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

Optimal. Leaf size=107 \[ \frac{\left (1-\sqrt{2}\right ) \log \left (-e^{2 c (a+b x)}+3-2 \sqrt{2}\right )}{2 b c}+\frac{\left (1+\sqrt{2}\right ) \log \left (-e^{2 c (a+b x)}+3+2 \sqrt{2}\right )}{2 b c}+\frac{e^{a c+b c x} \coth ^{-1}(\sinh (c (a+b x)))}{b c} \]

[Out]

(E^(a*c + b*c*x)*ArcCoth[Sinh[c*(a + b*x)]])/(b*c) + ((1 - Sqrt[2])*Log[3 - 2*Sqrt[2] - E^(2*c*(a + b*x))])/(2
*b*c) + ((1 + Sqrt[2])*Log[3 + 2*Sqrt[2] - E^(2*c*(a + b*x))])/(2*b*c)

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Rubi [A]  time = 0.1546, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {2194, 6276, 2282, 12, 1247, 632, 31} \[ \frac{\left (1-\sqrt{2}\right ) \log \left (-e^{2 c (a+b x)}+3-2 \sqrt{2}\right )}{2 b c}+\frac{\left (1+\sqrt{2}\right ) \log \left (-e^{2 c (a+b x)}+3+2 \sqrt{2}\right )}{2 b c}+\frac{e^{a c+b c x} \coth ^{-1}(\sinh (c (a+b x)))}{b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*ArcCoth[Sinh[a*c + b*c*x]],x]

[Out]

(E^(a*c + b*c*x)*ArcCoth[Sinh[c*(a + b*x)]])/(b*c) + ((1 - Sqrt[2])*Log[3 - 2*Sqrt[2] - E^(2*c*(a + b*x))])/(2
*b*c) + ((1 + Sqrt[2])*Log[3 + 2*Sqrt[2] - E^(2*c*(a + b*x))])/(2*b*c)

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 6276

Int[((a_.) + ArcCoth[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcCoth[u], w, x] - Di
st[b, Int[SimplifyIntegrand[(w*D[u, x])/(1 - u^2), x], x], x] /; InverseFunctionFreeQ[w, x]] /; FreeQ[{a, b},
x] && InverseFunctionFreeQ[u, x] &&  !MatchQ[v, ((c_.) + (d_.)*x)^(m_.) /; FreeQ[{c, d, m}, x]] && FalseQ[Func
tionOfLinear[v*(a + b*ArcCoth[u]), x]]

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 1247

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[
Int[(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x]

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int e^{c (a+b x)} \coth ^{-1}(\sinh (a c+b c x)) \, dx &=\frac{\operatorname{Subst}\left (\int e^x \coth ^{-1}(\sinh (x)) \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \coth ^{-1}(\sinh (c (a+b x)))}{b c}-\frac{\operatorname{Subst}\left (\int \frac{e^x \cosh (x)}{1-\sinh ^2(x)} \, dx,x,a c+b c x\right )}{b c}\\ &=\frac{e^{a c+b c x} \coth ^{-1}(\sinh (c (a+b x)))}{b c}-\frac{\operatorname{Subst}\left (\int \frac{2 x \left (-1-x^2\right )}{1-6 x^2+x^4} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \coth ^{-1}(\sinh (c (a+b x)))}{b c}-\frac{2 \operatorname{Subst}\left (\int \frac{x \left (-1-x^2\right )}{1-6 x^2+x^4} \, dx,x,e^{a c+b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \coth ^{-1}(\sinh (c (a+b x)))}{b c}-\frac{\operatorname{Subst}\left (\int \frac{-1-x}{1-6 x+x^2} \, dx,x,e^{2 a c+2 b c x}\right )}{b c}\\ &=\frac{e^{a c+b c x} \coth ^{-1}(\sinh (c (a+b x)))}{b c}+\frac{\left (1-\sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{-3+2 \sqrt{2}+x} \, dx,x,e^{2 a c+2 b c x}\right )}{2 b c}+\frac{\left (1+\sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-2 \sqrt{2}+x} \, dx,x,e^{2 a c+2 b c x}\right )}{2 b c}\\ &=\frac{e^{a c+b c x} \coth ^{-1}(\sinh (c (a+b x)))}{b c}+\frac{\left (1-\sqrt{2}\right ) \log \left (3-2 \sqrt{2}-e^{2 a c+2 b c x}\right )}{2 b c}+\frac{\left (1+\sqrt{2}\right ) \log \left (3+2 \sqrt{2}-e^{2 a c+2 b c x}\right )}{2 b c}\\ \end{align*}

