3.38 \(\int \frac{1}{x^4 (a+b \text{csch}^{-1}(c x))} \, dx\)

Optimal. Leaf size=117 \[ \frac{c^3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )}{4 b}-\frac{c^3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \text{csch}^{-1}(c x)\right )}{4 b}-\frac{c^3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )}{4 b}+\frac{c^3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \text{csch}^{-1}(c x)\right )}{4 b} \]

[Out]

(c^3*Cosh[a/b]*CoshIntegral[a/b + ArcCsch[c*x]])/(4*b) - (c^3*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcCsch[c
*x]])/(4*b) - (c^3*Sinh[a/b]*SinhIntegral[a/b + ArcCsch[c*x]])/(4*b) + (c^3*Sinh[(3*a)/b]*SinhIntegral[(3*a)/b
+ 3*ArcCsch[c*x]])/(4*b)

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Rubi [A]  time = 0.240291, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.357, Rules used = {6286, 5448, 3303, 3298, 3301} \[ \frac{c^3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )}{4 b}-\frac{c^3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \text{csch}^{-1}(c x)\right )}{4 b}-\frac{c^3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )}{4 b}+\frac{c^3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \text{csch}^{-1}(c x)\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*ArcCsch[c*x])),x]

[Out]

(c^3*Cosh[a/b]*CoshIntegral[a/b + ArcCsch[c*x]])/(4*b) - (c^3*Cosh[(3*a)/b]*CoshIntegral[(3*a)/b + 3*ArcCsch[c
*x]])/(4*b) - (c^3*Sinh[a/b]*SinhIntegral[a/b + ArcCsch[c*x]])/(4*b) + (c^3*Sinh[(3*a)/b]*SinhIntegral[(3*a)/b
+ 3*ArcCsch[c*x]])/(4*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 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^4 \left (a+b \text{csch}^{-1}(c x)\right )} \, dx &=-\left (c^3 \operatorname{Subst}\left (\int \frac{\cosh (x) \sinh ^2(x)}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )\right )\\ &=-\left (c^3 \operatorname{Subst}\left (\int \left (-\frac{\cosh (x)}{4 (a+b x)}+\frac{\cosh (3 x)}{4 (a+b x)}\right ) \, dx,x,\text{csch}^{-1}(c x)\right )\right )\\ &=\frac{1}{4} c^3 \operatorname{Subst}\left (\int \frac{\cosh (x)}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )-\frac{1}{4} c^3 \operatorname{Subst}\left (\int \frac{\cosh (3 x)}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=\frac{1}{4} \left (c^3 \cosh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )-\frac{1}{4} \left (c^3 \cosh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\cosh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )-\frac{1}{4} \left (c^3 \sinh \left (\frac{a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{a}{b}+x\right )}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )+\frac{1}{4} \left (c^3 \sinh \left (\frac{3 a}{b}\right )\right ) \operatorname{Subst}\left (\int \frac{\sinh \left (\frac{3 a}{b}+3 x\right )}{a+b x} \, dx,x,\text{csch}^{-1}(c x)\right )\\ &=\frac{c^3 \cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )}{4 b}-\frac{c^3 \cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (\frac{3 a}{b}+3 \text{csch}^{-1}(c x)\right )}{4 b}-\frac{c^3 \sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )}{4 b}+\frac{c^3 \sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (\frac{3 a}{b}+3 \text{csch}^{-1}(c x)\right )}{4 b}\\ \end{align*}

Mathematica [A]  time = 0.156946, size = 91, normalized size = 0.78 \[ -\frac{c^3 \left (-\cosh \left (\frac{a}{b}\right ) \text{Chi}\left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )+\cosh \left (\frac{3 a}{b}\right ) \text{Chi}\left (3 \left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )\right )+\sinh \left (\frac{a}{b}\right ) \text{Shi}\left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )-\sinh \left (\frac{3 a}{b}\right ) \text{Shi}\left (3 \left (\frac{a}{b}+\text{csch}^{-1}(c x)\right )\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*ArcCsch[c*x])),x]

[Out]

-(c^3*(-(Cosh[a/b]*CoshIntegral[a/b + ArcCsch[c*x]]) + Cosh[(3*a)/b]*CoshIntegral[3*(a/b + ArcCsch[c*x])] + Si
nh[a/b]*SinhIntegral[a/b + ArcCsch[c*x]] - Sinh[(3*a)/b]*SinhIntegral[3*(a/b + ArcCsch[c*x])]))/(4*b)

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Maple [F]  time = 0.179, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4} \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^4/(a+b*arccsch(c*x)),x)

[Out]

int(1/x^4/(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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arccsch(c*x)),x, algorithm="maxima")

[Out]

integrate(1/((b*arccsch(c*x) + a)*x^4), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{1}{b x^{4} \operatorname{arcsch}\left (c x\right ) + a x^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arccsch(c*x)),x, algorithm="fricas")

[Out]

integral(1/(b*x^4*arccsch(c*x) + a*x^4), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{4} \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**4/(a+b*acsch(c*x)),x)

[Out]

Integral(1/(x**4*(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^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arccsch(c*x)),x, algorithm="giac")

[Out]

integrate(1/((b*arccsch(c*x) + a)*x^4), x)