3.20 \(\int \frac{\cosh (c+d x)}{a+b \text{csch}(c+d x)} \, dx\)

Optimal. Leaf size=34 \[ \frac{\sinh (c+d x)}{a d}-\frac{b \log (a \sinh (c+d x)+b)}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.0890107, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3872, 2833, 12, 43} \[ \frac{\sinh (c+d x)}{a d}-\frac{b \log (a \sinh (c+d x)+b)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2833

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

Rule 12

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0217856, size = 30, normalized size = 0.88 \[ \frac{a \sinh (c+d x)-b \log (a \sinh (c+d x)+b)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.001, size = 52, normalized size = 1.5 \begin{align*}{\frac{1}{da{\rm csch} \left (dx+c\right )}}+{\frac{b\ln \left ({\rm csch} \left (dx+c\right ) \right ) }{d{a}^{2}}}-{\frac{b\ln \left ( a+b{\rm csch} \left (dx+c\right ) \right ) }{d{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.23365, size = 112, normalized size = 3.29 \begin{align*} -\frac{{\left (d x + c\right )} b}{a^{2} d} + \frac{e^{\left (d x + c\right )}}{2 \, a d} - \frac{e^{\left (-d x - c\right )}}{2 \, a d} - \frac{b \log \left (-2 \, b e^{\left (-d x - c\right )} + a e^{\left (-2 \, d x - 2 \, c\right )} - a\right )}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.65089, size = 354, normalized size = 10.41 \begin{align*} \frac{2 \, b d x \cosh \left (d x + c\right ) + a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} - 2 \,{\left (b \cosh \left (d x + c\right ) + b \sinh \left (d x + c\right )\right )} \log \left (\frac{2 \,{\left (a \sinh \left (d x + c\right ) + b\right )}}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 2 \,{\left (b d x + a \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) - a}{2 \,{\left (a^{2} d \cosh \left (d x + c\right ) + a^{2} d \sinh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*b*d*x*cosh(d*x + c) + a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 - 2*(b*cosh(d*x + c) + b*sinh(d*x + c))*log
(2*(a*sinh(d*x + c) + b)/(cosh(d*x + c) - sinh(d*x + c))) + 2*(b*d*x + a*cosh(d*x + c))*sinh(d*x + c) - a)/(a^
2*d*cosh(d*x + c) + a^2*d*sinh(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cosh{\left (c + d x \right )}}{a + b \operatorname{csch}{\left (c + d x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cosh(c + d*x)/(a + b*csch(c + d*x)), x)

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Giac [A]  time = 1.1507, size = 84, normalized size = 2.47 \begin{align*} \frac{e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}}{2 \, a d} - \frac{b \log \left ({\left | a{\left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )} + 2 \, b \right |}\right )}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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