\(\int \text {sech}^6(c+d x) (a+b \sinh ^2(c+d x))^3 \, dx\) [277]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=b^3 x+\frac {\left (a^3-b^3\right ) \tanh (c+d x)}{d}-\frac {(a-b)^2 (2 a+b) \tanh ^3(c+d x)}{3 d}+\frac {(a-b)^3 \tanh ^5(c+d x)}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 1.30 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.39 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {b^3 (c+d x)}{d}+\frac {(a-b) \left (64 a^2+22 a b+49 b^2+12 \left (4 a^2+7 a b+4 b^2\right ) \cosh (2 (c+d x))+\left (8 a^2+14 a b+23 b^2\right ) \cosh (4 (c+d x))\right ) \text {sech}^4(c+d x) \tanh (c+d x)}{120 d} \] Input:

Integrate[Sech[c + d*x]^6*(a + b*Sinh[c + d*x]^2)^3,x]
 

Output:

(b^3*(c + d*x))/d + ((a - b)*(64*a^2 + 22*a*b + 49*b^2 + 12*(4*a^2 + 7*a*b 
 + 4*b^2)*Cosh[2*(c + d*x)] + (8*a^2 + 14*a*b + 23*b^2)*Cosh[4*(c + d*x)]) 
*Sech[c + d*x]^4*Tanh[c + d*x])/(120*d)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3670, 300, 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.

\(\displaystyle \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a-b \sin (i c+i d x)^2\right )^3}{\cos (i c+i d x)^6}dx\)

\(\Big \downarrow \) 3670

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

\(\Big \downarrow \) 300

\(\displaystyle \frac {\int \left ((a-b)^3 \tanh ^4(c+d x)-(a-b)^2 (2 a+b) \tanh ^2(c+d x)+a^3-b^3+\frac {b^3}{1-\tanh ^2(c+d x)}\right )d\tanh (c+d x)}{d}\)

\(\Big \downarrow \) 2009

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

Input:

Int[Sech[c + d*x]^6*(a + b*Sinh[c + d*x]^2)^3,x]
 

Output:

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

Defintions of rubi rules used

rule 300

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 3670

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( 
p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f   Su 
bst[Int[(a + (a + b)*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + p + 1), x], x, Tan[e 
 + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(198\) vs. \(2(70)=140\).

Time = 0.16 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.69

\[\frac {a^{3} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )+3 a^{2} b \left (-\frac {\sinh \left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{4}\right )+3 b^{2} a \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )+b^{3} \left (d x +c -\tanh \left (d x +c \right )-\frac {\tanh \left (d x +c \right )^{3}}{3}-\frac {\tanh \left (d x +c \right )^{5}}{5}\right )}{d}\]

Input:

int(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x)
 

Output:

1/d*(a^3*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)^2)*tanh(d*x+c)+3*a^2*b*( 
-1/4*sinh(d*x+c)/cosh(d*x+c)^5+1/4*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c 
)^2)*tanh(d*x+c))+3*b^2*a*(-1/2*sinh(d*x+c)^3/cosh(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+4/15*sech(d*x+c)^2)*tanh(d*x+c 
))+b^3*(d*x+c-tanh(d*x+c)-1/3*tanh(d*x+c)^3-1/5*tanh(d*x+c)^5))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 530 vs. \(2 (70) = 140\).

Time = 0.09 (sec) , antiderivative size = 530, normalized size of antiderivative = 7.16 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {{\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{5} + 5 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + {\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (8 \, a^{3} + 6 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3} + 2 \, {\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (2 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, {\left (15 \, b^{3} d x - 8 \, a^{3} - 6 \, a^{2} b - 9 \, a b^{2} + 23 \, b^{3}\right )} \cosh \left (d x + c\right ) + 5 \, {\left ({\left (8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )} \cosh \left (d x + c\right )^{4} + 16 \, a^{3} - 24 \, a^{2} b + 18 \, a b^{2} - 10 \, b^{3} + 3 \, {\left (8 \, a^{3} + 6 \, a^{2} b - 9 \, a b^{2} - 5 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{15 \, {\left (d \cosh \left (d x + c\right )^{5} + 5 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 5 \, d \cosh \left (d x + c\right )^{3} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right )\right )}} \] Input:

