3.348 \(\int \sinh ^{-1}(c e^{a+b x}) \, dx\)

Optimal. Leaf size=76 \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac{\sinh ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sinh ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.0756375, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {2282, 5659, 3716, 2190, 2279, 2391} \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac{\sinh ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sinh ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[c*E^(a + b*x)],x]

[Out]

-ArcSinh[c*E^(a + b*x)]^2/(2*b) + (ArcSinh[c*E^(a + b*x)]*Log[1 - E^(2*ArcSinh[c*E^(a + b*x)])])/b + PolyLog[2
, E^(2*ArcSinh[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 5659

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Subst[Int[(a + b*x)^n/Tanh[x], x], x, ArcSinh
[c*x]] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rule 3716

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && 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 \sinh ^{-1}\left (c e^{a+b x}\right ) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(c x)}{x} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac{\sinh ^{-1}\left (c e^{a+b x}\right )^2}{2 b}-\frac{2 \operatorname{Subst}\left (\int \frac{e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac{\sinh ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sinh ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{b}-\frac{\operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (c e^{a+b x}\right )\right )}{b}\\ &=-\frac{\sinh ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sinh ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{b}-\frac{\operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}\\ &=-\frac{\sinh ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sinh ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{b}+\frac{\text{Li}_2\left (e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}\\ \end{align*}

Mathematica [A]  time = 0.491021, size = 76, normalized size = 1. \[ \frac{\text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{2 b}-\frac{\sinh ^{-1}\left (c e^{a+b x}\right )^2}{2 b}+\frac{\sinh ^{-1}\left (c e^{a+b x}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (c e^{a+b x}\right )}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[c*E^(a + b*x)],x]

[Out]

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

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Maple [A]  time = 0.023, size = 166, normalized size = 2.2 \begin{align*} -{\frac{ \left ({\it Arcsinh} \left ( c{{\rm e}^{bx+a}} \right ) \right ) ^{2}}{2\,b}}+{\frac{{\it Arcsinh} \left ( c{{\rm e}^{bx+a}} \right ) }{b}\ln \left ( 1+c{{\rm e}^{bx+a}}+\sqrt{1+{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}} \right ) }+{\frac{1}{b}{\it polylog} \left ( 2,-c{{\rm e}^{bx+a}}-\sqrt{1+{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}} \right ) }+{\frac{{\it Arcsinh} \left ( c{{\rm e}^{bx+a}} \right ) }{b}\ln \left ( 1-c{{\rm e}^{bx+a}}-\sqrt{1+{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}} \right ) }+{\frac{1}{b}{\it polylog} \left ( 2,c{{\rm e}^{bx+a}}+\sqrt{1+{c}^{2} \left ({{\rm e}^{bx+a}} \right ) ^{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -b c \int \frac{x e^{\left (b x + a\right )}}{c^{3} e^{\left (3 \, b x + 3 \, a\right )} + c e^{\left (b x + a\right )} +{\left (c^{2} e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}^{\frac{3}{2}}}\,{d x} + x \log \left (c e^{\left (b x + a\right )} + \sqrt{c^{2} e^{\left (2 \, b x + 2 \, a\right )} + 1}\right ) - \frac{2 \, b x \log \left (c^{2} e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) +{\rm Li}_2\left (-c^{2} e^{\left (2 \, b x + 2 \, a\right )}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*c*integrate(x*e^(b*x + a)/(c^3*e^(3*b*x + 3*a) + c*e^(b*x + a) + (c^2*e^(2*b*x + 2*a) + 1)^(3/2)), x) + x*l
og(c*e^(b*x + a) + sqrt(c^2*e^(2*b*x + 2*a) + 1)) - 1/4*(2*b*x*log(c^2*e^(2*b*x + 2*a) + 1) + dilog(-c^2*e^(2*
b*x + 2*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(arcsinh(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{asinh}{\left (c e^{a + b x} \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(asinh(c*exp(a + b*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \operatorname{arsinh}\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(arcsinh(c*exp(b*x+a)),x, algorithm="giac")

[Out]

integrate(arcsinh(c*e^(b*x + a)), x)