3.37 \(\int \frac{1}{x^3 (a+b \text{csch}^{-1}(c x))} \, dx\)

Optimal. Leaf size=63 \[ \frac{c^2 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \text{csch}^{-1}(c x)\right )}{2 b}-\frac{c^2 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \text{csch}^{-1}(c x)\right )}{2 b} \]

[Out]

(c^2*CoshIntegral[(2*a)/b + 2*ArcCsch[c*x]]*Sinh[(2*a)/b])/(2*b) - (c^2*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2
*ArcCsch[c*x]])/(2*b)

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Rubi [A]  time = 0.140565, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {6286, 5448, 12, 3303, 3298, 3301} \[ \frac{c^2 \sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \text{csch}^{-1}(c x)\right )}{2 b}-\frac{c^2 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \text{csch}^{-1}(c x)\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(a + b*ArcCsch[c*x])),x]

[Out]

(c^2*CoshIntegral[(2*a)/b + 2*ArcCsch[c*x]]*Sinh[(2*a)/b])/(2*b) - (c^2*Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2
*ArcCsch[c*x]])/(2*b)

Rule 6286

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

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
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 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 3298

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(I*SinhIntegral[(c*f*fz)
/d + f*fz*x])/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*e - c*f*fz*I, 0]

Rule 3301

Int[sin[(e_.) + (Complex[0, fz_])*(f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CoshIntegral[(c*f*fz)/d
+ f*fz*x]/d, x] /; FreeQ[{c, d, e, f, fz}, x] && EqQ[d*(e - Pi/2) - c*f*fz*I, 0]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (a+b \text{csch}^{-1}(c x)\right )} \, dx &=-\left (c^2 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh (x)}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )\right )\\ &=-\left (c^2 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{2 (a+b x)} \, dx,x,\text{csch}^{-1}(c x)\right )\right )\\ &=-\left (\frac{1}{2} c^2 \operatorname{Subst}\left (\int \frac{\sinh (2 x)}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )\right )\\ &=-\left (\frac{1}{2} \left (c^2 \cosh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )\right )+\frac{1}{2} \left (c^2 \sinh \left (\frac{2 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{2 a}{b}+2 x\right )}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=\frac{c^2 \text{Chi}\left (\frac{2 a}{b}+2 \text{csch}^{-1}(c x)\right ) \sinh \left (\frac{2 a}{b}\right )}{2 b}-\frac{c^2 \cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \text{csch}^{-1}(c x)\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.0715245, size = 56, normalized size = 0.89 \[ \frac{c^2 \left (\sinh \left (\frac{2 a}{b}\right ) \text{Chi}\left (\frac{2 a}{b}+2 \text{csch}^{-1}(c x)\right )-\cosh \left (\frac{2 a}{b}\right ) \text{Shi}\left (\frac{2 a}{b}+2 \text{csch}^{-1}(c x)\right )\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(a + b*ArcCsch[c*x])),x]

[Out]

(c^2*(CoshIntegral[(2*a)/b + 2*ArcCsch[c*x]]*Sinh[(2*a)/b] - Cosh[(2*a)/b]*SinhIntegral[(2*a)/b + 2*ArcCsch[c*
x]]))/(2*b)

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Maple [F]  time = 0.19, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3} \left ( a+b{\rm arccsch} \left (cx\right ) \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^3/(a+b*arccsch(c*x)),x)

[Out]

int(1/x^3/(a+b*arccsch(c*x)),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arccsch(c*x)),x, algorithm="maxima")

[Out]

integrate(1/((b*arccsch(c*x) + a)*x^3), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b x^{3} \operatorname{arcsch}\left (c x\right ) + a x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arccsch(c*x)),x, algorithm="fricas")

[Out]

integral(1/(b*x^3*arccsch(c*x) + a*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (a + b \operatorname{acsch}{\left (c x \right )}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**3/(a+b*acsch(c*x)),x)

[Out]

Integral(1/(x**3*(a + b*acsch(c*x))), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^3/(a+b*arccsch(c*x)),x, algorithm="giac")

[Out]

integrate(1/((b*arccsch(c*x) + a)*x^3), x)