3.4.60 \(\int e^{c (a+b x)} \tanh ^{-1}(\text {csch}(a c+b c x)) \, dx\) [360]

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

[Out]

exp(b*c*x+a*c)*arctanh(csch(c*(b*x+a)))/b/c+1/2*ln(3-exp(2*c*(b*x+a))-2*2^(1/2))*(1-2^(1/2))/b/c+1/2*ln(3-exp(
2*c*(b*x+a))+2*2^(1/2))*(1+2^(1/2))/b/c

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Rubi [A]
time = 0.12, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {2225, 6410, 2320, 12, 1261, 646, 31} \begin {gather*} \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} \tanh ^{-1}(\text {csch}(c (a+b x)))}{b c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*ArcTanh[Csch[a*c + b*c*x]],x]

[Out]

(E^(a*c + b*c*x)*ArcTanh[Csch[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 12

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

Rule 31

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

Rule 646

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 1261

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 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 6410

Int[((a_.) + ArcTanh[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[a + b*ArcTanh[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*ArcTanh[u]), x]]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 150, normalized size = 1.40 \begin {gather*} \frac {-2 \sqrt {2} \tanh ^{-1}\left (\frac {-1+e^{c (a+b x)}}{\sqrt {2}}\right )+2 \sqrt {2} \tanh ^{-1}\left (\frac {1+e^{c (a+b x)}}{\sqrt {2}}\right )+2 e^{c (a+b x)} \tanh ^{-1}\left (\frac {2 e^{c (a+b x)}}{-1+e^{2 c (a+b x)}}\right )+\log \left (1-2 e^{c (a+b x)}-e^{2 c (a+b x)}\right )+\log \left (1+2 e^{c (a+b x)}-e^{2 c (a+b x)}\right )}{2 b c} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))*ArcTanh[Csch[a*c + b*c*x]],x]

[Out]

(-2*Sqrt[2]*ArcTanh[(-1 + E^(c*(a + b*x)))/Sqrt[2]] + 2*Sqrt[2]*ArcTanh[(1 + E^(c*(a + b*x)))/Sqrt[2]] + 2*E^(
c*(a + b*x))*ArcTanh[(2*E^(c*(a + b*x)))/(-1 + E^(2*c*(a + b*x)))] + 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] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.32, size = 842, normalized size = 7.87

method	result	size
risch	\(\frac {{\mathrm e}^{c \left (b x +a \right )} \ln \left ({\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}-1\right )}{2 c b}+\frac {i \pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}+1\right )\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}+1\right )}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right ) {\mathrm e}^{c \left (b x +a \right )}}{4 c b}+\frac {i \pi \,\mathrm {csgn}\left (i \left (-{\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}+1\right )\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}+1\right )}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{2} {\mathrm e}^{c \left (b x +a \right )}}{4 c b}-\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right ) \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}+1\right )}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{2} {\mathrm e}^{c \left (b x +a \right )}}{4 c b}-\frac {i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}-1\right )\right ) \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}-1\right )}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right ) {\mathrm e}^{c \left (b x +a \right )}}{4 c b}+\frac {i \pi \,\mathrm {csgn}\left (\frac {i}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}-1\right )}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{2} {\mathrm e}^{c \left (b x +a \right )}}{4 c b}-\frac {i \pi \mathrm {csgn}\left (\frac {i \left (-{\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}+1\right )}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{3} {\mathrm e}^{c \left (b x +a \right )}}{4 c b}+\frac {i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}-1\right )\right ) \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}-1\right )}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{2} {\mathrm e}^{c \left (b x +a \right )}}{4 c b}-\frac {i \pi \mathrm {csgn}\left (\frac {i \left ({\mathrm e}^{2 c \left (b x +a \right )}+2 \,{\mathrm e}^{c \left (b x +a \right )}-1\right )}{{\mathrm e}^{2 c \left (b x +a \right )}-1}\right )^{3} {\mathrm e}^{c \left (b x +a \right )}}{4 c b}-\frac {{\mathrm e}^{c \left (b x +a \right )} \ln \left ({\mathrm e}^{2 c \left (b x +a \right )}-2 \,{\mathrm e}^{c \left (b x +a \right )}-1\right )}{2 c b}+\frac {\ln \left ({\mathrm e}^{2 c \left (b x +a \right )}-\left (1+\sqrt {2}\right )^{2}\right ) \sqrt {2}}{2 c b}-\frac {\ln \left ({\mathrm e}^{2 c \left (b x +a \right )}-\left (\sqrt {2}-1\right )^{2}\right ) \sqrt {2}}{2 c b}-\frac {2 a}{b}+\frac {\ln \left ({\mathrm e}^{2 c \left (b x +a \right )}-\left (1+\sqrt {2}\right )^{2}\right )}{2 c b}+\frac {\ln \left ({\mathrm e}^{2 c \left (b x +a \right )}-\left (\sqrt {2}-1\right )^{2}\right )}{2 c b}\)	\(842\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*arctanh(csch(b*c*x+a*c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (90) = 180\).
time = 0.47, size = 184, normalized size = 1.72 \begin {gather*} \frac {\operatorname {artanh}\left (\operatorname {csch}\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctanh(csch(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] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (90) = 180\).
time = 0.38, size = 233, normalized size = 2.18 \begin {gather*} \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*arctanh(csch(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} e^{a c} \int e^{b c x} \operatorname {atanh}{\left (\operatorname {csch}{\left (a c + b c x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*atanh(csch(b*c*x+a*c)),x)

