\(\int \frac {\tanh ^5(c+d x)}{(a+b \text {sech}^2(c+d x))^3} \, dx\) [159]

Optimal result

Integrand size = 23, antiderivative size = 77 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=-\frac {(a+b)^2}{4 a^3 d \left (b+a \cosh ^2(c+d x)\right )^2}+\frac {a+b}{a^3 d \left (b+a \cosh ^2(c+d x)\right )}+\frac {\log \left (b+a \cosh ^2(c+d x)\right )}{2 a^3 d} \] Output:

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

Mathematica [A] (verified)

Time = 1.85 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.77 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {2 \left (a^2+4 a b+3 b^2\right )+(a+2 b)^2 \log (a+2 b+a \cosh (2 (c+d x)))+a^2 \cosh ^2(2 (c+d x)) \log (a+2 b+a \cosh (2 (c+d x)))+2 a \cosh (2 (c+d x)) (2 (a+b)+(a+2 b) \log (a+2 b+a \cosh (2 (c+d x))))}{2 a^3 d (a+2 b+a \cosh (2 (c+d x)))^2} \] Input:

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

Output:

(2*(a^2 + 4*a*b + 3*b^2) + (a + 2*b)^2*Log[a + 2*b + a*Cosh[2*(c + d*x)]] 
+ a^2*Cosh[2*(c + d*x)]^2*Log[a + 2*b + a*Cosh[2*(c + d*x)]] + 2*a*Cosh[2* 
(c + d*x)]*(2*(a + b) + (a + 2*b)*Log[a + 2*b + a*Cosh[2*(c + d*x)]]))/(2* 
a^3*d*(a + 2*b + a*Cosh[2*(c + d*x)])^2)
 

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 26, 4626, 353, 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.

\(\displaystyle \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\frac {i \tan (i c+i d x)^5}{\left (a+b \sec (i c+i d x)^2\right )^3}dx\)

\(\Big \downarrow \) 26

\(\displaystyle -i \int \frac {\tan (i c+i d x)^5}{\left (b \sec (i c+i d x)^2+a\right )^3}dx\)

\(\Big \downarrow \) 4626

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 49

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(-1/2*(a + b)^2/(a^3*(b + a*Cosh[c + d*x]^2)^2) + (2*(a + b))/(a^3*(b + a* 
Cosh[c + d*x]^2)) + Log[b + a*Cosh[c + d*x]^2]/a^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 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(183\) vs. \(2(73)=146\).

Time = 136.53 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.39

method	result	size
risch	\(-\frac {x}{a^{3}}-\frac {2 c}{a^{3} d}+\frac {4 \,{\mathrm e}^{2 d x +2 c} \left (a^{2} {\mathrm e}^{4 d x +4 c}+a b \,{\mathrm e}^{4 d x +4 c}+a^{2} {\mathrm e}^{2 d x +2 c}+4 a b \,{\mathrm e}^{2 d x +2 c}+3 b^{2} {\mathrm e}^{2 d x +2 c}+a^{2}+a b \right )}{a^{3} d \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )^{2}}+\frac {\ln \left ({\mathrm e}^{4 d x +4 c}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 d x +2 c}}{a}+1\right )}{2 a^{3} d}\)	\(184\)
derivativedivides	\(\frac {\frac {\frac {\left (-2 a^{2}-2 a b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 \left (2 a -b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a +b \right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}{2}}{a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\)	\(226\)
default	\(\frac {\frac {\frac {\left (-2 a^{2}-2 a b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-4 \left (2 a -b \right ) a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a +b \right )}{\left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a +\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b +2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -2 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +a +b \right )}{2}}{a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3}}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3}}}{d}\)	\(226\)

Input:

int(tanh(d*x+c)^5/(a+b*sech(d*x+c)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1741 vs. \(2 (73) = 146\).

Time = 0.25 (sec) , antiderivative size = 1741, normalized size of antiderivative = 22.61 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Too large to display} \] Input:

integrate(tanh(d*x+c)^5/(a+b*sech(d*x+c)^2)^3,x, algorithm="fricas")
 

Output:

