\(\int (a+b \text {sech}^2(c+d x))^3 \tanh (c+d x) \, dx\) [127]

Optimal result

Integrand size = 21, antiderivative size = 71 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \tanh (c+d x) \, dx=\frac {a^3 \log (\cosh (c+d x))}{d}-\frac {3 a^2 b \text {sech}^2(c+d x)}{2 d}-\frac {3 a b^2 \text {sech}^4(c+d x)}{4 d}-\frac {b^3 \text {sech}^6(c+d x)}{6 d} \] Output:

a^3*ln(cosh(d*x+c))/d-3/2*a^2*b*sech(d*x+c)^2/d-3/4*a*b^2*sech(d*x+c)^4/d- 
1/6*b^3*sech(d*x+c)^6/d
 

Mathematica [A] (verified)

Time = 0.84 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \tanh (c+d x) \, dx=\frac {\cosh ^6(c+d x) \left (a+b \text {sech}^2(c+d x)\right )^3 \left (8 a^3 \log (\cosh (c+d x))-12 a^2 b \text {sech}^2(c+d x)-6 a b^2 \text {sech}^4(c+d x)-\frac {4}{3} b^3 \text {sech}^6(c+d x)\right )}{d (a+2 b+a \cosh (2 c+2 d x))^3} \] Input:

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

Output:

(Cosh[c + d*x]^6*(a + b*Sech[c + d*x]^2)^3*(8*a^3*Log[Cosh[c + d*x]] - 12* 
a^2*b*Sech[c + d*x]^2 - 6*a*b^2*Sech[c + d*x]^4 - (4*b^3*Sech[c + d*x]^6)/ 
3))/(d*(a + 2*b + a*Cosh[2*c + 2*d*x])^3)
 

Rubi [A] (warning: unable to verify)

Time = 0.30 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 26, 4626, 243, 49, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int -i \tan (i c+i d x) \left (a+b \sec (i c+i d x)^2\right )^3dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \left (b \sec (i c+i d x)^2+a\right )^3 \tan (i c+i d x)dx\)

\(\Big \downarrow \) 4626

\(\displaystyle \frac {\int \left (a \cosh ^2(c+d x)+b\right )^3 \text {sech}^7(c+d x)d\cosh (c+d x)}{d}\)

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {a^3 \log \left (\cosh ^2(c+d x)\right )-3 a^2 b \text {sech}(c+d x)-\frac {3}{2} a b^2 \text {sech}^2(c+d x)-\frac {1}{3} b^3 \text {sech}^3(c+d x)}{2 d}\)

Input:

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

Output:

(a^3*Log[Cosh[c + d*x]^2] - 3*a^2*b*Sech[c + d*x] - (3*a*b^2*Sech[c + d*x] 
^2)/2 - (b^3*Sech[c + d*x]^3)/3)/(2*d)
 

Defintions of rubi rules used

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

rule 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

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 4626

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

Maple [A] (warning: unable to verify)