[Out]

exp(a*c)*Integral(exp(b*c*x)*atanh(csch(a*c + b*c*x)), x)

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Giac [A]
time = 0.53, size = 167, normalized size = 1.56 \begin {gather*} \frac {e^{\left ({\left (b x + a\right )} c\right )} \log \left (-\frac {\frac {2}{e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}} + 1}{\frac {2}{e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )}} - 1}\right )}{2 \, b c} + \frac {\sqrt {2} \log \left (\frac {{\left | -4 \, \sqrt {2} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} - 6 \right |}}{{\left | 4 \, \sqrt {2} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} - 6 \right |}}\right ) + \log \left ({\left | e^{\left (4 \, b c x + 4 \, a c\right )} - 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1 \right |}\right )}{2 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*e^((b*x + a)*c)*log(-(2/(e^(b*c*x + a*c) - e^(-b*c*x - a*c)) + 1)/(2/(e^(b*c*x + a*c) - e^(-b*c*x - a*c))
- 1))/(b*c) + 1/2*(sqrt(2)*log(abs(-4*sqrt(2) + 2*e^(2*b*c*x + 2*a*c) - 6)/abs(4*sqrt(2) + 2*e^(2*b*c*x + 2*a*
c) - 6)) + log(abs(e^(4*b*c*x + 4*a*c) - 6*e^(2*b*c*x + 2*a*c) + 1)))/(b*c)

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Mupad [B]
time = 1.62, size = 187, normalized size = 1.75 \begin {gather*} \frac {\ln \left (6\,\sqrt {2}\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}-2\,\sqrt {2}-8\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}\right )\,\left (\sqrt {2}+1\right )}{2\,b\,c}-\frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\ln \left (1-\frac {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 {\ln \left (2\,\sqrt {2}-8\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}-6\,\sqrt {2}\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}\right )\,\left (\sqrt {2}-1\right )}{2\,b\,c}+\frac {\ln \left (\frac {1}{\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}-\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}}+1\right )\,{\mathrm {e}}^{a\,c+b\,c\,x}}{2\,b\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(1/sinh(a*c + b*c*x))*exp(c*(a + b*x)),x)

[Out]

(log(6*2^(1/2)*exp(2*c*(a + b*x)) - 2*2^(1/2) - 8*exp(2*c*(a + b*x)))*(2^(1/2) + 1))/(2*b*c) - (exp(a*c + b*c*
x)*log(1 - 1/((exp(b*c*x)*exp(a*c))/2 - (exp(-b*c*x)*exp(-a*c))/2)))/(2*b*c) - (log(2*2^(1/2) - 8*exp(2*c*(a +
b*x)) - 6*2^(1/2)*exp(2*c*(a + b*x)))*(2^(1/2) - 1))/(2*b*c) + (log(1/((exp(b*c*x)*exp(a*c))/2 - (exp(-b*c*x)
*exp(-a*c))/2) + 1)*exp(a*c + b*c*x))/(2*b*c)

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