3.923 \(\int e^{2 (a+b x)} \coth (a+b x) \text {csch}(a+b x) \, dx\)

Optimal. Leaf size=54 \[ \frac {2 e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {4 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

2*exp(b*x+a)/b+2*exp(b*x+a)/b/(1-exp(2*b*x+2*a))-4*arctanh(exp(b*x+a))/b

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Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2282, 12, 455, 388, 206} \[ \frac {2 e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {4 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*E^(a + b*x))/b + (2*E^(a + b*x))/(b*(1 - E^(2*a + 2*b*x))) - (4*ArcTanh[E^(a + b*x)])/b

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

Rule 455

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] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[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])

Rule 2282

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*} \int e^{2 (a+b x)} \coth (a+b x) \text {csch}(a+b x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {2 x^2 \left (1+x^2\right )}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {2 \operatorname {Subst}\left (\int \frac {x^2 \left (1+x^2\right )}{\left (1-x^2\right )^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {\operatorname {Subst}\left (\int \frac {2+2 x^2}{1-x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {2 e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {4 \operatorname {Subst}\left (\int \frac {1}{1-x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {2 e^{a+b x}}{b}+\frac {2 e^{a+b x}}{b \left (1-e^{2 a+2 b x}\right )}-\frac {4 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 62, normalized size = 1.15 \[ \frac {2 \left (\frac {e^{a+b x} \left (e^{2 (a+b x)}-2\right )}{e^{2 (a+b x)}-1}+\log \left (1-e^{a+b x}\right )-\log \left (e^{a+b x}+1\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.48, size = 200, normalized size = 3.70 \[ \frac {2 \, {\left (\cosh \left (b x + a\right )^{3} + 3 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} + \sinh \left (b x + a\right )^{3} - {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + {\left (\cosh \left (b x + a\right )^{2} + 2 \, \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + \sinh \left (b x + a\right )^{2} - 1\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + {\left (3 \, \cosh \left (b x + a\right )^{2} - 2\right )} \sinh \left (b x + a\right ) - 2 \, \cosh \left (b x + a\right )\right )}}{b \cosh \left (b x + a\right )^{2} + 2 \, b \cosh \left (b x + a\right ) \sinh \left (b x + a\right ) + b \sinh \left (b x + a\right )^{2} - b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(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)*si
nh(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)
*sinh(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 - 2)*sinh(b*
x + a) - 2*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)

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giac [A]  time = 0.15, size = 55, normalized size = 1.02 \[ -\frac {2 \, {\left (\frac {e^{\left (b x + a\right )}}{e^{\left (2 \, b x + 2 \, a\right )} - 1} - e^{\left (b x + a\right )} + \log \left (e^{\left (b x + a\right )} + 1\right ) - \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )\right )}}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)*csch(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

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maple [A]  time = 0.22, size = 65, normalized size = 1.20 \[ \frac {2 \,{\mathrm e}^{b x +a}}{b}-\frac {2 \,{\mathrm e}^{b x +a}}{b \left ({\mathrm e}^{2 b x +2 a}-1\right )}+\frac {2 \ln \left ({\mathrm e}^{b x +a}-1\right )}{b}-\frac {2 \ln \left (1+{\mathrm e}^{b x +a}\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(2*b*x+2*a)*cosh(b*x+a)*csch(b*x+a)^2,x)

[Out]

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

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maxima [A]  time = 0.32, size = 76, normalized size = 1.41 \[ -\frac {2 \, \log \left (e^{\left (-b x - a\right )} + 1\right )}{b} + \frac {2 \, \log \left (e^{\left (-b x - a\right )} - 1\right )}{b} - \frac {2 \, {\left (2 \, e^{\left (-2 \, b x - 2 \, a\right )} - 1\right )}}{b {\left (e^{\left (-b x - a\right )} - e^{\left (-3 \, b x - 3 \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*log(e^(-b*x - a) + 1)/b + 2*log(e^(-b*x - a) - 1)/b - 2*(2*e^(-2*b*x - 2*a) - 1)/(b*(e^(-b*x - a) - e^(-3*b
*x - 3*a)))

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mupad [B]  time = 1.76, size = 63, normalized size = 1.17 \[ \frac {2\,{\mathrm {e}}^{a+b\,x}}{b}-\frac {4\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{b\,x}\,{\mathrm {e}}^a\,\sqrt {-b^2}}{b}\right )}{\sqrt {-b^2}}-\frac {2\,{\mathrm {e}}^{a+b\,x}}{b\,\left ({\mathrm {e}}^{2\,a+2\,b\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)*exp(2*a + 2*b*x))/sinh(a + b*x)^2,x)

[Out]

(2*exp(a + b*x))/b - (4*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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(2*b*x+2*a)*cosh(b*x+a)*csch(b*x+a)**2,x)

[Out]

Timed out

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