Mathematica [A]  time = 0.164186, size = 153, normalized size = 1.43 \[ \frac{\log \left (-2 e^{c (a+b x)}-e^{2 c (a+b x)}+1\right )+\log \left (2 e^{c (a+b x)}-e^{2 c (a+b x)}+1\right )-2 \sqrt{2} \tanh ^{-1}\left (\frac{e^{c (a+b x)}-1}{\sqrt{2}}\right )+2 \sqrt{2} \tanh ^{-1}\left (\frac{e^{c (a+b x)}+1}{\sqrt{2}}\right )-2 e^{c (a+b x)} \coth ^{-1}\left (\frac{1}{2} e^{-c (a+b x)}-\frac{1}{2} e^{c (a+b x)}\right )}{2 b c} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^(c*(a + b*x))*ArcCoth[Sinh[a*c + b*c*x]],x]

[Out]

(-2*E^(c*(a + b*x))*ArcCoth[1/(2*E^(c*(a + b*x))) - E^(c*(a + b*x))/2] - 2*Sqrt[2]*ArcTanh[(-1 + E^(c*(a + b*x
)))/Sqrt[2]] + 2*Sqrt[2]*ArcTanh[(1 + E^(c*(a + b*x)))/Sqrt[2]] + Log[1 - 2*E^(c*(a + b*x)) - E^(2*c*(a + b*x)
)] + Log[1 + 2*E^(c*(a + b*x)) - E^(2*c*(a + b*x))])/(2*b*c)

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Maple [C]  time = 0.329, size = 794, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*arccoth(sinh(b*c*x+a*c)),x)

[Out]

1/2/b/c*exp(c*(b*x+a))*ln(exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-1)-1/4*I/b/c*Pi*csgn(I*exp(-c*(b*x+a))*(-exp(2*c*(
b*x+a))+2*exp(c*(b*x+a))+1))^3*exp(c*(b*x+a))-1/4*I/b/c*Pi*csgn(I*exp(-c*(b*x+a)))*csgn(I*exp(-c*(b*x+a))*(-ex
p(2*c*(b*x+a))+2*exp(c*(b*x+a))+1))^2*exp(c*(b*x+a))+1/4*I/b/c*Pi*csgn(I*(-exp(2*c*(b*x+a))+2*exp(c*(b*x+a))+1
))*csgn(I*exp(-c*(b*x+a))*(-exp(2*c*(b*x+a))+2*exp(c*(b*x+a))+1))^2*exp(c*(b*x+a))+1/4*I/b/c*Pi*csgn(I*(-exp(2
*c*(b*x+a))+2*exp(c*(b*x+a))+1))*csgn(I*exp(-c*(b*x+a)))*csgn(I*exp(-c*(b*x+a))*(-exp(2*c*(b*x+a))+2*exp(c*(b*
x+a))+1))*exp(c*(b*x+a))+1/4*I/b/c*Pi*csgn(I*(exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-1))*csgn(I*exp(-c*(b*x+a))*(ex
p(2*c*(b*x+a))+2*exp(c*(b*x+a))-1))^2*exp(c*(b*x+a))-1/4*I/b/c*Pi*csgn(I*(exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-1)
)*csgn(I*exp(-c*(b*x+a)))*csgn(I*exp(-c*(b*x+a))*(exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-1))*exp(c*(b*x+a))-1/4*I/b
/c*Pi*csgn(I*exp(-c*(b*x+a))*(exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-1))^3*exp(c*(b*x+a))+1/4*I/b/c*Pi*csgn(I*exp(-
c*(b*x+a)))*csgn(I*exp(-c*(b*x+a))*(exp(2*c*(b*x+a))+2*exp(c*(b*x+a))-1))^2*exp(c*(b*x+a))-1/2/b/c*exp(c*(b*x+
a))*ln(exp(2*c*(b*x+a))-2*exp(c*(b*x+a))-1)+1/2/b/c*ln(exp(2*c*(b*x+a))-(1+2^(1/2))^2)*2^(1/2)-1/2/b/c*ln(exp(
2*c*(b*x+a))-(2^(1/2)-1)^2)*2^(1/2)-2*a/b+1/2/b/c*ln(exp(2*c*(b*x+a))-(1+2^(1/2))^2)+1/2/b/c*ln(exp(2*c*(b*x+a
))-(2^(1/2)-1)^2)

