\(\int \frac {\sinh (c+d x)}{(a+b \text {sech}^2(c+d x))^2} \, dx\) [36]

Optimal result

Integrand size = 21, antiderivative size = 80 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=-\frac {3 \sqrt {b} \arctan \left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {b}}\right )}{2 a^{5/2} d}+\frac {\cosh (c+d x)}{a^2 d}+\frac {b \cosh (c+d x)}{2 a^2 d \left (b+a \cosh ^2(c+d x)\right )} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.82 (sec) , antiderivative size = 479, normalized size of antiderivative = 5.99 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {(a+2 b+a \cosh (2 (c+d x)))^2 \text {sech}^4(c+d x) \left (\frac {32 \cosh (c) \cosh (d x)}{a^2}+\frac {32 b \cosh (c+d x)}{a^2 (a+2 b+a \cosh (2 (c+d x)))}+\frac {2 \left (-\left (\left (a^2+24 b^2\right ) \arctan \left (\frac {\left (\sqrt {a}-i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{2}\right )+\cosh (c) \left (\sqrt {a}-i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2} \tanh \left (\frac {d x}{2}\right )\right )}{\sqrt {b}}\right )\right )-a^2 \arctan \left (\frac {\left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{2}\right )+\cosh (c) \left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2} \tanh \left (\frac {d x}{2}\right )\right )}{\sqrt {b}}\right )-24 b^2 \arctan \left (\frac {\left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2}\right ) \sinh (c) \tanh \left (\frac {d x}{2}\right )+\cosh (c) \left (\sqrt {a}+i \sqrt {a+b} \sqrt {(\cosh (c)-\sinh (c))^2} \tanh \left (\frac {d x}{2}\right )\right )}{\sqrt {b}}\right )+a^2 \arctan \left (\frac {\sqrt {a}-i \sqrt {a+b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )+a^2 \arctan \left (\frac {\sqrt {a}+i \sqrt {a+b} \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {b}}\right )+16 \sqrt {a} b^{3/2} \sinh (c) \sinh (d x)\right )}{a^{5/2} b^{3/2}}\right )}{128 d \left (a+b \text {sech}^2(c+d x)\right )^2} \] Input:

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

Output:

((a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x]^4*((32*Cosh[c]*Cosh[d*x]) 
/a^2 + (32*b*Cosh[c + d*x])/(a^2*(a + 2*b + a*Cosh[2*(c + d*x)])) + (2*(-( 
(a^2 + 24*b^2)*ArcTan[((Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2 
])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cosh[c] 
- Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]]) - a^2*ArcTan[((Sqrt[a] + I*Sqrt[a 
+ b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c]*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] 
 + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2]*Tanh[(d*x)/2]))/Sqrt[b]] - 24 
*b^2*ArcTan[((Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c])^2])*Sinh[c] 
*Tanh[(d*x)/2] + Cosh[c]*(Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cosh[c] - Sinh[c]) 
^2]*Tanh[(d*x)/2]))/Sqrt[b]] + a^2*ArcTan[(Sqrt[a] - I*Sqrt[a + b]*Tanh[(c 
 + d*x)/2])/Sqrt[b]] + a^2*ArcTan[(Sqrt[a] + I*Sqrt[a + b]*Tanh[(c + d*x)/ 
2])/Sqrt[b]] + 16*Sqrt[a]*b^(3/2)*Sinh[c]*Sinh[d*x]))/(a^(5/2)*b^(3/2))))/ 
(128*d*(a + b*Sech[c + d*x]^2)^2)
 

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 26, 4621, 252, 262, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 4621

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

\(\Big \downarrow \) 252

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

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

((3*(-((Sqrt[b]*ArcTan[(Sqrt[a]*Cosh[c + d*x])/Sqrt[b]])/a^(3/2)) + Cosh[c 
 + d*x]/a))/(2*a) - Cosh[c + d*x]^3/(2*a*(b + a*Cosh[c + d*x]^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 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4621

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

Maple [A] (verified)

Time = 31.79 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(-\frac {-\frac {1}{a^{2} \operatorname {sech}\left (d x +c \right )}-\frac {b \left (\frac {\operatorname {sech}\left (d x +c \right )}{2 a +2 b \operatorname {sech}\left (d x +c \right )^{2}}+\frac {3 \arctan \left (\frac {b \,\operatorname {sech}\left (d x +c \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}}{d}\)	\(70\)
default	\(-\frac {-\frac {1}{a^{2} \operatorname {sech}\left (d x +c \right )}-\frac {b \left (\frac {\operatorname {sech}\left (d x +c \right )}{2 a +2 b \operatorname {sech}\left (d x +c \right )^{2}}+\frac {3 \arctan \left (\frac {b \,\operatorname {sech}\left (d x +c \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{a^{2}}}{d}\)	\(70\)
risch	\(\frac {{\mathrm e}^{d x +c}}{2 a^{2} d}+\frac {{\mathrm e}^{-d x -c}}{2 a^{2} d}+\frac {{\mathrm e}^{d x +c} b \left ({\mathrm e}^{2 d x +2 c}+1\right )}{a^{2} d \left ({\mathrm e}^{4 d x +4 c} a +2 a \,{\mathrm e}^{2 d x +2 c}+4 b \,{\mathrm e}^{2 d x +2 c}+a \right )}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}-\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{4 a^{3} d}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 d x +2 c}+\frac {2 \sqrt {-a b}\, {\mathrm e}^{d x +c}}{a}+1\right )}{4 a^{3} d}\)	\(183\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 877 vs. \(2 (68) = 136\).

