### 3.10 $$\int (a+b \text{csch}^{-1}(c+d x))^2 \, dx$$

Optimal. Leaf size=85 $\frac{2 b^2 \text{PolyLog}\left (2,-e^{\text{csch}^{-1}(c+d x)}\right )}{d}-\frac{2 b^2 \text{PolyLog}\left (2,e^{\text{csch}^{-1}(c+d x)}\right )}{d}+\frac{(c+d x) \left (a+b \text{csch}^{-1}(c+d x)\right )^2}{d}+\frac{4 b \tanh ^{-1}\left (e^{\text{csch}^{-1}(c+d x)}\right ) \left (a+b \text{csch}^{-1}(c+d x)\right )}{d}$

[Out]

((c + d*x)*(a + b*ArcCsch[c + d*x])^2)/d + (4*b*(a + b*ArcCsch[c + d*x])*ArcTanh[E^ArcCsch[c + d*x]])/d + (2*b
^2*PolyLog[2, -E^ArcCsch[c + d*x]])/d - (2*b^2*PolyLog[2, E^ArcCsch[c + d*x]])/d

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Rubi [A]  time = 0.0831275, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 12, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.5, Rules used = {6316, 6280, 5452, 4182, 2279, 2391} $\frac{2 b^2 \text{PolyLog}\left (2,-e^{\text{csch}^{-1}(c+d x)}\right )}{d}-\frac{2 b^2 \text{PolyLog}\left (2,e^{\text{csch}^{-1}(c+d x)}\right )}{d}+\frac{(c+d x) \left (a+b \text{csch}^{-1}(c+d x)\right )^2}{d}+\frac{4 b \tanh ^{-1}\left (e^{\text{csch}^{-1}(c+d x)}\right ) \left (a+b \text{csch}^{-1}(c+d x)\right )}{d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(a + b*ArcCsch[c + d*x])^2,x]

[Out]

((c + d*x)*(a + b*ArcCsch[c + d*x])^2)/d + (4*b*(a + b*ArcCsch[c + d*x])*ArcTanh[E^ArcCsch[c + d*x]])/d + (2*b
^2*PolyLog[2, -E^ArcCsch[c + d*x]])/d - (2*b^2*PolyLog[2, E^ArcCsch[c + d*x]])/d

Rule 6316

Int[((a_.) + ArcCsch[(c_) + (d_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Dist[1/d, Subst[Int[(a + b*ArcCsch[x])^p, x
], x, c + d*x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[p, 0]

Rule 6280

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Dist[c^(-1), Subst[Int[(a + b*x)^n*Csch[x]*Coth[x]
, x], x, ArcCsch[c*x]], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[n, 0]

Rule 5452

Int[Coth[(a_.) + (b_.)*(x_)]^(p_.)*Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Si
mp[((c + d*x)^m*Csch[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Csch[a + b*x]^n, x], x] /
; FreeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

