Integrand size = 20, antiderivative size = 112 \[ \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx=\frac {e^{a c+b c x} \text {arctanh}(\sinh (c (a+b x)))}{b c}+\frac {\left (1-\sqrt {2}\right ) \log \left (3-2 \sqrt {2}-e^{2 c (a+b x)}\right )}{2 b c}+\frac {\left (1+\sqrt {2}\right ) \log \left (4+3 \sqrt {2}-\sqrt {2} e^{2 c (a+b x)}\right )}{2 b c} \] Output:
exp(b*c*x+a*c)*arctanh(sinh(c*(b*x+a)))/b/c+1/2*(1-2^(1/2))*ln(3-2*2^(1/2) -exp(2*c*(b*x+a)))/b/c+1/2*(1+2^(1/2))*ln(4+3*2^(1/2)-2^(1/2)*exp(2*c*(b*x +a)))/b/c
Time = 0.10 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.37 \[ \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx=\frac {-2 e^{c (a+b x)} \text {arctanh}\left (\frac {1}{2} e^{-c (a+b x)}-\frac {1}{2} e^{c (a+b x)}\right )-2 \sqrt {2} \text {arctanh}\left (\frac {-1+e^{c (a+b x)}}{\sqrt {2}}\right )+2 \sqrt {2} \text {arctanh}\left (\frac {1+e^{c (a+b x)}}{\sqrt {2}}\right )+\log \left (1-2 e^{c (a+b x)}-e^{2 c (a+b x)}\right )+\log \left (1+2 e^{c (a+b x)}-e^{2 c (a+b x)}\right )}{2 b c} \] Input:
Integrate[E^(c*(a + b*x))*ArcTanh[Sinh[a*c + b*c*x]],x]
Output:
(-2*E^(c*(a + b*x))*ArcTanh[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)
Time = 0.49 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.81, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {7281, 6829, 2720, 27, 1576, 1141, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle \frac {\int e^{a c+b x c} \text {arctanh}(\sinh (a c+b x c))d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 6829 |
\(\displaystyle \frac {e^{a c+b c x} \text {arctanh}(\sinh (a c+b c x))-\int \frac {e^{a c+b x c} \cosh (a c+b x c)}{1-\sinh ^2(a c+b x c)}d(a c+b x c)}{b c}\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {e^{a c+b c x} \text {arctanh}(\sinh (a c+b c x))-\int -\frac {2 e^{a c+b x c} \left (1+e^{2 a c+2 b x c}\right )}{1-6 e^{2 a c+2 b x c}+e^{4 a c+4 b x c}}de^{a c+b x c}}{b c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {e^{a c+b x c} \left (1+e^{2 a c+2 b x c}\right )}{1-6 e^{2 a c+2 b x c}+e^{4 a c+4 b x c}}de^{a c+b x c}+e^{a c+b c x} \text {arctanh}(\sinh (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 1576 |
\(\displaystyle \frac {\int \frac {1+e^{2 a c+2 b x c}}{1-5 e^{2 a c+2 b x c}}de^{2 a c+2 b x c}+e^{a c+b c x} \text {arctanh}(\sinh (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 1141 |
\(\displaystyle \frac {\int \left (-\frac {1+\sqrt {2}}{2 \left (-a c-b x c+2 \sqrt {2}+3\right )}-\frac {1-\sqrt {2}}{2 \left (-a c-b x c-2 \sqrt {2}+3\right )}\right )de^{2 a c+2 b x c}+e^{a c+b c x} \text {arctanh}(\sinh (a c+b c x))}{b c}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^{a c+b c x} \text {arctanh}(\sinh (a c+b c x))+\frac {1}{2} \left (1-\sqrt {2}\right ) \log \left (-a c-b c x-2 \sqrt {2}+3\right )+\frac {1}{2} \left (1+\sqrt {2}\right ) \log \left (-a c-b c x+2 \sqrt {2}+3\right )}{b c}\) |
Input:
Int[E^(c*(a + b*x))*ArcTanh[Sinh[a*c + b*c*x]],x]
Output:
(E^(a*c + b*c*x)*ArcTanh[Sinh[a*c + b*c*x]] + ((1 - Sqrt[2])*Log[3 - 2*Sqr t[2] - a*c - b*c*x])/2 + ((1 + Sqrt[2])*Log[3 + 2*Sqrt[2] - a*c - b*c*x])/ 2)/(b*c)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_ Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[1/c^p Int[ExpandIntegrand[ (d + e*x)^m*(b/2 - q/2 + c*x)^p*(b/2 + q/2 + c*x)^p, x], x], x] /; EqQ[p, - 1] || !FractionalPowerFactorQ[q]] /; FreeQ[{a, b, c, d, e}, x] && ILtQ[p, 0] && IntegerQ[m] && NiceSqrtQ[b^2 - 4*a*c]
Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^( p_.), x_Symbol] :> Simp[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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Int[((a_.) + ArcTanh[u_]*(b_.))*(v_), x_Symbol] :> With[{w = IntHide[v, x]} , Simp[(a + b*ArcTanh[u]) w, x] - Simp[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_.) /; F reeQ[{c, d, m}, x]] && FalseQ[FunctionOfLinear[v*(a + b*ArcTanh[u]), x]]
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 5.08 (sec) , antiderivative size = 868, normalized size of antiderivative = 7.75
Input:
int(exp(c*(b*x+a))*arctanh(sinh(b*c*x+a*c)),x,method=_RETURNVERBOSE)
Output:
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*P i*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(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))*(-exp(2*c*(b*x+a))+2*exp(c*(b*x+a))+1))^2*exp(c*(b*x+a))+1/2*I/b/c* Pi*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))-1/2*I/b/c*exp(c*(b*x+a))*Pi-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))+1/4*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*ex p(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)
Leaf count of result is larger than twice the leaf count of optimal. 234 vs. \(2 (93) = 186\).
