\(\int \text {sech}^3(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\) [103]

Optimal result

Integrand size = 23, antiderivative size = 172 \[ \int \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {\left (64 a^3+48 a^2 b+24 a b^2+5 b^3\right ) \arctan (\sinh (c+d x))}{128 d}+\frac {\left (64 a^3+48 a^2 b+24 a b^2+5 b^3\right ) \text {sech}(c+d x) \tanh (c+d x)}{128 d}-\frac {b \left (144 a^2+168 a b+59 b^2\right ) \text {sech}^3(c+d x) \tanh (c+d x)}{192 d}+\frac {b^2 (24 a+17 b) \text {sech}^5(c+d x) \tanh (c+d x)}{48 d}-\frac {b^3 \text {sech}^7(c+d x) \tanh (c+d x)}{8 d} \] Output:

1/128*(64*a^3+48*a^2*b+24*a*b^2+5*b^3)*arctan(sinh(d*x+c))/d+1/128*(64*a^3 
+48*a^2*b+24*a*b^2+5*b^3)*sech(d*x+c)*tanh(d*x+c)/d-1/192*b*(144*a^2+168*a 
*b+59*b^2)*sech(d*x+c)^3*tanh(d*x+c)/d+1/48*b^2*(24*a+17*b)*sech(d*x+c)^5* 
tanh(d*x+c)/d-1/8*b^3*sech(d*x+c)^7*tanh(d*x+c)/d
 

Mathematica [A] (verified)

Time = 12.14 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.92 \[ \int \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {6 \left (64 a^3+48 a^2 b+24 a b^2+5 b^3\right ) \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )+3 \left (64 a^3+48 a^2 b+24 a b^2+5 b^3\right ) \text {sech}(c+d x) \tanh (c+d x)-2 b \left (144 a^2+168 a b+59 b^2\right ) \text {sech}^3(c+d x) \tanh (c+d x)+8 b^2 (24 a+17 b) \text {sech}^5(c+d x) \tanh (c+d x)-48 b^3 \text {sech}^7(c+d x) \tanh (c+d x)}{384 d} \] Input:

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

Output:

(6*(64*a^3 + 48*a^2*b + 24*a*b^2 + 5*b^3)*ArcTan[Tanh[(c + d*x)/2]] + 3*(6 
4*a^3 + 48*a^2*b + 24*a*b^2 + 5*b^3)*Sech[c + d*x]*Tanh[c + d*x] - 2*b*(14 
4*a^2 + 168*a*b + 59*b^2)*Sech[c + d*x]^3*Tanh[c + d*x] + 8*b^2*(24*a + 17 
*b)*Sech[c + d*x]^5*Tanh[c + d*x] - 48*b^3*Sech[c + d*x]^7*Tanh[c + d*x])/ 
(384*d)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4159, 315, 401, 25, 298, 215, 216}

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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sec (i c+i d x)^3 \left (a-b \tan (i c+i d x)^2\right )^3dx\)

\(\Big \downarrow \) 4159

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

\(\Big \downarrow \) 315

\(\displaystyle \frac {\frac {1}{8} \int \frac {\left ((a+b) \sinh ^2(c+d x)+a\right ) \left ((a+b) (8 a+5 b) \sinh ^2(c+d x)+a (8 a+b)\right )}{\left (\sinh ^2(c+d x)+1\right )^4}d\sinh (c+d x)-\frac {b \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}{8 \left (\sinh ^2(c+d x)+1\right )^4}}{d}\)