Rule 4182

Int[csc[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*Ar
cTanh[E^(-(I*e) + f*fz*x)])/(f*fz*I), x] + (-Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 - E^(-(I*e) + f*
fz*x)], x], x] + Dist[(d*m)/(f*fz*I), Int[(c + d*x)^(m - 1)*Log[1 + E^(-(I*e) + f*fz*x)], x], x]) /; FreeQ[{c,
d, e, f, fz}, 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 \left (a+b \text{csch}^{-1}(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \left (a+b \text{csch}^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int (a+b x)^2 \coth (x) \text{csch}(x) \, dx,x,\text{csch}^{-1}(c+d x)\right )}{d}\\ &=\frac{(c+d x) \left (a+b \text{csch}^{-1}(c+d x)\right )^2}{d}-\frac{(2 b) \operatorname{Subst}\left (\int (a+b x) \text{csch}(x) \, dx,x,\text{csch}^{-1}(c+d x)\right )}{d}\\ &=\frac{(c+d x) \left (a+b \text{csch}^{-1}(c+d x)\right )^2}{d}+\frac{4 b \left (a+b \text{csch}^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\text{csch}^{-1}(c+d x)}\right )}{d}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \log \left (1-e^x\right ) \, dx,x,\text{csch}^{-1}(c+d x)\right )}{d}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \log \left (1+e^x\right ) \, dx,x,\text{csch}^{-1}(c+d x)\right )}{d}\\ &=\frac{(c+d x) \left (a+b \text{csch}^{-1}(c+d x)\right )^2}{d}+\frac{4 b \left (a+b \text{csch}^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\text{csch}^{-1}(c+d x)}\right )}{d}+\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{\text{csch}^{-1}(c+d x)}\right )}{d}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{\text{csch}^{-1}(c+d x)}\right )}{d}\\ &=\frac{(c+d x) \left (a+b \text{csch}^{-1}(c+d x)\right )^2}{d}+\frac{4 b \left (a+b \text{csch}^{-1}(c+d x)\right ) \tanh ^{-1}\left (e^{\text{csch}^{-1}(c+d x)}\right )}{d}+\frac{2 b^2 \text{Li}_2\left (-e^{\text{csch}^{-1}(c+d x)}\right )}{d}-\frac{2 b^2 \text{Li}_2\left (e^{\text{csch}^{-1}(c+d x)}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 0.248536, size = 160, normalized size = 1.88 $\frac{-2 b^2 \text{PolyLog}\left (2,-e^{-\text{csch}^{-1}(c+d x)}\right )+2 b^2 \text{PolyLog}\left (2,e^{-\text{csch}^{-1}(c+d x)}\right )+a^2 c+a^2 d x+2 a b (c+d x) \text{csch}^{-1}(c+d x)-2 a b \log \left (\tanh \left (\frac{1}{2} \text{csch}^{-1}(c+d x)\right )\right )+b^2 c \text{csch}^{-1}(c+d x)^2+b^2 d x \text{csch}^{-1}(c+d x)^2-2 b^2 \text{csch}^{-1}(c+d x) \log \left (1-e^{-\text{csch}^{-1}(c+d x)}\right )+2 b^2 \text{csch}^{-1}(c+d x) \log \left (e^{-\text{csch}^{-1}(c+d x)}+1\right )}{d}$

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCsch[c + d*x])^2,x]

[Out]

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

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Maple [F]  time = 0.218, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b{\rm arccsch} \left (dx+c\right ) \right ) ^{2}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsch(d*x+c))^2,x)

[Out]

int((a+b*arccsch(d*x+c))^2,x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(d*x+c))^2,x, algorithm="maxima")

[Out]

(x*log(sqrt(d^2*x^2 + 2*c*d*x + c^2 + 1) + 1)^2 - integrate(-((d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2)*log(d*x + c)
^2 + (d^2*x^2 + 2*c*d*x + c^2 + 1)*log(d*x + c)^2 - 2*((d^2*x^2 + 2*c*d*x + c^2 + 1)*log(d*x + c) + sqrt(d^2*x
^2 + 2*c*d*x + c^2 + 1)*(d^2*x^2 + c*d*x + (d^2*x^2 + 2*c*d*x + c^2 + 1)*log(d*x + c)))*log(sqrt(d^2*x^2 + 2*c
*d*x + c^2 + 1) + 1))/(d^2*x^2 + 2*c*d*x + c^2 + (d^2*x^2 + 2*c*d*x + c^2 + 1)^(3/2) + 1), x))*b^2 + a^2*x + (
2*(d*x + c)*arccsch(d*x + c) + log(sqrt(1/(d*x + c)^2 + 1) + 1) - log(sqrt(1/(d*x + c)^2 + 1) - 1))*a*b/d

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b^{2} \operatorname{arcsch}\left (d x + c\right )^{2} + 2 \, a b \operatorname{arcsch}\left (d x + c\right ) + a^{2}, x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(b^2*arccsch(d*x + c)^2 + 2*a*b*arccsch(d*x + c) + a^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{acsch}{\left (c + d x \right )}\right )^{2}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsch(d*x+c))**2,x)

[Out]

Integral((a + b*acsch(c + d*x))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \operatorname{arcsch}\left (d x + c\right ) + a\right )}^{2}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsch(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((b*arccsch(d*x + c) + a)^2, x)