integrate(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

1/15*((15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 23*b^3)*cosh(d*x + c)^5 + 
5*(15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 23*b^3)*cosh(d*x + c)*sinh(d*x 
 + c)^4 + (8*a^3 + 6*a^2*b + 9*a*b^2 - 23*b^3)*sinh(d*x + c)^5 + 5*(15*b^3 
*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 23*b^3)*cosh(d*x + c)^3 + 5*(8*a^3 + 6* 
a^2*b - 9*a*b^2 - 5*b^3 + 2*(8*a^3 + 6*a^2*b + 9*a*b^2 - 23*b^3)*cosh(d*x 
+ c)^2)*sinh(d*x + c)^3 + 5*(2*(15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 2 
3*b^3)*cosh(d*x + c)^3 + 3*(15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a*b^2 + 23*b^ 
3)*cosh(d*x + c))*sinh(d*x + c)^2 + 10*(15*b^3*d*x - 8*a^3 - 6*a^2*b - 9*a 
*b^2 + 23*b^3)*cosh(d*x + c) + 5*((8*a^3 + 6*a^2*b + 9*a*b^2 - 23*b^3)*cos 
h(d*x + c)^4 + 16*a^3 - 24*a^2*b + 18*a*b^2 - 10*b^3 + 3*(8*a^3 + 6*a^2*b 
- 9*a*b^2 - 5*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x + c)^5 + 5* 
d*cosh(d*x + c)*sinh(d*x + c)^4 + 5*d*cosh(d*x + c)^3 + 5*(2*d*cosh(d*x + 
c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^2 + 10*d*cosh(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:

integrate(sech(d*x+c)**6*(a+b*sinh(d*x+c)**2)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 824 vs. \(2 (70) = 140\).

Time = 0.06 (sec) , antiderivative size = 824, normalized size of antiderivative = 11.14 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \] Input:

integrate(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

1/15*b^3*(15*x + 15*c/d - 2*(70*e^(-2*d*x - 2*c) + 140*e^(-4*d*x - 4*c) + 
90*e^(-6*d*x - 6*c) + 45*e^(-8*d*x - 8*c) + 23)/(d*(5*e^(-2*d*x - 2*c) + 1 
0*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))) + 16/15*a^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 1 
0*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)) + 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)) + 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))) + 4/5*a^2*b*(5*e^(-2*d*x 
 - 2*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^(-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)) + 15*e^(-6*d*x - 6*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))) + 
6/5*a*b^2*(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...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 213 vs. \(2 (70) = 140\).

Time = 0.18 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.88 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {15 \, {\left (d x + c\right )} b^{3} - \frac {2 \, {\left (45 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} - 45 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} + 90 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 90 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 80 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} - 30 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 140 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 30 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 70 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 8 \, a^{3} + 6 \, a^{2} b + 9 \, a b^{2} - 23 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \] Input:

integrate(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

1/15*(15*(d*x + c)*b^3 - 2*(45*a*b^2*e^(8*d*x + 8*c) - 45*b^3*e^(8*d*x + 8 
*c) + 90*a^2*b*e^(6*d*x + 6*c) - 90*b^3*e^(6*d*x + 6*c) + 80*a^3*e^(4*d*x 
+ 4*c) - 30*a^2*b*e^(4*d*x + 4*c) + 90*a*b^2*e^(4*d*x + 4*c) - 140*b^3*e^( 
4*d*x + 4*c) + 40*a^3*e^(2*d*x + 2*c) + 30*a^2*b*e^(2*d*x + 2*c) - 70*b^3* 
e^(2*d*x + 2*c) + 8*a^3 + 6*a^2*b + 9*a*b^2 - 23*b^3)/(e^(2*d*x + 2*c) + 1 
)^5)/d
 

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 563, normalized size of antiderivative = 7.61 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=b^3\,x-\frac {\frac {2\,\left (8\,a^3-12\,a^2\,b+9\,a\,b^2-5\,b^3\right )}{15\,d}-\frac {12\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a\,b^2-b^3\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}+\frac {\frac {6\,\left (a\,b^2-a^2\,b\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b^2-b^3\right )}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}+\frac {\frac {6\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {18\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}-\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (8\,a^3-12\,a^2\,b+9\,a\,b^2-5\,b^3\right )}{5\,d}-\frac {6\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a\,b^2-b^3\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {6\,\left (a\,b^2-b^3\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}-\frac {24\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a\,b^2-a^2\,b\right )}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (8\,a^3-12\,a^2\,b+9\,a\,b^2-5\,b^3\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{8\,c+8\,d\,x}\,\left (a\,b^2-b^3\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {6\,\left (a\,b^2-b^3\right )}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:

int((a + b*sinh(c + d*x)^2)^3/cosh(c + d*x)^6,x)
 