\(\Big \downarrow \) 401

\(\displaystyle \frac {\frac {1}{8} \left (-\frac {1}{6} \int -\frac {3 (a+b) \left (16 a^2+14 b a+5 b^2\right ) \sinh ^2(c+d x)+a \left (48 a^2+18 b a+5 b^2\right )}{\left (\sinh ^2(c+d x)+1\right )^3}d\sinh (c+d x)-\frac {b (12 a+5 b) \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{6 \left (\sinh ^2(c+d x)+1\right )^3}\right )-\frac {b \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}{8 \left (\sinh ^2(c+d x)+1\right )^4}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \int \frac {3 (a+b) \left (16 a^2+14 b a+5 b^2\right ) \sinh ^2(c+d x)+a \left (48 a^2+18 b a+5 b^2\right )}{\left (\sinh ^2(c+d x)+1\right )^3}d\sinh (c+d x)-\frac {b (12 a+5 b) \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{6 \left (\sinh ^2(c+d x)+1\right )^3}\right )-\frac {b \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}{8 \left (\sinh ^2(c+d x)+1\right )^4}}{d}\)

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} \left (64 a^3+48 a^2 b+24 a b^2+5 b^3\right ) \int \frac {1}{\left (\sinh ^2(c+d x)+1\right )^2}d\sinh (c+d x)-\frac {b \left (72 a^2+52 a b+15 b^2\right ) \sinh (c+d x)}{4 \left (\sinh ^2(c+d x)+1\right )^2}\right )-\frac {b (12 a+5 b) \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{6 \left (\sinh ^2(c+d x)+1\right )^3}\right )-\frac {b \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}{8 \left (\sinh ^2(c+d x)+1\right )^4}}{d}\)

\(\Big \downarrow \) 215

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} \left (64 a^3+48 a^2 b+24 a b^2+5 b^3\right ) \left (\frac {1}{2} \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)+\frac {\sinh (c+d x)}{2 \left (\sinh ^2(c+d x)+1\right )}\right )-\frac {b \left (72 a^2+52 a b+15 b^2\right ) \sinh (c+d x)}{4 \left (\sinh ^2(c+d x)+1\right )^2}\right )-\frac {b (12 a+5 b) \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{6 \left (\sinh ^2(c+d x)+1\right )^3}\right )-\frac {b \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}{8 \left (\sinh ^2(c+d x)+1\right )^4}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{8} \left (\frac {1}{6} \left (\frac {3}{4} \left (64 a^3+48 a^2 b+24 a b^2+5 b^3\right ) \left (\frac {1}{2} \arctan (\sinh (c+d x))+\frac {\sinh (c+d x)}{2 \left (\sinh ^2(c+d x)+1\right )}\right )-\frac {b \left (72 a^2+52 a b+15 b^2\right ) \sinh (c+d x)}{4 \left (\sinh ^2(c+d x)+1\right )^2}\right )-\frac {b (12 a+5 b) \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{6 \left (\sinh ^2(c+d x)+1\right )^3}\right )-\frac {b \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )^2}{8 \left (\sinh ^2(c+d x)+1\right )^4}}{d}\)

Input:

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

Output:

(-1/8*(b*Sinh[c + d*x]*(a + (a + b)*Sinh[c + d*x]^2)^2)/(1 + Sinh[c + d*x] 
^2)^4 + (-1/6*(b*(12*a + 5*b)*Sinh[c + d*x]*(a + (a + b)*Sinh[c + d*x]^2)) 
/(1 + Sinh[c + d*x]^2)^3 + (-1/4*(b*(72*a^2 + 52*a*b + 15*b^2)*Sinh[c + d* 
x])/(1 + Sinh[c + d*x]^2)^2 + (3*(64*a^3 + 48*a^2*b + 24*a*b^2 + 5*b^3)*(A 
rcTan[Sinh[c + d*x]]/2 + Sinh[c + d*x]/(2*(1 + Sinh[c + d*x]^2))))/4)/6)/8 
)/d
 

