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

Optimal. Leaf size=90 \[ -\frac{b^2 \log (a+b \sinh (c+d x))}{a d \left (a^2+b^2\right )}-\frac{b \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )}-\frac{a \log (\cosh (c+d x))}{d \left (a^2+b^2\right )}+\frac{\log (\sinh (c+d x))}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.169638, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2837, 12, 894, 635, 203, 260} \[ -\frac{b^2 \log (a+b \sinh (c+d x))}{a d \left (a^2+b^2\right )}-\frac{b \tan ^{-1}(\sinh (c+d x))}{d \left (a^2+b^2\right )}-\frac{a \log (\cosh (c+d x))}{d \left (a^2+b^2\right )}+\frac{\log (\sinh (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2837

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 894

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && IntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 203

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{\text{csch}(c+d x) \text{sech}(c+d x)}{a+b \sinh (c+d x)} \, dx &=-\frac{b \operatorname{Subst}\left (\int \frac{b}{x (a+x) \left (-b^2-x^2\right )} \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{x (a+x) \left (-b^2-x^2\right )} \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=-\frac{b^2 \operatorname{Subst}\left (\int \left (-\frac{1}{a b^2 x}+\frac{1}{a \left (a^2+b^2\right ) (a+x)}+\frac{b^2+a x}{b^2 \left (a^2+b^2\right ) \left (b^2+x^2\right )}\right ) \, dx,x,b \sinh (c+d x)\right )}{d}\\ &=\frac{\log (\sinh (c+d x))}{a d}-\frac{b^2 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right ) d}-\frac{\operatorname{Subst}\left (\int \frac{b^2+a x}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=\frac{\log (\sinh (c+d x))}{a d}-\frac{b^2 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right ) d}-\frac{a \operatorname{Subst}\left (\int \frac{x}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right ) d}-\frac{b^2 \operatorname{Subst}\left (\int \frac{1}{b^2+x^2} \, dx,x,b \sinh (c+d x)\right )}{\left (a^2+b^2\right ) d}\\ &=-\frac{b \tan ^{-1}(\sinh (c+d x))}{\left (a^2+b^2\right ) d}-\frac{a \log (\cosh (c+d x))}{\left (a^2+b^2\right ) d}+\frac{\log (\sinh (c+d x))}{a d}-\frac{b^2 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 0.133677, size = 92, normalized size = 1.02 \[ -\frac{\frac{2 b^2 \log (a+b \sinh (c+d x))}{a \left (a^2+b^2\right )}+\frac{\log (-\sinh (c+d x)+i)}{a+i b}+\frac{\log (\sinh (c+d x)+i)}{a-i b}-\frac{2 \log (\sinh (c+d x))}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Log[I - Sinh[c + d*x]]/(a + I*b) - (2*Log[Sinh[c + d*x]])/a + Log[I + Sinh[c + d*x]]/(a - I*b) + (2*b^2*Log[
a + b*Sinh[c + d*x]])/(a*(a^2 + b^2)))/(2*d)

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 123, normalized size = 1.4 \begin{align*}{\frac{1}{da}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }-{\frac{{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) a}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a-2\,\tanh \left ( 1/2\,dx+c/2 \right ) b-a \right ) }-{\frac{a}{d \left ({a}^{2}+{b}^{2} \right ) }\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+1 \right ) }-2\,{\frac{b\arctan \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d/a*ln(tanh(1/2*d*x+1/2*c))-1/d*b^2/(a^2+b^2)/a*ln(tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)*b-a)-1/d/(a
^2+b^2)*a*ln(tanh(1/2*d*x+1/2*c)^2+1)-2/d/(a^2+b^2)*b*arctan(tanh(1/2*d*x+1/2*c))

________________________________________________________________________________________

Maxima [A]  time = 1.72629, size = 186, normalized size = 2.07 \begin{align*} -\frac{b^{2} \log \left (-2 \, a e^{\left (-d x - c\right )} + b e^{\left (-2 \, d x - 2 \, c\right )} - b\right )}{{\left (a^{3} + a b^{2}\right )} d} + \frac{2 \, b \arctan \left (e^{\left (-d x - c\right )}\right )}{{\left (a^{2} + b^{2}\right )} d} - \frac{a \log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{{\left (a^{2} + b^{2}\right )} d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{a d} + \frac{\log \left (e^{\left (-d x - c\right )} - 1\right )}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.6949, size = 350, normalized size = 3.89 \begin{align*} -\frac{2 \, a b \arctan \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right )\right ) + b^{2} \log \left (\frac{2 \,{\left (b \sinh \left (d x + c\right ) + a\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + a^{2} \log \left (\frac{2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) -{\left (a^{2} + b^{2}\right )} \log \left (\frac{2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right )}{{\left (a^{3} + a b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.27236, size = 211, normalized size = 2.34 \begin{align*} -\frac{b^{3} \log \left ({\left | b{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, a \right |}\right )}{a^{3} b d + a b^{3} d} - \frac{{\left (\pi + 2 \, \arctan \left (\frac{1}{2} \,{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )} e^{\left (-d x - c\right )}\right )\right )} b}{2 \,{\left (a^{2} d + b^{2} d\right )}} - \frac{a \log \left ({\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}^{2} + 4\right )}{2 \,{\left (a^{2} d + b^{2} d\right )}} + \frac{\log \left ({\left | e^{\left (d x + c\right )} - e^{\left (-d x - c\right )} \right |}\right )}{a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^3*log(abs(b*(e^(d*x + c) - e^(-d*x - c)) + 2*a))/(a^3*b*d + a*b^3*d) - 1/2*(pi + 2*arctan(1/2*(e^(2*d*x + 2
*c) - 1)*e^(-d*x - c)))*b/(a^2*d + b^2*d) - 1/2*a*log((e^(d*x + c) - e^(-d*x - c))^2 + 4)/(a^2*d + b^2*d) + lo
g(abs(e^(d*x + c) - e^(-d*x - c)))/(a*d)