\(\int \text {sech}^4(c+d x) (a+b \sinh ^2(c+d x))^3 \, dx\) [275]

Optimal result

Integrand size = 23, antiderivative size = 82 \[ \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {1}{2} (6 a-5 b) b^2 x+\frac {b^3 \cosh (c+d x) \sinh (c+d x)}{2 d}+\frac {(a-b)^2 (a+2 b) \tanh (c+d x)}{d}-\frac {(a-b)^3 \tanh ^3(c+d x)}{3 d} \] Output:

1/2*(6*a-5*b)*b^2*x+1/2*b^3*cosh(d*x+c)*sinh(d*x+c)/d+(a-b)^2*(a+2*b)*tanh 
(d*x+c)/d-1/3*(a-b)^3*tanh(d*x+c)^3/d
 

Mathematica [A] (verified)

Time = 1.93 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.02 \[ \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {6 (6 a-5 b) b^2 (c+d x)+3 b^3 \sinh (2 (c+d x))+2 (a-b)^2 (4 a+5 b+(2 a+7 b) \cosh (2 (c+d x))) \text {sech}^2(c+d x) \tanh (c+d x)}{12 d} \] Input:

Integrate[Sech[c + d*x]^4*(a + b*Sinh[c + d*x]^2)^3,x]
 

Output:

(6*(6*a - 5*b)*b^2*(c + d*x) + 3*b^3*Sinh[2*(c + d*x)] + 2*(a - b)^2*(4*a 
+ 5*b + (2*a + 7*b)*Cosh[2*(c + d*x)])*Sech[c + d*x]^2*Tanh[c + d*x])/(12* 
d)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3670, 300, 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.

\(\displaystyle \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (a-b \sin (i c+i d x)^2\right )^3}{\cos (i c+i d x)^4}dx\)

\(\Big \downarrow \) 3670

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

\(\Big \downarrow \) 300

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{2} b^2 (6 a-5 b) \text {arctanh}(\tanh (c+d x))-\frac {1}{3} (a-b)^3 \tanh ^3(c+d x)+(a-b)^2 (a+2 b) \tanh (c+d x)+\frac {b^3 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\)

Input:

Int[Sech[c + d*x]^4*(a + b*Sinh[c + d*x]^2)^3,x]
 

Output:

(((6*a - 5*b)*b^2*ArcTanh[Tanh[c + d*x]])/2 + (a - b)^2*(a + 2*b)*Tanh[c + 
 d*x] - ((a - b)^3*Tanh[c + d*x]^3)/3 + (b^3*Tanh[c + d*x])/(2*(1 - Tanh[c 
 + d*x]^2)))/d
 

Defintions of rubi rules used

rule 300

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(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] && ILtQ[q, 0] && GeQ[p, -q]
 

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 = 0.20 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.80

\[\frac {a^{3} \left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )+3 a^{2} b \left (-\frac {\sinh \left (d x +c \right )}{2 \cosh \left (d x +c \right )^{3}}+\frac {\left (\frac {2}{3}+\frac {\operatorname {sech}\left (d x +c \right )^{2}}{3}\right ) \tanh \left (d x +c \right )}{2}\right )+3 b^{2} a \left (d x +c -\tanh \left (d x +c \right )-\frac {\tanh \left (d x +c \right )^{3}}{3}\right )+b^{3} \left (\frac {\sinh \left (d x +c \right )^{5}}{2 \cosh \left (d x +c \right )^{3}}-\frac {5 d x}{2}-\frac {5 c}{2}+\frac {5 \tanh \left (d x +c \right )}{2}+\frac {5 \tanh \left (d x +c \right )^{3}}{6}\right )}{d}\]

Input:

int(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^3,x)
 

Output:

1/d*(a^3*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c)+3*a^2*b*(-1/2*sinh(d*x+c)/cos 
h(d*x+c)^3+1/2*(2/3+1/3*sech(d*x+c)^2)*tanh(d*x+c))+3*b^2*a*(d*x+c-tanh(d* 
x+c)-1/3*tanh(d*x+c)^3)+b^3*(1/2*sinh(d*x+c)^5/cosh(d*x+c)^3-5/2*d*x-5/2*c 
+5/2*tanh(d*x+c)+5/6*tanh(d*x+c)^3))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 321 vs. \(2 (76) = 152\).