Defintions of rubi rules used

rule 25

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

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 298

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

rule 315

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 401

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
q/(a*b*2*(p + 1))), x] + Simp[1/(a*b*2*(p + 1))   Int[(a + b*x^2)^(p + 1)*( 
c + d*x^2)^(q - 1)*Simp[c*(b*e*2*(p + 1) + b*e - a*f) + d*(b*e*2*(p + 1) + 
(b*e - a*f)*(2*q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && L 
tQ[p, -1] && GtQ[q, 0]
 

rule 3042

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

rule 4159

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_ 
))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff/f 
  Subst[Int[ExpandToSum[b*(ff*x)^n + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2 
*x^2)^((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f} 
, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2] && IntegerQ[p]
 

Maple [A] (verified)

Time = 80.24 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.69

method	result	size
derivativedivides	\(\frac {a^{3} \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+3 a^{2} b \left (-\frac {\sinh \left (d x +c \right )}{3 \cosh \left (d x +c \right )^{4}}+\frac {\left (\frac {\operatorname {sech}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (d x +c \right )}{8}\right ) \tanh \left (d x +c \right )}{3}+\frac {\arctan \left ({\mathrm e}^{d x +c}\right )}{4}\right )+3 b^{2} a \left (-\frac {\sinh \left (d x +c \right )^{3}}{3 \cosh \left (d x +c \right )^{6}}-\frac {\sinh \left (d x +c \right )}{5 \cosh \left (d x +c \right )^{6}}+\frac {\left (\frac {\operatorname {sech}\left (d x +c \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (d x +c \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (d x +c \right )}{16}\right ) \tanh \left (d x +c \right )}{5}+\frac {\arctan \left ({\mathrm e}^{d x +c}\right )}{8}\right )+b^{3} \left (-\frac {\sinh \left (d x +c \right )^{5}}{3 \cosh \left (d x +c \right )^{8}}-\frac {\sinh \left (d x +c \right )^{3}}{3 \cosh \left (d x +c \right )^{8}}-\frac {\sinh \left (d x +c \right )}{7 \cosh \left (d x +c \right )^{8}}+\frac {\left (\frac {\operatorname {sech}\left (d x +c \right )^{7}}{8}+\frac {7 \operatorname {sech}\left (d x +c \right )^{5}}{48}+\frac {35 \operatorname {sech}\left (d x +c \right )^{3}}{192}+\frac {35 \,\operatorname {sech}\left (d x +c \right )}{128}\right ) \tanh \left (d x +c \right )}{7}+\frac {5 \arctan \left ({\mathrm e}^{d x +c}\right )}{64}\right )}{d}\)	\(291\)
