\(\int e^{a+b x} \coth ^2(a+b x) \, dx\) [211]

Optimal result

Integrand size = 16, antiderivative size = 53 \[ \int e^{a+b x} \coth ^2(a+b x) \, dx=\frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 \text {arctanh}\left (e^{a+b x}\right )}{b} \]

[Out]

exp(b*x+a)/b+2*exp(b*x+a)/b/(1-exp(2*b*x+2*a))-2*arctanh(exp(b*x+a))/b

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2320, 398, 294, 212} \[ \int e^{a+b x} \coth ^2(a+b x) \, dx=-\frac {2 \text {arctanh}\left (e^{a+b x}\right )}{b}+\frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )} \]

[In]

Int[E^(a + b*x)*Coth[a + b*x]^2,x]

[Out]

E^(a + b*x)/b + (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (2*ArcTanh[E^(a + b*x)])/b

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 398

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Int[PolynomialDivide[(a + b*x^n)
^p, (c + d*x^n)^(-q), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && IGtQ[p, 0] && ILt
Q[q, 0] && GeQ[p, -q]

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]]]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {\text {Subst}\left (\int \left (1+\frac {4 x^2}{\left (1-x^2\right )^2}\right ) \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}+\frac {4 \text {Subst}\left (\int \frac {x^2}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 \text {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^{a+b x}\right )}{b} \\ & = \frac {e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {2 \text {arctanh}\left (e^{a+b x}\right )}{b} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 2.33 (sec) , antiderivative size = 179, normalized size of antiderivative = 3.38 \[ \int e^{a+b x} \coth ^2(a+b x) \, dx=\frac {e^{a+b x} \left (\frac {1}{48} e^{-4 (a+b x)} \left (-375-713 e^{2 (a+b x)}-181 e^{4 (a+b x)}+61 e^{6 (a+b x)}+\frac {3 \left (125+196 e^{2 (a+b x)}-14 e^{4 (a+b x)}-52 e^{6 (a+b x)}+e^{8 (a+b x)}\right ) \text {arctanh}\left (\sqrt {e^{2 (a+b x)}}\right )}{\sqrt {e^{2 (a+b x)}}}\right )+\frac {4}{105} \left (e^{a+b x}+e^{3 (a+b x)}\right )^2 \, _4F_3\left (\frac {3}{2},2,2,2;1,1,\frac {9}{2};e^{2 (a+b x)}\right )\right )}{b} \]

[In]

Integrate[E^(a + b*x)*Coth[a + b*x]^2,x]

[Out]

(E^(a + b*x)*((-375 - 713*E^(2*(a + b*x)) - 181*E^(4*(a + b*x)) + 61*E^(6*(a + b*x)) + (3*(125 + 196*E^(2*(a +
b*x)) - 14*E^(4*(a + b*x)) - 52*E^(6*(a + b*x)) + E^(8*(a + b*x)))*ArcTanh[Sqrt[E^(2*(a + b*x))]])/Sqrt[E^(2*
(a + b*x))])/(48*E^(4*(a + b*x))) + (4*(E^(a + b*x) + E^(3*(a + b*x)))^2*HypergeometricPFQ[{3/2, 2, 2, 2}, {1,
1, 9/2}, E^(2*(a + b*x))])/105))/b

Maple [A] (verified)

Time = 0.09 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\cosh \left (b x +a \right )-2 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )+\frac {\cosh \left (b x +a \right )^{2}}{\sinh \left (b x +a \right )}-\frac {2}{\sinh \left (b x +a \right )}}{b}\)	\(48\)
default	\(\frac {\cosh \left (b x +a \right )-2 \,\operatorname {arctanh}\left ({\mathrm e}^{b x +a}\right )+\frac {\cosh \left (b x +a \right )^{2}}{\sinh \left (b x +a \right )}-\frac {2}{\sinh \left (b x +a \right )}}{b}\)	\(48\)
risch	\(\frac {{\mathrm e}^{b x +a}}{b}-\frac {2 \,{\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}+\frac {\ln \left ({\mathrm e}^{b x +a}-1\right )}{b}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right )}{b}\)	\(63\)

[In]

int(exp(b*x+a)*coth(b*x+a)^2,x,method=_RETURNVERBOSE)

[Out]

1/b*(cosh(b*x+a)-2*arctanh(exp(b*x+a))+cosh(b*x+a)^2/sinh(b*x+a)-2/sinh(b*x+a))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (47) = 94\).