Time = 0.11 (sec) , antiderivative size = 321, normalized size of antiderivative = 3.91 \[ \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {3 \, b^{3} \sinh \left (d x + c\right )^{5} - 4 \, {\left (4 \, a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (6 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{3} - 12 \, {\left (4 \, a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (6 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + {\left (30 \, b^{3} \cosh \left (d x + c\right )^{2} + 16 \, a^{3} + 24 \, a^{2} b - 96 \, a b^{2} + 65 \, b^{3}\right )} \sinh \left (d x + c\right )^{3} - 12 \, {\left (4 \, a^{3} + 6 \, a^{2} b - 24 \, a b^{2} + 14 \, b^{3} - 3 \, {\left (6 \, a b^{2} - 5 \, b^{3}\right )} d x\right )} \cosh \left (d x + c\right ) + 3 \, {\left (5 \, b^{3} \cosh \left (d x + c\right )^{4} + 16 \, a^{3} - 24 \, a^{2} b + 10 \, b^{3} + {\left (16 \, a^{3} + 24 \, a^{2} b - 96 \, a b^{2} + 65 \, b^{3}\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{24 \, {\left (d \cosh \left (d x + c\right )^{3} + 3 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} + 3 \, d \cosh \left (d x + c\right )\right )}} \] Input:

integrate(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

1/24*(3*b^3*sinh(d*x + c)^5 - 4*(4*a^3 + 6*a^2*b - 24*a*b^2 + 14*b^3 - 3*( 
6*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c)^3 - 12*(4*a^3 + 6*a^2*b - 24*a*b^2 + 1 
4*b^3 - 3*(6*a*b^2 - 5*b^3)*d*x)*cosh(d*x + c)*sinh(d*x + c)^2 + (30*b^3*c 
osh(d*x + c)^2 + 16*a^3 + 24*a^2*b - 96*a*b^2 + 65*b^3)*sinh(d*x + c)^3 - 
12*(4*a^3 + 6*a^2*b - 24*a*b^2 + 14*b^3 - 3*(6*a*b^2 - 5*b^3)*d*x)*cosh(d* 
x + c) + 3*(5*b^3*cosh(d*x + c)^4 + 16*a^3 - 24*a^2*b + 10*b^3 + (16*a^3 + 
 24*a^2*b - 96*a*b^2 + 65*b^3)*cosh(d*x + c)^2)*sinh(d*x + c))/(d*cosh(d*x 
 + c)^3 + 3*d*cosh(d*x + c)*sinh(d*x + c)^2 + 3*d*cosh(d*x + c))
 

Sympy [F(-1)]

Timed out. \[ \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\text {Timed out} \] Input:

integrate(sech(d*x+c)**4*(a+b*sinh(d*x+c)**2)**3,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 382 vs. \(2 (76) = 152\).

Time = 0.06 (sec) , antiderivative size = 382, normalized size of antiderivative = 4.66 \[ \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=a b^{2} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} - \frac {1}{24} \, b^{3} {\left (\frac {60 \, {\left (d x + c\right )}}{d} + \frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {121 \, e^{\left (-2 \, d x - 2 \, c\right )} + 201 \, e^{\left (-4 \, d x - 4 \, c\right )} + 147 \, e^{\left (-6 \, d x - 6 \, c\right )} + 3}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 3 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )}\right )}}\right )} + \frac {4}{3} \, a^{3} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + 2 \, a^{2} b {\left (\frac {3 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}} + \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} \] Input:

integrate(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

a*b^2*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(d*(3 
*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c) + 1))) - 1/24*b^ 
3*(60*(d*x + c)/d + 3*e^(-2*d*x - 2*c)/d - (121*e^(-2*d*x - 2*c) + 201*e^( 
-4*d*x - 4*c) + 147*e^(-6*d*x - 6*c) + 3)/(d*(e^(-2*d*x - 2*c) + 3*e^(-4*d 
*x - 4*c) + 3*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c)))) + 4/3*a^3*(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*b*(3*e^(-4*d*x - 4*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)))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 208 vs. \(2 (76) = 152\).