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Maxima [B]  time = 1.61425, size = 248, normalized size = 2.32 \begin{align*} \frac{\operatorname{arcoth}\left (\sinh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac{\sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (b c x + a c\right )} + 1}{\sqrt{2} + e^{\left (b c x + a c\right )} - 1}\right )}{2 \, b c} - \frac{\sqrt{2} \log \left (-\frac{\sqrt{2} - e^{\left (b c x + a c\right )} - 1}{\sqrt{2} + e^{\left (b c x + a c\right )} + 1}\right )}{2 \, b c} + \frac{\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )} - 1\right )}{2 \, b c} + \frac{\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} - 2 \, e^{\left (b c x + a c\right )} - 1\right )}{2 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arccoth(sinh(b*c*x + a*c))*e^((b*x + a)*c)/(b*c) + 1/2*sqrt(2)*log(-(sqrt(2) - e^(b*c*x + a*c) + 1)/(sqrt(2) +
e^(b*c*x + a*c) - 1))/(b*c) - 1/2*sqrt(2)*log(-(sqrt(2) - e^(b*c*x + a*c) - 1)/(sqrt(2) + e^(b*c*x + a*c) + 1
))/(b*c) + 1/2*log(e^(2*b*c*x + 2*a*c) + 2*e^(b*c*x + a*c) - 1)/(b*c) + 1/2*log(e^(2*b*c*x + 2*a*c) - 2*e^(b*c
*x + a*c) - 1)/(b*c)

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Fricas [B]  time = 1.74446, size = 624, normalized size = 5.83 \begin{align*} \frac{{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \log \left (\frac{\sinh \left (b c x + a c\right ) + 1}{\sinh \left (b c x + a c\right ) - 1}\right ) + \sqrt{2} \log \left (\frac{3 \,{\left (2 \, \sqrt{2} + 3\right )} \cosh \left (b c x + a c\right )^{2} - 4 \,{\left (3 \, \sqrt{2} + 4\right )} \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + 3 \,{\left (2 \, \sqrt{2} + 3\right )} \sinh \left (b c x + a c\right )^{2} - 2 \, \sqrt{2} - 3}{\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} - 3}\right ) + \log \left (\frac{2 \,{\left (\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} - 3\right )}}{\cosh \left (b c x + a c\right )^{2} - 2 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )^{2}}\right )}{2 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((cosh(b*c*x + a*c) + sinh(b*c*x + a*c))*log((sinh(b*c*x + a*c) + 1)/(sinh(b*c*x + a*c) - 1)) + sqrt(2)*lo
g((3*(2*sqrt(2) + 3)*cosh(b*c*x + a*c)^2 - 4*(3*sqrt(2) + 4)*cosh(b*c*x + a*c)*sinh(b*c*x + a*c) + 3*(2*sqrt(2
) + 3)*sinh(b*c*x + a*c)^2 - 2*sqrt(2) - 3)/(cosh(b*c*x + a*c)^2 + sinh(b*c*x + a*c)^2 - 3)) + log(2*(cosh(b*c
*x + a*c)^2 + sinh(b*c*x + a*c)^2 - 3)/(cosh(b*c*x + a*c)^2 - 2*cosh(b*c*x + a*c)*sinh(b*c*x + a*c) + sinh(b*c
*x + a*c)^2)))/(b*c)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*acoth(sinh(b*c*x+a*c)),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(arccoth(sinh(b*c*x + a*c))*e^((b*x + a)*c), x)