Time = 0.25 (sec) , antiderivative size = 198, normalized size of antiderivative = 3.74 \[ \int e^{a+b x} \coth ^2(a+b x) \, dx=\frac {\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} - {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 3 \, {\left (\cosh \left (b x + a\right )^{2} - 1\right )} \sinh \left (b x + a\right ) - 3 \, \cosh \left (b x + a\right )}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b} \]

[In]

integrate(exp(b*x+a)*coth(b*x+a)^2,x, algorithm="fricas")

[Out]

(cosh(b*x + a)^3 + 3*cosh(b*x + a)*sinh(b*x + a)^2 + sinh(b*x + a)^3 - (cosh(b*x + a)^2 + 2*cosh(b*x + a)*sinh
(b*x + a) + sinh(b*x + a)^2 - 1)*log(cosh(b*x + a) + sinh(b*x + a) + 1) + (cosh(b*x + a)^2 + 2*cosh(b*x + a)*s
inh(b*x + a) + sinh(b*x + a)^2 - 1)*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 3*(cosh(b*x + a)^2 - 1)*sinh(b*x
+ a) - 3*cosh(b*x + a))/(b*cosh(b*x + a)^2 + 2*b*cosh(b*x + a)*sinh(b*x + a) + b*sinh(b*x + a)^2 - b)

Sympy [F]

\[ \int e^{a+b x} \coth ^2(a+b x) \, dx=e^{a} \int e^{b x} \coth ^{2}{\left (a + b x \right )}\, dx \]

[In]

integrate(exp(b*x+a)*coth(b*x+a)**2,x)

[Out]

exp(a)*Integral(exp(b*x)*coth(a + b*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.17 \[ \int e^{a+b x} \coth ^2(a+b x) \, dx=\frac {e^{\left (b x + a\right )}}{b} - \frac {\log \left (e^{\left (b x + a\right )} + 1\right )}{b} + \frac {\log \left (e^{\left (b x + a\right )} - 1\right )}{b} - \frac {2 \, e^{\left (b x + a\right )}}{b {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )}} \]

[In]

integrate(exp(b*x+a)*coth(b*x+a)^2,x, algorithm="maxima")

[Out]

e^(b*x + a)/b - log(e^(b*x + a) + 1)/b + log(e^(b*x + a) - 1)/b - 2*e^(b*x + a)/(b*(e^(2*b*x + 2*a) - 1))

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int e^{a+b x} \coth ^2(a+b x) \, dx=-\frac {\frac {2 \, e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - e^{\left (b x + a\right )} + \log \left (e^{\left (b x + a\right )} + 1\right ) - \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{b} \]

[In]

integrate(exp(b*x+a)*coth(b*x+a)^2,x, algorithm="giac")

[Out]

-(2*e^(b*x + a)/(e^(2*b*x + 2*a) - 1) - e^(b*x + a) + log(e^(b*x + a) + 1) - log(abs(e^(b*x + a) - 1)))/b

Mupad [B] (verification not implemented)

Time = 1.76 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.17 \[ \int e^{a+b x} \coth ^2(a+b x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}}{b}-\frac {2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

[In]

int(coth(a + b*x)^2*exp(a + b*x),x)

[Out]

exp(a + b*x)/b - (2*atan((exp(b*x)*exp(a)*(-b^2)^(1/2))/b))/(-b^2)^(1/2) - (2*exp(a + b*x))/(b*(exp(2*a + 2*b*
x) - 1))