Time = 0.34 (sec) , antiderivative size = 1780, normalized size of antiderivative = 22.25 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/4*(2*a*cosh(d*x + c)^6 + 12*a*cosh(d*x + c)*sinh(d*x + c)^5 + 2*a*sinh( 
d*x + c)^6 + 6*(a + 2*b)*cosh(d*x + c)^4 + 6*(5*a*cosh(d*x + c)^2 + a + 2* 
b)*sinh(d*x + c)^4 + 8*(5*a*cosh(d*x + c)^3 + 3*(a + 2*b)*cosh(d*x + c))*s 
inh(d*x + c)^3 + 6*(a + 2*b)*cosh(d*x + c)^2 + 6*(5*a*cosh(d*x + c)^4 + 6* 
(a + 2*b)*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 3*(a*cosh(d*x + c)^ 
5 + 5*a*cosh(d*x + c)*sinh(d*x + c)^4 + a*sinh(d*x + c)^5 + 2*(a + 2*b)*co 
sh(d*x + c)^3 + 2*(5*a*cosh(d*x + c)^2 + a + 2*b)*sinh(d*x + c)^3 + 2*(5*a 
*cosh(d*x + c)^3 + 3*(a + 2*b)*cosh(d*x + c))*sinh(d*x + c)^2 + a*cosh(d*x 
 + c) + (5*a*cosh(d*x + c)^4 + 6*(a + 2*b)*cosh(d*x + c)^2 + a)*sinh(d*x + 
 c))*sqrt(-b/a)*log((a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x + c)^3 
 + a*sinh(d*x + c)^4 + 2*(a - 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d*x + c)^ 
2 + a - 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a - 2*b)*cosh(d*x + 
 c))*sinh(d*x + c) - 4*(a*cosh(d*x + c)^3 + 3*a*cosh(d*x + c)*sinh(d*x + c 
)^2 + a*sinh(d*x + c)^3 + a*cosh(d*x + c) + (3*a*cosh(d*x + c)^2 + a)*sinh 
(d*x + c))*sqrt(-b/a) + a)/(a*cosh(d*x + c)^4 + 4*a*cosh(d*x + c)*sinh(d*x 
 + c)^3 + a*sinh(d*x + c)^4 + 2*(a + 2*b)*cosh(d*x + c)^2 + 2*(3*a*cosh(d* 
x + c)^2 + a + 2*b)*sinh(d*x + c)^2 + 4*(a*cosh(d*x + c)^3 + (a + 2*b)*cos 
h(d*x + c))*sinh(d*x + c) + a)) + 12*(a*cosh(d*x + c)^5 + 2*(a + 2*b)*cosh 
(d*x + c)^3 + (a + 2*b)*cosh(d*x + c))*sinh(d*x + c) + 2*a)/(a^3*d*cosh(d* 
x + c)^5 + 5*a^3*d*cosh(d*x + c)*sinh(d*x + c)^4 + a^3*d*sinh(d*x + c)^...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate(sinh(d*x+c)/(a+b*sech(d*x+c)^2)^2,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 [B] (verification not implemented)

Time = 2.57 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.89 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx=\frac {b\,\mathrm {cosh}\left (c+d\,x\right )}{2\,\left (d\,a^3\,{\mathrm {cosh}\left (c+d\,x\right )}^2+b\,d\,a^2\right )}+\frac {\mathrm {cosh}\left (c+d\,x\right )}{a^2\,d}-\frac {3\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\mathrm {cosh}\left (c+d\,x\right )}{\sqrt {b}}\right )}{2\,a^{5/2}\,d} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 1643, normalized size of antiderivative = 20.54 \[ \int \frac {\sinh (c+d x)}{\left (a+b \text {sech}^2(c+d x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 6*e**(5*c + 5*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + 
 b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + 
 a + 2*b)))*a - 12*e**(3*c + 3*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sqrt(2*sqr 
t(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)* 
sqrt(a + b) + a + 2*b)))*a - 24*e**(3*c + 3*d*x)*sqrt(b)*sqrt(a)*sqrt(a + 
b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sq 
rt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*b - 6*e**(c + d*x)*sqrt(b)*sqrt(a)*s 
qrt(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sq 
rt(a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a + 6*e**(5*c + 5*d*x)*sqrt( 
a)*sqrt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sq 
rt(2*sqrt(b)*sqrt(a + b) + a + 2*b)))*a*b + 12*e**(3*c + 3*d*x)*sqrt(a)*sq 
rt(2*sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2* 
sqrt(b)*sqrt(a + b) + a + 2*b)))*a*b + 24*e**(3*c + 3*d*x)*sqrt(a)*sqrt(2* 
sqrt(b)*sqrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt( 
b)*sqrt(a + b) + a + 2*b)))*b**2 + 6*e**(c + d*x)*sqrt(a)*sqrt(2*sqrt(b)*s 
qrt(a + b) + a + 2*b)*atan((e**(c + d*x)*a)/(sqrt(a)*sqrt(2*sqrt(b)*sqrt(a 
 + b) + a + 2*b)))*a*b - 3*e**(5*c + 5*d*x)*sqrt(b)*sqrt(a)*sqrt(a + b)*sq 
rt(2*sqrt(b)*sqrt(a + b) - a - 2*b)*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a 
- 2*b) + e**(c + d*x)*sqrt(a))*a + 3*e**(5*c + 5*d*x)*sqrt(b)*sqrt(a)*sqrt 
(a + b)*sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b)*log(sqrt(2*sqrt(b)*sqrt(a...