\(\int \text {sech}(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\) [93]

Optimal result

Integrand size = 21, antiderivative size = 81 \[ \int \text {sech}(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {\left (8 a^2+8 a b+3 b^2\right ) \arctan (\sinh (c+d x))}{8 d}-\frac {b (8 a+5 b) \text {sech}(c+d x) \tanh (c+d x)}{8 d}+\frac {b^2 \text {sech}^3(c+d x) \tanh (c+d x)}{4 d} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 6.80 (sec) , antiderivative size = 427, normalized size of antiderivative = 5.27 \[ \int \text {sech}(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {\text {csch}^3(c+d x) \left (128 \, _5F_4\left (\frac {3}{2},2,2,2,2;1,1,1,\frac {9}{2};-\sinh ^2(c+d x)\right ) \sinh ^6(c+d x) \left (a+a \sinh ^2(c+d x)+b \sinh ^2(c+d x)\right )^2+128 \, _4F_3\left (\frac {3}{2},2,2,2;1,1,\frac {9}{2};-\sinh ^2(c+d x)\right ) \sinh ^6(c+d x) \left (5 b^2 \sinh ^4(c+d x)+2 a b \sinh ^2(c+d x) \left (6+5 \sinh ^2(c+d x)\right )+a^2 \left (7+12 \sinh ^2(c+d x)+5 \sinh ^4(c+d x)\right )\right )+35 \left (b^2 \sinh ^4(c+d x) \left (1947+485 \sinh ^2(c+d x)\right )+2 a b \sinh ^2(c+d x) \left (2625+2554 \sinh ^2(c+d x)+485 \sinh ^4(c+d x)\right )+a^2 \left (3375+5907 \sinh ^2(c+d x)+3161 \sinh ^4(c+d x)+485 \sinh ^6(c+d x)\right )\right )-\frac {105 \text {arctanh}\left (\sqrt {-\sinh ^2(c+d x)}\right ) \left (b^2 \sinh ^4(c+d x) \left (649+378 \sinh ^2(c+d x)+9 \sinh ^4(c+d x)\right )+2 a b \sinh ^2(c+d x) \left (875+1143 \sinh ^2(c+d x)+389 \sinh ^4(c+d x)+9 \sinh ^6(c+d x)\right )+a^2 \left (1125+2344 \sinh ^2(c+d x)+1674 \sinh ^4(c+d x)+400 \sinh ^6(c+d x)+9 \sinh ^8(c+d x)\right )\right )}{\sqrt {-\sinh ^2(c+d x)}}\right )}{6720 d} \] Input:

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

Output:

-1/6720*(Csch[c + d*x]^3*(128*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 
1, 9/2}, -Sinh[c + d*x]^2]*Sinh[c + d*x]^6*(a + a*Sinh[c + d*x]^2 + b*Sinh 
[c + d*x]^2)^2 + 128*HypergeometricPFQ[{3/2, 2, 2, 2}, {1, 1, 9/2}, -Sinh[ 
c + d*x]^2]*Sinh[c + d*x]^6*(5*b^2*Sinh[c + d*x]^4 + 2*a*b*Sinh[c + d*x]^2 
*(6 + 5*Sinh[c + d*x]^2) + a^2*(7 + 12*Sinh[c + d*x]^2 + 5*Sinh[c + d*x]^4 
)) + 35*(b^2*Sinh[c + d*x]^4*(1947 + 485*Sinh[c + d*x]^2) + 2*a*b*Sinh[c + 
 d*x]^2*(2625 + 2554*Sinh[c + d*x]^2 + 485*Sinh[c + d*x]^4) + a^2*(3375 + 
5907*Sinh[c + d*x]^2 + 3161*Sinh[c + d*x]^4 + 485*Sinh[c + d*x]^6)) - (105 
*ArcTanh[Sqrt[-Sinh[c + d*x]^2]]*(b^2*Sinh[c + d*x]^4*(649 + 378*Sinh[c + 
d*x]^2 + 9*Sinh[c + d*x]^4) + 2*a*b*Sinh[c + d*x]^2*(875 + 1143*Sinh[c + d 
*x]^2 + 389*Sinh[c + d*x]^4 + 9*Sinh[c + d*x]^6) + a^2*(1125 + 2344*Sinh[c 
 + d*x]^2 + 1674*Sinh[c + d*x]^4 + 400*Sinh[c + d*x]^6 + 9*Sinh[c + d*x]^8 
)))/Sqrt[-Sinh[c + d*x]^2]))/d
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.25, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3042, 4159, 315, 298, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4159

