### 3.25 $$\int \text{csch}^{-1}(c e^{a+b x}) \, dx$$

Optimal. Leaf size=77 $-\frac{\text{PolyLog}\left (2,e^{2 \text{csch}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac{\text{csch}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{\text{csch}^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 \text{csch}^{-1}\left (c e^{a+b x}\right )}\right )}{b}$

[Out]

ArcCsch[c*E^(a + b*x)]^2/(2*b) - (ArcCsch[c*E^(a + b*x)]*Log[1 - E^(2*ArcCsch[c*E^(a + b*x)])])/b - PolyLog[2,
E^(2*ArcCsch[c*E^(a + b*x)])]/(2*b)

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Rubi [A]  time = 0.0879159, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.7, Rules used = {2282, 6282, 5659, 3716, 2190, 2279, 2391} $-\frac{\text{PolyLog}\left (2,e^{2 \text{csch}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac{\text{csch}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{\text{csch}^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 \text{csch}^{-1}\left (c e^{a+b x}\right )}\right )}{b}$

Antiderivative was successfully veriﬁed.

[In]

Int[ArcCsch[c*E^(a + b*x)],x]

[Out]

ArcCsch[c*E^(a + b*x)]^2/(2*b) - (ArcCsch[c*E^(a + b*x)]*Log[1 - E^(2*ArcCsch[c*E^(a + b*x)])])/b - PolyLog[2,
E^(2*ArcCsch[c*E^(a + b*x)])]/(2*b)

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6282

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcSinh[x/c])/x, x], x, 1/x] /; F
reeQ[{a, b, c}, x]

Rule 5659

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tanh[x], x], x, ArcSinh
[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> -Simp[(I*(c
+ d*x)^(m + 1))/(d*(m + 1)), x] + Dist[2*I, Int[((c + d*x)^m*E^(2*(-(I*e) + f*fz*x)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \text{csch}^{-1}\left (c e^{a+b x}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\text{csch}^{-1}(c x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}\left (\frac{x}{c}\right )}{x} \, dx,x,e^{-a-b x}\right )}{b}\\ &=-\frac{\operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}+\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}+\frac{\operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )\right )}{b}\\ &=\frac{\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}+\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{2 b}\\ &=\frac{\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )^2}{2 b}-\frac{\sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{b}-\frac{\text{Li}_2\left (e^{2 \sinh ^{-1}\left (\frac{e^{-a-b x}}{c}\right )}\right )}{2 b}\\ \end{align*}

Mathematica [B]  time = 0.568507, size = 236, normalized size = 3.06 $\frac{e^{-a-b x} \sqrt{c^2 e^{2 (a+b x)}+1} \left (-4 \text{PolyLog}\left (2,\frac{1}{2} \left (1-\sqrt{c^2 e^{2 (a+b x)}+1}\right )\right )+\log ^2\left (-c^2 e^{2 (a+b x)}\right )+2 \log ^2\left (\frac{1}{2} \left (\sqrt{c^2 e^{2 (a+b x)}+1}+1\right )\right )-4 \log \left (\frac{1}{2} \left (\sqrt{c^2 e^{2 (a+b x)}+1}+1\right )\right ) \log \left (-c^2 e^{2 (a+b x)}\right )+\left (4 \log \left (-c^2 e^{2 (a+b x)}\right )-8 b x\right ) \tanh ^{-1}\left (\sqrt{c^2 e^{2 (a+b x)}+1}\right )\right )}{8 b c \sqrt{\frac{e^{-2 (a+b x)}}{c^2}+1}}+x \text{csch}^{-1}\left (c e^{a+b x}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[ArcCsch[c*E^(a + b*x)],x]

[Out]

x*ArcCsch[c*E^(a + b*x)] + (E^(-a - b*x)*Sqrt[1 + c^2*E^(2*(a + b*x))]*(Log[-(c^2*E^(2*(a + b*x)))]^2 + ArcTan
h[Sqrt[1 + c^2*E^(2*(a + b*x))]]*(-8*b*x + 4*Log[-(c^2*E^(2*(a + b*x)))]) - 4*Log[-(c^2*E^(2*(a + b*x)))]*Log[
(1 + Sqrt[1 + c^2*E^(2*(a + b*x))])/2] + 2*Log[(1 + Sqrt[1 + c^2*E^(2*(a + b*x))])/2]^2 - 4*PolyLog[2, (1 - Sq
rt[1 + c^2*E^(2*(a + b*x))])/2]))/(8*b*c*Sqrt[1 + 1/(c^2*E^(2*(a + b*x)))])

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Maple [F]  time = 0.434, size = 0, normalized size = 0. \begin{align*} \int{\rm arccsch} \left (c{{\rm e}^{bx+a}}\right )\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(arccsch(c*exp(b*x+a)),x)

[Out]

int(arccsch(c*exp(b*x+a)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b c^{2} \int \frac{x e^{\left (2 \, b x + 2 \, a\right )}}{c^{2} e^{\left (2 \, b x + 2 \, a\right )} +{\left (c^{2} e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{\frac{3}{2}} + 1}\,{d x} - \frac{1}{2} \, b x^{2} -{\left (a + \log \left (c\right )\right )} x + x \log \left (\sqrt{c^{2} e^{\left (2 \, b x + 2 \, a\right )} + 1} + 1\right ) - \frac{2 \, b x \log \left (c^{2} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-c^{2} e^{\left (2 \, b x + 2 \, a\right )}\right )}{4 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(c*exp(b*x+a)),x, algorithm="maxima")

[Out]

b*c^2*integrate(x*e^(2*b*x + 2*a)/(c^2*e^(2*b*x + 2*a) + (c^2*e^(2*b*x + 2*a) + 1)^(3/2) + 1), x) - 1/2*b*x^2
- (a + log(c))*x + x*log(sqrt(c^2*e^(2*b*x + 2*a) + 1) + 1) - 1/4*(2*b*x*log(c^2*e^(2*b*x + 2*a) + 1) + dilog(
-c^2*e^(2*b*x + 2*a)))/b

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(c*exp(b*x+a)),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsch(c*exp(b*x+a)),x)

[Out]

Integral(acsch(c*exp(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arcsch}\left (c e^{\left (b x + a\right )}\right )\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(c*exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(arccsch(c*e^(b*x + a)), x)