-1/2*(2*a^2*d*x*cosh(d*x + c)^8 + 16*a^2*d*x*cosh(d*x + c)*sinh(d*x + c)^7 
 + 2*a^2*d*x*sinh(d*x + c)^8 + 8*((a^2 + 2*a*b)*d*x - a^2 - a*b)*cosh(d*x 
+ c)^6 + 8*(7*a^2*d*x*cosh(d*x + c)^2 + (a^2 + 2*a*b)*d*x - a^2 - a*b)*sin 
h(d*x + c)^6 + 16*(7*a^2*d*x*cosh(d*x + c)^3 + 3*((a^2 + 2*a*b)*d*x - a^2 
- a*b)*cosh(d*x + c))*sinh(d*x + c)^5 + 4*((3*a^2 + 8*a*b + 8*b^2)*d*x - 2 
*a^2 - 8*a*b - 6*b^2)*cosh(d*x + c)^4 + 4*(35*a^2*d*x*cosh(d*x + c)^4 + (3 
*a^2 + 8*a*b + 8*b^2)*d*x + 30*((a^2 + 2*a*b)*d*x - a^2 - a*b)*cosh(d*x + 
c)^2 - 2*a^2 - 8*a*b - 6*b^2)*sinh(d*x + c)^4 + 2*a^2*d*x + 16*(7*a^2*d*x* 
cosh(d*x + c)^5 + 10*((a^2 + 2*a*b)*d*x - a^2 - a*b)*cosh(d*x + c)^3 + ((3 
*a^2 + 8*a*b + 8*b^2)*d*x - 2*a^2 - 8*a*b - 6*b^2)*cosh(d*x + c))*sinh(d*x 
 + c)^3 + 8*((a^2 + 2*a*b)*d*x - a^2 - a*b)*cosh(d*x + c)^2 + 8*(7*a^2*d*x 
*cosh(d*x + c)^6 + 15*((a^2 + 2*a*b)*d*x - a^2 - a*b)*cosh(d*x + c)^4 + (a 
^2 + 2*a*b)*d*x + 3*((3*a^2 + 8*a*b + 8*b^2)*d*x - 2*a^2 - 8*a*b - 6*b^2)* 
cosh(d*x + c)^2 - a^2 - a*b)*sinh(d*x + c)^2 - (a^2*cosh(d*x + c)^8 + 8*a^ 
2*cosh(d*x + c)*sinh(d*x + c)^7 + a^2*sinh(d*x + c)^8 + 4*(a^2 + 2*a*b)*co 
sh(d*x + c)^6 + 4*(7*a^2*cosh(d*x + c)^2 + a^2 + 2*a*b)*sinh(d*x + c)^6 + 
8*(7*a^2*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b)*cosh(d*x + c))*sinh(d*x + c)^5 
+ 2*(3*a^2 + 8*a*b + 8*b^2)*cosh(d*x + c)^4 + 2*(35*a^2*cosh(d*x + c)^4 + 
30*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 3*a^2 + 8*a*b + 8*b^2)*sinh(d*x + c)^4 
+ 8*(7*a^2*cosh(d*x + c)^5 + 10*(a^2 + 2*a*b)*cosh(d*x + c)^3 + (3*a^2 ...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Timed out} \] Input:

integrate(tanh(d*x+c)**5/(a+b*sech(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. 206 vs. \(2 (73) = 146\).

Time = 0.06 (sec) , antiderivative size = 206, normalized size of antiderivative = 2.68 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\frac {4 \, {\left ({\left (a^{2} + a b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + {\left (a^{2} + 4 \, a b + 3 \, b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + {\left (a^{2} + a b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} e^{\left (-8 \, d x - 8 \, c\right )} + a^{5} + 4 \, {\left (a^{5} + 2 \, a^{4} b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \, {\left (3 \, a^{5} + 8 \, a^{4} b + 8 \, a^{3} b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, {\left (a^{5} + 2 \, a^{4} b\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )} d} + \frac {d x + c}{a^{3} d} + \frac {\log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, a^{3} d} \] Input:

integrate(tanh(d*x+c)^5/(a+b*sech(d*x+c)^2)^3,x, algorithm="maxima")
 

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tanh(d*x+c)^5/(a+b*sech(d*x+c)^2)^3,x, algorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Limit: Max order reached or unable 
to make series expansion Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx=\int \frac {{\mathrm {cosh}\left (c+d\,x\right )}^6\,{\mathrm {tanh}\left (c+d\,x\right )}^5}{{\left (a\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\right )}^3} \,d x \] Input:

int(tanh(c + d*x)^5/(a + b/cosh(c + d*x)^2)^3,x)
 

Output:

int((cosh(c + d*x)^6*tanh(c + d*x)^5)/(b + a*cosh(c + d*x)^2)^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 2301, normalized size of antiderivative = 29.88 \[ \int \frac {\tanh ^5(c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(tanh(d*x+c)^5/(a+b*sech(d*x+c)^2)^3,x)
 

Output:

(e**(8*c + 8*d*x)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d 
*x)*sqrt(a))*a**3 + 2*e**(8*c + 8*d*x)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - 
 a - 2*b) + e**(c + d*x)*sqrt(a))*a**2*b + e**(8*c + 8*d*x)*log(sqrt(2*sqr 
t(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**3 + 2*e**(8*c + 8*d 
*x)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a**2 
*b + e**(8*c + 8*d*x)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 
 2*b)*a**3 + 2*e**(8*c + 8*d*x)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d* 
x)*a + a + 2*b)*a**2*b - 2*e**(8*c + 8*d*x)*a**3*d*x - 2*e**(8*c + 8*d*x)* 
a**3 - 4*e**(8*c + 8*d*x)*a**2*b*d*x - 2*e**(8*c + 8*d*x)*a**2*b + 4*e**(6 
*c + 6*d*x)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sq 
rt(a))*a**3 + 16*e**(6*c + 6*d*x)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 
2*b) + e**(c + d*x)*sqrt(a))*a**2*b + 16*e**(6*c + 6*d*x)*log( - sqrt(2*sq 
rt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a*b**2 + 4*e**(6*c + 
6*d*x)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a 
**3 + 16*e**(6*c + 6*d*x)*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + e**( 
c + d*x)*sqrt(a))*a**2*b + 16*e**(6*c + 6*d*x)*log(sqrt(2*sqrt(b)*sqrt(a + 
 b) - a - 2*b) + e**(c + d*x)*sqrt(a))*a*b**2 + 4*e**(6*c + 6*d*x)*log(2*s 
qrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*a**3 + 16*e**(6*c + 6*d 
*x)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2*b)*a**2*b + 16* 
e**(6*c + 6*d*x)*log(2*sqrt(b)*sqrt(a + b) + e**(2*c + 2*d*x)*a + a + 2...