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

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

[Out]

(-2*(a + b*ArcSinh[c*x])*ArcTanh[E^(2*ArcSinh[c*x])])/d - (b*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*d) + (b*PolyL
og[2, E^(2*ArcSinh[c*x])])/(2*d)

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Rubi [A]  time = 0.116704, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {5720, 5461, 4182, 2279, 2391} \[ -\frac{b \text{PolyLog}\left (2,-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{2 \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*ArcSinh[c*x])*ArcTanh[E^(2*ArcSinh[c*x])])/d - (b*PolyLog[2, -E^(2*ArcSinh[c*x])])/(2*d) + (b*PolyL
og[2, E^(2*ArcSinh[c*x])])/(2*d)

Rule 5720

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Dist[1/d, Subst[Int[(
a + b*x)^n/(Cosh[x]*Sinh[x]), x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && 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 \sinh ^{-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,\sinh ^{-1}(c x)\right )}{d}\\ &=\frac{2 \operatorname{Subst}\left (\int (a+b x) \text{csch}(2 x) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac{b \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}+\frac{b \operatorname{Subst}\left (\int \log \left (1+e^{2 x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{d}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac{b \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}+\frac{b \operatorname{Subst}\left (\int \frac{\log (1+x)}{x} \, dx,x,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}\\ &=-\frac{2 \left (a+b \sinh ^{-1}(c x)\right ) \tanh ^{-1}\left (e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac{b \text{Li}_2\left (-e^{2 \sinh ^{-1}(c x)}\right )}{2 d}+\frac{b \text{Li}_2\left (e^{2 \sinh ^{-1}(c x)}\right )}{2 d}\\ \end{align*}

Mathematica [B]  time = 0.0903523, size = 207, normalized size = 3.39 \[ -\frac{b \text{PolyLog}\left (2,-\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{d}-\frac{b \text{PolyLog}\left (2,\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{d}+\frac{b \text{PolyLog}\left (2,e^{2 \sinh ^{-1}(c x)}\right )}{2 d}-\frac{a \log \left (c^2 x^2+1\right )}{2 d}-\frac{a \sinh ^{-1}(c x)}{d}+\frac{a \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{d}-\frac{b \sinh ^{-1}(c x) \log \left (1-\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}\right )}{d}-\frac{b \sinh ^{-1}(c x) \log \left (\frac{\sqrt{-c^2} e^{\sinh ^{-1}(c x)}}{c}+1\right )}{d}+\frac{b \sinh ^{-1}(c x) \log \left (1-e^{2 \sinh ^{-1}(c x)}\right )}{d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((a*ArcSinh[c*x])/d) - (b*ArcSinh[c*x]*Log[1 - (Sqrt[-c^2]*E^ArcSinh[c*x])/c])/d - (b*ArcSinh[c*x]*Log[1 + (S
qrt[-c^2]*E^ArcSinh[c*x])/c])/d + (a*Log[1 - E^(2*ArcSinh[c*x])])/d + (b*ArcSinh[c*x]*Log[1 - E^(2*ArcSinh[c*x
])])/d - (a*Log[1 + c^2*x^2])/(2*d) - (b*PolyLog[2, -((Sqrt[-c^2]*E^ArcSinh[c*x])/c)])/d - (b*PolyLog[2, (Sqrt
[-c^2]*E^ArcSinh[c*x])/c])/d + (b*PolyLog[2, E^(2*ArcSinh[c*x])])/(2*d)

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Maple [A]  time = 0.042, size = 74, normalized size = 1.2 \begin{align*}{\frac{a\ln \left ( cx \right ) }{d}}-{\frac{a\ln \left ({c}^{2}{x}^{2}+1 \right ) }{2\,d}}+{\frac{b}{d}{\it dilog} \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{-2} \right ) }-{\frac{b}{4\,d}{\it dilog} \left ( \left ( cx+\sqrt{{c}^{2}{x}^{2}+1} \right ) ^{-4} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a/d*ln(c*x)-1/2*a/d*ln(c^2*x^2+1)+b/d*dilog(1/(c*x+(c^2*x^2+1)^(1/2))^2)-1/4*b/d*dilog(1/(c*x+(c^2*x^2+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^{2} x^{2} + 1\right )}{d} - \frac{2 \, \log \left (x\right )}{d}\right )} + b \int \frac{\log \left (c x + \sqrt{c^{2} x^{2} + 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*arcsinh(c*x))/x/(c^2*d*x^2+d),x, algorithm="maxima")

[Out]

-1/2*a*(log(c^2*x^2 + 1)/d - 2*log(x)/d) + b*integrate(log(c*x + sqrt(c^2*x^2 + 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{arsinh}\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*arcsinh(c*x))/x/(c^2*d*x^2+d),x, algorithm="fricas")

[Out]

integral((b*arcsinh(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{asinh}{\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*asinh(c*x))/x/(c**2*d*x**2+d),x)

[Out]

(Integral(a/(c**2*x**3 + x), x) + Integral(b*asinh(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{arsinh}\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*arcsinh(c*x))/x/(c^2*d*x^2+d),x, algorithm="giac")

[Out]

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