3.215 \(\int \frac{e^{c (a+b x)}}{\coth ^2(a c+b c x)^{3/2}} \, dx\)

Optimal. Leaf size=193 \[ \frac{e^{c (a+b x)} \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}}+\frac{3 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right ) \sqrt{\coth ^2(a c+b c x)}}-\frac{2 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^2 \sqrt{\coth ^2(a c+b c x)}}-\frac{3 \tan ^{-1}\left (e^{c (a+b x)}\right ) \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}} \]

[Out]

(E^(c*(a + b*x))*Coth[a*c + b*c*x])/(b*c*Sqrt[Coth[a*c + b*c*x]^2]) - (2*E^(c*(a + b*x))*Coth[a*c + b*c*x])/(b
*c*(1 + E^(2*c*(a + b*x)))^2*Sqrt[Coth[a*c + b*c*x]^2]) + (3*E^(c*(a + b*x))*Coth[a*c + b*c*x])/(b*c*(1 + E^(2
*c*(a + b*x)))*Sqrt[Coth[a*c + b*c*x]^2]) - (3*ArcTan[E^(c*(a + b*x))]*Coth[a*c + b*c*x])/(b*c*Sqrt[Coth[a*c +
b*c*x]^2])

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Rubi [A]  time = 0.86382, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {6720, 2282, 390, 1158, 12, 288, 203} \[ \frac{e^{c (a+b x)} \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}}+\frac{3 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right ) \sqrt{\coth ^2(a c+b c x)}}-\frac{2 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^2 \sqrt{\coth ^2(a c+b c x)}}-\frac{3 \tan ^{-1}\left (e^{c (a+b x)}\right ) \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))/(Coth[a*c + b*c*x]^2)^(3/2),x]

[Out]

(E^(c*(a + b*x))*Coth[a*c + b*c*x])/(b*c*Sqrt[Coth[a*c + b*c*x]^2]) - (2*E^(c*(a + b*x))*Coth[a*c + b*c*x])/(b
*c*(1 + E^(2*c*(a + b*x)))^2*Sqrt[Coth[a*c + b*c*x]^2]) + (3*E^(c*(a + b*x))*Coth[a*c + b*c*x])/(b*c*(1 + E^(2
*c*(a + b*x)))*Sqrt[Coth[a*c + b*c*x]^2]) - (3*ArcTan[E^(c*(a + b*x))]*Coth[a*c + b*c*x])/(b*c*Sqrt[Coth[a*c +
b*c*x]^2])

Rule 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

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]]]

Rule 390

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 1158

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + c*
x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + c*x^4)^p, d + e*x^2, x], x, 0]}, -Simp[(R*x*(d + e*x
^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*ExpandToSum[2*d*(q + 1)*Qx + R*
(2*q + 3), x], x], x]] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 12

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

Rule 288

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 203

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

Rubi steps

\begin{align*} \int \frac{e^{c (a+b x)}}{\coth ^2(a c+b c x)^{3/2}} \, dx &=\frac{\coth (a c+b c x) \int e^{c (a+b x)} \tanh ^3(a c+b c x) \, dx}{\sqrt{\coth ^2(a c+b c x)}}\\ &=\frac{\coth (a c+b c x) \operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^3}{\left (1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\coth ^2(a c+b c x)}}\\ &=\frac{\coth (a c+b c x) \operatorname{Subst}\left (\int \left (1-\frac{2 \left (1+3 x^4\right )}{\left (1+x^2\right )^3}\right ) \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\coth ^2(a c+b c x)}}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}}-\frac{(2 \coth (a c+b c x)) \operatorname{Subst}\left (\int \frac{1+3 x^4}{\left (1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\coth ^2(a c+b c x)}}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}}-\frac{2 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (1+e^{2 c (a+b x)}\right )^2 \sqrt{\coth ^2(a c+b c x)}}+\frac{\coth (a c+b c x) \operatorname{Subst}\left (\int -\frac{12 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{2 b c \sqrt{\coth ^2(a c+b c x)}}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}}-\frac{2 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (1+e^{2 c (a+b x)}\right )^2 \sqrt{\coth ^2(a c+b c x)}}-\frac{(6 \coth (a c+b c x)) \operatorname{Subst}\left (\int \frac{x^2}{\left (1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\coth ^2(a c+b c x)}}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}}-\frac{2 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (1+e^{2 c (a+b x)}\right )^2 \sqrt{\coth ^2(a c+b c x)}}+\frac{3 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (1+e^{2 c (a+b x)}\right ) \sqrt{\coth ^2(a c+b c x)}}-\frac{(3 \coth (a c+b c x)) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^{c (a+b x)}\right )}{b c \sqrt{\coth ^2(a c+b c x)}}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}}-\frac{2 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (1+e^{2 c (a+b x)}\right )^2 \sqrt{\coth ^2(a c+b c x)}}+\frac{3 e^{c (a+b x)} \coth (a c+b c x)}{b c \left (1+e^{2 c (a+b x)}\right ) \sqrt{\coth ^2(a c+b c x)}}-\frac{3 \tan ^{-1}\left (e^{c (a+b x)}\right ) \coth (a c+b c x)}{b c \sqrt{\coth ^2(a c+b c x)}}\\ \end{align*}