default	\(\frac {a^{3} \left (\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+3 a^{2} b \left (-\frac {\sinh \left (d x +c \right )}{3 \cosh \left (d x +c \right )^{4}}+\frac {\left (\frac {\operatorname {sech}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (d x +c \right )}{8}\right ) \tanh \left (d x +c \right )}{3}+\frac {\arctan \left ({\mathrm e}^{d x +c}\right )}{4}\right )+3 b^{2} a \left (-\frac {\sinh \left (d x +c \right )^{3}}{3 \cosh \left (d x +c \right )^{6}}-\frac {\sinh \left (d x +c \right )}{5 \cosh \left (d x +c \right )^{6}}+\frac {\left (\frac {\operatorname {sech}\left (d x +c \right )^{5}}{6}+\frac {5 \operatorname {sech}\left (d x +c \right )^{3}}{24}+\frac {5 \,\operatorname {sech}\left (d x +c \right )}{16}\right ) \tanh \left (d x +c \right )}{5}+\frac {\arctan \left ({\mathrm e}^{d x +c}\right )}{8}\right )+b^{3} \left (-\frac {\sinh \left (d x +c \right )^{5}}{3 \cosh \left (d x +c \right )^{8}}-\frac {\sinh \left (d x +c \right )^{3}}{3 \cosh \left (d x +c \right )^{8}}-\frac {\sinh \left (d x +c \right )}{7 \cosh \left (d x +c \right )^{8}}+\frac {\left (\frac {\operatorname {sech}\left (d x +c \right )^{7}}{8}+\frac {7 \operatorname {sech}\left (d x +c \right )^{5}}{48}+\frac {35 \operatorname {sech}\left (d x +c \right )^{3}}{192}+\frac {35 \,\operatorname {sech}\left (d x +c \right )}{128}\right ) \tanh \left (d x +c \right )}{7}+\frac {5 \arctan \left ({\mathrm e}^{d x +c}\right )}{64}\right )}{d}\)	\(291\)
risch	\(\frac {{\mathrm e}^{d x +c} \left (-1765 b^{3} {\mathrm e}^{8 d x +8 c}+1765 b^{3} {\mathrm e}^{6 d x +6 c}-895 b^{3} {\mathrm e}^{4 d x +4 c}-960 \,{\mathrm e}^{2 d x +2 c} a^{3}-72 b^{2} a -192 a^{3}-144 a^{2} b -432 a^{2} b \,{\mathrm e}^{12 d x +12 c}-2160 a^{2} b \,{\mathrm e}^{10 d x +10 c}-312 a \,b^{2} {\mathrm e}^{10 d x +10 c}-1584 a^{2} b \,{\mathrm e}^{8 d x +8 c}+744 a \,b^{2} {\mathrm e}^{8 d x +8 c}+1584 a^{2} b \,{\mathrm e}^{6 d x +6 c}-744 a \,b^{2} {\mathrm e}^{6 d x +6 c}+2160 a^{2} b \,{\mathrm e}^{4 d x +4 c}+312 a \,b^{2} {\mathrm e}^{4 d x +4 c}+432 a^{2} b \,{\mathrm e}^{2 d x +2 c}+984 a \,b^{2} {\mathrm e}^{2 d x +2 c}-15 b^{3}-960 a^{3} {\mathrm e}^{6 d x +6 c}+895 b^{3} {\mathrm e}^{10 d x +10 c}+960 a^{3} {\mathrm e}^{8 d x +8 c}-397 b^{3} {\mathrm e}^{12 d x +12 c}+397 b^{3} {\mathrm e}^{2 d x +2 c}+192 a^{3} {\mathrm e}^{14 d x +14 c}-984 a \,b^{2} {\mathrm e}^{12 d x +12 c}+144 a^{2} b \,{\mathrm e}^{14 d x +14 c}+72 a \,b^{2} {\mathrm e}^{14 d x +14 c}+15 b^{3} {\mathrm e}^{14 d x +14 c}+960 a^{3} {\mathrm e}^{12 d x +12 c}-1728 \,{\mathrm e}^{4 d x +4 c} a^{3}+1728 \,{\mathrm e}^{10 d x +10 c} a^{3}\right )}{192 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{8}}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{3}}{2 d}+\frac {3 i b \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{2}}{8 d}+\frac {3 i b^{2} \ln \left ({\mathrm e}^{d x +c}+i\right ) a}{16 d}+\frac {5 i \ln \left ({\mathrm e}^{d x +c}+i\right ) b^{3}}{128 d}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{3}}{2 d}-\frac {3 i b \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{2}}{8 d}-\frac {3 i b^{2} \ln \left ({\mathrm e}^{d x +c}-i\right ) a}{16 d}-\frac {5 i b^{3} \ln \left ({\mathrm e}^{d x +c}-i\right )}{128 d}\)	\(582\)