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

\(\Big \downarrow \) 315

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

\(\Big \downarrow \) 298

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^2+8 a b+3 b^2\right ) \int \frac {1}{\sinh ^2(c+d x)+1}d\sinh (c+d x)-\frac {3 b (2 a+b) \sinh (c+d x)}{2 \left (\sinh ^2(c+d x)+1\right )}\right )-\frac {b \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{4 \left (\sinh ^2(c+d x)+1\right )^2}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (8 a^2+8 a b+3 b^2\right ) \arctan (\sinh (c+d x))-\frac {3 b (2 a+b) \sinh (c+d x)}{2 \left (\sinh ^2(c+d x)+1\right )}\right )-\frac {b \sinh (c+d x) \left ((a+b) \sinh ^2(c+d x)+a\right )}{4 \left (\sinh ^2(c+d x)+1\right )^2}}{d}\)

Input:

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

Output:

(-1/4*(b*Sinh[c + d*x]*(a + (a + b)*Sinh[c + d*x]^2))/(1 + Sinh[c + d*x]^2 
)^2 + (((8*a^2 + 8*a*b + 3*b^2)*ArcTan[Sinh[c + d*x]])/2 - (3*b*(2*a + b)* 
Sinh[c + d*x])/(2*(1 + Sinh[c + d*x]^2)))/4)/d
 

Defintions of rubi rules used

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 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 = 4.56 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.65

method	result	size
derivativedivides	\(\frac {2 a^{2} \arctan \left ({\mathrm e}^{d x +c}\right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{2}}+\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{\cosh \left (d x +c \right )^{4}}-\frac {\sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{4}}+\left (\frac {\operatorname {sech}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (d x +c \right )}{8}\right ) \tanh \left (d x +c \right )+\frac {3 \arctan \left ({\mathrm e}^{d x +c}\right )}{4}\right )}{d}\)	\(134\)
default	\(\frac {2 a^{2} \arctan \left ({\mathrm e}^{d x +c}\right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{2}}+\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{\cosh \left (d x +c \right )^{4}}-\frac {\sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{4}}+\left (\frac {\operatorname {sech}\left (d x +c \right )^{3}}{4}+\frac {3 \,\operatorname {sech}\left (d x +c \right )}{8}\right ) \tanh \left (d x +c \right )+\frac {3 \arctan \left ({\mathrm e}^{d x +c}\right )}{4}\right )}{d}\)	\(134\)
risch	\(-\frac {b \,{\mathrm e}^{d x +c} \left (8 \,{\mathrm e}^{6 d x +6 c} a +5 \,{\mathrm e}^{6 d x +6 c} b +8 \,{\mathrm e}^{4 d x +4 c} a -3 b \,{\mathrm e}^{4 d x +4 c}-8 \,{\mathrm e}^{2 d x +2 c} a +3 \,{\mathrm e}^{2 d x +2 c} b -8 a -5 b \right )}{4 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{4}}+\frac {i \ln \left ({\mathrm e}^{d x +c}+i\right ) a^{2}}{d}+\frac {i b a \ln \left ({\mathrm e}^{d x +c}+i\right )}{d}+\frac {3 i b^{2} \ln \left ({\mathrm e}^{d x +c}+i\right )}{8 d}-\frac {i \ln \left ({\mathrm e}^{d x +c}-i\right ) a^{2}}{d}-\frac {i b a \ln \left ({\mathrm e}^{d x +c}-i\right )}{d}-\frac {3 i b^{2} \ln \left ({\mathrm e}^{d x +c}-i\right )}{8 d}\)	\(218\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1373 vs. \(2 (75) = 150\).

Time = 0.11 (sec) , antiderivative size = 1373, normalized size of antiderivative = 16.95 \[ \int \text {sech}(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\text {Too large to display} \] Input:

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

Output:

-1/4*((8*a*b + 5*b^2)*cosh(d*x + c)^7 + 7*(8*a*b + 5*b^2)*cosh(d*x + c)*si 
nh(d*x + c)^6 + (8*a*b + 5*b^2)*sinh(d*x + c)^7 + (8*a*b - 3*b^2)*cosh(d*x 
 + c)^5 + (21*(8*a*b + 5*b^2)*cosh(d*x + c)^2 + 8*a*b - 3*b^2)*sinh(d*x + 
c)^5 + 5*(7*(8*a*b + 5*b^2)*cosh(d*x + c)^3 + (8*a*b - 3*b^2)*cosh(d*x + c 
))*sinh(d*x + c)^4 - (8*a*b - 3*b^2)*cosh(d*x + c)^3 + (35*(8*a*b + 5*b^2) 
*cosh(d*x + c)^4 + 10*(8*a*b - 3*b^2)*cosh(d*x + c)^2 - 8*a*b + 3*b^2)*sin 
h(d*x + c)^3 + (21*(8*a*b + 5*b^2)*cosh(d*x + c)^5 + 10*(8*a*b - 3*b^2)*co 
sh(d*x + c)^3 - 3*(8*a*b - 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - ((8*a^2 
 + 8*a*b + 3*b^2)*cosh(d*x + c)^8 + 8*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c 
)*sinh(d*x + c)^7 + (8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^8 + 4*(8*a^2 + 8 
*a*b + 3*b^2)*cosh(d*x + c)^6 + 4*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c) 
^2 + 8*a^2 + 8*a*b + 3*b^2)*sinh(d*x + c)^6 + 8*(7*(8*a^2 + 8*a*b + 3*b^2) 
*cosh(d*x + c)^3 + 3*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c))*sinh(d*x + c)^ 
5 + 6*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^4 + 2*(35*(8*a^2 + 8*a*b + 3*b 
^2)*cosh(d*x + c)^4 + 30*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2 + 24*a^2 
+ 24*a*b + 9*b^2)*sinh(d*x + c)^4 + 8*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x 
+ c)^5 + 10*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^3 + 3*(8*a^2 + 8*a*b + 3 
*b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x 
+ c)^2 + 4*(7*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^6 + 15*(8*a^2 + 8*a*b 
+ 3*b^2)*cosh(d*x + c)^4 + 9*(8*a^2 + 8*a*b + 3*b^2)*cosh(d*x + c)^2 + ...
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 199 vs. \(2 (75) = 150\).

Time = 0.12 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.46 \[ \int \text {sech}(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {1}{4} \, b^{2} {\left (\frac {3 \, \arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac {5 \, e^{\left (-d x - c\right )} - 3 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )} - 5 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} - 2 \, a b {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + \frac {a^{2} \arctan \left (\sinh \left (d x + c\right )\right )}{d} \] Input:

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

Output:

-1/4*b^2*(3*arctan(e^(-d*x - c))/d + (5*e^(-d*x - c) - 3*e^(-3*d*x - 3*c) 
+ 3*e^(-5*d*x - 5*c) - 5*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))) - 2*a*b*(arcta 
n(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))) + a^2*arctan(sinh(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (75) = 150\).

Time = 0.15 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.10 \[ \int \text {sech}(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {{\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} {\left (8 \, a^{2} + 8 \, a b + 3 \, b^{2}\right )} - \frac {4 \, {\left (8 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 5 \, b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{3} + 32 \, a b {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 12 \, b^{2} {\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}\right )}}{{\left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}^{2}}}{16 \, d} \] Input:

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

Output:

1/16*((pi + 2*arctan(1/2*(e^(2*d*x + 2*c) - 1)*e^(-d*x - c)))*(8*a^2 + 8*a 
*b + 3*b^2) - 4*(8*a*b*(e^(d*x + c) - e^(-d*x - c))^3 + 5*b^2*(e^(d*x + c) 
 - e^(-d*x - c))^3 + 32*a*b*(e^(d*x + c) - e^(-d*x - c)) + 12*b^2*(e^(d*x 
+ c) - e^(-d*x - c)))/((e^(d*x + c) - e^(-d*x - c))^2 + 4)^2)/d
 

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 303, normalized size of antiderivative = 3.74 \[ \int \text {sech}(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\,\left (8\,a^2\,\sqrt {d^2}+3\,b^2\,\sqrt {d^2}+8\,a\,b\,\sqrt {d^2}\right )}{d\,\sqrt {64\,a^4+128\,a^3\,b+112\,a^2\,b^2+48\,a\,b^3+9\,b^4}}\right )\,\sqrt {64\,a^4+128\,a^3\,b+112\,a^2\,b^2+48\,a\,b^3+9\,b^4}}{4\,\sqrt {d^2}}-\frac {6\,b^2\,{\mathrm {e}}^{c+d\,x}}{d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}+\frac {4\,b^2\,{\mathrm {e}}^{c+d\,x}}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-\frac {{\mathrm {e}}^{c+d\,x}\,\left (5\,b^2+8\,a\,b\right )}{4\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {{\mathrm {e}}^{c+d\,x}\,\left (9\,b^2+8\,a\,b\right )}{2\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \] Input:

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

Output:

(atan((exp(d*x)*exp(c)*(8*a^2*(d^2)^(1/2) + 3*b^2*(d^2)^(1/2) + 8*a*b*(d^2 
)^(1/2)))/(d*(48*a*b^3 + 128*a^3*b + 64*a^4 + 9*b^4 + 112*a^2*b^2)^(1/2))) 
*(48*a*b^3 + 128*a^3*b + 64*a^4 + 9*b^4 + 112*a^2*b^2)^(1/2))/(4*(d^2)^(1/ 
2)) - (6*b^2*exp(c + d*x))/(d*(3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + e 
xp(6*c + 6*d*x) + 1)) + (4*b^2*exp(c + d*x))/(d*(4*exp(2*c + 2*d*x) + 6*ex 
p(4*c + 4*d*x) + 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1)) - (exp(c + d* 
x)*(8*a*b + 5*b^2))/(4*d*(exp(2*c + 2*d*x) + 1)) + (exp(c + d*x)*(8*a*b + 
9*b^2))/(2*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 476, normalized size of antiderivative = 5.88 \[ \int \text {sech}(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {8 e^{8 d x +8 c} \mathit {atan} \left (e^{d x +c}\right ) a^{2}+8 e^{8 d x +8 c} \mathit {atan} \left (e^{d x +c}\right ) a b +3 e^{8 d x +8 c} \mathit {atan} \left (e^{d x +c}\right ) b^{2}+32 e^{6 d x +6 c} \mathit {atan} \left (e^{d x +c}\right ) a^{2}+32 e^{6 d x +6 c} \mathit {atan} \left (e^{d x +c}\right ) a b +12 e^{6 d x +6 c} \mathit {atan} \left (e^{d x +c}\right ) b^{2}+48 e^{4 d x +4 c} \mathit {atan} \left (e^{d x +c}\right ) a^{2}+48 e^{4 d x +4 c} \mathit {atan} \left (e^{d x +c}\right ) a b +18 e^{4 d x +4 c} \mathit {atan} \left (e^{d x +c}\right ) b^{2}+32 e^{2 d x +2 c} \mathit {atan} \left (e^{d x +c}\right ) a^{2}+32 e^{2 d x +2 c} \mathit {atan} \left (e^{d x +c}\right ) a b +12 e^{2 d x +2 c} \mathit {atan} \left (e^{d x +c}\right ) b^{2}+8 \mathit {atan} \left (e^{d x +c}\right ) a^{2}+8 \mathit {atan} \left (e^{d x +c}\right ) a b +3 \mathit {atan} \left (e^{d x +c}\right ) b^{2}-8 e^{7 d x +7 c} a b -5 e^{7 d x +7 c} b^{2}-8 e^{5 d x +5 c} a b +3 e^{5 d x +5 c} b^{2}+8 e^{3 d x +3 c} a b -3 e^{3 d x +3 c} b^{2}+8 e^{d x +c} a b +5 e^{d x +c} b^{2}}{4 d \left (e^{8 d x +8 c}+4 e^{6 d x +6 c}+6 e^{4 d x +4 c}+4 e^{2 d x +2 c}+1\right )} \] Input:

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

Output:

(8*e**(8*c + 8*d*x)*atan(e**(c + d*x))*a**2 + 8*e**(8*c + 8*d*x)*atan(e**( 
c + d*x))*a*b + 3*e**(8*c + 8*d*x)*atan(e**(c + d*x))*b**2 + 32*e**(6*c + 
6*d*x)*atan(e**(c + d*x))*a**2 + 32*e**(6*c + 6*d*x)*atan(e**(c + d*x))*a* 
b + 12*e**(6*c + 6*d*x)*atan(e**(c + d*x))*b**2 + 48*e**(4*c + 4*d*x)*atan 
(e**(c + d*x))*a**2 + 48*e**(4*c + 4*d*x)*atan(e**(c + d*x))*a*b + 18*e**( 
4*c + 4*d*x)*atan(e**(c + d*x))*b**2 + 32*e**(2*c + 2*d*x)*atan(e**(c + d* 
x))*a**2 + 32*e**(2*c + 2*d*x)*atan(e**(c + d*x))*a*b + 12*e**(2*c + 2*d*x 
)*atan(e**(c + d*x))*b**2 + 8*atan(e**(c + d*x))*a**2 + 8*atan(e**(c + d*x 
))*a*b + 3*atan(e**(c + d*x))*b**2 - 8*e**(7*c + 7*d*x)*a*b - 5*e**(7*c + 
7*d*x)*b**2 - 8*e**(5*c + 5*d*x)*a*b + 3*e**(5*c + 5*d*x)*b**2 + 8*e**(3*c 
 + 3*d*x)*a*b - 3*e**(3*c + 3*d*x)*b**2 + 8*e**(c + d*x)*a*b + 5*e**(c + d 
*x)*b**2)/(4*d*(e**(8*c + 8*d*x) + 4*e**(6*c + 6*d*x) + 6*e**(4*c + 4*d*x) 
 + 4*e**(2*c + 2*d*x) + 1))