\(\int \sinh ^2(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\) [11]

Optimal result

Integrand size = 23, antiderivative size = 71 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {1}{2} (a+b) (a+5 b) x+\frac {(a+b)^2 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {2 b (a+b) \tanh (c+d x)}{d}+\frac {b^2 \tanh ^3(c+d x)}{3 d} \] Output:

-1/2*(a+b)*(a+5*b)*x+1/2*(a+b)^2*cosh(d*x+c)*sinh(d*x+c)/d+2*b*(a+b)*tanh( 
d*x+c)/d+1/3*b^2*tanh(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.99 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {-6 \left (a^2+6 a b+5 b^2\right ) (c+d x)+3 (a+b)^2 \sinh (2 (c+d x))+4 b \left (6 a+7 b-b \text {sech}^2(c+d x)\right ) \tanh (c+d x)}{12 d} \] Input:

Integrate[Sinh[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^2,x]
 

Output:

(-6*(a^2 + 6*a*b + 5*b^2)*(c + d*x) + 3*(a + b)^2*Sinh[2*(c + d*x)] + 4*b* 
(6*a + 7*b - b*Sech[c + d*x]^2)*Tanh[c + d*x])/(12*d)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 4146, 366, 363, 262, 219}

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.

\(\displaystyle \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\sin (i c+i d x)^2 \left (a-b \tan (i c+i d x)^2\right )^2dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \sin (i c+i d x)^2 \left (a-b \tan (i c+i d x)^2\right )^2dx\)

\(\Big \downarrow \) 4146

\(\displaystyle \frac {\int \frac {\tanh ^2(c+d x) \left (b \tanh ^2(c+d x)+a\right )^2}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 366

\(\displaystyle \frac {\frac {(a+b)^2 \tanh ^3(c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}-\frac {1}{2} \int \frac {\tanh ^2(c+d x) \left (a^2+6 b a+3 b^2+2 b^2 \tanh ^2(c+d x)\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 363

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2}{3} b^2 \tanh ^3(c+d x)-(a+b) (a+5 b) \int \frac {\tanh ^2(c+d x)}{1-\tanh ^2(c+d x)}d\tanh (c+d x)\right )+\frac {(a+b)^2 \tanh ^3(c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 262

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2}{3} b^2 \tanh ^3(c+d x)-(a+b) (a+5 b) \left (\int \frac {1}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\tanh (c+d x)\right )\right )+\frac {(a+b)^2 \tanh ^3(c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {1}{2} \left (\frac {2}{3} b^2 \tanh ^3(c+d x)-(a+b) (a+5 b) (\text {arctanh}(\tanh (c+d x))-\tanh (c+d x))\right )+\frac {(a+b)^2 \tanh ^3(c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\)

Input:

Int[Sinh[c + d*x]^2*(a + b*Tanh[c + d*x]^2)^2,x]
 

Output:

(((a + b)^2*Tanh[c + d*x]^3)/(2*(1 - Tanh[c + d*x]^2)) + (-((a + b)*(a + 5 
*b)*(ArcTanh[Tanh[c + d*x]] - Tanh[c + d*x])) + (2*b^2*Tanh[c + d*x]^3)/3) 
/2)/d
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 363

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x 
_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), 
 x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3))   Int[(e*x)^ 
m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d 
, 0] && NeQ[m + 2*p + 3, 0]
 

rule 366

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, 
x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* 
b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1))   Int[(e*x)^m*(a + b*x^2)^(p 
 + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 
2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p 
, -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4146

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]
 

Maple [A] (verified)

Time = 2.09 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.66

method	result	size
derivativedivides	\(\frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (\frac {\sinh \left (d x +c \right )^{5}}{2 \cosh \left (d x +c \right )^{3}}-\frac {5 d x}{2}-\frac {5 c}{2}+\frac {5 \tanh \left (d x +c \right )}{2}+\frac {5 \tanh \left (d x +c \right )^{3}}{6}\right )}{d}\)	\(118\)
default	\(\frac {a^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )+2 a b \left (\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )}-\frac {3 d x}{2}-\frac {3 c}{2}+\frac {3 \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (\frac {\sinh \left (d x +c \right )^{5}}{2 \cosh \left (d x +c \right )^{3}}-\frac {5 d x}{2}-\frac {5 c}{2}+\frac {5 \tanh \left (d x +c \right )}{2}+\frac {5 \tanh \left (d x +c \right )^{3}}{6}\right )}{d}\)	\(118\)
risch	\(-\frac {a^{2} x}{2}-3 a b x -\frac {5 b^{2} x}{2}+\frac {{\mathrm e}^{2 d x +2 c} a^{2}}{8 d}+\frac {{\mathrm e}^{2 d x +2 c} a b}{4 d}+\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} a^{2}}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} a b}{4 d}-\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}-\frac {2 b \left (6 \,{\mathrm e}^{4 d x +4 c} a +9 b \,{\mathrm e}^{4 d x +4 c}+12 \,{\mathrm e}^{2 d x +2 c} a +12 \,{\mathrm e}^{2 d x +2 c} b +6 a +7 b \right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{3}}\)	\(193\)

Input:

int(sinh(d*x+c)^2*(a+tanh(d*x+c)^2*b)^2,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^2*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+2*a*b*(1/2*sinh(d*x+c 
)^3/cosh(d*x+c)-3/2*d*x-3/2*c+3/2*tanh(d*x+c))+b^2*(1/2*sinh(d*x+c)^5/cosh 
(d*x+c)^3-5/2*d*x-5/2*c+5/2*tanh(d*x+c)+5/6*tanh(d*x+c)^3))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 291 vs. \(2 (65) = 130\).

