\(\int \text {sech}^6(c+d x) (a+b \sinh ^2(c+d x)) \, dx\) [256]

Optimal result

Integrand size = 21, antiderivative size = 54 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {a \tanh (c+d x)}{d}-\frac {(2 a-b) \tanh ^3(c+d x)}{3 d}+\frac {(a-b) \tanh ^5(c+d x)}{5 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.89 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {a \tanh (c+d x)}{d}+\frac {2 b \tanh (c+d x)}{15 d}+\frac {b \text {sech}^2(c+d x) \tanh (c+d x)}{15 d}-\frac {b \text {sech}^4(c+d x) \tanh (c+d x)}{5 d}-\frac {2 a \tanh ^3(c+d x)}{3 d}+\frac {a \tanh ^5(c+d x)}{5 d} \] Input:

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

Output:

(a*Tanh[c + d*x])/d + (2*b*Tanh[c + d*x])/(15*d) + (b*Sech[c + d*x]^2*Tanh 
[c + d*x])/(15*d) - (b*Sech[c + d*x]^4*Tanh[c + d*x])/(5*d) - (2*a*Tanh[c 
+ d*x]^3)/(3*d) + (a*Tanh[c + d*x]^5)/(5*d)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3670, 290, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3670

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

\(\Big \downarrow \) 290

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 290

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] :> I 
nt[ExpandIntegrand[(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] && IGtQ[q, 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 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 [A] (verified)

Time = 73.43 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.56

method	result	size
risch	\(-\frac {4 \left (15 \,{\mathrm e}^{6 d x +6 c} b +40 \,{\mathrm e}^{4 d x +4 c} a -5 b \,{\mathrm e}^{4 d x +4 c}+20 \,{\mathrm e}^{2 d x +2 c} a +5 \,{\mathrm e}^{2 d x +2 c} b +4 a +b \right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}\)	\(84\)
derivativedivides	\(\frac {a \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )+b \left (-\frac {\sinh \left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{4}\right )}{d}\)	\(85\)
default	\(\frac {a \left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )+b \left (-\frac {\sinh \left (d x +c \right )}{4 \cosh \left (d x +c \right )^{5}}+\frac {\left (\frac {8}{15}+\frac {\operatorname {sech}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {sech}\left (d x +c \right )^{2}}{15}\right ) \tanh \left (d x +c \right )}{4}\right )}{d}\)	\(85\)

Input:

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

Output:

-4/15*(15*exp(6*d*x+6*c)*b+40*exp(4*d*x+4*c)*a-5*b*exp(4*d*x+4*c)+20*exp(2 
*d*x+2*c)*a+5*exp(2*d*x+2*c)*b+4*a+b)/d/(exp(2*d*x+2*c)+1)^5
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 343 vs. \(2 (50) = 100\).

Time = 0.09 (sec) , antiderivative size = 343, normalized size of antiderivative = 6.35 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=-\frac {8 \, {\left (2 \, {\left (a + 4 \, b\right )} \cosh \left (d x + c\right )^{3} + 6 \, {\left (a + 4 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} - {\left (2 \, a - 7 \, b\right )} \sinh \left (d x + c\right )^{3} + 30 \, a \cosh \left (d x + c\right ) - {\left (3 \, {\left (2 \, a - 7 \, b\right )} \cosh \left (d x + c\right )^{2} - 10 \, a + 5 \, b\right )} \sinh \left (d x + c\right )\right )}}{15 \, {\left (d \cosh \left (d x + c\right )^{7} + 7 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} + d \sinh \left (d x + c\right )^{7} + 5 \, d \cosh \left (d x + c\right )^{5} + {\left (21 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (7 \, d \cosh \left (d x + c\right )^{3} + 5 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 11 \, d \cosh \left (d x + c\right )^{3} + {\left (35 \, d \cosh \left (d x + c\right )^{4} + 50 \, d \cosh \left (d x + c\right )^{2} + 9 \, d\right )} \sinh \left (d x + c\right )^{3} + {\left (21 \, d \cosh \left (d x + c\right )^{5} + 50 \, d \cosh \left (d x + c\right )^{3} + 33 \, d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} + 15 \, d \cosh \left (d x + c\right ) + {\left (7 \, d \cosh \left (d x + c\right )^{6} + 25 \, d \cosh \left (d x + c\right )^{4} + 27 \, d \cosh \left (d x + c\right )^{2} + 5 \, d\right )} \sinh \left (d x + c\right )\right )}} \] Input:

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

Output:

