3.1.62 \(\int \text {sech}^2(c+d x) (a+b \text {sech}^2(c+d x))^2 \, dx\) [62]

3.1.62.1 Optimal result

Integrand size = 23, antiderivative size = 53 \[ \int \text {sech}^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {(a+b)^2 \tanh (c+d x)}{d}-\frac {2 b (a+b) \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

output

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

3.1.62.2 Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.75 \[ \int \text {sech}^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {a^2 \tanh (c+d x)}{d}+\frac {2 a b \tanh (c+d x)}{d}+\frac {b^2 \tanh (c+d x)}{d}-\frac {2 a b \tanh ^3(c+d x)}{3 d}-\frac {2 b^2 \tanh ^3(c+d x)}{3 d}+\frac {b^2 \tanh ^5(c+d x)}{5 d} \]

input

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

output

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

3.1.62.3 Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4634, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 210

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.1.62.3.1 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 4634

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

3.1.62.4 Maple [A] (verified)

Time = 1.30 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.32

method	result	size
derivativedivides	\(\frac {a^{2} \tanh \left (d x +c \right )+2 a b \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+b^{2} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{d}\)	\(70\)
default	\(\frac {a^{2} \tanh \left (d x +c \right )+2 a b \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+b^{2} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{d}\)	\(70\)
parts	\(\frac {a^{2} \tanh \left (d x +c \right )}{d}+\frac {b^{2} \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{d}+\frac {2 a b \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{d}\)	\(75\)
parallelrisch	\(\frac {\left (45 a^{2}+100 a b +40 b^{2}\right ) \sinh \left (3 d x +3 c \right )+\left (15 a^{2}+20 a b +8 b^{2}\right ) \sinh \left (5 d x +5 c \right )+30 \left (a^{2}+\frac {8}{3} a b +\frac {8}{3} b^{2}\right ) \sinh \left (d x +c \right )}{15 d \left (\cosh \left (5 d x +5 c \right )+5 \cosh \left (3 d x +3 c \right )+10 \cosh \left (d x +c \right )\right )}\)	\(109\)
risch	\(-\frac {2 \left (15 a^{2} {\mathrm e}^{8 d x +8 c}+60 a^{2} {\mathrm e}^{6 d x +6 c}+60 a b \,{\mathrm e}^{6 d x +6 c}+90 a^{2} {\mathrm e}^{4 d x +4 c}+140 a b \,{\mathrm e}^{4 d x +4 c}+80 \,{\mathrm e}^{4 d x +4 c} b^{2}+60 a^{2} {\mathrm e}^{2 d x +2 c}+100 a b \,{\mathrm e}^{2 d x +2 c}+40 \,{\mathrm e}^{2 d x +2 c} b^{2}+15 a^{2}+20 a b +8 b^{2}\right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\)	\(157\)

input

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

output

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

3.1.62.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 404 vs. \(2 (49) = 98\).

Time = 0.24 (sec) , antiderivative size = 404, normalized size of antiderivative = 7.62 \[ \int \text {sech}^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {4 \, {\left ({\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{4} - 8 \, {\left (5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \sinh \left (d x + c\right )^{4} + 20 \, {\left (3 \, a^{2} + 4 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (15 \, a^{2} + 10 \, a b + 4 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 30 \, a^{2} + 40 \, a b + 10 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} + 45 \, a^{2} + 70 \, a b + 40 \, b^{2} - 8 \, {\left ({\left (5 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 5 \, {\left (a b + b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{5} + d \sinh \left (d x + c\right )^{6} + 6 \, d \cosh \left (d x + c\right )^{4} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )^{4} + 4 \, {\left (5 \, d \cosh \left (d x + c\right )^{3} + 4 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{3} + 15 \, d \cosh \left (d x + c\right )^{2} + 3 \, {\left (5 \, d \cosh \left (d x + c\right )^{4} + 12 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{5} + 8 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + 10 \, d\right )}} \]

input

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

output

-4/15*((15*a^2 + 10*a*b + 4*b^2)*cosh(d*x + c)^4 - 8*(5*a*b + 2*b^2)*cosh( 
d*x + c)*sinh(d*x + c)^3 + (15*a^2 + 10*a*b + 4*b^2)*sinh(d*x + c)^4 + 20* 
(3*a^2 + 4*a*b + b^2)*cosh(d*x + c)^2 + 2*(3*(15*a^2 + 10*a*b + 4*b^2)*cos 
h(d*x + c)^2 + 30*a^2 + 40*a*b + 10*b^2)*sinh(d*x + c)^2 + 45*a^2 + 70*a*b 
 + 40*b^2 - 8*((5*a*b + 2*b^2)*cosh(d*x + c)^3 + 5*(a*b + b^2)*cosh(d*x + 
c))*sinh(d*x + c))/(d*cosh(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)
 

3.1.62.6 Sympy [F]

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

input

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

output

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

3.1.62.7 Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 324 vs. \(2 (49) = 98\).

Time = 0.21 (sec) , antiderivative size = 324, normalized size of antiderivative = 6.11 \[ \int \text {sech}^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=\frac {16}{15} \, b^{2} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {8}{3} \, a b {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + \frac {2 \, a^{2}}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}} \]

input

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

output

16/15*b^2*(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)) + 1 
0*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))) + 8/3*a*b*(3*e^(-2*d*x - 2*c)/(d*(3*e^(-2* 
d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1)) + 1/(d*(3*e^(-2*d 
*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1))) + 2*a^2/(d*(e^(-2 
*d*x - 2*c) + 1))
 

3.1.62.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 156 vs. \(2 (49) = 98\).

Time = 0.30 (sec) , antiderivative size = 156, normalized size of antiderivative = 2.94 \[ \int \text {sech}^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {2 \, {\left (15 \, a^{2} e^{\left (8 \, d x + 8 \, c\right )} + 60 \, a^{2} e^{\left (6 \, d x + 6 \, c\right )} + 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 90 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 140 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 80 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 60 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 100 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 40 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 15 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \]

input

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

output

-2/15*(15*a^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) + 140*a*b*e^(4*d*x + 4*c) + 80*b^2*e^(4*d*x 
 + 4*c) + 60*a^2*e^(2*d*x + 2*c) + 100*a*b*e^(2*d*x + 2*c) + 40*b^2*e^(2*d 
*x + 2*c) + 15*a^2 + 20*a*b + 8*b^2)/(d*(e^(2*d*x + 2*c) + 1)^5)
 

3.1.62.9 Mupad [B] (verification not implemented)

Time = 2.03 (sec) , antiderivative size = 452, normalized size of antiderivative = 8.53 \[ \int \text {sech}^2(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^2 \, dx=-\frac {\frac {2\,a\,\left (a+2\,b\right )}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {\frac {2\,a^2}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}+\frac {4\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{5\,d}+\frac {8\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{5\,d}+\frac {8\,a\,{\mathrm {e}}^{6\,c+6\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,a\,\left (a+2\,b\right )}{5\,d}+\frac {2\,a^2\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{5\,d}+\frac {6\,a\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {2\,\left (3\,a^2+8\,a\,b+8\,b^2\right )}{15\,d}+\frac {2\,a^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}+\frac {4\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,a^2}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \]

input

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

output

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