3.5 $$\int \frac{\text{csch}^{-1}(a+b x)}{x^2} \, dx$$

Optimal. Leaf size=63 $\frac{2 b \tanh ^{-1}\left (\frac{\tanh \left (\frac{1}{2} \text{csch}^{-1}(a+b x)\right )+a}{\sqrt{a^2+1}}\right )}{a \sqrt{a^2+1}}-\frac{b \text{csch}^{-1}(a+b x)}{a}-\frac{\text{csch}^{-1}(a+b x)}{x}$

[Out]

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

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Rubi [A]  time = 0.101801, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.6, Rules used = {6322, 5469, 3783, 2660, 618, 206} $\frac{2 b \tanh ^{-1}\left (\frac{\tanh \left (\frac{1}{2} \text{csch}^{-1}(a+b x)\right )+a}{\sqrt{a^2+1}}\right )}{a \sqrt{a^2+1}}-\frac{b \text{csch}^{-1}(a+b x)}{a}-\frac{\text{csch}^{-1}(a+b x)}{x}$

Antiderivative was successfully veriﬁed.

[In]

Int[ArcCsch[a + b*x]/x^2,x]

[Out]

-((b*ArcCsch[a + b*x])/a) - ArcCsch[a + b*x]/x + (2*b*ArcTanh[(a + Tanh[ArcCsch[a + b*x]/2])/Sqrt[1 + a^2]])/(
a*Sqrt[1 + a^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 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\text{csch}^{-1}(a+b x)}{x^2} \, dx &=-\left (b \operatorname{Subst}\left (\int \frac{x \coth (x) \text{csch}(x)}{(-a+\text{csch}(x))^2} \, dx,x,\text{csch}^{-1}(a+b x)\right )\right )\\ &=-\frac{\text{csch}^{-1}(a+b x)}{x}+b \operatorname{Subst}\left (\int \frac{1}{-a+\text{csch}(x)} \, dx,x,\text{csch}^{-1}(a+b x)\right )\\ &=-\frac{b \text{csch}^{-1}(a+b x)}{a}-\frac{\text{csch}^{-1}(a+b x)}{x}+\frac{b \operatorname{Subst}\left (\int \frac{1}{1-a \sinh (x)} \, dx,x,\text{csch}^{-1}(a+b x)\right )}{a}\\ &=-\frac{b \text{csch}^{-1}(a+b x)}{a}-\frac{\text{csch}^{-1}(a+b x)}{x}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{1-2 a x-x^2} \, dx,x,\tanh \left (\frac{1}{2} \text{csch}^{-1}(a+b x)\right )\right )}{a}\\ &=-\frac{b \text{csch}^{-1}(a+b x)}{a}-\frac{\text{csch}^{-1}(a+b x)}{x}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{1}{4 \left (1+a^2\right )-x^2} \, dx,x,-2 a-2 \tanh \left (\frac{1}{2} \text{csch}^{-1}(a+b x)\right )\right )}{a}\\ &=-\frac{b \text{csch}^{-1}(a+b x)}{a}-\frac{\text{csch}^{-1}(a+b x)}{x}+\frac{2 b \tanh ^{-1}\left (\frac{a+\tanh \left (\frac{1}{2} \text{csch}^{-1}(a+b x)\right )}{\sqrt{1+a^2}}\right )}{a \sqrt{1+a^2}}\\ \end{align*}

Mathematica [B]  time = 0.162538, size = 141, normalized size = 2.24 $-\frac{b \left (-\log \left (\sqrt{a^2+1} a \sqrt{\frac{a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}+\sqrt{a^2+1} b x \sqrt{\frac{a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}+a^2+a b x+1\right )+\sqrt{a^2+1} \sinh ^{-1}\left (\frac{1}{a+b x}\right )+\log (x)\right )}{a \sqrt{a^2+1}}-\frac{\text{csch}^{-1}(a+b x)}{x}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[ArcCsch[a + b*x]/x^2,x]

[Out]

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

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Maple [B]  time = 0.233, size = 154, normalized size = 2.4 \begin{align*} -{\frac{{\rm arccsch} \left (bx+a\right )}{x}}-{\frac{b}{a \left ( bx+a \right ) }\sqrt{1+ \left ( bx+a \right ) ^{2}}{\it Artanh} \left ({\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}} \right ){\frac{1}{\sqrt{{\frac{1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}}+{\frac{b}{a \left ( bx+a \right ) }\sqrt{1+ \left ( bx+a \right ) ^{2}}\ln \left ( 2\,{\frac{\sqrt{{a}^{2}+1}\sqrt{1+ \left ( bx+a \right ) ^{2}}+a \left ( bx+a \right ) +1}{bx}} \right ){\frac{1}{\sqrt{{\frac{1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}{\frac{1}{\sqrt{{a}^{2}+1}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.79782, size = 792, normalized size = 12.57 \begin{align*} -\frac{{\left (a^{2} + 1\right )} b x \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 ) -{\left (a^{2} + 1\right )} b x \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 ) - \sqrt{a^{2} + 1} b x \log \left (-\frac{a^{2} b x + a^{3} +{\left (a b x + a^{2} +{\left (a b x + a^{2}\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\right )} \sqrt{a^{2} + 1} +{\left (a^{3} +{\left (a^{2} + 1\right )} 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}{x}\right ) +{\left (a^{3} + a\right )} \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 )}{{\left (a^{3} + a\right )} x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((a^2 + 1)*b*x*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 + 1)*b*x*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) - sq
rt(a^2 + 1)*b*x*log(-(a^2*b*x + a^3 + (a*b*x + a^2 + (a*b*x + a^2)*sqrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2
+ 2*a*b*x + a^2)) + 1)*sqrt(a^2 + 1) + (a^3 + (a^2 + 1)*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)/x) + (a^3 + a)*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^3 + a)*x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsch(b*x+a)/x**2,x)

[Out]

Integral(acsch(a + b*x)/x**2, x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccsch(b*x + a)/x^2, x)