Time = 58.92 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(-\frac {\frac {\operatorname {sech}\left (d x +c \right )^{6} b^{3}}{6}+\frac {3 \operatorname {sech}\left (d x +c \right )^{4} a \,b^{2}}{4}+\frac {3 \operatorname {sech}\left (d x +c \right )^{2} a^{2} b}{2}+a^{3} \ln \left (\operatorname {sech}\left (d x +c \right )\right )}{d}\)	\(59\)
default	\(-\frac {\frac {\operatorname {sech}\left (d x +c \right )^{6} b^{3}}{6}+\frac {3 \operatorname {sech}\left (d x +c \right )^{4} a \,b^{2}}{4}+\frac {3 \operatorname {sech}\left (d x +c \right )^{2} a^{2} b}{2}+a^{3} \ln \left (\operatorname {sech}\left (d x +c \right )\right )}{d}\)	\(59\)
parts	\(\frac {a^{3} \ln \left (\cosh \left (d x +c \right )\right )}{d}-\frac {b^{3} \operatorname {sech}\left (d x +c \right )^{6}}{6 d}+\frac {3 a^{2} b \tanh \left (d x +c \right )^{2}}{2 d}-\frac {3 a \,b^{2} \operatorname {sech}\left (d x +c \right )^{4}}{4 d}\)	\(66\)
risch	\(-a^{3} x -\frac {2 a^{3} c}{d}-\frac {2 b \,{\mathrm e}^{2 d x +2 c} \left (9 a^{2} {\mathrm e}^{8 d x +8 c}+36 a^{2} {\mathrm e}^{6 d x +6 c}+18 a b \,{\mathrm e}^{6 d x +6 c}+54 a^{2} {\mathrm e}^{4 d x +4 c}+36 a b \,{\mathrm e}^{4 d x +4 c}+16 b^{2} {\mathrm e}^{4 d x +4 c}+36 a^{2} {\mathrm e}^{2 d x +2 c}+18 a b \,{\mathrm e}^{2 d x +2 c}+9 a^{2}\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{6}}+\frac {a^{3} \ln \left ({\mathrm e}^{2 d x +2 c}+1\right )}{d}\)	\(179\)

Input:

int((a+b*sech(d*x+c)^2)^3*tanh(d*x+c),x,method=_RETURNVERBOSE)
 

Output:

-1/d*(1/6*sech(d*x+c)^6*b^3+3/4*sech(d*x+c)^4*a*b^2+3/2*sech(d*x+c)^2*a^2* 
b+a^3*ln(sech(d*x+c)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2519 vs. \(2 (65) = 130\).

Time = 0.37 (sec) , antiderivative size = 2519, normalized size of antiderivative = 35.48 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \tanh (c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/3*(3*a^3*d*x*cosh(d*x + c)^12 + 36*a^3*d*x*cosh(d*x + c)*sinh(d*x + c)^ 
11 + 3*a^3*d*x*sinh(d*x + c)^12 + 18*(a^3*d*x + a^2*b)*cosh(d*x + c)^10 + 
18*(11*a^3*d*x*cosh(d*x + c)^2 + a^3*d*x + a^2*b)*sinh(d*x + c)^10 + 60*(1 
1*a^3*d*x*cosh(d*x + c)^3 + 3*(a^3*d*x + a^2*b)*cosh(d*x + c))*sinh(d*x + 
c)^9 + 9*(5*a^3*d*x + 8*a^2*b + 4*a*b^2)*cosh(d*x + c)^8 + 9*(165*a^3*d*x* 
cosh(d*x + c)^4 + 5*a^3*d*x + 8*a^2*b + 4*a*b^2 + 90*(a^3*d*x + a^2*b)*cos 
h(d*x + c)^2)*sinh(d*x + c)^8 + 72*(33*a^3*d*x*cosh(d*x + c)^5 + 30*(a^3*d 
*x + a^2*b)*cosh(d*x + c)^3 + (5*a^3*d*x + 8*a^2*b + 4*a*b^2)*cosh(d*x + c 
))*sinh(d*x + c)^7 + 4*(15*a^3*d*x + 27*a^2*b + 18*a*b^2 + 8*b^3)*cosh(d*x 
 + c)^6 + 4*(693*a^3*d*x*cosh(d*x + c)^6 + 15*a^3*d*x + 945*(a^3*d*x + a^2 
*b)*cosh(d*x + c)^4 + 27*a^2*b + 18*a*b^2 + 8*b^3 + 63*(5*a^3*d*x + 8*a^2* 
b + 4*a*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 24*(99*a^3*d*x*cosh(d*x + 
c)^7 + 189*(a^3*d*x + a^2*b)*cosh(d*x + c)^5 + 21*(5*a^3*d*x + 8*a^2*b + 4 
*a*b^2)*cosh(d*x + c)^3 + (15*a^3*d*x + 27*a^2*b + 18*a*b^2 + 8*b^3)*cosh( 
d*x + c))*sinh(d*x + c)^5 + 3*a^3*d*x + 9*(5*a^3*d*x + 8*a^2*b + 4*a*b^2)* 
cosh(d*x + c)^4 + 3*(495*a^3*d*x*cosh(d*x + c)^8 + 1260*(a^3*d*x + a^2*b)* 
cosh(d*x + c)^6 + 15*a^3*d*x + 210*(5*a^3*d*x + 8*a^2*b + 4*a*b^2)*cosh(d* 
x + c)^4 + 24*a^2*b + 12*a*b^2 + 20*(15*a^3*d*x + 27*a^2*b + 18*a*b^2 + 8* 
b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 4*(165*a^3*d*x*cosh(d*x + c)^9 + 5 
40*(a^3*d*x + a^2*b)*cosh(d*x + c)^7 + 126*(5*a^3*d*x + 8*a^2*b + 4*a*b...
 

