Integrand size = 22, antiderivative size = 70 \[ \int e^{a+b x} \coth (a+b x) \text {csch}^2(a+b x) \, dx=-\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )^2}+\frac {3 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {\text {arctanh}\left (e^{a+b x}\right )}{b} \]
-2*exp(b*x+a)/b/(1-exp(2*b*x+2*a))^2+3*exp(b*x+a)/b/(1-exp(2*b*x+2*a))-arc tanh(exp(b*x+a))/b
Time = 0.06 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.84 \[ \int e^{a+b x} \coth (a+b x) \text {csch}^2(a+b x) \, dx=\frac {e^{a+b x}-3 e^{3 (a+b x)}-\left (-1+e^{2 (a+b x)}\right )^2 \text {arctanh}\left (e^{a+b x}\right )}{b \left (-1+e^{2 (a+b x)}\right )^2} \]
(E^(a + b*x) - 3*E^(3*(a + b*x)) - (-1 + E^(2*(a + b*x)))^2*ArcTanh[E^(a + b*x)])/(b*(-1 + E^(2*(a + b*x)))^2)
Time = 0.21 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2720, 27, 360, 27, 298, 219}
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.
\(\displaystyle \int e^{a+b x} \coth (a+b x) \text {csch}^2(a+b x) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {4 e^{2 a+2 b x} \left (1+e^{2 a+2 b x}\right )}{\left (1-e^{2 a+2 b x}\right )^3}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 \int \frac {e^{2 a+2 b x} \left (1+e^{2 a+2 b x}\right )}{\left (1-e^{2 a+2 b x}\right )^3}de^{a+b x}}{b}\) |
\(\Big \downarrow \) 360 |
\(\displaystyle -\frac {4 \left (\frac {e^{a+b x}}{2 \left (1-e^{2 a+2 b x}\right )^2}-\frac {1}{4} \int \frac {2 \left (1+2 e^{2 a+2 b x}\right )}{\left (1-e^{2 a+2 b x}\right )^2}de^{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 \left (\frac {e^{a+b x}}{2 \left (1-e^{2 a+2 b x}\right )^2}-\frac {1}{2} \int \frac {1+2 e^{2 a+2 b x}}{\left (1-e^{2 a+2 b x}\right )^2}de^{a+b x}\right )}{b}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle -\frac {4 \left (\frac {1}{2} \left (\frac {1}{2} \int \frac {1}{1-e^{2 a+2 b x}}de^{a+b x}-\frac {3 e^{a+b x}}{2 \left (1-e^{2 a+2 b x}\right )}\right )+\frac {e^{a+b x}}{2 \left (1-e^{2 a+2 b x}\right )^2}\right )}{b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {4 \left (\frac {1}{2} \left (\frac {1}{2} \text {arctanh}\left (e^{a+b x}\right )-\frac {3 e^{a+b x}}{2 \left (1-e^{2 a+2 b x}\right )}\right )+\frac {e^{a+b x}}{2 \left (1-e^{2 a+2 b x}\right )^2}\right )}{b}\) |
(-4*(E^(a + b*x)/(2*(1 - E^(2*a + 2*b*x))^2) + ((-3*E^(a + b*x))/(2*(1 - E ^(2*a + 2*b*x))) + ArcTanh[E^(a + b*x)]/2)/2))/b
3.10.6.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
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]]]
Time = 0.44 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(\frac {-\frac {1}{\sinh \left (b x +a \right )}-\frac {\cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {\coth \left (b x +a \right ) \operatorname {csch}\left (b x +a \right )}{2}-\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )}{b}\) | \(55\) |
default | \(\frac {-\frac {1}{\sinh \left (b x +a \right )}-\frac {\cosh \left (b x +a \right )}{\sinh \left (b x +a \right )^{2}}+\frac {\coth \left (b x +a \right ) \operatorname {csch}\left (b x +a \right )}{2}-\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )}{b}\) | \(55\) |
risch | \(-\frac {{\mathrm e}^{b x +a} \left (3 \,{\mathrm e}^{2 b x +2 a}-1\right )}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right )}{2 b}+\frac {\ln \left ({\mathrm e}^{b x +a}-1\right )}{2 b}\) | \(67\) |
1/b*(-1/sinh(b*x+a)-cosh(b*x+a)/sinh(b*x+a)^2+1/2*coth(b*x+a)*csch(b*x+a)- arctanh(exp(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 387 vs. \(2 (61) = 122\).
