3.3.99 \(\int e^{c (a+b x)} \coth ^{-1}(\text {sech}(a c+b c x)) \, dx\) [299]

Optimal. Leaf size=49 \[ \frac {e^{a c+b c x} \coth ^{-1}(\text {sech}(c (a+b x)))}{b c}+\frac {\log \left (1-e^{2 c (a+b x)}\right )}{b c} \]

[Out]

exp(b*c*x+a*c)*arccoth(sech(c*(b*x+a)))/b/c+ln(1-exp(2*c*(b*x+a)))/b/c

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Rubi [A]
time = 0.05, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2225, 6411, 2320, 12, 266} \begin {gather*} \frac {\log \left (1-e^{2 c (a+b x)}\right )}{b c}+\frac {e^{a c+b c x} \coth ^{-1}(\text {sech}(c (a+b x)))}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 266

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

Rule 2225

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 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 6411

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]]

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 59, normalized size = 1.20 \begin {gather*} \frac {e^{c (a+b x)} \coth ^{-1}\left (\frac {2 e^{c (a+b x)}}{1+e^{2 c (a+b x)}}\right )+\log \left (1-e^{2 c (a+b x)}\right )}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.15, size = 939, normalized size = 19.16

method	result	size
risch	\(\text {Expression too large to display}\)	\(939\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*arccoth(sech(b*c*x+a*c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 64, normalized size = 1.31 \begin {gather*} \frac {\operatorname {arcoth}\left (\operatorname {sech}\left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} + 1\right )}{b c} + \frac {\log \left (e^{\left (b c x + a c\right )} - 1\right )}{b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arccoth(sech(b*c*x + a*c))*e^((b*x + a)*c)/(b*c) + log(e^(b*c*x + a*c) + 1)/(b*c) + log(e^(b*c*x + a*c) - 1)/(
b*c)

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Fricas [A]
time = 0.35, size = 93, normalized size = 1.90 \begin {gather*} \frac {{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \log \left (-\frac {\cosh \left (b c x + a c\right ) + 1}{\cosh \left (b c x + a c\right ) - 1}\right ) + 2 \, \log \left (\frac {2 \, \sinh \left (b c x + a c\right )}{\cosh \left (b c x + a c\right ) - \sinh \left (b c x + a c\right )}\right )}{2 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((cosh(b*c*x + a*c) + sinh(b*c*x + a*c))*log(-(cosh(b*c*x + a*c) + 1)/(cosh(b*c*x + a*c) - 1)) + 2*log(2*s
inh(b*c*x + a*c)/(cosh(b*c*x + a*c) - sinh(b*c*x + a*c))))/(b*c)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 147 vs. \(2 (47) = 94\).
time = 0.47, size = 147, normalized size = 3.00 \begin {gather*} \frac {{\left (e^{\left (b c x\right )} \log \left (-\frac {e^{\left (2 \, b c x + 2 \, a c\right )}}{e^{\left (2 \, b c x + 2 \, a c\right )} - 2 \, e^{\left (b c x + a c\right )} + 1} - \frac {2 \, e^{\left (b c x + a c\right )}}{e^{\left (2 \, b c x + 2 \, a c\right )} - 2 \, e^{\left (b c x + a c\right )} + 1} - \frac {1}{e^{\left (2 \, b c x + 2 \, a c\right )} - 2 \, e^{\left (b c x + a c\right )} + 1}\right ) + 2 \, e^{\left (-a c\right )} \log \left ({\left | e^{\left (2 \, b c x + 2 \, a c\right )} - 1 \right |}\right )\right )} e^{\left (a c\right )}}{2 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(e^(b*c*x)*log(-e^(2*b*c*x + 2*a*c)/(e^(2*b*c*x + 2*a*c) - 2*e^(b*c*x + a*c) + 1) - 2*e^(b*c*x + a*c)/(e^(
2*b*c*x + 2*a*c) - 2*e^(b*c*x + a*c) + 1) - 1/(e^(2*b*c*x + 2*a*c) - 2*e^(b*c*x + a*c) + 1)) + 2*e^(-a*c)*log(
abs(e^(2*b*c*x + 2*a*c) - 1)))*e^(a*c)/(b*c)

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Mupad [B]
time = 1.44, size = 111, normalized size = 2.27 \begin {gather*} \frac {\ln \left ({\mathrm {e}}^{2\,b\,c\,x}\,{\mathrm {e}}^{2\,a\,c}-1\right )}{b\,c}-\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,\ln \left (1-\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}-\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}\right )}{2\,b\,c}+\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}\,\ln \left (\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}+\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}+1\right )}{2\,b\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(1/cosh(a*c + b*c*x))*exp(c*(a + b*x)),x)

[Out]

log(exp(2*b*c*x)*exp(2*a*c) - 1)/(b*c) - (exp(b*c*x)*exp(a*c)*log(1 - (exp(-b*c*x)*exp(-a*c))/2 - (exp(b*c*x)*
exp(a*c))/2))/(2*b*c) + (exp(b*c*x)*exp(a*c)*log((exp(b*c*x)*exp(a*c))/2 + (exp(-b*c*x)*exp(-a*c))/2 + 1))/(2*
b*c)

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