Integrand size = 16, antiderivative size = 60 \[ \int e^{a+b x} \coth ^2(d+b x) \, dx=\frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 d+2 b x}\right )}-\frac {2 e^{a-d} \text {arctanh}\left (e^{d+b x}\right )}{b} \] Output:
exp(b*x+a)/b+2*exp(b*x+a)/b/(1-exp(2*b*x+2*d))-2*exp(a-d)*arctanh(exp(b*x+ d))/b
Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.65 \[ \int e^{a+b x} \coth ^2(d+b x) \, dx=\frac {e^a \left (e^{b x} \left (\left (-3+e^{2 b x}\right ) \cosh (d)+\left (3+e^{2 b x}\right ) \sinh (d)\right )-2 \text {arctanh}\left (e^{b x} (\cosh (d)+\sinh (d))\right ) \left (e^{2 b x}-\cosh ^2(d)-\sinh ^2(d)+\sinh (2 d)\right )\right )}{b \left (\left (-1+e^{2 b x}\right ) \cosh (d)+\left (1+e^{2 b x}\right ) \sinh (d)\right )} \] Input:
Integrate[E^(a + b*x)*Coth[d + b*x]^2,x]
Output:
(E^a*(E^(b*x)*((-3 + E^(2*b*x))*Cosh[d] + (3 + E^(2*b*x))*Sinh[d]) - 2*Arc Tanh[E^(b*x)*(Cosh[d] + Sinh[d])]*(E^(2*b*x) - Cosh[d]^2 - Sinh[d]^2 + Sin h[2*d])))/(b*((-1 + E^(2*b*x))*Cosh[d] + (1 + E^(2*b*x))*Sinh[d]))
Time = 0.22 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.67, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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.
\(\displaystyle \int e^{a+b x} \coth ^2(b x+d) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {\int \frac {e^a \left (1+e^{2 b x}\right )^2}{\left (1-e^{2 b x}\right )^2}de^{b x}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e^a \int \frac {\left (1+e^{2 b x}\right )^2}{\left (1-e^{2 b x}\right )^2}de^{b x}}{b}\) |
\(\Big \downarrow \) 300 |
\(\displaystyle \frac {e^a \int \left (1+\frac {4 e^{2 b x}}{\left (1-e^{2 b x}\right )^2}\right )de^{b x}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e^a \left (-2 \text {arctanh}\left (e^{b x}\right )+e^{b x}+\frac {2 e^{b x}}{1-e^{2 b x}}\right )}{b}\) |
Input:
Int[E^(a + b*x)*Coth[d + b*x]^2,x]
Output:
(E^a*(E^(b*x) + (2*E^(b*x))/(1 - E^(2*b*x)) - 2*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 = 0.24 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.62
method | result | size |
risch | \(\frac {{\mathrm e}^{b x +a}}{b}+\frac {2 \,{\mathrm e}^{b x +3 a}}{\left (-{\mathrm e}^{2 b x +2 a +2 d}+{\mathrm e}^{2 a}\right ) b}+\frac {\ln \left ({\mathrm e}^{b x +a}-{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{b}-\frac {\ln \left ({\mathrm e}^{b x +a}+{\mathrm e}^{a -d}\right ) {\mathrm e}^{a -d}}{b}\) | \(97\) |
Input:
int(exp(b*x+a)*coth(b*x+d)^2,x,method=_RETURNVERBOSE)
Output:
exp(b*x+a)/b+2/(-exp(2*b*x+2*a+2*d)+exp(2*a))/b*exp(b*x+3*a)+ln(exp(b*x+a) -exp(a-d))/b*exp(a-d)-ln(exp(b*x+a)+exp(a-d))/b*exp(a-d)
Leaf count of result is larger than twice the leaf count of optimal. 446 vs. \(2 (53) = 106\).