Time = 0.25 (sec) , antiderivative size = 387, normalized size of antiderivative = 5.53 \[ \int e^{a+b x} \coth (a+b x) \text {csch}^2(a+b x) \, dx=-\frac {6 \, \cosh \left (b x + a\right )^{3} + 18 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + 6 \, \sinh \left (b x + a\right )^{3} + {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) - {\left (\cosh \left (b x + a\right )^{4} + 4 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + \sinh \left (b x + a\right )^{4} + 2 \, {\left (3 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right )^{2} - 2 \, \cosh \left (b x + a\right )^{2} + 4 \, {\left (\cosh \left (b x + a\right )^{3} - \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 2 \, {\left (9 \, \cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 2 \, \cosh \left (b x + a\right )}{2 \, {\left (b \cosh \left (b x + a\right )^{4} + 4 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{3} + b \sinh \left (b x + a\right )^{4} - 2 \, b \cosh \left (b x + a\right )^{2} + 2 \, {\left (3 \, b \cosh \left (b x + a\right )^{2} - b\right )} \sinh \left (b x + a\right )^{2} + 4 \, {\left (b \cosh \left (b x + a\right )^{3} - b \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + b\right )}} \]
-1/2*(6*cosh(b*x + a)^3 + 18*cosh(b*x + a)*sinh(b*x + a)^2 + 6*sinh(b*x + a)^3 + (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^ 4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cos h(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a) + sinh( b*x + a) + 1) - (cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh( b*x + a)^4 + 2*(3*cosh(b*x + a)^2 - 1)*sinh(b*x + a)^2 - 2*cosh(b*x + a)^2 + 4*(cosh(b*x + a)^3 - cosh(b*x + a))*sinh(b*x + a) + 1)*log(cosh(b*x + a ) + sinh(b*x + a) - 1) + 2*(9*cosh(b*x + a)^2 - 1)*sinh(b*x + a) - 2*cosh( b*x + a))/(b*cosh(b*x + a)^4 + 4*b*cosh(b*x + a)*sinh(b*x + a)^3 + b*sinh( b*x + a)^4 - 2*b*cosh(b*x + a)^2 + 2*(3*b*cosh(b*x + a)^2 - b)*sinh(b*x + a)^2 + 4*(b*cosh(b*x + a)^3 - b*cosh(b*x + a))*sinh(b*x + a) + b)
Timed out. \[ \int e^{a+b x} \coth (a+b x) \text {csch}^2(a+b x) \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.11 \[ \int e^{a+b x} \coth (a+b x) \text {csch}^2(a+b x) \, dx=-\frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{2 \, b} + \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{2 \, b} - \frac {3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}}{b {\left (e^{\left (4 \, b x + 4 \, a\right )} - 2 \, e^{\left (2 \, b x + 2 \, a\right )} + 1\right )}} \]
-1/2*log(e^(b*x + a) + 1)/b + 1/2*log(e^(b*x + a) - 1)/b - (3*e^(3*b*x + 3 *a) - e^(b*x + a))/(b*(e^(4*b*x + 4*a) - 2*e^(2*b*x + 2*a) + 1))
Time = 0.26 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.89 \[ \int e^{a+b x} \coth (a+b x) \text {csch}^2(a+b x) \, dx=-\frac {\frac {2 \, {\left (3 \, e^{\left (3 \, b x + 3 \, a\right )} - e^{\left (b x + a\right )}\right )}}{{\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}^{2}} + \log \left (e^{\left (b x + a\right )} + 1\right ) - \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{2 \, b} \]
-1/2*(2*(3*e^(3*b*x + 3*a) - e^(b*x + a))/(e^(2*b*x + 2*a) - 1)^2 + log(e^ (b*x + a) + 1) - log(abs(e^(b*x + a) - 1)))/b
Time = 0.15 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.46 \[ \int e^{a+b x} \coth (a+b x) \text {csch}^2(a+b x) \, dx=-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {\frac {{\mathrm {e}}^{a+b\,x}}{b}+\frac {{\mathrm {e}}^{3\,a+3\,b\,x}}{b}}{{\mathrm {e}}^{4\,a+4\,b\,x}-2\,{\mathrm {e}}^{2\,a+2\,b\,x}+1}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]