Sympy [A] (verification not implemented)

Time = 0.61 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.23 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \tanh (c+d x) \, dx=\begin {cases} a^{3} x - \frac {a^{3} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {3 a^{2} b \operatorname {sech}^{2}{\left (c + d x \right )}}{2 d} - \frac {3 a b^{2} \operatorname {sech}^{4}{\left (c + d x \right )}}{4 d} - \frac {b^{3} \operatorname {sech}^{6}{\left (c + d x \right )}}{6 d} & \text {for}\: d \neq 0 \\x \left (a + b \operatorname {sech}^{2}{\left (c \right )}\right )^{3} \tanh {\left (c \right )} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((a**3*x - a**3*log(tanh(c + d*x) + 1)/d - 3*a**2*b*sech(c + d*x) 
**2/(2*d) - 3*a*b**2*sech(c + d*x)**4/(4*d) - b**3*sech(c + d*x)**6/(6*d), 
 Ne(d, 0)), (x*(a + b*sech(c)**2)**3*tanh(c), True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \tanh (c+d x) \, dx=\frac {3 \, a^{2} b \tanh \left (d x + c\right )^{2}}{2 \, d} + \frac {a^{3} \log \left (\cosh \left (d x + c\right )\right )}{d} - \frac {12 \, a b^{2}}{d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{4}} - \frac {32 \, b^{3}}{3 \, d {\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{6}} \] Input:

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

Output:

3/2*a^2*b*tanh(d*x + c)^2/d + a^3*log(cosh(d*x + c))/d - 12*a*b^2/(d*(e^(d 
*x + c) + e^(-d*x - c))^4) - 32/3*b^3/(d*(e^(d*x + c) + e^(-d*x - c))^6)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 271 vs. \(2 (65) = 130\).

Time = 0.15 (sec) , antiderivative size = 271, normalized size of antiderivative = 3.82 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \tanh (c+d x) \, dx=-\frac {60 \, {\left (d x + c\right )} a^{3} - 60 \, a^{3} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac {147 \, a^{3} e^{\left (12 \, d x + 12 \, c\right )} + 882 \, a^{3} e^{\left (10 \, d x + 10 \, c\right )} + 360 \, a^{2} b e^{\left (10 \, d x + 10 \, c\right )} + 2205 \, a^{3} e^{\left (8 \, d x + 8 \, c\right )} + 1440 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 720 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 2940 \, a^{3} e^{\left (6 \, d x + 6 \, c\right )} + 2160 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} + 1440 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} + 640 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 2205 \, a^{3} e^{\left (4 \, d x + 4 \, c\right )} + 1440 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 720 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 882 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} + 360 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} + 147 \, a^{3}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{6}}}{60 \, d} \] Input:

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

Output:

-1/60*(60*(d*x + c)*a^3 - 60*a^3*log(e^(2*d*x + 2*c) + 1) + (147*a^3*e^(12 
*d*x + 12*c) + 882*a^3*e^(10*d*x + 10*c) + 360*a^2*b*e^(10*d*x + 10*c) + 2 
205*a^3*e^(8*d*x + 8*c) + 1440*a^2*b*e^(8*d*x + 8*c) + 720*a*b^2*e^(8*d*x 
+ 8*c) + 2940*a^3*e^(6*d*x + 6*c) + 2160*a^2*b*e^(6*d*x + 6*c) + 1440*a*b^ 
2*e^(6*d*x + 6*c) + 640*b^3*e^(6*d*x + 6*c) + 2205*a^3*e^(4*d*x + 4*c) + 1 
440*a^2*b*e^(4*d*x + 4*c) + 720*a*b^2*e^(4*d*x + 4*c) + 882*a^3*e^(2*d*x + 
 2*c) + 360*a^2*b*e^(2*d*x + 2*c) + 147*a^3)/(e^(2*d*x + 2*c) + 1)^6)/d
 

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 347, normalized size of antiderivative = 4.89 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \tanh (c+d x) \, dx=\frac {32\,b^3}{3\,d\,\left (6\,{\mathrm {e}}^{2\,c+2\,d\,x}+15\,{\mathrm {e}}^{4\,c+4\,d\,x}+20\,{\mathrm {e}}^{6\,c+6\,d\,x}+15\,{\mathrm {e}}^{8\,c+8\,d\,x}+6\,{\mathrm {e}}^{10\,c+10\,d\,x}+{\mathrm {e}}^{12\,c+12\,d\,x}+1\right )}-\frac {4\,\left (3\,a\,b^2-8\,b^3\right )}{d\,\left (4\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1\right )}-a^3\,x-\frac {6\,\left (2\,a\,b^2-a^2\,b\right )}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}+\frac {a^3\,\ln \left ({\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+1\right )}{d}+\frac {8\,\left (9\,a\,b^2-4\,b^3\right )}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}+1\right )}-\frac {32\,b^3}{d\,\left (5\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}+1\right )}-\frac {6\,a^2\,b}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )} \] Input:

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

Output:

(32*b^3)/(3*d*(6*exp(2*c + 2*d*x) + 15*exp(4*c + 4*d*x) + 20*exp(6*c + 6*d 
*x) + 15*exp(8*c + 8*d*x) + 6*exp(10*c + 10*d*x) + exp(12*c + 12*d*x) + 1) 
) - (4*(3*a*b^2 - 8*b^3))/(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)) - a^3*x - (6*(2*a*b^2 - a^2*b))/ 
(d*(2*exp(2*c + 2*d*x) + exp(4*c + 4*d*x) + 1)) + (a^3*log(exp(2*c)*exp(2* 
d*x) + 1))/d + (8*(9*a*b^2 - 4*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)) - (32*b^3)/(d*(5*exp(2*c + 2*d*x) + 10*e 
xp(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)) - (6*a^2*b)/(d*(exp(2*c + 2*d*x) + 1))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.01 \[ \int \left (a+b \text {sech}^2(c+d x)\right )^3 \tanh (c+d x) \, dx=\frac {12 \,\mathrm {log}\left (e^{2 d x +2 c}+1\right ) a^{3}-2 \mathrm {sech}\left (d x +c \right )^{6} b^{3}-9 \mathrm {sech}\left (d x +c \right )^{4} a \,b^{2}-18 \mathrm {sech}\left (d x +c \right )^{2} a^{2} b -12 a^{3} d x}{12 d} \] Input:

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

Output:

(12*log(e**(2*c + 2*d*x) + 1)*a**3 - 2*sech(c + d*x)**6*b**3 - 9*sech(c + 
d*x)**4*a*b**2 - 18*sech(c + d*x)**2*a**2*b - 12*a**3*d*x)/(12*d)