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

Optimal. Leaf size=170 \[ \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\frac{1}{2} \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\text{sech}^{-1}(a+b x) \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right ) \]

[Out]

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

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Rubi [A]  time = 0.289761, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {6321, 5595, 5570, 3718, 2190, 2279, 2391, 5562} \[ \text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{PolyLog}\left (2,\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\frac{1}{2} \text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{1-\sqrt{1-a^2}}\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{a e^{\text{sech}^{-1}(a+b x)}}{\sqrt{1-a^2}+1}\right )-\text{sech}^{-1}(a+b x) \log \left (e^{2 \text{sech}^{-1}(a+b x)}+1\right ) \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[a + b*x]/x,x]

[Out]

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

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

Mathematica [C]  time = 0.191604, size = 332, normalized size = 1.95 \[ -\text{PolyLog}\left (2,-\frac{\left (\sqrt{1-a^2}-1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}\right )-\text{PolyLog}\left (2,\frac{\left (\sqrt{1-a^2}+1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}\right )+\frac{1}{2} \text{PolyLog}\left (2,-e^{-2 \text{sech}^{-1}(a+b x)}\right )+\text{sech}^{-1}(a+b x) \log \left (\frac{\left (\sqrt{1-a^2}-1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}+1\right )+\text{sech}^{-1}(a+b x) \log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}\right )+2 i \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \log \left (\frac{\left (\sqrt{1-a^2}-1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}+1\right )-2 i \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \log \left (1-\frac{\left (\sqrt{1-a^2}+1\right ) e^{-\text{sech}^{-1}(a+b x)}}{a}\right )-4 i \sin ^{-1}\left (\frac{\sqrt{\frac{a-1}{a}}}{\sqrt{2}}\right ) \tanh ^{-1}\left (\frac{(a+1) \tanh \left (\frac{1}{2} \text{sech}^{-1}(a+b x)\right )}{\sqrt{1-a^2}}\right )-\text{sech}^{-1}(a+b x) \log \left (e^{-2 \text{sech}^{-1}(a+b x)}+1\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[ArcSech[a + b*x]/x,x]

[Out]

(-4*I)*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*ArcTanh[((1 + a)*Tanh[ArcSech[a + b*x]/2])/Sqrt[1 - a^2]] - ArcSech[a
+ b*x]*Log[1 + E^(-2*ArcSech[a + b*x])] + ArcSech[a + b*x]*Log[1 + (-1 + Sqrt[1 - a^2])/(a*E^ArcSech[a + b*x])
] + (2*I)*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 + (-1 + Sqrt[1 - a^2])/(a*E^ArcSech[a + b*x])] + ArcSech[a +
b*x]*Log[1 - (1 + Sqrt[1 - a^2])/(a*E^ArcSech[a + b*x])] - (2*I)*ArcSin[Sqrt[(-1 + a)/a]/Sqrt[2]]*Log[1 - (1 +
Sqrt[1 - a^2])/(a*E^ArcSech[a + b*x])] + PolyLog[2, -E^(-2*ArcSech[a + b*x])]/2 - PolyLog[2, -((-1 + Sqrt[1 -
a^2])/(a*E^ArcSech[a + b*x]))] - PolyLog[2, (1 + Sqrt[1 - a^2])/(a*E^ArcSech[a + b*x])]

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Maple [C]  time = 0.352, size = 886, normalized size = 5.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arcsech(b*x+a)*ln((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)+1)/(1+(-a^2+1)^(1
/2)))+1/2*arcsech(b*x+a)*ln((a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+(-a^2
+1)^(1/2)))-1/2*(-a^2+1)^(1/2)/(a^2-1)*arcsech(b*x+a)*ln((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2
))+(-a^2+1)^(1/2)+1)/(1+(-a^2+1)^(1/2)))+1/2*(-a^2+1)^(1/2)/(a^2-1)*arcsech(b*x+a)*ln((a*(1/(b*x+a)+(1/(b*x+a)
-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+(-a^2+1)^(1/2)))+dilog((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)
*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)+1)/(1+(-a^2+1)^(1/2)))+dilog((a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)
+1)^(1/2))+(-a^2+1)^(1/2)-1)/(-1+(-a^2+1)^(1/2)))-arcsech(b*x+a)*ln(1+I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x
+a)+1)^(1/2)))-arcsech(b*x+a)*ln(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))-dilog(1+I*(1/(b*x+a)
+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))-dilog(1-I*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2)))+1/2*
(a^2-1-(-a^2+1)^(1/2))/a^2/(a^2-1)*arcsech(b*x+a)*(ln((a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(
-a^2+1)^(1/2)-1)/(-1+(-a^2+1)^(1/2)))*a^2+ln((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^
(1/2)+1)/(1+(-a^2+1)^(1/2)))*a^2-2*ln((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)+1
)/(1+(-a^2+1)^(1/2)))+2*ln((-a*(1/(b*x+a)+(1/(b*x+a)-1)^(1/2)*(1/(b*x+a)+1)^(1/2))+(-a^2+1)^(1/2)+1)/(1+(-a^2+
1)^(1/2)))*(-a^2+1)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(b*x + a)/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 )}{x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(arcsech(b*x + a)/x, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asech}{\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)/x,x)

[Out]

Integral(asech(a + b*x)/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 )}{x}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(b*x + a)/x, x)