### 3.3 $$\int x \text{csch}^{-1}(a+b x) \, dx$$

Optimal. Leaf size=75 $-\frac{a^2 \text{csch}^{-1}(a+b x)}{2 b^2}+\frac{(a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{2 b^2}-\frac{a \tanh ^{-1}\left (\sqrt{\frac{1}{(a+b x)^2}+1}\right )}{b^2}+\frac{1}{2} x^2 \text{csch}^{-1}(a+b x)$

[Out]

((a + b*x)*Sqrt[1 + (a + b*x)^(-2)])/(2*b^2) - (a^2*ArcCsch[a + b*x])/(2*b^2) + (x^2*ArcCsch[a + b*x])/2 - (a*
ArcTanh[Sqrt[1 + (a + b*x)^(-2)]])/b^2

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Rubi [A]  time = 0.0605459, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 8, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.75, Rules used = {6322, 5469, 3773, 3770, 3767, 8} $-\frac{a^2 \text{csch}^{-1}(a+b x)}{2 b^2}+\frac{(a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{2 b^2}-\frac{a \tanh ^{-1}\left (\sqrt{\frac{1}{(a+b x)^2}+1}\right )}{b^2}+\frac{1}{2} x^2 \text{csch}^{-1}(a+b x)$

Antiderivative was successfully veriﬁed.

[In]

Int[x*ArcCsch[a + b*x],x]

[Out]

((a + b*x)*Sqrt[1 + (a + b*x)^(-2)])/(2*b^2) - (a^2*ArcCsch[a + b*x])/(2*b^2) + (x^2*ArcCsch[a + b*x])/2 - (a*
ArcTanh[Sqrt[1 + (a + b*x)^(-2)]])/b^2

Rule 6322

Int[((a_.) + ArcCsch[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Dist[(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Csch[x]*Coth[x]*(d*e - c*f + f*Csch[x])^m, x], x, ArcCsch[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rule 5469

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

Rule 3773

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Simp[a^2*x, x] + (Dist[2*a*b, Int[Csc[c + d*x], x],
x] + Dist[b^2, Int[Csc[c + d*x]^2, x], x]) /; FreeQ[{a, b, c, d}, x]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x \text{csch}^{-1}(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x \coth (x) \text{csch}(x) (-a+\text{csch}(x)) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{b^2}\\ &=\frac{1}{2} x^2 \text{csch}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\text{csch}(x))^2 \, dx,x,\text{csch}^{-1}(a+b x)\right )}{2 b^2}\\ &=-\frac{a^2 \text{csch}^{-1}(a+b x)}{2 b^2}+\frac{1}{2} x^2 \text{csch}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \text{csch}^2(x) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{2 b^2}+\frac{a \operatorname{Subst}\left (\int \text{csch}(x) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{b^2}\\ &=-\frac{a^2 \text{csch}^{-1}(a+b x)}{2 b^2}+\frac{1}{2} x^2 \text{csch}^{-1}(a+b x)-\frac{a \tanh ^{-1}\left (\sqrt{1+\frac{1}{(a+b x)^2}}\right )}{b^2}+\frac{i \operatorname{Subst}\left (\int 1 \, dx,x,-i (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}\right )}{2 b^2}\\ &=\frac{(a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{2 b^2}-\frac{a^2 \text{csch}^{-1}(a+b x)}{2 b^2}+\frac{1}{2} x^2 \text{csch}^{-1}(a+b x)-\frac{a \tanh ^{-1}\left (\sqrt{1+\frac{1}{(a+b x)^2}}\right )}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.085407, size = 110, normalized size = 1.47 $\frac{(a+b x) \sqrt{\frac{a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}-2 a \log \left ((a+b x) \left (\sqrt{\frac{a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}+1\right )\right )+a^2 \left (-\sinh ^{-1}\left (\frac{1}{a+b x}\right )\right )+b^2 x^2 \text{csch}^{-1}(a+b x)}{2 b^2}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*ArcCsch[a + b*x],x]

[Out]

((a + b*x)*Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2] + b^2*x^2*ArcCsch[a + b*x] - a^2*ArcSinh[(a + b*x)^
(-1)] - 2*a*Log[(a + b*x)*(1 + Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2])])/(2*b^2)

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Maple [A]  time = 0.234, size = 97, normalized size = 1.3 \begin{align*}{\frac{1}{{b}^{2}} \left ({\frac{{\rm arccsch} \left (bx+a\right ) \left ( bx+a \right ) ^{2}}{2}}-{\rm arccsch} \left (bx+a\right )a \left ( bx+a \right ) -{\frac{1}{2\,bx+2\,a}\sqrt{1+ \left ( bx+a \right ) ^{2}} \left ( 2\,a{\it Arcsinh} \left ( bx+a \right ) -\sqrt{1+ \left ( bx+a \right ) ^{2}} \right ){\frac{1}{\sqrt{{\frac{1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/b^2*(1/2*arccsch(b*x+a)*(b*x+a)^2-arccsch(b*x+a)*a*(b*x+a)-1/2*(1+(b*x+a)^2)^(1/2)*(2*a*arcsinh(b*x+a)-(1+(b
*x+a)^2)^(1/2))/((1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*I*a*(log(I*(b^2*x + a*b)/b + 1) - log(-I*(b^2*x + a*b)/b + 1))/b^2 + 1/4*(2*b^2*x^2*log(sqrt(b^2*x^2 + 2*a
*b*x + a^2 + 1) + 1) - (a^2 - 1)*log(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*(b^2*x^2 - a^2)*log(b*x + a))/b^2 + inte
grate(1/2*(b^2*x^3 + a*b*x^2)/(b^2*x^2 + 2*a*b*x + a^2 + (b^2*x^2 + 2*a*b*x + a^2 + 1)^(3/2) + 1), x)

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Fricas [B]  time = 3.15985, size = 643, normalized size = 8.57 \begin{align*} \frac{b^{2} x^{2} \log \left (\frac{{\left (b x + a\right )} \sqrt{\frac{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right ) - a^{2} \log \left (-b x +{\left (b x + a\right )} \sqrt{\frac{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a + 1\right ) + a^{2} \log \left (-b x +{\left (b x + a\right )} \sqrt{\frac{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a - 1\right ) + 2 \, a \log \left (-b x +{\left (b x + a\right )} \sqrt{\frac{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a\right ) +{\left (b x + a\right )} \sqrt{\frac{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, b^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b^2*x^2*log(((b*x + a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)) + 1)/(b*x + a)) - a^
2*log(-b*x + (b*x + a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)) - a + 1) + a^2*log(-b*x +
(b*x + a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)) - a - 1) + 2*a*log(-b*x + (b*x + a)*s
qrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)) - a) + (b*x + a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 +
1)/(b^2*x^2 + 2*a*b*x + a^2)))/b^2

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*acsch(a + b*x), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x*arccsch(b*x + a), x)