3.31 \(\int \text{sech}^{-1}(c e^{a+b x}) \, dx\)

Optimal. Leaf size=77 \[ -\frac{\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac{\text{sech}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{\text{sech}^{-1}\left (c e^{a+b x}\right ) \log \left (e^{2 \text{sech}^{-1}\left (c e^{a+b x}\right )}+1\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.101172, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {2282, 6281, 5660, 3718, 2190, 2279, 2391} \[ -\frac{\text{PolyLog}\left (2,-e^{2 \text{sech}^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}+\frac{\text{sech}^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{\text{sech}^{-1}\left (c e^{a+b x}\right ) \log \left (e^{2 \text{sech}^{-1}\left (c e^{a+b x}\right )}+1\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSech[c*E^(a + b*x)],x]

[Out]

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

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 6281

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> -Subst[Int[(a + b*ArcCosh[x/c])/x, x], x, 1/x] /; F
reeQ[{a, b, c}, x]

Rule 5660

Int[((a_.) + ArcCosh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Coth[x], x], x, ArcCosh
[c*x]] /; FreeQ[{a, b, c}, x] && 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]

Rubi steps

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

Mathematica [B]  time = 1.17138, size = 249, normalized size = 3.23 \[ x \text{sech}^{-1}\left (c e^{a+b x}\right )-\frac{\sqrt{\frac{1-c e^{a+b x}}{c e^{a+b x}+1}} \sqrt{c e^{a+b x}+1} \left (4 \text{PolyLog}\left (2,\frac{1}{2} \left (1-\sqrt{1-c^2 e^{2 (a+b x)}}\right )\right )-\log ^2\left (c^2 e^{2 (a+b x)}\right )-2 \log ^2\left (\frac{1}{2} \left (\sqrt{1-c^2 e^{2 (a+b x)}}+1\right )\right )+4 \log \left (\frac{1}{2} \left (\sqrt{1-c^2 e^{2 (a+b x)}}+1\right )\right ) \log \left (c^2 e^{2 (a+b x)}\right )+\left (8 b x-4 \log \left (c^2 e^{2 (a+b x)}\right )\right ) \tanh ^{-1}\left (\sqrt{1-c^2 e^{2 (a+b x)}}\right )\right )}{8 b \sqrt{1-c e^{a+b x}}} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSech[c*E^(a + b*x)],x]

[Out]

x*ArcSech[c*E^(a + b*x)] - (Sqrt[(1 - c*E^(a + b*x))/(1 + c*E^(a + b*x))]*Sqrt[1 + c*E^(a + b*x)]*(ArcTanh[Sqr
t[1 - c^2*E^(2*(a + b*x))]]*(8*b*x - 4*Log[c^2*E^(2*(a + b*x))]) - Log[c^2*E^(2*(a + b*x))]^2 + 4*Log[c^2*E^(2
*(a + b*x))]*Log[(1 + Sqrt[1 - c^2*E^(2*(a + b*x))])/2] - 2*Log[(1 + Sqrt[1 - c^2*E^(2*(a + b*x))])/2]^2 + 4*P
olyLog[2, (1 - Sqrt[1 - c^2*E^(2*(a + b*x))])/2]))/(8*b*Sqrt[1 - c*E^(a + b*x)])

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Maple [A]  time = 0.361, size = 140, normalized size = 1.8 \begin{align*}{\frac{ \left ({\rm arcsech} \left (c{{\rm e}^{bx+a}}\right ) \right ) ^{2}}{2\,b}}-{\frac{{\rm arcsech} \left (c{{\rm e}^{bx+a}}\right )}{b}\ln \left ( 1+ \left ({\frac{1}{c{{\rm e}^{bx+a}}}}+\sqrt{{\frac{1}{c{{\rm e}^{bx+a}}}}-1}\sqrt{{\frac{1}{c{{\rm e}^{bx+a}}}}+1} \right ) ^{2} \right ) }-{\frac{1}{2\,b}{\it polylog} \left ( 2,- \left ({\frac{1}{c{{\rm e}^{bx+a}}}}+\sqrt{{\frac{1}{c{{\rm e}^{bx+a}}}}-1}\sqrt{{\frac{1}{c{{\rm e}^{bx+a}}}}+1} \right ) ^{2} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsech(c*exp(b*x+a)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} b c^{2} \int \frac{x e^{\left (2 \, b x + 2 \, a\right )}}{c^{2} e^{\left (2 \, b x + 2 \, a\right )} +{\left (c^{2} e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (\frac{1}{2} \, \log \left (c e^{\left (b x + a\right )} + 1\right ) + \frac{1}{2} \, \log \left (-c e^{\left (b x + a\right )} + 1\right )\right )} - 1}\,{d x} - \frac{1}{2} \, b x^{2} -{\left (a + \log \left (c\right )\right )} x + x \log \left (\sqrt{c e^{\left (b x + a\right )} + 1} \sqrt{-c e^{\left (b x + a\right )} + 1} + 1\right ) - \frac{b x \log \left (c e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (-c e^{\left (b x + a\right )}\right )}{2 \, b} - \frac{b x \log \left (-c e^{\left (b x + a\right )} + 1\right ) +{\rm Li}_2\left (c e^{\left (b x + a\right )}\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asech(c*exp(b*x+a)),x)

[Out]

Integral(asech(c*exp(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arcsech(c*e^(b*x + a)), x)