Integrand size = 22, antiderivative size = 60 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth (d+b x) \, dx=\frac {3 e^{2 a-d+b x}}{2 b}+\frac {e^{2 a+d+3 b x}}{6 b}-\frac {2 e^{2 a-2 d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
3/2*exp(b*x+2*a-d)/b+1/6*exp(3*b*x+2*a+d)/b-2*exp(2*a-2*d)*arctanh(exp(b*x +d))/b
Leaf count is larger than twice the leaf count of optimal. \(183\) vs. \(2(60)=120\).
Time = 0.43 (sec) , antiderivative size = 183, normalized size of antiderivative = 3.05 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth (d+b x) \, dx=\frac {e^{2 a} \left (-6 \cosh (2 d) \left (\log \left (\left (1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (-1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right )-\log \left (\left (-1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right )\right )+e^{b x} \left (-9+e^{2 b x}\right ) \sinh (d)+\cosh (d) \left (e^{b x} \left (9+e^{2 b x}\right )+12 \left (\log \left (\left (1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (-1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right )-\log \left (\left (-1+e^{b x}\right ) \cosh \left (\frac {d}{2}\right )+\left (1+e^{b x}\right ) \sinh \left (\frac {d}{2}\right )\right )\right ) \sinh (d)\right )\right )}{6 b} \] Input:
Integrate[E^(2*(a + b*x))*Cosh[d + b*x]*Coth[d + b*x],x]
Output:
(E^(2*a)*(-6*Cosh[2*d]*(Log[(1 + E^(b*x))*Cosh[d/2] + (-1 + E^(b*x))*Sinh[ d/2]] - Log[(-1 + E^(b*x))*Cosh[d/2] + (1 + E^(b*x))*Sinh[d/2]]) + E^(b*x) *(-9 + E^(2*b*x))*Sinh[d] + Cosh[d]*(E^(b*x)*(9 + E^(2*b*x)) + 12*(Log[(1 + E^(b*x))*Cosh[d/2] + (-1 + E^(b*x))*Sinh[d/2]] - Log[(-1 + E^(b*x))*Cosh [d/2] + (1 + E^(b*x))*Sinh[d/2]])*Sinh[d])))/(6*b)
Time = 0.21 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.63, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2720, 27, 300, 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.
\(\displaystyle \int e^{2 (a+b x)} \cosh (b x+d) \coth (b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int -\frac {e^{2 a} \left (1+e^{2 b x}\right )^2}{2 \left (1-e^{2 b x}\right )}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {e^{2 a} \int \frac {\left (1+e^{2 b x}\right )^2}{1-e^{2 b x}}de^{b x}}{2 b}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle -\frac {e^{2 a} \int \left (-e^{2 b x}-3+\frac {4}{1-e^{2 b x}}\right )de^{b x}}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e^{2 a} \left (4 \text {arctanh}\left (e^{b x}\right )-3 e^{b x}-\frac {1}{3} e^{3 b x}\right )}{2 b}\) |
Input:
Int[E^(2*(a + b*x))*Cosh[d + b*x]*Coth[d + b*x],x]
Output:
-1/2*(E^(2*a)*(-3*E^(b*x) - E^(3*b*x)/3 + 4*ArcTanh[E^(b*x)]))/b
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)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int [PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c , d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
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 = 0.37 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.47
method | result | size |
risch | \(\frac {{\mathrm e}^{3 b x +2 a +d}}{6 b}+\frac {3 \,{\mathrm e}^{b x +2 a -d}}{2 b}+\frac {\ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{b}-\frac {\ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{2 a -2 d}}{b}\) | \(88\) |
Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d),x,method=_RETURNVERBOSE)
Output:
1/6*exp(3*b*x+2*a+d)/b+3/2*exp(b*x+2*a-d)/b+ln(exp(b*x+a)-exp(a-d))/b*exp( 2*a-2*d)-ln(exp(b*x+a)+exp(a-d))/b*exp(2*a-2*d)
Leaf count of result is larger than twice the leaf count of optimal. 264 vs. \(2 (52) = 104\).
