3.10.16 \(\int e^{a+b x} \cosh ^2(a+b x) \coth (a+b x) \, dx\) [916]

Optimal. Leaf size=59 \[ \frac {e^{-a-b x}}{4 b}+\frac {e^{a+b x}}{b}+\frac {e^{3 a+3 b x}}{12 b}-\frac {2 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \]

[Out]

1/4*exp(-b*x-a)/b+exp(b*x+a)/b+1/12*exp(3*b*x+3*a)/b-2*arctanh(exp(b*x+a))/b

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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2320, 12, 472, 213} \begin {gather*} \frac {e^{-a-b x}}{4 b}+\frac {e^{a+b x}}{b}+\frac {e^{3 a+3 b x}}{12 b}-\frac {2 \tanh ^{-1}\left (e^{a+b x}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

E^(-a - b*x)/(4*b) + E^(a + b*x)/b + E^(3*a + 3*b*x)/(12*b) - (2*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 213

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

Rule 472

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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*} \int e^{a+b x} \cosh ^2(a+b x) \coth (a+b x) \, dx &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{4 x^2 \left (-1+x^2\right )} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {\text {Subst}\left (\int \frac {\left (1+x^2\right )^3}{x^2 \left (-1+x^2\right )} \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac {\text {Subst}\left (\int \left (4-\frac {1}{x^2}+x^2+\frac {8}{-1+x^2}\right ) \, dx,x,e^{a+b x}\right )}{4 b}\\ &=\frac {e^{-a-b x}}{4 b}+\frac {e^{a+b x}}{b}+\frac {e^{3 a+3 b x}}{12 b}+\frac {2 \text {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{a+b x}\right )}{b}\\ &=\frac {e^{-a-b x}}{4 b}+\frac {e^{a+b x}}{b}+\frac {e^{3 a+3 b x}}{12 b}-\frac {2 \tanh ^{-1}\left (e^{a+b x}\right )}{b}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 68, normalized size = 1.15 \begin {gather*} \frac {e^{-a-b x} \left (3+12 e^{2 (a+b x)}+e^{4 (a+b x)}-24 \sqrt {e^{2 (a+b x)}} \tanh ^{-1}\left (\sqrt {e^{2 (a+b x)}}\right )\right )}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^(-a - b*x)*(3 + 12*E^(2*(a + b*x)) + E^(4*(a + b*x)) - 24*Sqrt[E^(2*(a + b*x))]*ArcTanh[Sqrt[E^(2*(a + b*x)
)]]))/(12*b)

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Maple [A]
time = 2.34, size = 57, normalized size = 0.97

method	result	size
default	\(\frac {3 \sinh \left (b x +a \right )}{4 b}+\frac {\sinh \left (3 b x +3 a \right )}{12 b}+\frac {\frac {\left (\cosh ^{3}\left (b x +a \right )\right )}{3}+\cosh \left (b x +a \right )-2 \arctanh \left ({\mathrm e}^{b x +a}\right )}{b}\)	\(57\)
risch	\(\frac {{\mathrm e}^{3 b x +3 a}}{12 b}+\frac {{\mathrm e}^{b x +a}}{b}+\frac {{\mathrm e}^{-b x -a}}{4 b}-\frac {\ln \left ({\mathrm e}^{b x +a}+1\right )}{b}+\frac {\ln \left ({\mathrm e}^{b x +a}-1\right )}{b}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(b*x+a)*cosh(b*x+a)^3*csch(b*x+a),x,method=_RETURNVERBOSE)

[Out]

3/4*sinh(b*x+a)/b+1/12/b*sinh(3*b*x+3*a)+1/b*(1/3*cosh(b*x+a)^3+cosh(b*x+a)-2*arctanh(exp(b*x+a)))

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Maxima [A]
time = 0.28, size = 65, normalized size = 1.10 \begin {gather*} \frac {e^{\left (3 \, b x + 3 \, a\right )} + 12 \, e^{\left (b x + a\right )}}{12 \, b} + \frac {e^{\left (-b x - a\right )}}{4 \, 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3*csch(b*x+a),x, algorithm="maxima")

[Out]

1/12*(e^(3*b*x + 3*a) + 12*e^(b*x + a))/b + 1/4*e^(-b*x - a)/b - log(e^(b*x + a) + 1)/b + log(e^(b*x + a) - 1)
/b

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (51) = 102\).
time = 0.36, size = 170, normalized size = 2.88 \begin {gather*} \frac {\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} + 6 \, {\left (\cosh \left (b x + a\right )^{2} + 2\right )} \sinh \left (b x + a\right )^{2} + 12 \, \cosh \left (b x + a\right )^{2} - 12 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) + 1\right ) + 12 \, {\left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right )\right )} \log \left (\cosh \left (b x + a\right ) + \sinh \left (b x + a\right ) - 1\right ) + 4 \, {\left (\cosh \left (b x + a\right )^{3} + 6 \, \cosh \left (b x + a\right )\right )} \sinh \left (b x + a\right ) + 3}{12 \, {\left (b \cosh \left (b x + a\right ) + b \sinh \left (b x + a\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3*csch(b*x+a),x, algorithm="fricas")

[Out]

1/12*(cosh(b*x + a)^4 + 4*cosh(b*x + a)*sinh(b*x + a)^3 + sinh(b*x + a)^4 + 6*(cosh(b*x + a)^2 + 2)*sinh(b*x +
a)^2 + 12*cosh(b*x + a)^2 - 12*(cosh(b*x + a) + sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) + 1) + 12*(c
osh(b*x + a) + sinh(b*x + a))*log(cosh(b*x + a) + sinh(b*x + a) - 1) + 4*(cosh(b*x + a)^3 + 6*cosh(b*x + a))*s
inh(b*x + a) + 3)/(b*cosh(b*x + a) + b*sinh(b*x + a))

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)**3*csch(b*x+a),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 6436 deep

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Giac [A]
time = 0.42, size = 57, normalized size = 0.97 \begin {gather*} \frac {e^{\left (3 \, b x + 3 \, a\right )} + 12 \, e^{\left (b x + a\right )} + 3 \, e^{\left (-b x - a\right )} - 12 \, \log \left (e^{\left (b x + a\right )} + 1\right ) + 12 \, \log \left ({\left | e^{\left (b x + a\right )} - 1 \right |}\right )}{12 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(b*x+a)*cosh(b*x+a)^3*csch(b*x+a),x, algorithm="giac")

[Out]

1/12*(e^(3*b*x + 3*a) + 12*e^(b*x + a) + 3*e^(-b*x - a) - 12*log(e^(b*x + a) + 1) + 12*log(abs(e^(b*x + a) - 1
)))/b

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Mupad [B]
time = 0.11, size = 66, normalized size = 1.12 \begin {gather*} \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 {{\mathrm {e}}^{-a-b\,x}}{4\,b}+\frac {{\mathrm {e}}^{3\,a+3\,b\,x}}{12\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cosh(a + b*x)^3*exp(a + b*x))/sinh(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) + exp(- a - b*x)/(4*b) + exp(3*a + 3*
b*x)/(12*b)

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