3.1.8 \(\int \frac {a+b \text {csch}^{-1}(c x)}{x} \, dx\) [8]

Optimal. Leaf size=56 \[ -\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{2 b}-\left (a+b \text {csch}^{-1}(c x)\right ) \log \left (1-e^{-2 \text {csch}^{-1}(c x)}\right )+\frac {1}{2} b \text {PolyLog}\left (2,e^{-2 \text {csch}^{-1}(c x)}\right ) \]

[Out]

-1/2*(a+b*arccsch(c*x))^2/b-(a+b*arccsch(c*x))*ln(1-1/(1/c/x+(1+1/c^2/x^2)^(1/2))^2)+1/2*b*polylog(2,1/(1/c/x+
(1+1/c^2/x^2)^(1/2))^2)

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Rubi [A]
time = 0.10, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6417, 5775, 3797, 2221, 2317, 2438} \begin {gather*} -\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{2 b}-\log \left (1-e^{-2 \text {csch}^{-1}(c x)}\right ) \left (a+b \text {csch}^{-1}(c x)\right )+\frac {1}{2} b \text {Li}_2\left (e^{-2 \text {csch}^{-1}(c x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsch[c*x])/x,x]

[Out]

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

Rule 2221

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

Rule 2317

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 2438

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]

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

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

Rule 6417

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcSinh[x/c])/x, x], x, 1/x] /; F
reeQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {a+b \text {csch}^{-1}(c x)}{x} \, dx &=-\text {Subst}\left (\int \frac {a+b \sinh ^{-1}\left (\frac {x}{c}\right )}{x} \, dx,x,\frac {1}{x}\right )\\ &=-\text {Subst}\left (\int (a+b x) \coth (x) \, dx,x,\text {csch}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{2 b}+2 \text {Subst}\left (\int \frac {e^{2 x} (a+b x)}{1-e^{2 x}} \, dx,x,\text {csch}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{2 b}-\left (a+b \text {csch}^{-1}(c x)\right ) \log \left (1-e^{2 \text {csch}^{-1}(c x)}\right )+b \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\text {csch}^{-1}(c x)\right )\\ &=\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{2 b}-\left (a+b \text {csch}^{-1}(c x)\right ) \log \left (1-e^{2 \text {csch}^{-1}(c x)}\right )+\frac {1}{2} b \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \text {csch}^{-1}(c x)}\right )\\ &=\frac {\left (a+b \text {csch}^{-1}(c x)\right )^2}{2 b}-\left (a+b \text {csch}^{-1}(c x)\right ) \log \left (1-e^{2 \text {csch}^{-1}(c x)}\right )-\frac {1}{2} b \text {Li}_2\left (e^{2 \text {csch}^{-1}(c x)}\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 47, normalized size = 0.84 \begin {gather*} a \log (x)+\frac {1}{2} b \left (-\text {csch}^{-1}(c x) \left (\text {csch}^{-1}(c x)+2 \log \left (1-e^{-2 \text {csch}^{-1}(c x)}\right )\right )+\text {PolyLog}\left (2,e^{-2 \text {csch}^{-1}(c x)}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsch[c*x])/x,x]

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {a +b \,\mathrm {arccsch}\left (c x \right )}{x}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*c^2*integrate(x^2*log(x)/(c^2*x^3 + x), x) - 2*c^2*integrate(x*log(x)/(c^2*x^2 + (c^2*x^2 + 1)^(3/2) +
1), x) - (log(c^2*x^2 + 1) - 2*log(x))*log(c) + log(c^2*x^2 + 1)*log(c) - 2*log(x)*log(sqrt(c^2*x^2 + 1) + 1)
+ 2*integrate(log(x)/(c^2*x^3 + x), x))*b + a*log(x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arccsch(c*x) + a)/x, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b \operatorname {acsch}{\left (c x \right )}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*acsch(c*x))/x, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccsch(c*x) + a)/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asinh(1/(c*x)))/x,x)

[Out]

int((a + b*asinh(1/(c*x)))/x, x)

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