOSE)
Output:
1/d*(-coth(d*x+c)*a^3+3*tanh(d*x+c)*a^2*b+3*b^2*a*(-1/2*sinh(d*x+c)/cosh(d *x+c)^3+1/2*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c))+b^3*(-1/2*sinh(d*x+c)^3/c osh(d*x+c)^5-3/8*sinh(d*x+c)/cosh(d*x+c)^5+3/8*(8/15+1/5*sech(d*x+c)^4+4/1 5*sech(d*x+c)^2)*tanh(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 572 vs. \(2 (62) = 124\).
Time = 0.09 (sec) , antiderivative size = 572, normalized size of antiderivative = 8.94 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")
Output:
-4/5*((5*a^3 + 5*a*b^2 + 2*b^3)*cosh(d*x + c)^5 + 5*(5*a^3 + 5*a*b^2 + 2*b ^3)*cosh(d*x + c)*sinh(d*x + c)^4 + (15*a^2*b + 10*a*b^2 + 3*b^3)*sinh(d*x + c)^5 + (25*a^3 + 5*a*b^2 - 2*b^3)*cosh(d*x + c)^3 + (45*a^2*b + 10*a*b^ 2 - 3*b^3 + 10*(15*a^2*b + 10*a*b^2 + 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c )^3 + (10*(5*a^3 + 5*a*b^2 + 2*b^3)*cosh(d*x + c)^3 + 3*(25*a^3 + 5*a*b^2 - 2*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(5*a^3 - a*b^2)*cosh(d*x + c) + (5*(15*a^2*b + 10*a*b^2 + 3*b^3)*cosh(d*x + c)^4 + 30*a^2*b + 10*b^3 + 3*(45*a^2*b + 10*a*b^2 - 3*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d* x + c)^7 + 7*d*cosh(d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 3*d*cos h(d*x + c)^5 + (21*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^5 + 5*(7*d*cosh( d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^4 + d*cosh(d*x + c)^3 + (35* d*cosh(d*x + c)^4 + 50*d*cosh(d*x + c)^2 + 9*d)*sinh(d*x + c)^3 + 3*(7*d*c osh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c)^2 - 5*d*cosh(d*x + c) + (7*d*cosh(d*x + c)^6 + 25*d*cosh(d*x + c)^4 + 27*d*co sh(d*x + c)^2 + 5*d)*sinh(d*x + c))
\[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \operatorname {csch}^{2}{\left (c + d x \right )}\, dx \] Input:
integrate(csch(d*x+c)**2*(a+b*tanh(d*x+c)**2)**3,x)
Output:
Integral((a + b*tanh(c + d*x)**2)**3*csch(c + d*x)**2, x)
Leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (62) = 124\).
Time = 0.05 (sec) , antiderivative size = 348, normalized size of antiderivative = 5.44 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {2}{5} \, b^{3} {\left (\frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {5 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + 2 \, a b^{2} {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {6 \, a^{2} b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} + \frac {2 \, a^{3}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")
Output:
2/5*b^3*(10*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 5* e^(-8*d*x - 8*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d *x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2* d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c ) + e^(-10*d*x - 10*c) + 1))) + 2*a*b^2*(3*e^(-4*d*x - 4*c)/(d*(3*e^(-2*d* x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1))) + 6*a^2*b/(d*(e^(-2 *d*x - 2*c) + 1)) + 2*a^3/(d*(e^(-2*d*x - 2*c) - 1))
Leaf count of result is larger than twice the leaf count of optimal. 202 vs. \(2 (62) = 124\).
Time = 0.22 (sec) , antiderivative size = 202, normalized size of antiderivative = 3.16 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {2 \, {\left (\frac {5 \, a^{3}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} + \frac {15 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 15 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 5 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 30 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 20 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 10 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} b + 5 \, a b^{2} + b^{3}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}\right )}}{5 \, d} \] Input:
integrate(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")
Output:
-2/5*(5*a^3/(e^(2*d*x + 2*c) - 1) + (15*a^2*b*e^(8*d*x + 8*c) + 15*a*b^2*e ^(8*d*x + 8*c) + 5*b^3*e^(8*d*x + 8*c) + 60*a^2*b*e^(6*d*x + 6*c) + 30*a*b ^2*e^(6*d*x + 6*c) + 90*a^2*b*e^(4*d*x + 4*c) + 20*a*b^2*e^(4*d*x + 4*c) + 10*b^3*e^(4*d*x + 4*c) + 60*a^2*b*e^(2*d*x + 2*c) + 10*a*b^2*e^(2*d*x + 2 *c) + 15*a^2*b + 5*a*b^2 + b^3)/(e^(2*d*x + 2*c) + 1)^5)/d
Time = 1.48 (sec) , antiderivative size = 590, normalized size of antiderivative = 9.22 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=-\frac {\frac {2\,\left (3\,a^2\,b-b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2\,b-a\,b^2+b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b-b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2\,b-b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b-a\,b^2+b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b-b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {12\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2\,b-a\,b^2+b^3\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2\,b-b^3\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (3\,a^2\,b-b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {2\,a^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {2\,\left (3\,a^2\,b+3\,a\,b^2+b^3\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:
int((a + b*tanh(c + d*x)^2)^3/sinh(c + d*x)^2,x)
Output:
- ((2*(3*a^2*b - b^3))/(5*d) + (2*exp(2*c + 2*d*x)*(3*a*b^2 + 3*a^2*b + b^ 3))/(5*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2*(3*a^2*b - a* b^2 + b^3))/(5*d) + (2*exp(4*c + 4*d*x)*(3*a*b^2 + 3*a^2*b + b^3))/(5*d) + (4*exp(2*c + 2*d*x)*(3*a^2*b - b^3))/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4 *c + 4*d*x) + exp(6*c + 6*d*x) + 1) - ((2*(3*a^2*b - b^3))/(5*d) + (6*exp( 2*c + 2*d*x)*(3*a^2*b - a*b^2 + b^3))/(5*d) + (2*exp(6*c + 6*d*x)*(3*a*b^2 + 3*a^2*b + b^3))/(5*d) + (6*exp(4*c + 4*d*x)*(3*a^2*b - b^3))/(5*d))/(4* exp(2*c + 2*d*x) + 6*exp(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d *x) + 1) - ((2*(3*a*b^2 + 3*a^2*b + b^3))/(5*d) + (12*exp(4*c + 4*d*x)*(3* a^2*b - a*b^2 + b^3))/(5*d) + (2*exp(8*c + 8*d*x)*(3*a*b^2 + 3*a^2*b + b^3 ))/(5*d) + (8*exp(2*c + 2*d*x)*(3*a^2*b - b^3))/(5*d) + (8*exp(6*c + 6*d*x )*(3*a^2*b - b^3))/(5*d))/(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x) + 10*e xp(6*c + 6*d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) - (2*a^3)/( d*(exp(2*c + 2*d*x) - 1)) - (2*(3*a*b^2 + 3*a^2*b + b^3))/(5*d*(exp(2*c + 2*d*x) + 1))
Time = 0.27 (sec) , antiderivative size = 400, normalized size of antiderivative = 6.25 \[ \int \text {csch}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {5 e^{12 d x +12 c} a^{3}+15 e^{12 d x +12 c} a^{2} b +15 e^{12 d x +12 c} a \,b^{2}+5 e^{12 d x +12 c} b^{3}-75 e^{8 d x +8 c} a^{3}-105 e^{8 d x +8 c} a^{2} b +15 e^{8 d x +8 c} a \,b^{2}+45 e^{8 d x +8 c} b^{3}-200 e^{6 d x +6 c} a^{3}-120 e^{6 d x +6 c} a^{2} b +40 e^{6 d x +6 c} a \,b^{2}-40 e^{6 d x +6 c} b^{3}-225 e^{4 d x +4 c} a^{3}+45 e^{4 d x +4 c} a^{2} b -35 e^{4 d x +4 c} a \,b^{2}+15 e^{4 d x +4 c} b^{3}-120 e^{2 d x +2 c} a^{3}+120 e^{2 d x +2 c} a^{2} b -40 e^{2 d x +2 c} a \,b^{2}-24 e^{2 d x +2 c} b^{3}-25 a^{3}+45 a^{2} b +5 a \,b^{2}-b^{3}}{10 d \left (e^{12 d x +12 c}+4 e^{10 d x +10 c}+5 e^{8 d x +8 c}-5 e^{4 d x +4 c}-4 e^{2 d x +2 c}-1\right )} \] Input:
int(csch(d*x+c)^2*(a+b*tanh(d*x+c)^2)^3,x)
Output:
(5*e**(12*c + 12*d*x)*a**3 + 15*e**(12*c + 12*d*x)*a**2*b + 15*e**(12*c + 12*d*x)*a*b**2 + 5*e**(12*c + 12*d*x)*b**3 - 75*e**(8*c + 8*d*x)*a**3 - 10 5*e**(8*c + 8*d*x)*a**2*b + 15*e**(8*c + 8*d*x)*a*b**2 + 45*e**(8*c + 8*d* x)*b**3 - 200*e**(6*c + 6*d*x)*a**3 - 120*e**(6*c + 6*d*x)*a**2*b + 40*e** (6*c + 6*d*x)*a*b**2 - 40*e**(6*c + 6*d*x)*b**3 - 225*e**(4*c + 4*d*x)*a** 3 + 45*e**(4*c + 4*d*x)*a**2*b - 35*e**(4*c + 4*d*x)*a*b**2 + 15*e**(4*c + 4*d*x)*b**3 - 120*e**(2*c + 2*d*x)*a**3 + 120*e**(2*c + 2*d*x)*a**2*b - 4 0*e**(2*c + 2*d*x)*a*b**2 - 24*e**(2*c + 2*d*x)*b**3 - 25*a**3 + 45*a**2*b + 5*a*b**2 - b**3)/(10*d*(e**(12*c + 12*d*x) + 4*e**(10*c + 10*d*x) + 5*e **(8*c + 8*d*x) - 5*e**(4*c + 4*d*x) - 4*e**(2*c + 2*d*x) - 1))