Time = 0.11 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.09 \[ \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx=\frac {{\left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right )} \log \left (-\frac {\sinh \left (b c x + a c\right ) + 1}{\sinh \left (b c x + a c\right ) - 1}\right ) + \sqrt {2} \log \left (\frac {3 \, {\left (2 \, \sqrt {2} + 3\right )} \cosh \left (b c x + a c\right )^{2} - 4 \, {\left (3 \, \sqrt {2} + 4\right )} \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + 3 \, {\left (2 \, \sqrt {2} + 3\right )} \sinh \left (b c x + a c\right )^{2} - 2 \, \sqrt {2} - 3}{\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} - 3}\right ) + \log \left (\frac {2 \, {\left (\cosh \left (b c x + a c\right )^{2} + \sinh \left (b c x + a c\right )^{2} - 3\right )}}{\cosh \left (b c x + a c\right )^{2} - 2 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )^{2}}\right )}{2 \, b c} \] Input:
integrate(exp(c*(b*x+a))*arctanh(sinh(b*c*x+a*c)),x, algorithm="fricas")
Output:
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)*log((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)
Timed out. \[ \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx=\text {Timed out} \] Input:
integrate(exp(c*(b*x+a))*atanh(sinh(b*c*x+a*c)),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.64 \[ \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx=\frac {\operatorname {artanh}\left (\sinh \left (b c x + a c\right )\right ) e^{\left ({\left (b x + a\right )} c\right )}}{b c} + \frac {\sqrt {2} \log \left (-\frac {\sqrt {2} - e^{\left (b c x + a c\right )} + 1}{\sqrt {2} + e^{\left (b c x + a c\right )} - 1}\right )}{2 \, b c} - \frac {\sqrt {2} \log \left (-\frac {\sqrt {2} - e^{\left (b c x + a c\right )} - 1}{\sqrt {2} + e^{\left (b c x + a c\right )} + 1}\right )}{2 \, b c} + \frac {\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )} - 1\right )}{2 \, b c} + \frac {\log \left (e^{\left (2 \, b c x + 2 \, a c\right )} - 2 \, e^{\left (b c x + a c\right )} - 1\right )}{2 \, b c} \] Input:
integrate(exp(c*(b*x+a))*arctanh(sinh(b*c*x+a*c)),x, algorithm="maxima")
Output:
arctanh(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*sqr t(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)
Time = 0.28 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.40 \[ \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx=\frac {e^{\left ({\left (b x + a\right )} c\right )} \log \left (-\frac {e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )} + 2}{e^{\left (b c x + a c\right )} - e^{\left (-b c x - a c\right )} - 2}\right )}{2 \, b c} + \frac {\sqrt {2} \log \left (\frac {{\left | -4 \, \sqrt {2} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} - 6 \right |}}{{\left | 4 \, \sqrt {2} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} - 6 \right |}}\right ) + \log \left ({\left | e^{\left (4 \, b c x + 4 \, a c\right )} - 6 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1 \right |}\right )}{2 \, b c} \] Input:
integrate(exp(c*(b*x+a))*arctanh(sinh(b*c*x+a*c)),x, algorithm="giac")
Output:
1/2*e^((b*x + a)*c)*log(-(e^(b*c*x + a*c) - e^(-b*c*x - a*c) + 2)/(e^(b*c* x + a*c) - e^(-b*c*x - a*c) - 2))/(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)
Time = 4.10 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.60 \[ \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx=\frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\ln \left (\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}-\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}+1\right )}{2\,b\,c}-\frac {{\mathrm {e}}^{a\,c+b\,c\,x}\,\ln \left (\frac {{\mathrm {e}}^{-b\,c\,x}\,{\mathrm {e}}^{-a\,c}}{2}-\frac {{\mathrm {e}}^{b\,c\,x}\,{\mathrm {e}}^{a\,c}}{2}+1\right )}{2\,b\,c}+\frac {\ln \left (6\,\sqrt {2}\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}-2\,\sqrt {2}-8\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}\right )\,\left (\sqrt {2}+1\right )}{2\,b\,c}-\frac {\ln \left (2\,\sqrt {2}-8\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}-6\,\sqrt {2}\,{\mathrm {e}}^{2\,c\,\left (a+b\,x\right )}\right )\,\left (\sqrt {2}-1\right )}{2\,b\,c} \] Input:
int(exp(c*(a + b*x))*atanh(sinh(a*c + b*c*x)),x)
Output:
(exp(a*c + b*c*x)*log((exp(b*c*x)*exp(a*c))/2 - (exp(-b*c*x)*exp(-a*c))/2 + 1))/(2*b*c) - (exp(a*c + b*c*x)*log((exp(-b*c*x)*exp(-a*c))/2 - (exp(b*c *x)*exp(a*c))/2 + 1))/(2*b*c) + (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) - (log(2*2^(1/2) - 8*ex p(2*c*(a + b*x)) - 6*2^(1/2)*exp(2*c*(a + b*x)))*(2^(1/2) - 1))/(2*b*c)
\[ \int e^{c (a+b x)} \text {arctanh}(\sinh (a c+b c x)) \, dx=e^{a c} \left (\int e^{b c x} \mathit {atanh} \left (\sinh \left (b c x +a c \right )\right )d x \right ) \] Input:
int(exp(c*(b*x+a))*atanh(sinh(b*c*x+a*c)),x)
Output:
e**(a*c)*int(e**(b*c*x)*atanh(sinh(a*c + b*c*x)),x)