3.12 \(\int \frac{\text{sech}^{-1}(a+b x)^2}{x} \, dx\)

Optimal. Leaf size=274 \[ 2 \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+2 \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-2 \text{PolyLog}\left (3,\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-2 \text{PolyLog}\left (3,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\text{sech}^{-1}(a+b x)^2 \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right ) \]

[Out]

ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])] + ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSe
ch[a + b*x])/(1 + Sqrt[1 - a^2])] - ArcSech[a + b*x]^2*Log[1 + E^(2*ArcSech[a + b*x])] + 2*ArcSech[a + b*x]*Po
lyLog[2, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])] + 2*ArcSech[a + b*x]*PolyLog[2, (a*E^ArcSech[a + b*x])/(1
+ Sqrt[1 - a^2])] - ArcSech[a + b*x]*PolyLog[2, -E^(2*ArcSech[a + b*x])] - 2*PolyLog[3, (a*E^ArcSech[a + b*x]
)/(1 - Sqrt[1 - a^2])] - 2*PolyLog[3, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])] + PolyLog[3, -E^(2*ArcSech[a
+ b*x])]/2

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Rubi [A]  time = 0.463482, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 9, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.75, Rules used = {6321, 5595, 5570, 3718, 2190, 2531, 2282, 6589, 5562} \[ 2 \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+2 \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-2 \text{PolyLog}\left (3,\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-2 \text{PolyLog}\left (3,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\text{sech}^{-1}(a+b x)^2 \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[a + b*x]^2/x,x]

[Out]

ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])] + ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSe
ch[a + b*x])/(1 + Sqrt[1 - a^2])] - ArcSech[a + b*x]^2*Log[1 + E^(2*ArcSech[a + b*x])] + 2*ArcSech[a + b*x]*Po
lyLog[2, (a*E^ArcSech[a + b*x])/(1 - Sqrt[1 - a^2])] + 2*ArcSech[a + b*x]*PolyLog[2, (a*E^ArcSech[a + b*x])/(1
+ Sqrt[1 - a^2])] - ArcSech[a + b*x]*PolyLog[2, -E^(2*ArcSech[a + b*x])] - 2*PolyLog[3, (a*E^ArcSech[a + b*x]
)/(1 - Sqrt[1 - a^2])] - 2*PolyLog[3, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])] + PolyLog[3, -E^(2*ArcSech[a
+ b*x])]/2

Rule 6321

Int[((a_.) + ArcSech[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Dist[(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Sech[x]*Tanh[x]*(d*e - c*f + f*Sech[x])^m, x], x, ArcSech[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rule 5595

Int[(((e_.) + (f_.)*(x_))^(m_.)*(F_)[(c_.) + (d_.)*(x_)]^(n_.)*(G_)[(c_.) + (d_.)*(x_)]^(p_.))/((a_) + (b_.)*S
ech[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[((e + f*x)^m*Cosh[c + d*x]*F[c + d*x]^n*G[c + d*x]^p)/(b + a*Cosh[c
+ d*x]), x] /; FreeQ[{a, b, c, d, e, f}, x] && HyperbolicQ[F] && HyperbolicQ[G] && IntegersQ[m, n, p]

Rule 5570

Int[(((e_.) + (f_.)*(x_))^(m_.)*Tanh[(c_.) + (d_.)*(x_)]^(n_.))/(Cosh[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Sym
bol] :> Dist[1/a, Int[(e + f*x)^m*Tanh[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Sinh[c + d*x]*Tanh[c +
d*x]^(n - 1))/(a + b*Cosh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3718

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (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))), x],
x] /; FreeQ[{c, d, e, f, fz}, x] && IGtQ[m, 0]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rule 5562