Input:

int(sech(d*x+c)^3*(a+tanh(d*x+c)^2*b)^3,x,method=_RETURNVERBOSE)
 

Output:

1/d*(a^3*(1/2*sech(d*x+c)*tanh(d*x+c)+arctan(exp(d*x+c)))+3*a^2*b*(-1/3/co 
sh(d*x+c)^4*sinh(d*x+c)+1/3*(1/4*sech(d*x+c)^3+3/8*sech(d*x+c))*tanh(d*x+c 
)+1/4*arctan(exp(d*x+c)))+3*b^2*a*(-1/3*sinh(d*x+c)^3/cosh(d*x+c)^6-1/5*si 
nh(d*x+c)/cosh(d*x+c)^6+1/5*(1/6*sech(d*x+c)^5+5/24*sech(d*x+c)^3+5/16*sec 
h(d*x+c))*tanh(d*x+c)+1/8*arctan(exp(d*x+c)))+b^3*(-1/3*sinh(d*x+c)^5/cosh 
(d*x+c)^8-1/3*sinh(d*x+c)^3/cosh(d*x+c)^8-1/7*sinh(d*x+c)/cosh(d*x+c)^8+1/ 
7*(1/8*sech(d*x+c)^7+7/48*sech(d*x+c)^5+35/192*sech(d*x+c)^3+35/128*sech(d 
*x+c))*tanh(d*x+c)+5/64*arctan(exp(d*x+c))))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 6114 vs. \(2 (162) = 324\).

Time = 0.12 (sec) , antiderivative size = 6114, normalized size of antiderivative = 35.55 \[ \int \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

\[ \int \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \operatorname {sech}^{3}{\left (c + d x \right )}\, dx \] Input:

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

Output:

Integral((a + b*tanh(c + d*x)**2)**3*sech(c + d*x)**3, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 553 vs. \(2 (162) = 324\).

Time = 0.14 (sec) , antiderivative size = 553, normalized size of antiderivative = 3.22 \[ \int \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:

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

Output:

-1/192*b^3*(15*arctan(e^(-d*x - c))/d - (15*e^(-d*x - c) - 397*e^(-3*d*x - 
 3*c) + 895*e^(-5*d*x - 5*c) - 1765*e^(-7*d*x - 7*c) + 1765*e^(-9*d*x - 9* 
c) - 895*e^(-11*d*x - 11*c) + 397*e^(-13*d*x - 13*c) - 15*e^(-15*d*x - 15* 
c))/(d*(8*e^(-2*d*x - 2*c) + 28*e^(-4*d*x - 4*c) + 56*e^(-6*d*x - 6*c) + 7 
0*e^(-8*d*x - 8*c) + 56*e^(-10*d*x - 10*c) + 28*e^(-12*d*x - 12*c) + 8*e^( 
-14*d*x - 14*c) + e^(-16*d*x - 16*c) + 1))) - 1/8*a*b^2*(3*arctan(e^(-d*x 
- c))/d - (3*e^(-d*x - c) - 47*e^(-3*d*x - 3*c) + 78*e^(-5*d*x - 5*c) - 78 
*e^(-7*d*x - 7*c) + 47*e^(-9*d*x - 9*c) - 3*e^(-11*d*x - 11*c))/(d*(6*e^(- 
2*d*x - 2*c) + 15*e^(-4*d*x - 4*c) + 20*e^(-6*d*x - 6*c) + 15*e^(-8*d*x - 
8*c) + 6*e^(-10*d*x - 10*c) + e^(-12*d*x - 12*c) + 1))) - 3/4*a^2*b*(arcta 
n(e^(-d*x - c))/d - (e^(-d*x - c) - 7*e^(-3*d*x - 3*c) + 7*e^(-5*d*x - 5*c 
) - e^(-7*d*x - 7*c))/(d*(4*e^(-2*d*x - 2*c) + 6*e^(-4*d*x - 4*c) + 4*e^(- 
6*d*x - 6*c) + e^(-8*d*x - 8*c) + 1))) - a^3*(arctan(e^(-d*x - c))/d - (e^ 
(-d*x - c) - e^(-3*d*x - 3*c))/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 
 1)))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 485 vs. \(2 (162) = 324\).