Time = 0.18 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.54 \[ \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {3 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, {\left (6 \, a b^{2} - 5 \, b^{3}\right )} {\left (d x + c\right )} - 3 \, {\left (12 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - \frac {16 \, {\left (9 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 18 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 18 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 12 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{3} + 3 \, a^{2} b - 12 \, a b^{2} + 7 \, b^{3}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{3}}}{24 \, d} \] Input:

integrate(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

1/24*(3*b^3*e^(2*d*x + 2*c) + 12*(6*a*b^2 - 5*b^3)*(d*x + c) - 3*(12*a*b^2 
*e^(2*d*x + 2*c) - 10*b^3*e^(2*d*x + 2*c) + b^3)*e^(-2*d*x - 2*c) - 16*(9* 
a^2*b*e^(4*d*x + 4*c) - 18*a*b^2*e^(4*d*x + 4*c) + 9*b^3*e^(4*d*x + 4*c) + 
 6*a^3*e^(2*d*x + 2*c) - 18*a*b^2*e^(2*d*x + 2*c) + 12*b^3*e^(2*d*x + 2*c) 
 + 2*a^3 + 3*a^2*b - 12*a*b^2 + 7*b^3)/(e^(2*d*x + 2*c) + 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 273, normalized size of antiderivative = 3.33 \[ \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {b^2\,x\,\left (6\,a-5\,b\right )}{2}-\frac {\frac {2\,\left (a^2\,b-2\,a\,b^2+b^3\right )}{d}+\frac {2\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (a^2\,b-2\,a\,b^2+b^3\right )}{d}+\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (2\,a^3-3\,a^2\,b+b^3\right )}{3\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1}-\frac {\frac {2\,\left (2\,a^3-3\,a^2\,b+b^3\right )}{3\,d}+\frac {2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a^2\,b-2\,a\,b^2+b^3\right )}{d}}{2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1}-\frac {b^3\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}+\frac {b^3\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {2\,\left (a^2\,b-2\,a\,b^2+b^3\right )}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:

int((a + b*sinh(c + d*x)^2)^3/cosh(c + d*x)^4,x)
 

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 355, normalized size of antiderivative = 4.33 \[ \int \text {sech}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx=\frac {3 e^{10 d x +10 c} b^{3}+48 e^{8 d x +8 c} a^{2} b +72 e^{8 d x +8 c} a \,b^{2} d x -96 e^{8 d x +8 c} a \,b^{2}-60 e^{8 d x +8 c} b^{3} d x +55 e^{8 d x +8 c} b^{3}+216 e^{6 d x +6 c} a \,b^{2} d x -180 e^{6 d x +6 c} b^{3} d x -96 e^{4 d x +4 c} a^{3}+144 e^{4 d x +4 c} a^{2} b +216 e^{4 d x +4 c} a \,b^{2} d x -180 e^{4 d x +4 c} b^{3} d x -60 e^{4 d x +4 c} b^{3}-32 e^{2 d x +2 c} a^{3}+72 e^{2 d x +2 c} a \,b^{2} d x +96 e^{2 d x +2 c} a \,b^{2}-60 e^{2 d x +2 c} b^{3} d x -75 e^{2 d x +2 c} b^{3}-3 b^{3}}{24 e^{2 d x +2 c} d \left (e^{6 d x +6 c}+3 e^{4 d x +4 c}+3 e^{2 d x +2 c}+1\right )} \] Input:

int(sech(d*x+c)^4*(a+b*sinh(d*x+c)^2)^3,x)
                                                                                    
                                                                                    
 

Output:

(3*e**(10*c + 10*d*x)*b**3 + 48*e**(8*c + 8*d*x)*a**2*b + 72*e**(8*c + 8*d 
*x)*a*b**2*d*x - 96*e**(8*c + 8*d*x)*a*b**2 - 60*e**(8*c + 8*d*x)*b**3*d*x 
 + 55*e**(8*c + 8*d*x)*b**3 + 216*e**(6*c + 6*d*x)*a*b**2*d*x - 180*e**(6* 
c + 6*d*x)*b**3*d*x - 96*e**(4*c + 4*d*x)*a**3 + 144*e**(4*c + 4*d*x)*a**2 
*b + 216*e**(4*c + 4*d*x)*a*b**2*d*x - 180*e**(4*c + 4*d*x)*b**3*d*x - 60* 
e**(4*c + 4*d*x)*b**3 - 32*e**(2*c + 2*d*x)*a**3 + 72*e**(2*c + 2*d*x)*a*b 
**2*d*x + 96*e**(2*c + 2*d*x)*a*b**2 - 60*e**(2*c + 2*d*x)*b**3*d*x - 75*e 
**(2*c + 2*d*x)*b**3 - 3*b**3)/(24*e**(2*c + 2*d*x)*d*(e**(6*c + 6*d*x) + 
3*e**(4*c + 4*d*x) + 3*e**(2*c + 2*d*x) + 1))