Time = 0.09 (sec) , antiderivative size = 291, normalized size of antiderivative = 4.10 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{5} - 4 \, {\left (3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} d x + 12 \, a b + 14 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} - 12 \, {\left (3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} d x + 12 \, a b + 14 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (30 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 9 \, a^{2} + 66 \, a b + 65 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} - 12 \, {\left (3 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} d x + 12 \, a b + 14 \, b^{2}\right )} \cosh \left (d x + c\right ) + 3 \, {\left (5 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} + {\left (9 \, a^{2} + 66 \, a b + 65 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, a^{2} + 20 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \] Input:

integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="fricas")
 

Output:

1/24*(3*(a^2 + 2*a*b + b^2)*sinh(d*x + c)^5 - 4*(3*(a^2 + 6*a*b + 5*b^2)*d 
*x + 12*a*b + 14*b^2)*cosh(d*x + c)^3 - 12*(3*(a^2 + 6*a*b + 5*b^2)*d*x + 
12*a*b + 14*b^2)*cosh(d*x + c)*sinh(d*x + c)^2 + (30*(a^2 + 2*a*b + b^2)*c 
osh(d*x + c)^2 + 9*a^2 + 66*a*b + 65*b^2)*sinh(d*x + c)^3 - 12*(3*(a^2 + 6 
*a*b + 5*b^2)*d*x + 12*a*b + 14*b^2)*cosh(d*x + c) + 3*(5*(a^2 + 2*a*b + b 
^2)*cosh(d*x + c)^4 + (9*a^2 + 66*a*b + 65*b^2)*cosh(d*x + c)^2 + 2*a^2 + 
20*a*b + 10*b^2)*sinh(d*x + c))/(d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c)*sin 
h(d*x + c)^2 + 3*d*cosh(d*x + c))
 

Sympy [F]

\[ \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{2} \sinh ^{2}{\left (c + d x \right )}\, dx \] Input:

integrate(sinh(d*x+c)**2*(a+b*tanh(d*x+c)**2)**2,x)
 

Output:

Integral((a + b*tanh(c + d*x)**2)**2*sinh(c + d*x)**2, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 217 vs. \(2 (65) = 130\).

Time = 0.04 (sec) , antiderivative size = 217, normalized size of antiderivative = 3.06 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {1}{8} \, a^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {1}{24} \, b^{2} {\left (\frac {60 \, {\left (d x + c\right )}}{d} + \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} - \frac {1}{4} \, a b {\left (\frac {12 \, {\left (d x + c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {17 \, e^{\left (-2 \, d x - 2 \, c\right )} + 1}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )}\right )}}\right )} \] Input:

integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="maxima")
 

Output:

-1/8*a^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/24*b^2*(60*(d* 
x + c)/d + 3*e^(-2*d*x - 2*c)/d - (121*e^(-2*d*x - 2*c) + 201*e^(-4*d*x - 
4*c) + 147*e^(-6*d*x - 6*c) + 3)/(d*(e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) 
 + 3*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c)))) - 1/4*a*b*(12*(d*x + c)/d + e^ 
(-2*d*x - 2*c)/d - (17*e^(-2*d*x - 2*c) + 1)/(d*(e^(-2*d*x - 2*c) + e^(-4* 
d*x - 4*c))))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (65) = 130\).

Time = 0.17 (sec) , antiderivative size = 213, normalized size of antiderivative = 3.00 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 12 \, {\left (a^{2} + 6 \, a b + 5 \, b^{2}\right )} {\left (d x + c\right )} + 3 \, {\left (2 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 10 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2} - 2 \, a b - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - \frac {16 \, {\left (6 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b + 7 \, b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \] Input:

integrate(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x, algorithm="giac")
 

Output:

1/24*(3*a^2*e^(2*d*x + 2*c) + 6*a*b*e^(2*d*x + 2*c) + 3*b^2*e^(2*d*x + 2*c 
) - 12*(a^2 + 6*a*b + 5*b^2)*(d*x + c) + 3*(2*a^2*e^(2*d*x + 2*c) + 12*a*b 
*e^(2*d*x + 2*c) + 10*b^2*e^(2*d*x + 2*c) - a^2 - 2*a*b - b^2)*e^(-2*d*x - 
 2*c) - 16*(6*a*b*e^(4*d*x + 4*c) + 9*b^2*e^(4*d*x + 4*c) + 12*a*b*e^(2*d* 
x + 2*c) + 12*b^2*e^(2*d*x + 2*c) + 6*a*b + 7*b^2)/(e^(2*d*x + 2*c) + 1)^3 
)/d
 

Mupad [B] (verification not implemented)