Time = 0.22 (sec) , antiderivative size = 485, normalized size of antiderivative = 2.82 \[ \int \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx=\frac {3 \, {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (64 \, a^{3} + 48 \, a^{2} b + 24 \, a b^{2} + 5 \, b^{3}\right )} + \frac {4 \, {\left (192 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{7} + 144 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{7} + 72 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{7} + 15 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{7} + 2304 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} + 576 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} - 480 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} - 292 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{5} + 9216 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 2304 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 4224 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} - 880 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 12288 \, a^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 9216 \, a^{2} b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 4608 \, a b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} - 960 \, b^{3} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}^{4}}}{768 \, d} \] Input:

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

Output:

1/768*(3*(pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(64*a^3 + 
 48*a^2*b + 24*a*b^2 + 5*b^3) + 4*(192*a^3*(e^(d*x + c) - e^(-d*x - c))^7 
+ 144*a^2*b*(e^(d*x + c) - e^(-d*x - c))^7 + 72*a*b^2*(e^(d*x + c) - e^(-d 
*x - c))^7 + 15*b^3*(e^(d*x + c) - e^(-d*x - c))^7 + 2304*a^3*(e^(d*x + c) 
 - e^(-d*x - c))^5 + 576*a^2*b*(e^(d*x + c) - e^(-d*x - c))^5 - 480*a*b^2* 
(e^(d*x + c) - e^(-d*x - c))^5 - 292*b^3*(e^(d*x + c) - e^(-d*x - c))^5 + 
9216*a^3*(e^(d*x + c) - e^(-d*x - c))^3 - 2304*a^2*b*(e^(d*x + c) - e^(-d* 
x - c))^3 - 4224*a*b^2*(e^(d*x + c) - e^(-d*x - c))^3 - 880*b^3*(e^(d*x + 
c) - e^(-d*x - c))^3 + 12288*a^3*(e^(d*x + c) - e^(-d*x - c)) - 9216*a^2*b 
*(e^(d*x + c) - e^(-d*x - c)) - 4608*a*b^2*(e^(d*x + c) - e^(-d*x - c)) - 
960*b^3*(e^(d*x + c) - e^(-d*x - c)))/((e^(d*x + c) - e^(-d*x - c))^2 + 4) 
^4)/d
 

Mupad [B] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 951, normalized size of antiderivative = 5.53 \[ \int \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:

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

Output:

(atan((exp(d*x)*exp(c)*(64*a^3*(d^2)^(1/2) + 5*b^3*(d^2)^(1/2) + 24*a*b^2* 
(d^2)^(1/2) + 48*a^2*b*(d^2)^(1/2)))/(d*(240*a*b^5 + 6144*a^5*b + 4096*a^6 
 + 25*b^6 + 1056*a^2*b^4 + 2944*a^3*b^3 + 5376*a^4*b^2)^(1/2)))*(240*a*b^5 
 + 6144*a^5*b + 4096*a^6 + 25*b^6 + 1056*a^2*b^4 + 2944*a^3*b^3 + 5376*a^4 
*b^2)^(1/2))/(64*(d^2)^(1/2)) - ((exp(c + d*x)*(a + b)^3)/(2*d) + (exp(13* 
c + 13*d*x)*(a + b)^3)/(2*d) - (3*exp(5*c + 5*d*x)*(a*b^2 + a^2*b - 5*a^3 
- 5*b^3))/(2*d) - (3*exp(9*c + 9*d*x)*(a*b^2 + a^2*b - 5*a^3 - 5*b^3))/(2* 
d) + (2*exp(7*c + 7*d*x)*(3*a*b^2 - 3*a^2*b + 5*a^3 - 5*b^3))/d + (3*exp(3 
*c + 3*d*x)*(a + b)^2*(a - b))/d + (3*exp(11*c + 11*d*x)*(a + b)^2*(a - b) 
)/d)/(8*exp(2*c + 2*d*x) + 28*exp(4*c + 4*d*x) + 56*exp(6*c + 6*d*x) + 70* 
exp(8*c + 8*d*x) + 56*exp(10*c + 10*d*x) + 28*exp(12*c + 12*d*x) + 8*exp(1 
4*c + 14*d*x) + exp(16*c + 16*d*x) + 1) + (2*exp(c + d*x)*(48*a*b^2 + 85*b 
^3))/(3*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)) + (16*b^3*exp(c + d*x))/(d 
*(7*exp(2*c + 2*d*x) + 21*exp(4*c + 4*d*x) + 35*exp(6*c + 6*d*x) + 35*exp( 
8*c + 8*d*x) + 21*exp(10*c + 10*d*x) + 7*exp(12*c + 12*d*x) + exp(14*c + 1 
4*d*x) + 1)) + (exp(c + d*x)*(24*a*b^2 + 48*a^2*b + 64*a^3 + 5*b^3))/(64*d 
*(exp(2*c + 2*d*x) + 1)) - (4*exp(c + d*x)*(6*a*b^2 + 35*b^3))/(3*d*(6*exp 
(2*c + 2*d*x) + 15*exp(4*c + 4*d*x) + 20*exp(6*c + 6*d*x) + 15*exp(8*c + 8 
*d*x) + 6*exp(10*c + 10*d*x) + exp(12*c + 12*d*x) + 1)) - (exp(c + d*x)...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1394, normalized size of antiderivative = 8.10 \[ \int \text {sech}^3(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx =\text {Too large to display} \] Input:

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

Output:

(192*e**(16*c + 16*d*x)*atan(e**(c + d*x))*a**3 + 144*e**(16*c + 16*d*x)*a 
tan(e**(c + d*x))*a**2*b + 72*e**(16*c + 16*d*x)*atan(e**(c + d*x))*a*b**2 
 + 15*e**(16*c + 16*d*x)*atan(e**(c + d*x))*b**3 + 1536*e**(14*c + 14*d*x) 
*atan(e**(c + d*x))*a**3 + 1152*e**(14*c + 14*d*x)*atan(e**(c + d*x))*a**2 
*b + 576*e**(14*c + 14*d*x)*atan(e**(c + d*x))*a*b**2 + 120*e**(14*c + 14* 
d*x)*atan(e**(c + d*x))*b**3 + 5376*e**(12*c + 12*d*x)*atan(e**(c + d*x))* 
a**3 + 4032*e**(12*c + 12*d*x)*atan(e**(c + d*x))*a**2*b + 2016*e**(12*c + 
 12*d*x)*atan(e**(c + d*x))*a*b**2 + 420*e**(12*c + 12*d*x)*atan(e**(c + d 
*x))*b**3 + 10752*e**(10*c + 10*d*x)*atan(e**(c + d*x))*a**3 + 8064*e**(10 
*c + 10*d*x)*atan(e**(c + d*x))*a**2*b + 4032*e**(10*c + 10*d*x)*atan(e**( 
c + d*x))*a*b**2 + 840*e**(10*c + 10*d*x)*atan(e**(c + d*x))*b**3 + 13440* 
e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**3 + 10080*e**(8*c + 8*d*x)*atan(e** 
(c + d*x))*a**2*b + 5040*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a*b**2 + 1050 
*e**(8*c + 8*d*x)*atan(e**(c + d*x))*b**3 + 10752*e**(6*c + 6*d*x)*atan(e* 
*(c + d*x))*a**3 + 8064*e**(6*c + 6*d*x)*atan(e**(c + d*x))*a**2*b + 4032* 
e**(6*c + 6*d*x)*atan(e**(c + d*x))*a*b**2 + 840*e**(6*c + 6*d*x)*atan(e** 
(c + d*x))*b**3 + 5376*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a**3 + 4032*e** 
(4*c + 4*d*x)*atan(e**(c + d*x))*a**2*b + 2016*e**(4*c + 4*d*x)*atan(e**(c 
 + d*x))*a*b**2 + 420*e**(4*c + 4*d*x)*atan(e**(c + d*x))*b**3 + 1536*e**( 
2*c + 2*d*x)*atan(e**(c + d*x))*a**3 + 1152*e**(2*c + 2*d*x)*atan(e**(c...