### 3.69 $$\int \frac{\text{csch}^{-1}(a+b x)}{\frac{a d}{b}+d x} \, dx$$

Optimal. Leaf size=61 $-\frac{\text{PolyLog}\left (2,e^{2 \text{csch}^{-1}(a+b x)}\right )}{2 d}+\frac{\text{csch}^{-1}(a+b x)^2}{2 d}-\frac{\text{csch}^{-1}(a+b x) \log \left (1-e^{2 \text{csch}^{-1}(a+b x)}\right )}{d}$

[Out]

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

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Rubi [A]  time = 0.0977974, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 19, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.421, Rules used = {6320, 12, 6282, 5659, 3716, 2190, 2279, 2391} $-\frac{\text{PolyLog}\left (2,e^{2 \text{csch}^{-1}(a+b x)}\right )}{2 d}+\frac{\text{csch}^{-1}(a+b x)^2}{2 d}-\frac{\text{csch}^{-1}(a+b x) \log \left (1-e^{2 \text{csch}^{-1}(a+b x)}\right )}{d}$

Antiderivative was successfully veriﬁed.

[In]

Int[ArcCsch[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

Rule 6320

Int[((a_.) + ArcCsch[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((f*x)/d)^m*(a + b*ArcCsch[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[d*e - c*f,
0] && IGtQ[p, 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 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 \frac{\text{csch}^{-1}(a+b x)}{\frac{a d}{b}+d x} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{b \text{csch}^{-1}(x)}{d x} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\text{csch}^{-1}(x)}{x} \, dx,x,a+b x\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{x} \, dx,x,\frac{1}{a+b x}\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{\sinh ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}+\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{\sinh ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\sinh ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}+\frac{\operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (\frac{1}{a+b x}\right )\right )}{d}\\ &=\frac{\sinh ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\sinh ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}+\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{2 d}\\ &=\frac{\sinh ^{-1}\left (\frac{1}{a+b x}\right )^2}{2 d}-\frac{\sinh ^{-1}\left (\frac{1}{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{d}-\frac{\text{Li}_2\left (e^{2 \sinh ^{-1}\left (\frac{1}{a+b x}\right )}\right )}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.058268, size = 52, normalized size = 0.85 $\frac{\text{PolyLog}\left (2,e^{-2 \text{csch}^{-1}(a+b x)}\right )-\text{csch}^{-1}(a+b x) \left (\text{csch}^{-1}(a+b x)+2 \log \left (1-e^{-2 \text{csch}^{-1}(a+b x)}\right )\right )}{2 d}$

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcCsch[a + b*x]/((a*d)/b + d*x),x]

[Out]

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

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Maple [F]  time = 0.509, size = 0, normalized size = 0. \begin{align*} \int{{\rm arccsch} \left (bx+a\right ) \left ({\frac{ad}{b}}+dx \right ) ^{-1}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(arccsch(b*x+a)/(a*d/b+d*x),x)

[Out]

int(arccsch(b*x+a)/(a*d/b+d*x),x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(b*x+a)/(a*d/b+d*x),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \operatorname{arcsch}\left (b x + a\right )}{b d x + a d}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(b*x+a)/(a*d/b+d*x),x, algorithm="fricas")

[Out]

integral(b*arccsch(b*x + a)/(b*d*x + a*d), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsch(b*x+a)/(a*d/b+d*x),x)

[Out]

b*Integral(acsch(a + b*x)/(a + b*x), x)/d

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(b*x+a)/(a*d/b+d*x),x, algorithm="giac")

[Out]

integrate(arccsch(b*x + a)/(d*x + a*d/b), x)