\(\int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx\) [356]

Optimal result

Integrand size = 25, antiderivative size = 78 \[ \int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {2 a b \text {arctanh}\left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{3/2} d}-\frac {\text {sech}(c+d x) (a-b \sinh (c+d x))}{\left (a^2+b^2\right ) d} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.33 \[ \int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {-2 a b \arctan \left (\frac {b-a \tanh \left (\frac {1}{2} (c+d x)\right )}{\sqrt {-a^2-b^2}}\right )-a \sqrt {-a^2-b^2} \text {sech}(c+d x)+b \sqrt {-a^2-b^2} \tanh (c+d x)}{\left (-a^2-b^2\right )^{3/2} d} \] Input:

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

Output:

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

Rubi [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 0.41 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 26, 3345, 26, 27, 3042, 3139, 1083, 217}

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.

\(\displaystyle \int \frac {\tanh (c+d x) \text {sech}(c+d x)}{a+b \sinh (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 3345

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 27

\(\displaystyle -i \left (-\frac {i a b \int \frac {1}{a+b \sinh (c+d x)}dx}{a^2+b^2}-\frac {i \text {sech}(c+d x) (a-b \sinh (c+d x))}{d \left (a^2+b^2\right )}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle -i \left (-\frac {i a b \int \frac {1}{a-i b \sin (i c+i d x)}dx}{a^2+b^2}-\frac {i \text {sech}(c+d x) (a-b \sinh (c+d x))}{d \left (a^2+b^2\right )}\right )\)

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

\(\displaystyle -i \left (\frac {4 a b \int \frac {1}{\tanh ^2\left (\frac {1}{2} (c+d x)\right )-4 \left (a^2+b^2\right )}d\left (2 i a \tanh \left (\frac {1}{2} (c+d x)\right )-2 i b\right )}{d \left (a^2+b^2\right )}-\frac {i \text {sech}(c+d x) (a-b \sinh (c+d x))}{d \left (a^2+b^2\right )}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle -i \left (-\frac {2 i a b \text {arctanh}\left (\frac {\tanh \left (\frac {1}{2} (c+d x)\right )}{2 \sqrt {a^2+b^2}}\right )}{d \left (a^2+b^2\right )^{3/2}}-\frac {i \text {sech}(c+d x) (a-b \sinh (c+d x))}{d \left (a^2+b^2\right )}\right )\)

Input:

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

Output:

(-I)*(((-2*I)*a*b*ArcTanh[Tanh[(c + d*x)/2]/(2*Sqrt[a^2 + b^2])])/((a^2 + 
b^2)^(3/2)*d) - (I*Sech[c + d*x]*(a - b*Sinh[c + d*x]))/((a^2 + b^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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

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

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3345

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*Co 
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*((b*c - a*d - (a*c - b*d)* 
Sin[e + f*x])/(f*g*(a^2 - b^2)*(p + 1))), x] + Simp[1/(g^2*(a^2 - b^2)*(p + 
 1))   Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*Simp[c*(a^2*(p + 
 2) - b^2*(m + p + 2)) + a*b*d*m + b*(a*c - b*d)*(m + p + 3)*Sin[e + f*x], 
x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && Lt 
Q[p, -1] && IntegerQ[2*m]
 

Maple [A] (verified)

Time = 2.06 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.29

method	result	size
derivativedivides	\(\frac {\frac {2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a}{\left (a^{2}+b^{2}\right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {4 a b \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(101\)
default	\(\frac {\frac {2 b \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 a}{\left (a^{2}+b^{2}\right ) \left (1+\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}-\frac {4 a b \,\operatorname {arctanh}\left (\frac {2 a \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2 b}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (2 a^{2}+2 b^{2}\right ) \sqrt {a^{2}+b^{2}}}}{d}\)	\(101\)
risch	\(-\frac {2 \left (a \,{\mathrm e}^{d x +c}+b \right )}{d \left (a^{2}+b^{2}\right ) \left (1+{\mathrm e}^{2 d x +2 c}\right )}+\frac {b a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}+a^{4}+2 a^{2} b^{2}+b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}-\frac {b a \ln \left ({\mathrm e}^{d x +c}+\frac {a \left (a^{2}+b^{2}\right )^{\frac {3}{2}}-a^{4}-2 a^{2} b^{2}-b^{4}}{b \left (a^{2}+b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a^{2}+b^{2}\right )^{\frac {3}{2}} d}\)	\(168\)

Input:

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

Output:

1/d*(2/(a^2+b^2)*(b*tanh(1/2*d*x+1/2*c)-a)/(1+tanh(1/2*d*x+1/2*c)^2)-4*a*b 
/(2*a^2+2*b^2)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*a*tanh(1/2*d*x+1/2*c)-2*b)/( 
a^2+b^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 350 vs. \(2 (75) = 150\).

