3.5.91 \(\int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx\) [491]

3.5.91.1 Optimal result

Integrand size = 15, antiderivative size = 24 \[ \int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx=-\frac {\arctan (\sinh (a+b x))}{b}-\frac {\text {csch}(a+b x)}{b} \]

output

-arctan(sinh(b*x+a))/b-csch(b*x+a)/b
 

3.5.91.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21 \[ \int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx=-\frac {\text {csch}(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\sinh ^2(a+b x)\right )}{b} \]

input

Integrate[Csch[a + b*x]^2*Sech[a + b*x],x]
 

output

-((Csch[a + b*x]*Hypergeometric2F1[-1/2, 1, 1/2, -Sinh[a + b*x]^2])/b)
 

3.5.91.3 Rubi [C] (verified)

Result contains complex when optimal does not.

Time = 0.20 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.21, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 25, 3101, 25, 262, 219}

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.

\(\displaystyle \int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int -\csc (i a+i b x)^2 \sec (i a+i b x)dx\)

\(\Big \downarrow \) 25

\(\displaystyle -\int \csc (i a+i b x)^2 \sec (i a+i b x)dx\)

\(\Big \downarrow \) 3101

\(\displaystyle -\frac {i \int \frac {\text {csch}^2(a+b x)}{\text {csch}^2(a+b x)+1}d(-i \text {csch}(a+b x))}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {i \int -\frac {\text {csch}^2(a+b x)}{\text {csch}^2(a+b x)+1}d(-i \text {csch}(a+b x))}{b}\)

\(\Big \downarrow \) 262

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

\(\Big \downarrow \) 219

\(\displaystyle -\frac {i (i \arctan (\text {csch}(a+b x))-i \text {csch}(a+b x))}{b}\)

input

Int[Csch[a + b*x]^2*Sech[a + b*x],x]
 

output

((-I)*(I*ArcTan[Csch[a + b*x]] - I*Csch[a + b*x]))/b
 

3.5.91.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

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

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 3101

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S 
ymbol] :> Simp[-(f*a^n)^(-1)   Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 
 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n 
 + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
 

3.5.91.4 Maple [A] (verified)

Time = 0.62 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {-\frac {1}{\sinh \left (b x +a \right )}-2 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\)	\(25\)
default	\(\frac {-\frac {1}{\sinh \left (b x +a \right )}-2 \arctan \left ({\mathrm e}^{b x +a}\right )}{b}\)	\(25\)
risch	\(-\frac {2 \,{\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}+\frac {i \ln \left ({\mathrm e}^{b x +a}-i\right )}{b}-\frac {i \ln \left ({\mathrm e}^{b x +a}+i\right )}{b}\)	\(58\)

input

int(csch(b*x+a)^2*sech(b*x+a),x,method=_RETURNVERBOSE)
 

output

1/b*(-1/sinh(b*x+a)-2*arctan(exp(b*x+a)))
 

3.5.91.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 103 vs. \(2 (24) = 48\).

Time = 0.25 (sec) , antiderivative size = 103, normalized size of antiderivative = 4.29 \[ \int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx=-\frac {2 \, {\left ({\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \arctan \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right ) + \cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b} \]

input

integrate(csch(b*x+a)^2*sech(b*x+a),x, algorithm="fricas")
 

output

-2*((cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh(b*x + a) + sinh(b*x + a)^2 - 1 
)*arctan(cosh(b*x + a) + sinh(b*x + a)) + cosh(b*x + a) + sinh(b*x + a))/( 
b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 - 
b)
 

3.5.91.6 Sympy [F]

\[ \int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx=\int \operatorname {csch}^{2}{\left (a + b x \right )} \operatorname {sech}{\left (a + b x \right )}\, dx \]

input

integrate(csch(b*x+a)**2*sech(b*x+a),x)
 

output

Integral(csch(a + b*x)**2*sech(a + b*x), x)
 

3.5.91.7 Maxima [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.79 \[ \int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx=\frac {2 \, \arctan \left (e^{\left (-b x - a\right )}\right )}{b} + \frac {2 \, e^{\left (-b x - a\right )}}{b {\left (e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )}} \]

input

integrate(csch(b*x+a)^2*sech(b*x+a),x, algorithm="maxima")
 

output

2*arctan(e^(-b*x - a))/b + 2*e^(-b*x - a)/(b*(e^(-2*b*x - 2*a) - 1))
 

3.5.91.8 Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 54 vs. \(2 (24) = 48\).

Time = 0.26 (sec) , antiderivative size = 54, normalized size of antiderivative = 2.25 \[ \int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx=-\frac {\pi + \frac {4}{e^{\left (b x + a\right )} - e^{\left (-b x - a\right )}} + 2 \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right )}{2 \, b} \]

input

integrate(csch(b*x+a)^2*sech(b*x+a),x, algorithm="giac")
 

output

-1/2*(pi + 4/(e^(b*x + a) - e^(-b*x - a)) + 2*arctan(1/2*(e^(2*b*x + 2*a) 
- 1)*e^(-b*x - a)))/b
 

3.5.91.9 Mupad [B] (verification not implemented)

Time = 2.26 (sec) , antiderivative size = 48, normalized size of antiderivative = 2.00 \[ \int \text {csch}^2(a+b x) \text {sech}(a+b x) \, dx=-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {b^2}}{b}\right )}{\sqrt {b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

input

int(1/(cosh(a + b*x)*sinh(a + b*x)^2),x)
 

output

- (2*atan((exp(b*x)*exp(a)*(b^2)^(1/2))/b))/(b^2)^(1/2) - (2*exp(a + b*x)) 
/(b*(exp(2*a + 2*b*x) - 1))