Time = 2.45 (sec) , antiderivative size = 248, normalized size of antiderivative = 3.49 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2}{8\,d}-x\,\left (\frac {a^2}{2}+3\,a\,b+\frac {5\,b^2}{2}\right )-\frac {\frac {2\,\left (3\,b^2+2\,a\,b\right )}{3\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (b^2+2\,a\,b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,b^2+2\,a\,b\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,\left (3\,b^2+2\,a\,b\right )}{3\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{-2\,c-2\,d\,x}\,{\left (a+b\right )}^2}{8\,d}-\frac {\frac {2\,\left (b^2+2\,a\,b\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,b^2+2\,a\,b\right )}{3\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1} \] Input:

int(sinh(c + d*x)^2*(a + b*tanh(c + d*x)^2)^2,x)
 

Output:

(exp(2*c + 2*d*x)*(a + b)^2)/(8*d) - x*(3*a*b + a^2/2 + (5*b^2)/2) - ((2*( 
2*a*b + 3*b^2))/(3*d) + (4*exp(2*c + 2*d*x)*(2*a*b + b^2))/(3*d) + (2*exp( 
4*c + 4*d*x)*(2*a*b + 3*b^2))/(3*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d 
*x) + exp(6*c + 6*d*x) + 1) - (2*(2*a*b + 3*b^2))/(3*d*(exp(2*c + 2*d*x) + 
 1)) - (exp(- 2*c - 2*d*x)*(a + b)^2)/(8*d) - ((2*(2*a*b + b^2))/(3*d) + ( 
2*exp(2*c + 2*d*x)*(2*a*b + 3*b^2))/(3*d))/(2*exp(2*c + 2*d*x) + exp(4*c + 
 4*d*x) + 1)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 446, normalized size of antiderivative = 6.28 \[ \int \sinh ^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {46 e^{8 d x +8 c} a b +3 e^{10 d x +10 c} a^{2}+3 e^{10 d x +10 c} b^{2}-12 e^{2 d x +2 c} a^{2} d x -12 e^{8 d x +8 c} a^{2} d x -60 e^{8 d x +8 c} b^{2} d x -36 e^{6 d x +6 c} a^{2} d x -180 e^{6 d x +6 c} b^{2} d x -36 e^{4 d x +4 c} a^{2} d x -180 e^{4 d x +4 c} b^{2} d x -3 a^{2}-72 e^{2 d x +2 c} a b d x -72 e^{8 d x +8 c} a b d x -216 e^{6 d x +6 c} a b d x -216 e^{4 d x +4 c} a b d x -75 e^{2 d x +2 c} b^{2}-60 e^{2 d x +2 c} b^{2} d x -3 b^{2}+6 e^{10 d x +10 c} a b -120 e^{4 d x +4 c} a b -86 e^{2 d x +2 c} a b -6 a b +7 e^{8 d x +8 c} a^{2}+55 e^{8 d x +8 c} b^{2}-12 e^{4 d x +4 c} a^{2}-60 e^{4 d x +4 c} b^{2}-11 e^{2 d x +2 c} a^{2}}{24 e^{2 d x +2 c} d \left (e^{6 d x +6 c}+3 e^{4 d x +4 c}+3 e^{2 d x +2 c}+1\right )} \] Input:

int(sinh(d*x+c)^2*(a+b*tanh(d*x+c)^2)^2,x)
 

Output:

(3*e**(10*c + 10*d*x)*a**2 + 6*e**(10*c + 10*d*x)*a*b + 3*e**(10*c + 10*d* 
x)*b**2 - 12*e**(8*c + 8*d*x)*a**2*d*x + 7*e**(8*c + 8*d*x)*a**2 - 72*e**( 
8*c + 8*d*x)*a*b*d*x + 46*e**(8*c + 8*d*x)*a*b - 60*e**(8*c + 8*d*x)*b**2* 
d*x + 55*e**(8*c + 8*d*x)*b**2 - 36*e**(6*c + 6*d*x)*a**2*d*x - 216*e**(6* 
c + 6*d*x)*a*b*d*x - 180*e**(6*c + 6*d*x)*b**2*d*x - 36*e**(4*c + 4*d*x)*a 
**2*d*x - 12*e**(4*c + 4*d*x)*a**2 - 216*e**(4*c + 4*d*x)*a*b*d*x - 120*e* 
*(4*c + 4*d*x)*a*b - 180*e**(4*c + 4*d*x)*b**2*d*x - 60*e**(4*c + 4*d*x)*b 
**2 - 12*e**(2*c + 2*d*x)*a**2*d*x - 11*e**(2*c + 2*d*x)*a**2 - 72*e**(2*c 
 + 2*d*x)*a*b*d*x - 86*e**(2*c + 2*d*x)*a*b - 60*e**(2*c + 2*d*x)*b**2*d*x 
 - 75*e**(2*c + 2*d*x)*b**2 - 3*a**2 - 6*a*b - 3*b**2)/(24*e**(2*c + 2*d*x 
)*d*(e**(6*c + 6*d*x) + 3*e**(4*c + 4*d*x) + 3*e**(2*c + 2*d*x) + 1))