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)])/(Cosh[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :
> -Simp[(e + f*x)^(m + 1)/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(c + d*x))/(a - Rt[a^2 - b^2, 2] + b*E^(c +
d*x)), x] + Int[((e + f*x)^m*E^(c + d*x))/(a + Rt[a^2 - b^2, 2] + b*E^(c + d*x)), x]) /; FreeQ[{a, b, c, d, e,
f}, x] && IGtQ[m, 0] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{\text{sech}^{-1}(a+b x)^2}{x} \, dx &=-\operatorname{Subst}\left (\int \frac{x^2 \text{sech}(x) \tanh (x)}{-a+\text{sech}(x)} \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=-\operatorname{Subst}\left (\int \frac{x^2 \tanh (x)}{1-a \cosh (x)} \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=-\left (a \operatorname{Subst}\left (\int \frac{x^2 \sinh (x)}{1-a \cosh (x)} \, dx,x,\text{sech}^{-1}(a+b x)\right )\right )-\operatorname{Subst}\left (\int x^2 \tanh (x) \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=-\left (2 \operatorname{Subst}\left (\int \frac{e^{2 x} x^2}{1+e^{2 x}} \, dx,x,\text{sech}^{-1}(a+b x)\right )\right )-a \operatorname{Subst}\left (\int \frac{e^x x^2}{1-\sqrt{1-a^2}-a e^x} \, dx,x,\text{sech}^{-1}(a+b x)\right )-a \operatorname{Subst}\left (\int \frac{e^x x^2}{1+\sqrt{1-a^2}-a e^x} \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x)^2 \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )-2 \operatorname{Subst}\left (\int x \log \left (1-\frac{a e^x}{1-\sqrt{1-a^2}}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )-2 \operatorname{Subst}\left (\int x \log \left (1-\frac{a e^x}{1+\sqrt{1-a^2}}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )+2 \operatorname{Subst}\left (\int x \log \left (1+e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x)^2 \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )+2 \text{sech}^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+2 \text{sech}^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x) \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a+b x)}\right )-2 \operatorname{Subst}\left (\int \text{Li}_2\left (\frac{a e^x}{1-\sqrt{1-a^2}}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )-2 \operatorname{Subst}\left (\int \text{Li}_2\left (\frac{a e^x}{1+\sqrt{1-a^2}}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )+\operatorname{Subst}\left (\int \text{Li}_2\left (-e^{2 x}\right ) \, dx,x,\text{sech}^{-1}(a+b x)\right )\\ &=\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x)^2 \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )+2 \text{sech}^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+2 \text{sech}^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x) \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a+b x)}\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{\text{Li}_2(-x)}{x} \, dx,x,e^{2 \text{sech}^{-1}(a+b x)}\right )-2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{a x}{1-\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )-2 \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (\frac{a x}{1+\sqrt{1-a^2}}\right )}{x} \, dx,x,e^{\text{sech}^{-1}(a+b x)}\right )\\ &=\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x)^2 \log \left (1+e^{2 \text{sech}^{-1}(a+b x)}\right )+2 \text{sech}^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+2 \text{sech}^{-1}(a+b x) \text{Li}_2\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x) \text{Li}_2\left (-e^{2 \text{sech}^{-1}(a+b x)}\right )-2 \text{Li}_3\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )-2 \text{Li}_3\left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{1+\sqrt{1-a^2}}\right )+\frac{1}{2} \text{Li}_3\left (-e^{2 \text{sech}^{-1}(a+b x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.294566, size = 280, normalized size = 1.02 \[ 2 \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}-1}\right )+2 \text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-2 \text{PolyLog}\left (3,-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}-1}\right )-2 \text{PolyLog}\left (3,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )+\text{sech}^{-1}(a+b x) \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(a+b x)}\right )+\frac{1}{2} \text{PolyLog}\left (3,-e^{-2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x)^2 \log \left (\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}-1}+1\right )+\text{sech}^{-1}(a+b x)^2 \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\frac{2}{3} \text{sech}^{-1}(a+b x)^3-\text{sech}^{-1}(a+b x)^2 \log \left (e^{-2 \text{sech}^{-1}(a+b x)}+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSech[a + b*x]^2/x,x]

[Out]

(-2*ArcSech[a + b*x]^3)/3 - ArcSech[a + b*x]^2*Log[1 + E^(-2*ArcSech[a + b*x])] + ArcSech[a + b*x]^2*Log[1 + (
a*E^ArcSech[a + b*x])/(-1 + Sqrt[1 - a^2])] + ArcSech[a + b*x]^2*Log[1 - (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 -
a^2])] + ArcSech[a + b*x]*PolyLog[2, -E^(-2*ArcSech[a + b*x])] + 2*ArcSech[a + b*x]*PolyLog[2, -((a*E^ArcSech[
a + b*x])/(-1 + Sqrt[1 - a^2]))] + 2*ArcSech[a + b*x]*PolyLog[2, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])] +
PolyLog[3, -E^(-2*ArcSech[a + b*x])]/2 - 2*PolyLog[3, -((a*E^ArcSech[a + b*x])/(-1 + Sqrt[1 - a^2]))] - 2*Pol
yLog[3, (a*E^ArcSech[a + b*x])/(1 + Sqrt[1 - a^2])]

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Maple [F]  time = 0.392, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ({\rm arcsech} \left (bx+a\right ) \right ) ^{2}}{x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(b*x+a)^2/x,x)

[Out]

int(arcsech(b*x+a)^2/x,x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(b*x+a)^2/x,x, algorithm="maxima")

[Out]

integrate(arcsech(b*x + a)^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(b*x+a)^2/x,x, algorithm="fricas")

[Out]

integral(arcsech(b*x + a)^2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asech(b*x+a)**2/x,x)

[Out]

Integral(asech(a + b*x)**2/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsech(b*x+a)^2/x,x, algorithm="giac")

[Out]

integrate(arcsech(b*x + a)^2/x, x)