Time = 0.09 (sec) , antiderivative size = 446, normalized size of antiderivative = 7.43 \[ \int e^{a+b x} \coth ^2(d+b x) \, dx=\frac {\cosh \left (b x + d\right )^{3} \cosh \left (-a + d\right ) + {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{3} + 3 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} - 3 \, \cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - {\left (\cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) + {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} + 2 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{2} - 1\right )} \sinh \left (-a + d\right ) - \cosh \left (-a + d\right )\right )} \log \left (\cosh \left (b x + d\right ) + \sinh \left (b x + d\right ) + 1\right ) + {\left (\cosh \left (b x + d\right )^{2} \cosh \left (-a + d\right ) + {\left (\cosh \left (-a + d\right ) - \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right )^{2} + 2 \, {\left (\cosh \left (b x + d\right ) \cosh \left (-a + d\right ) - \cosh \left (b x + d\right ) \sinh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{2} - 1\right )} \sinh \left (-a + d\right ) - \cosh \left (-a + 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 (-a + d\right ) - {\left (\cosh \left (b x + d\right )^{2} - 1\right )} \sinh \left (-a + d\right ) - \cosh \left (-a + d\right )\right )} \sinh \left (b x + d\right ) - {\left (\cosh \left (b x + d\right )^{3} - 3 \, \cosh \left (b x + d\right )\right )} \sinh \left (-a + d\right )}{b \cosh \left (b x + d\right )^{2} + 2 \, b \cosh \left (b x + d\right ) \sinh \left (b x + d\right ) + b \sinh \left (b x + d\right )^{2} - b} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^2,x, algorithm="fricas")
Output:
(cosh(b*x + d)^3*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b*x + d )^3 + 3*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d)^2 - 3*cosh(b*x + d)*cosh(-a + d) - (cosh(b*x + d)^2*cosh(-a + d) + ( cosh(-a + d) - sinh(-a + d))*sinh(b*x + d)^2 + 2*(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^2 - 1)*sin h(-a + d) - cosh(-a + d))*log(cosh(b*x + d) + sinh(b*x + d) + 1) + (cosh(b *x + d)^2*cosh(-a + d) + (cosh(-a + d) - sinh(-a + d))*sinh(b*x + d)^2 + 2 *(cosh(b*x + d)*cosh(-a + d) - cosh(b*x + d)*sinh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^2 - 1)*sinh(-a + d) - cosh(-a + d))*log(cosh(b*x + d) + si nh(b*x + d) - 1) + 3*(cosh(b*x + d)^2*cosh(-a + d) - (cosh(b*x + d)^2 - 1) *sinh(-a + d) - cosh(-a + d))*sinh(b*x + d) - (cosh(b*x + d)^3 - 3*cosh(b* x + d))*sinh(-a + d))/(b*cosh(b*x + d)^2 + 2*b*cosh(b*x + d)*sinh(b*x + d) + b*sinh(b*x + d)^2 - b)
\[ \int e^{a+b x} \coth ^2(d+b x) \, dx=e^{a} \int e^{b x} \coth ^{2}{\left (b x + d \right )}\, dx \] Input:
integrate(exp(b*x+a)*coth(b*x+d)**2,x)
Output:
exp(a)*Integral(exp(b*x)*coth(b*x + d)**2, x)
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.50 \[ \int e^{a+b x} \coth ^2(d+b x) \, dx=-\frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right )}{b} + \frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} - e^{a}\right )}{b} + \frac {e^{\left (b x + a\right )}}{b} - \frac {2 \, e^{\left (b x + 3 \, a\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} - e^{\left (2 \, a\right )}\right )}} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^2,x, algorithm="maxima")
Output:
-e^(a - d)*log(e^(b*x + a + d) + e^a)/b + e^(a - d)*log(e^(b*x + a + d) - e^a)/b + e^(b*x + a)/b - 2*e^(b*x + 3*a)/(b*(e^(2*b*x + 2*a + 2*d) - e^(2* a)))
Time = 0.12 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.42 \[ \int e^{a+b x} \coth ^2(d+b x) \, dx=-\frac {e^{\left (a - d\right )} \log \left (e^{\left (b x + a + d\right )} + e^{a}\right ) - e^{\left (a - d\right )} \log \left ({\left | e^{\left (b x + a + d\right )} - e^{a} \right |}\right ) + \frac {2 \, e^{\left (b x + 3 \, a\right )}}{e^{\left (2 \, b x + 2 \, a + 2 \, d\right )} - e^{\left (2 \, a\right )}} - e^{\left (b x + a\right )}}{b} \] Input:
integrate(exp(b*x+a)*coth(b*x+d)^2,x, algorithm="giac")
Output:
-(e^(a - d)*log(e^(b*x + a + d) + e^a) - e^(a - d)*log(abs(e^(b*x + a + d) - e^a)) + 2*e^(b*x + 3*a)/(e^(2*b*x + 2*a + 2*d) - e^(2*a)) - e^(b*x + a) )/b
Time = 0.00 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.65 \[ \int e^{a+b x} \coth ^2(d+b x) \, dx=\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a}{b}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b\,\sqrt {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,d}}}\right )\,\sqrt {{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,d}}}{\sqrt {-b^2}}+\frac {2\,{\mathrm {e}}^{3\,a}\,{\mathrm {e}}^{-2\,d}\,{\mathrm {e}}^{b\,x}}{b\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{-2\,d}-b\,{\mathrm {e}}^{2\,a}\,{\mathrm {e}}^{2\,b\,x}} \] Input:
int(coth(d + b*x)^2*exp(a + b*x),x)
Output:
(exp(b*x)*exp(a))/b - (2*atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/(b*(exp(2*a)* exp(-2*d))^(1/2)))*(exp(2*a)*exp(-2*d))^(1/2))/(-b^2)^(1/2) + (2*exp(3*a)* exp(-2*d)*exp(b*x))/(b*exp(2*a)*exp(-2*d) - b*exp(2*a)*exp(2*b*x))
Time = 0.21 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.85 \[ \int e^{a+b x} \coth ^2(d+b x) \, dx=\frac {e^{a} \left (e^{3 b x +3 d}+e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}-1\right )-e^{2 b x +2 d} \mathrm {log}\left (e^{b x +d}+1\right )-3 e^{b x +d}-\mathrm {log}\left (e^{b x +d}-1\right )+\mathrm {log}\left (e^{b x +d}+1\right )\right )}{e^{d} b \left (e^{2 b x +2 d}-1\right )} \] Input:
int(exp(b*x+a)*coth(b*x+d)^2,x)
Output:
(e**a*(e**(3*b*x + 3*d) + e**(2*b*x + 2*d)*log(e**(b*x + d) - 1) - e**(2*b *x + 2*d)*log(e**(b*x + d) + 1) - 3*e**(b*x + d) - log(e**(b*x + d) - 1) + log(e**(b*x + d) + 1)))/(e**d*b*(e**(2*b*x + 2*d) - 1))