-8/15*(2*(a + 4*b)*cosh(d*x + c)^3 + 6*(a + 4*b)*cosh(d*x + c)*sinh(d*x + 
c)^2 - (2*a - 7*b)*sinh(d*x + c)^3 + 30*a*cosh(d*x + c) - (3*(2*a - 7*b)*c 
osh(d*x + c)^2 - 10*a + 5*b)*sinh(d*x + c))/(d*cosh(d*x + c)^7 + 7*d*cosh( 
d*x + c)*sinh(d*x + c)^6 + d*sinh(d*x + c)^7 + 5*d*cosh(d*x + c)^5 + (21*d 
*cosh(d*x + c)^2 + 5*d)*sinh(d*x + c)^5 + 5*(7*d*cosh(d*x + c)^3 + 5*d*cos 
h(d*x + c))*sinh(d*x + c)^4 + 11*d*cosh(d*x + c)^3 + (35*d*cosh(d*x + c)^4 
 + 50*d*cosh(d*x + c)^2 + 9*d)*sinh(d*x + c)^3 + (21*d*cosh(d*x + c)^5 + 5 
0*d*cosh(d*x + c)^3 + 33*d*cosh(d*x + c))*sinh(d*x + c)^2 + 15*d*cosh(d*x 
+ c) + (7*d*cosh(d*x + c)^6 + 25*d*cosh(d*x + c)^4 + 27*d*cosh(d*x + c)^2 
+ 5*d)*sinh(d*x + c))
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 486 vs. \(2 (50) = 100\).

Time = 0.04 (sec) , antiderivative size = 486, normalized size of antiderivative = 9.00 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {16}{15} \, a {\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 {4}{15} \, b {\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 {5 \, 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 {15 \, e^{\left (-6 \, d x - 6 \, 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 )} \] Input:

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

Output:

16/15*a*(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)) + 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))) + 4/15*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)))
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.54 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (15 \, b e^{\left (6 \, d x + 6 \, c\right )} + 40 \, a e^{\left (4 \, d x + 4 \, c\right )} - 5 \, b e^{\left (4 \, d x + 4 \, c\right )} + 20 \, a e^{\left (2 \, d x + 2 \, c\right )} + 5 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + b\right )}}{15 \, d {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}} \] Input:

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

Output:

-4/15*(15*b*e^(6*d*x + 6*c) + 40*a*e^(4*d*x + 4*c) - 5*b*e^(4*d*x + 4*c) + 
 20*a*e^(2*d*x + 2*c) + 5*b*e^(2*d*x + 2*c) + 4*a + b)/(d*(e^(2*d*x + 2*c) 
 + 1)^5)
 

Mupad [B] (verification not implemented)

Time = 1.99 (sec) , antiderivative size = 298, normalized size of antiderivative = 5.52 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=-\frac {\frac {8\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}+\frac {8\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {16\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1}-\frac {\frac {2\,b}{5\,d}+\frac {6\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}+\frac {8\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a-b\right )}{5\,d}}{4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {\frac {8\,\left (2\,a-b\right )}{15\,d}+\frac {4\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {2\,b}{5\,d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )} \] Input:

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

Output:

- ((8*b*exp(2*c + 2*d*x))/(5*d) + (8*b*exp(6*c + 6*d*x))/(5*d) + (16*exp(4 
*c + 4*d*x)*(2*a - 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*b 
)/(5*d) + (6*b*exp(4*c + 4*d*x))/(5*d) + (8*exp(2*c + 2*d*x)*(2*a - 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) - ((8*(2*a - b))/(15*d) + (4*b*exp(2*c + 2*d*x))/(5*d))/( 
3*exp(2*c + 2*d*x) + 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) + 1) - (2*b)/(5 
*d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1))
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 139, normalized size of antiderivative = 2.57 \[ \int \text {sech}^6(c+d x) \left (a+b \sinh ^2(c+d x)\right ) \, dx=\frac {-4 e^{6 d x +6 c} b -\frac {32 e^{4 d x +4 c} a}{3}+\frac {4 e^{4 d x +4 c} b}{3}-\frac {16 e^{2 d x +2 c} a}{3}-\frac {4 e^{2 d x +2 c} b}{3}-\frac {16 a}{15}-\frac {4 b}{15}}{d \left (e^{10 d x +10 c}+5 e^{8 d x +8 c}+10 e^{6 d x +6 c}+10 e^{4 d x +4 c}+5 e^{2 d x +2 c}+1\right )} \] Input:

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

Output:

(4*( - 15*e**(6*c + 6*d*x)*b - 40*e**(4*c + 4*d*x)*a + 5*e**(4*c + 4*d*x)* 
b - 20*e**(2*c + 2*d*x)*a - 5*e**(2*c + 2*d*x)*b - 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))