\(\int \text {sech}^2(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\) [94]

Optimal result

Integrand size = 23, antiderivative size = 49 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {a^2 \tanh (c+d x)}{d}+\frac {2 a b \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {a^2 \tanh (c+d x)}{d}+\frac {2 a b \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4158, 210, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4158

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

\(\Big \downarrow \) 210

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 210

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 
)^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
 

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 4158

Int[sec[(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[ff/(c^(m - 1)*f)   Subst[Int[(c^2 + ff^2*x^2)^(m/2 - 1)*(a + b*(ff*x)^n)^ 
p, x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && I 
ntegerQ[m/2] && (IntegersQ[n, p] || IGtQ[m, 0] || IGtQ[p, 0] || EqQ[n^2, 4] 
 || EqQ[n^2, 16])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(125\) vs. \(2(45)=90\).

Time = 13.82 (sec) , antiderivative size = 126, normalized size of antiderivative = 2.57

method	result	size
derivativedivides	\(\frac {a^{2} \tanh \left (d x +c \right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\)	\(126\)
default	\(\frac {a^{2} \tanh \left (d x +c \right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{3}}{2 \cosh \left (d x +c \right )^{5}}-\frac {3 \sinh \left (d x +c \right )}{8 \cosh \left (d x +c \right )^{5}}+\frac {3 \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{8}\right )}{d}\)	\(126\)
risch	\(-\frac {2 \left (15 \,{\mathrm e}^{8 d x +8 c} a^{2}+30 \,{\mathrm e}^{8 d x +8 c} a b +15 \,{\mathrm e}^{8 d x +8 c} b^{2}+60 \,{\mathrm e}^{6 d x +6 c} a^{2}+60 \,{\mathrm e}^{6 d x +6 c} a b +90 \,{\mathrm e}^{4 d x +4 c} a^{2}+40 \,{\mathrm e}^{4 d x +4 c} a b +30 \,{\mathrm e}^{4 d x +4 c} b^{2}+60 \,{\mathrm e}^{2 d x +2 c} a^{2}+20 \,{\mathrm e}^{2 d x +2 c} b a +15 a^{2}+10 a b +3 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\)	\(170\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 391 vs. \(2 (45) = 90\).

Time = 0.08 (sec) , antiderivative size = 391, normalized size of antiderivative = 7.98 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {4 \, {\left ({\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} + 8 \, {\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 20 \, {\left (3 \, a^{2} + 2 \, a b\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (15 \, a^{2} + 20 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 30 \, a^{2} + 20 \, a b\right )} \sinh \left (d x + c\right )^{2} + 45 \, a^{2} + 20 \, a b + 15 \, b^{2} + 8 \, {\left ({\left (5 \, a b + 3 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, a b \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} + 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{4} + 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} + 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 10 \, d\right )}} \] Input:

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

Output:

-4/15*((15*a^2 + 20*a*b + 9*b^2)*cosh(d*x + c)^4 + 8*(5*a*b + 3*b^2)*cosh( 
d*x + c)*sinh(d*x + c)^3 + (15*a^2 + 20*a*b + 9*b^2)*sinh(d*x + c)^4 + 20* 
(3*a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(15*a^2 + 20*a*b + 9*b^2)*cosh(d*x 
+ c)^2 + 30*a^2 + 20*a*b)*sinh(d*x + c)^2 + 45*a^2 + 20*a*b + 15*b^2 + 8*( 
(5*a*b + 3*b^2)*cosh(d*x + c)^3 + 5*a*b*cosh(d*x + c))*sinh(d*x + c))/(d*c 
osh(d*x + c)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 + 6 
*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^4 + 4*(5* 
d*cosh(d*x + c)^3 + 4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 15*d*cosh(d*x + c 
)^2 + 3*(5*d*cosh(d*x + c)^4 + 12*d*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^2 
 + 2*(3*d*cosh(d*x + c)^5 + 8*d*cosh(d*x + c)^3 + 5*d*cosh(d*x + c))*sinh( 
d*x + c) + 10*d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.08 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {b^{2} \tanh \left (d x + c\right )^{5}}{5 \, d} + \frac {2 \, a b \tanh \left (d x + c\right )^{3}}{3 \, d} + \frac {2 \, a^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 169 vs. \(2 (45) = 90\).