Output:

b^3*x - ((2*(9*a*b^2 - 12*a^2*b + 8*a^3 - 5*b^3))/(15*d) - (12*exp(2*c + 2 
*d*x)*(a*b^2 - a^2*b))/(5*d) + (6*exp(4*c + 4*d*x)*(a*b^2 - 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) + ((6*(a*b 
^2 - a^2*b))/(5*d) - (6*exp(2*c + 2*d*x)*(a*b^2 - b^3))/(5*d))/(2*exp(2*c 
+ 2*d*x) + exp(4*c + 4*d*x) + 1) + ((6*(a*b^2 - a^2*b))/(5*d) + (18*exp(4* 
c + 4*d*x)*(a*b^2 - a^2*b))/(5*d) - (2*exp(2*c + 2*d*x)*(9*a*b^2 - 12*a^2* 
b + 8*a^3 - 5*b^3))/(5*d) - (6*exp(6*c + 6*d*x)*(a*b^2 - b^3))/(5*d))/(4*e 
xp(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) - ((6*(a*b^2 - b^3))/(5*d) - (24*exp(2*c + 2*d*x)*(a*b^2 - a^2*b)) 
/(5*d) - (24*exp(6*c + 6*d*x)*(a*b^2 - a^2*b))/(5*d) + (4*exp(4*c + 4*d*x) 
*(9*a*b^2 - 12*a^2*b + 8*a^3 - 5*b^3))/(5*d) + (6*exp(8*c + 8*d*x)*(a*b^2 
- b^3))/(5*d))/(5*exp(2*c + 2*d*x) + 10*exp(4*c + 4*d*x) + 10*exp(6*c + 6* 
d*x) + 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) + 1) - (6*(a*b^2 - b^3))/(5 
*d*(exp(2*c + 2*d*x) + 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 347, normalized size of antiderivative = 4.69 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {18 e^{10 d x +10 c} a \,b^{2}+15 e^{10 d x +10 c} b^{3} d x -18 e^{10 d x +10 c} b^{3}+75 e^{8 d x +8 c} b^{3} d x -180 e^{6 d x +6 c} a^{2} b +180 e^{6 d x +6 c} a \,b^{2}+150 e^{6 d x +6 c} b^{3} d x -160 e^{4 d x +4 c} a^{3}+60 e^{4 d x +4 c} a^{2} b +150 e^{4 d x +4 c} b^{3} d x +100 e^{4 d x +4 c} b^{3}-80 e^{2 d x +2 c} a^{3}-60 e^{2 d x +2 c} a^{2} b +90 e^{2 d x +2 c} a \,b^{2}+75 e^{2 d x +2 c} b^{3} d x +50 e^{2 d x +2 c} b^{3}-16 a^{3}-12 a^{2} b +15 b^{3} d x +28 b^{3}}{15 d \left (e^{10 d x +10 c}+5 e^{8 d x +8 c}+10 e^{6 d x +6 c}+10 e^{4 d x +4 c}+5 e^{2 d x +2 c}+1\right )} \] Input:

int(sech(d*x+c)^6*(a+b*sinh(d*x+c)^2)^3,x)
 

Output:

(18*e**(10*c + 10*d*x)*a*b**2 + 15*e**(10*c + 10*d*x)*b**3*d*x - 18*e**(10 
*c + 10*d*x)*b**3 + 75*e**(8*c + 8*d*x)*b**3*d*x - 180*e**(6*c + 6*d*x)*a* 
*2*b + 180*e**(6*c + 6*d*x)*a*b**2 + 150*e**(6*c + 6*d*x)*b**3*d*x - 160*e 
**(4*c + 4*d*x)*a**3 + 60*e**(4*c + 4*d*x)*a**2*b + 150*e**(4*c + 4*d*x)*b 
**3*d*x + 100*e**(4*c + 4*d*x)*b**3 - 80*e**(2*c + 2*d*x)*a**3 - 60*e**(2* 
c + 2*d*x)*a**2*b + 90*e**(2*c + 2*d*x)*a*b**2 + 75*e**(2*c + 2*d*x)*b**3* 
d*x + 50*e**(2*c + 2*d*x)*b**3 - 16*a**3 - 12*a**2*b + 15*b**3*d*x + 28*b* 
*3)/(15*d*(e**(10*c + 10*d*x) + 5*e**(8*c + 8*d*x) + 10*e**(6*c + 6*d*x) + 
 10*e**(4*c + 4*d*x) + 5*e**(2*c + 2*d*x) + 1))