3.33 \(\int \frac{a+b \cosh ^{-1}(c x)}{x (d-c^2 d x^2)} \, dx\)

Optimal. Leaf size=61 \[ \frac{b \text{PolyLog}\left (2,-e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac{b \text{PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 d}+\frac{2 \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d} \]

[Out]

(2*(a + b*ArcCosh[c*x])*ArcTanh[E^(2*ArcCosh[c*x])])/d + (b*PolyLog[2, -E^(2*ArcCosh[c*x])])/(2*d) - (b*PolyLo
g[2, E^(2*ArcCosh[c*x])])/(2*d)

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Rubi [A]  time = 0.126603, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {5721, 5461, 4182, 2279, 2391} \[ \frac{b \text{PolyLog}\left (2,-e^{2 \cosh ^{-1}(c x)}\right )}{2 d}-\frac{b \text{PolyLog}\left (2,e^{2 \cosh ^{-1}(c x)}\right )}{2 d}+\frac{2 \tanh ^{-1}\left (e^{2 \cosh ^{-1}(c x)}\right ) \left (a+b \cosh ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCosh[c*x])/(x*(d - c^2*d*x^2)),x]

[Out]

(2*(a + b*ArcCosh[c*x])*ArcTanh[E^(2*ArcCosh[c*x])])/d + (b*PolyLog[2, -E^(2*ArcCosh[c*x])])/(2*d) - (b*PolyLo
g[2, E^(2*ArcCosh[c*x])])/(2*d)

Rule 5721

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> -Dist[d^(-1), Subst[I
nt[(a + b*x)^n/(Cosh[x]*Sinh[x]), x], x, ArcCosh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] &
& IGtQ[n, 0]

Rule 5461

Int[Csch[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.)*Sech[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Dis
t[2^n, Int[(c + d*x)^m*Csch[2*a + 2*b*x]^n, x], x] /; FreeQ[{a, b, c, d}, x] && RationalQ[m] && IntegerQ[n]

Rule 4182

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

Rule 2279

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 2391

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]

Rubi steps

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

Mathematica [A]  time = 0.148326, size = 93, normalized size = 1.52 \[ -\frac{b \left (\text{PolyLog}\left (2,-e^{-2 \cosh ^{-1}(c x)}\right )-\text{PolyLog}\left (2,e^{-2 \cosh ^{-1}(c x)}\right )+2 \cosh ^{-1}(c x) \left (\log \left (1-e^{-2 \cosh ^{-1}(c x)}\right )-\log \left (e^{-2 \cosh ^{-1}(c x)}+1\right )\right )\right )}{2 d}-\frac{a \log \left (1-c^2 x^2\right )}{2 d}+\frac{a \log (x)}{d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcCosh[c*x])/(x*(d - c^2*d*x^2)),x]

[Out]

(a*Log[x])/d - (a*Log[1 - c^2*x^2])/(2*d) - (b*(2*ArcCosh[c*x]*(Log[1 - E^(-2*ArcCosh[c*x])] - Log[1 + E^(-2*A
rcCosh[c*x])]) + PolyLog[2, -E^(-2*ArcCosh[c*x])] - PolyLog[2, E^(-2*ArcCosh[c*x])]))/(2*d)

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Maple [A]  time = 0.05, size = 91, normalized size = 1.5 \begin{align*} -{\frac{a\ln \left ( cx-1 \right ) }{2\,d}}+{\frac{a\ln \left ( cx \right ) }{d}}-{\frac{a\ln \left ( cx+1 \right ) }{2\,d}}-{\frac{b}{d}{\it dilog} \left ( \left ( cx+\sqrt{cx-1}\sqrt{cx+1} \right ) ^{-2} \right ) }+{\frac{b}{4\,d}{\it dilog} \left ( \left ( cx+\sqrt{cx-1}\sqrt{cx+1} \right ) ^{-4} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccosh(c*x))/x/(-c^2*d*x^2+d),x)

[Out]

-1/2*a/d*ln(c*x-1)+a/d*ln(c*x)-1/2*a/d*ln(c*x+1)-b/d*dilog(1/(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^2)+1/4*b/d*dilo
g(1/(c*x+(c*x-1)^(1/2)*(c*x+1)^(1/2))^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x/(-c^2*d*x^2+d),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b \operatorname{arcosh}\left (c x\right ) + a}{c^{2} d x^{3} - d x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x/(-c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral(-(b*arccosh(c*x) + a)/(c^2*d*x^3 - d*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{a}{c^{2} x^{3} - x}\, dx + \int \frac{b \operatorname{acosh}{\left (c x \right )}}{c^{2} x^{3} - x}\, dx}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acosh(c*x))/x/(-c**2*d*x**2+d),x)

[Out]

-(Integral(a/(c**2*x**3 - x), x) + Integral(b*acosh(c*x)/(c**2*x**3 - x), x))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccosh(c*x))/x/(-c^2*d*x^2+d),x, algorithm="giac")

[Out]

integrate(-(b*arccosh(c*x) + a)/((c^2*d*x^2 - d)*x), x)