Time = 0.08 (sec) , antiderivative size = 350, normalized size of antiderivative = 4.49 \[ \int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {2 \, a^{2} b + 2 \, b^{3} - {\left (a b \cosh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + a b \sinh \left (d x + c\right )^{2} + a b\right )} \sqrt {a^{2} + b^{2}} \log \left (\frac {b^{2} \cosh \left (d x + c\right )^{2} + b^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + 2 \, a^{2} + b^{2} + 2 \, {\left (b^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right ) + 2 \, \sqrt {a^{2} + b^{2}} {\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right ) + a\right )}}{b \cosh \left (d x + c\right )^{2} + b \sinh \left (d x + c\right )^{2} + 2 \, a \cosh \left (d x + c\right ) + 2 \, {\left (b \cosh \left (d x + c\right ) + a\right )} \sinh \left (d x + c\right ) - b}\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \cosh \left (d x + c\right ) + 2 \, {\left (a^{3} + a b^{2}\right )} \sinh \left (d x + c\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right )^{2} + 2 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d \sinh \left (d x + c\right )^{2} + {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d} \] Input:

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

Output:

-(2*a^2*b + 2*b^3 - (a*b*cosh(d*x + c)^2 + 2*a*b*cosh(d*x + c)*sinh(d*x + 
c) + a*b*sinh(d*x + c)^2 + a*b)*sqrt(a^2 + b^2)*log((b^2*cosh(d*x + c)^2 + 
 b^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + 2*a^2 + b^2 + 2*(b^2*cosh(d*x 
 + c) + a*b)*sinh(d*x + c) + 2*sqrt(a^2 + b^2)*(b*cosh(d*x + c) + b*sinh(d 
*x + c) + a))/(b*cosh(d*x + c)^2 + b*sinh(d*x + c)^2 + 2*a*cosh(d*x + c) + 
 2*(b*cosh(d*x + c) + a)*sinh(d*x + c) - b)) + 2*(a^3 + a*b^2)*cosh(d*x + 
c) + 2*(a^3 + a*b^2)*sinh(d*x + c))/((a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + 
c)^2 + 2*(a^4 + 2*a^2*b^2 + b^4)*d*cosh(d*x + c)*sinh(d*x + c) + (a^4 + 2* 
a^2*b^2 + b^4)*d*sinh(d*x + c)^2 + (a^4 + 2*a^2*b^2 + b^4)*d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.50 \[ \int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {a b \log \left (\frac {b e^{\left (-d x - c\right )} - a - \sqrt {a^{2} + b^{2}}}{b e^{\left (-d x - c\right )} - a + \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}} d} - \frac {2 \, {\left (a e^{\left (-d x - c\right )} - b\right )}}{{\left (a^{2} + b^{2} + {\left (a^{2} + b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )}\right )} d} \] Input:

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

Output:

-a*b*log((b*e^(-d*x - c) - a - sqrt(a^2 + b^2))/(b*e^(-d*x - c) - a + sqrt 
(a^2 + b^2)))/((a^2 + b^2)^(3/2)*d) - 2*(a*e^(-d*x - c) - b)/((a^2 + b^2 + 
 (a^2 + b^2)*e^(-2*d*x - 2*c))*d)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.36 \[ \int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=-\frac {\frac {a b \log \left (\frac {{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b e^{\left (d x + c\right )} + 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{2} + b^{2}\right )}^{\frac {3}{2}}} + \frac {2 \, {\left (a e^{\left (d x + c\right )} + b\right )}}{{\left (a^{2} + b^{2}\right )} {\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}}}{d} \] Input:

integrate(sech(d*x+c)*tanh(d*x+c)/(a+b*sinh(d*x+c)),x, algorithm="giac")
 

Output:

-(a*b*log(abs(2*b*e^(d*x + c) + 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*e^(d*x + 
c) + 2*a + 2*sqrt(a^2 + b^2)))/(a^2 + b^2)^(3/2) + 2*(a*e^(d*x + c) + b)/( 
(a^2 + b^2)*(e^(2*d*x + 2*c) + 1)))/d
 

Mupad [B] (verification not implemented)

Time = 1.44 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.18 \[ \int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {a\,b\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{a^2+b^2}+\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{{\left (a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\frac {a\,b\,\ln \left (\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{a^2+b^2}-\frac {2\,a\,\left (b-a\,{\mathrm {e}}^{c+d\,x}\right )}{{\left (a^2+b^2\right )}^{3/2}}\right )}{d\,{\left (a^2+b^2\right )}^{3/2}}-\frac {\frac {2\,b}{d\,\left (a^2+b^2\right )}+\frac {2\,a\,{\mathrm {e}}^{c+d\,x}}{d\,\left (a^2+b^2\right )}}{{\mathrm {e}}^{2\,c+2\,d\,x}+1} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.71 \[ \int \frac {\text {sech}(c+d x) \tanh (c+d x)}{a+b \sinh (c+d x)} \, dx=\frac {-2 e^{2 d x +2 c} \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a b i -2 \sqrt {a^{2}+b^{2}}\, \mathit {atan} \left (\frac {e^{d x +c} b i +a i}{\sqrt {a^{2}+b^{2}}}\right ) a b i +2 e^{2 d x +2 c} a^{2} b +2 e^{2 d x +2 c} b^{3}-2 e^{d x +c} a^{3}-2 e^{d x +c} a \,b^{2}}{d \left (e^{2 d x +2 c} a^{4}+2 e^{2 d x +2 c} a^{2} b^{2}+e^{2 d x +2 c} b^{4}+a^{4}+2 a^{2} b^{2}+b^{4}\right )} \] Input:

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

Output:

(2*( - e**(2*c + 2*d*x)*sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/sq 
rt(a**2 + b**2))*a*b*i - sqrt(a**2 + b**2)*atan((e**(c + d*x)*b*i + a*i)/s 
qrt(a**2 + b**2))*a*b*i + e**(2*c + 2*d*x)*a**2*b + e**(2*c + 2*d*x)*b**3 
- e**(c + d*x)*a**3 - e**(c + d*x)*a*b**2))/(d*(e**(2*c + 2*d*x)*a**4 + 2* 
e**(2*c + 2*d*x)*a**2*b**2 + e**(2*c + 2*d*x)*b**4 + a**4 + 2*a**2*b**2 + 
b**4))