Time = 0.16 (sec) , antiderivative size = 169, normalized size of antiderivative = 3.45 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {2 \, {\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 30 \, a b e^{\left (8 \, d x + 8 \, c\right )} + 15 \, b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 40 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 30 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 20 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} + 10 \, a b + 3 \, b^{2}\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \] Input:

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

Output:

-2/15*(15*a^2*e^(8*d*x + 8*c) + 30*a*b*e^(8*d*x + 8*c) + 15*b^2*e^(8*d*x + 
 8*c) + 60*a^2*e^(6*d*x + 6*c) + 60*a*b*e^(6*d*x + 6*c) + 90*a^2*e^(4*d*x 
+ 4*c) + 40*a*b*e^(4*d*x + 4*c) + 30*b^2*e^(4*d*x + 4*c) + 60*a^2*e^(2*d*x 
 + 2*c) + 20*a*b*e^(2*d*x + 2*c) + 15*a^2 + 10*a*b + 3*b^2)/(d*(e^(2*d*x + 
 2*c) + 1)^5)
 

Mupad [B] (verification not implemented)

Time = 2.39 (sec) , antiderivative size = 482, normalized size of antiderivative = 9.84 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=-\frac {\frac {2\,\left (a^2-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,{\left (a+b\right )}^2}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,\left (a^2-b^2\right )}{5\,d}+\frac {6\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{6\,c+6\,d\,x}\,{\left (a+b\right )}^2}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2-2\,a\,b+3\,b^2\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,{\left (a+b\right )}^2}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2-b^2\right )}{5\,d}+\frac {8\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a^2-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{8\,c+8\,d\,x}\,{\left (a+b\right )}^2}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2-2\,a\,b+3\,b^2\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2-2\,a\,b+3\,b^2\right )}{15\,d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2-b^2\right )}{5\,d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,{\left (a+b\right )}^2}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,{\left (a+b\right )}^2}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:

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

Output:

- ((2*(a^2 - b^2))/(5*d) + (2*exp(2*c + 2*d*x)*(a + b)^2)/(5*d))/(2*exp(2* 
c + 2*d*x) + exp(4*c + 4*d*x) + 1) - ((2*(a^2 - b^2))/(5*d) + (6*exp(4*c + 
 4*d*x)*(a^2 - b^2))/(5*d) + (2*exp(6*c + 6*d*x)*(a + b)^2)/(5*d) + (2*exp 
(2*c + 2*d*x)*(3*a^2 - 2*a*b + 3*b^2))/(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*(a + b)^2) 
/(5*d) + (8*exp(2*c + 2*d*x)*(a^2 - b^2))/(5*d) + (8*exp(6*c + 6*d*x)*(a^2 
 - b^2))/(5*d) + (2*exp(8*c + 8*d*x)*(a + b)^2)/(5*d) + (4*exp(4*c + 4*d*x 
)*(3*a^2 - 2*a*b + 3*b^2))/(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) - ( 
(2*(3*a^2 - 2*a*b + 3*b^2))/(15*d) + (4*exp(2*c + 2*d*x)*(a^2 - b^2))/(5*d 
) + (2*exp(4*c + 4*d*x)*(a + b)^2)/(5*d))/(3*exp(2*c + 2*d*x) + 3*exp(4*c 
+ 4*d*x) + exp(6*c + 6*d*x) + 1) - (2*(a + b)^2)/(5*d*(exp(2*c + 2*d*x) + 
1))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 224, normalized size of antiderivative = 4.57 \[ \int \text {sech}^2(c+d x) \left (a+b \tanh ^2(c+d x)\right )^2 \, dx=\frac {\frac {2 e^{10 d x +10 c} a^{2}}{5}+\frac {4 e^{10 d x +10 c} a b}{5}+\frac {2 e^{10 d x +10 c} b^{2}}{5}-4 e^{6 d x +6 c} a^{2}+4 e^{6 d x +6 c} b^{2}-8 e^{4 d x +4 c} a^{2}+\frac {8 e^{4 d x +4 c} a b}{3}-6 e^{2 d x +2 c} a^{2}+\frac {4 e^{2 d x +2 c} a b}{3}+2 e^{2 d x +2 c} b^{2}-\frac {8 a^{2}}{5}-\frac {8 a b}{15}}{d \left (e^{10 d x +10 c}+5 e^{8 d x +8 c}+10 e^{6 d x +6 c}+10 e^{4 d x +4 c}+5 e^{2 d x +2 c}+1\right )} \] Input:

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

Output:

(2*(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 - 30*e**(6*c + 6*d*x)*a**2 + 30*e**(6*c + 6*d*x)*b**2 - 60*e**( 
4*c + 4*d*x)*a**2 + 20*e**(4*c + 4*d*x)*a*b - 45*e**(2*c + 2*d*x)*a**2 + 1 
0*e**(2*c + 2*d*x)*a*b + 15*e**(2*c + 2*d*x)*b**2 - 12*a**2 - 4*a*b))/(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))