Time = 0.09 (sec) , antiderivative size = 264, normalized size of antiderivative = 4.40 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth (d+b x) \, dx=\frac {\cosh \left (b x + d\right )^{3} \cosh \left (-2 \, a + 2 \, d\right ) + {\left (\cosh \left (-2 \, a + 2 \, d\right ) - \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{3} + 3 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - \cosh \left (b x + d\right ) \sinh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right )^{2} + 9 \, \cosh \left (b x + d\right ) \cosh \left (-2 \, a + 2 \, d\right ) - 6 \, {\left (\cosh \left (-2 \, a + 2 \, d\right ) - \sinh \left (-2 \, a + 2 \, d\right )\right )} \log \left (\cosh \left (b x + d\right ) + \sinh \left (b x + d\right ) + 1\right ) + 6 \, {\left (\cosh \left (-2 \, a + 2 \, d\right ) - \sinh \left (-2 \, a + 2 \, d\right )\right )} \log \left (\cosh \left (b x + d\right ) + \sinh \left (b x + d\right ) - 1\right ) + 3 \, {\left (\cosh \left (b x + d\right )^{2} \cosh \left (-2 \, a + 2 \, d\right ) - {\left (\cosh \left (b x + d\right )^{2} + 3\right )} \sinh \left (-2 \, a + 2 \, d\right ) + 3 \, \cosh \left (-2 \, a + 2 \, d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{3} + 9 \, \cosh \left (b x + d\right )\right )} \sinh \left (-2 \, a + 2 \, d\right )}{6 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d),x, algorithm="fricas")
Output:
1/6*(cosh(b*x + d)^3*cosh(-2*a + 2*d) + (cosh(-2*a + 2*d) - sinh(-2*a + 2* d))*sinh(b*x + d)^3 + 3*(cosh(b*x + d)*cosh(-2*a + 2*d) - cosh(b*x + d)*si nh(-2*a + 2*d))*sinh(b*x + d)^2 + 9*cosh(b*x + d)*cosh(-2*a + 2*d) - 6*(co sh(-2*a + 2*d) - sinh(-2*a + 2*d))*log(cosh(b*x + d) + sinh(b*x + d) + 1) + 6*(cosh(-2*a + 2*d) - sinh(-2*a + 2*d))*log(cosh(b*x + d) + sinh(b*x + d ) - 1) + 3*(cosh(b*x + d)^2*cosh(-2*a + 2*d) - (cosh(b*x + d)^2 + 3)*sinh( -2*a + 2*d) + 3*cosh(-2*a + 2*d))*sinh(b*x + d) - (cosh(b*x + d)^3 + 9*cos h(b*x + d))*sinh(-2*a + 2*d))/b
\[ \int e^{2 (a+b x)} \cosh (d+b x) \coth (d+b x) \, dx=e^{2 a} \int e^{2 b x} \cosh {\left (b x + d \right )} \coth {\left (b x + d \right )}\, dx \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d),x)
Output:
exp(2*a)*Integral(exp(2*b*x)*cosh(b*x + d)*coth(b*x + d), x)
Time = 0.03 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.30 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth (d+b x) \, dx=\frac {{\left (9 \, e^{\left (-2 \, b x - 2 \, d\right )} + 1\right )} e^{\left (3 \, b x + 2 \, a + d\right )}}{6 \, b} - \frac {e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} + 1\right )}{b} + \frac {e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (-b x - d\right )} - 1\right )}{b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d),x, algorithm="maxima")
Output:
1/6*(9*e^(-2*b*x - 2*d) + 1)*e^(3*b*x + 2*a + d)/b - e^(2*a - 2*d)*log(e^( -b*x - d) + 1)/b + e^(2*a - 2*d)*log(e^(-b*x - d) - 1)/b
Time = 0.12 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.27 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth (d+b x) \, dx=\frac {{\left (e^{\left (3 \, b x + 2 \, a + 7 \, d\right )} + 9 \, e^{\left (b x + 2 \, a + 5 \, d\right )}\right )} e^{\left (-6 \, d\right )} - 6 \, e^{\left (2 \, a - 2 \, d\right )} \log \left (e^{\left (b x + d\right )} + 1\right ) + 6 \, e^{\left (2 \, a - 2 \, d\right )} \log \left ({\left | e^{\left (b x + d\right )} - 1 \right |}\right )}{6 \, b} \] Input:
integrate(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d),x, algorithm="giac")
Output:
1/6*((e^(3*b*x + 2*a + 7*d) + 9*e^(b*x + 2*a + 5*d))*e^(-6*d) - 6*e^(2*a - 2*d)*log(e^(b*x + d) + 1) + 6*e^(2*a - 2*d)*log(abs(e^(b*x + d) - 1)))/b
Time = 2.79 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.48 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth (d+b x) \, dx=\frac {3\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}}{2\,b}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-d}\,{\mathrm {e}}^{b\,x}\,\sqrt {-b^2}}{b\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}\right )\,\sqrt {{\mathrm {e}}^{4\,a}\,{\mathrm {e}}^{-4\,d}}}{\sqrt {-b^2}}+\frac {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{3\,b\,x}\,{\mathrm {e}}^d}{6\,b} \] Input:
int(cosh(d + b*x)*coth(d + b*x)*exp(2*a + 2*b*x),x)
Output:
(3*exp(2*a)*exp(-d)*exp(b*x))/(2*b) - (2*atan((exp(2*a)*exp(-d)*exp(b*x)*( -b^2)^(1/2))/(b*(exp(4*a)*exp(-4*d))^(1/2)))*(exp(4*a)*exp(-4*d))^(1/2))/( -b^2)^(1/2) + (exp(2*a)*exp(3*b*x)*exp(d))/(6*b)
Time = 0.22 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int e^{2 (a+b x)} \cosh (d+b x) \coth (d+b x) \, dx=\frac {e^{2 a} \left (e^{3 b x +3 d}+9 e^{b x +d}+6 \,\mathrm {log}\left (e^{b x +d}-1\right )-6 \,\mathrm {log}\left (e^{b x +d}+1\right )\right )}{6 e^{2 d} b} \] Input:
int(exp(2*b*x+2*a)*cosh(b*x+d)*coth(b*x+d),x)
Output:
(e**(2*a)*(e**(3*b*x + 3*d) + 9*e**(b*x + d) + 6*log(e**(b*x + d) - 1) - 6 *log(e**(b*x + d) + 1)))/(6*e**(2*d)*b)