3.1.12 \(\int \frac {\text {sech}^{-1}(a+b x)^2}{x} \, dx\) [12]

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

[Out]

-arcsech(b*x+a)^2*ln(1+(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))^2)+arcsech(b*x+a)^2*ln(1-a*(1/(b*x+
a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1-(-a^2+1)^(1/2)))+arcsech(b*x+a)^2*ln(1-a*(1/(b*x+a)+(1/(b*x+a)-
1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1+(-a^2+1)^(1/2)))-arcsech(b*x+a)*polylog(2,-(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/
(b*x+a)+1)^(1/2))^2)+2*arcsech(b*x+a)*polylog(2,a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1-(-a^2
+1)^(1/2)))+2*arcsech(b*x+a)*polylog(2,a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1+(-a^2+1)^(1/2)
))+1/2*polylog(3,-(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))^2)-2*polylog(3,a*(1/(b*x+a)+(1/(b*x+a)-1
)^(1/2)*(1/(b*x+a)+1)^(1/2))/(1-(-a^2+1)^(1/2)))-2*polylog(3,a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1
/2))/(1+(-a^2+1)^(1/2)))

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Rubi [A]
time = 0.31, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 9, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.750, Rules used = {6456, 5714, 5689, 3799, 2221, 2611, 2320, 6724, 5681} \begin {gather*} 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)}}{\sqrt {1-a^2}+1}\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)}}{\sqrt {1-a^2}+1}\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) \text {Li}_2\left (-e^{2 \text {sech}^{-1}(a+b x)}\right )+\frac {1}{2} \text {Li}_3\left (-e^{2 \text {sech}^{-1}(a+b x)}\right )-\text {sech}^{-1}(a+b x)^2 \log \left (e^{2 \text {sech}^{-1}(a+b x)}+1\right ) \end {gather*}

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 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 2320

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 2611

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 3799

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 5681

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]

Rule 5689

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 5714

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 6456

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 6724

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]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 280, normalized size = 1.02 \begin {gather*} -\frac {2}{3} \text {sech}^{-1}(a+b x)^3-\text {sech}^{-1}(a+b x)^2 \log \left (1+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)}}{1+\sqrt {1-a^2}}\right )+\text {sech}^{-1}(a+b x) \text {PolyLog}\left (2,-e^{-2 \text {sech}^{-1}(a+b x)}\right )+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)}}{1+\sqrt {1-a^2}}\right )+\frac {1}{2} \text {PolyLog}\left (3,-e^{-2 \text {sech}^{-1}(a+b x)}\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)}}{1+\sqrt {1-a^2}}\right ) \end {gather*}

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 = 0.42, size = 0, normalized size = 0.00 \[\int \frac {\mathrm {arcsech}\left (b x +a \right )^{2}}{x}\, dx\]

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 = 0.00, 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(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 = 0.00, 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(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {asech}^{2}{\left (a + b x \right )}}{x}\, dx \end {gather*}

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 = 0.00, 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(arcsech(b*x+a)^2/x,x, algorithm="giac")

[Out]

integrate(arcsech(b*x + a)^2/x, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\mathrm {acosh}\left (\frac {1}{a+b\,x}\right )}^2}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acosh(1/(a + b*x))^2/x,x)

[Out]

int(acosh(1/(a + b*x))^2/x, x)

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