Mathematica [A]  time = 0.297194, size = 104, normalized size = 0.54 \[ \frac{\left (e^{c (a+b x)} \left (5 e^{2 c (a+b x)}+e^{4 c (a+b x)}+2\right )-3 \left (e^{2 c (a+b x)}+1\right )^2 \tan ^{-1}\left (e^{c (a+b x)}\right )\right ) \coth (c (a+b x))}{b c \left (e^{2 c (a+b x)}+1\right )^2 \sqrt{\coth ^2(c (a+b x))}} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))/(Coth[a*c + b*c*x]^2)^(3/2),x]

[Out]

((E^(c*(a + b*x))*(2 + 5*E^(2*c*(a + b*x)) + E^(4*c*(a + b*x))) - 3*(1 + E^(2*c*(a + b*x)))^2*ArcTan[E^(c*(a +
b*x))])*Coth[c*(a + b*x)])/(b*c*(1 + E^(2*c*(a + b*x)))^2*Sqrt[Coth[c*(a + b*x)]^2])

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Maple [C]  time = 0.209, size = 301, normalized size = 1.6 \begin{align*}{\frac{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ){{\rm e}^{c \left ( bx+a \right ) }}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) cb}{\frac{1}{\sqrt{{\frac{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}}}}+{\frac{{{\rm e}^{c \left ( bx+a \right ) }} \left ( 3\,{{\rm e}^{2\,c \left ( bx+a \right ) }}+1 \right ) }{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) cb}{\frac{1}{\sqrt{{\frac{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}}}}+{\frac{{\frac{3\,i}{2}} \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) \ln \left ({{\rm e}^{c \left ( bx+a \right ) }}-i \right ) }{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) cb}{\frac{1}{\sqrt{{\frac{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}}}}-{\frac{{\frac{3\,i}{2}} \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) \ln \left ({{\rm e}^{c \left ( bx+a \right ) }}+i \right ) }{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) cb}{\frac{1}{\sqrt{{\frac{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))/(coth(b*c*x+a*c)^2)^(3/2),x)

[Out]

1/((1+exp(2*c*(b*x+a)))^2/(exp(2*c*(b*x+a))-1)^2)^(1/2)/(exp(2*c*(b*x+a))-1)*(1+exp(2*c*(b*x+a)))/c/b*exp(c*(b
*x+a))+1/(1+exp(2*c*(b*x+a)))/(exp(2*c*(b*x+a))-1)/((1+exp(2*c*(b*x+a)))^2/(exp(2*c*(b*x+a))-1)^2)^(1/2)*exp(c
*(b*x+a))*(3*exp(2*c*(b*x+a))+1)/c/b+3/2*I*(1+exp(2*c*(b*x+a)))/(exp(2*c*(b*x+a))-1)/((1+exp(2*c*(b*x+a)))^2/(
exp(2*c*(b*x+a))-1)^2)^(1/2)/c/b*ln(exp(c*(b*x+a))-I)-3/2*I*(1+exp(2*c*(b*x+a)))/(exp(2*c*(b*x+a))-1)/((1+exp(
2*c*(b*x+a)))^2/(exp(2*c*(b*x+a))-1)^2)^(1/2)/c/b*ln(exp(c*(b*x+a))+I)

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Maxima [A]  time = 1.66612, size = 122, normalized size = 0.63 \begin{align*} -\frac{3 \, \arctan \left (e^{\left (b c x + a c\right )}\right )}{b c} + \frac{e^{\left (5 \, b c x + 5 \, a c\right )} + 5 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 2 \, e^{\left (b c x + a c\right )}}{b c{\left (e^{\left (4 \, b c x + 4 \, a c\right )} + 2 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(coth(b*c*x+a*c)^2)^(3/2),x, algorithm="maxima")

[Out]

-3*arctan(e^(b*c*x + a*c))/(b*c) + (e^(5*b*c*x + 5*a*c) + 5*e^(3*b*c*x + 3*a*c) + 2*e^(b*c*x + a*c))/(b*c*(e^(
4*b*c*x + 4*a*c) + 2*e^(2*b*c*x + 2*a*c) + 1))

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Fricas [B]  time = 2.54231, size = 1175, normalized size = 6.09 \begin{align*} \frac{\cosh \left (b c x + a c\right )^{5} + 5 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{4} + \sinh \left (b c x + a c\right )^{5} + 5 \,{\left (2 \, \cosh \left (b c x + a c\right )^{2} + 1\right )} \sinh \left (b c x + a c\right )^{3} + 5 \, \cosh \left (b c x + a c\right )^{3} + 5 \,{\left (2 \, \cosh \left (b c x + a c\right )^{3} + 3 \, \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )^{2} - 3 \,{\left (\cosh \left (b c x + a c\right )^{4} + 4 \, \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + \sinh \left (b c x + a c\right )^{4} + 2 \,{\left (3 \, \cosh \left (b c x + a c\right )^{2} + 1\right )} \sinh \left (b c x + a c\right )^{2} + 2 \, \cosh \left (b c x + a c\right )^{2} + 4 \,{\left (\cosh \left (b c x + a c\right )^{3} + \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right ) + 1\right )} \arctan \left (\cosh \left (b c x + a c\right ) + \sinh \left (b c x + a c\right )\right ) +{\left (5 \, \cosh \left (b c x + a c\right )^{4} + 15 \, \cosh \left (b c x + a c\right )^{2} + 2\right )} \sinh \left (b c x + a c\right ) + 2 \, \cosh \left (b c x + a c\right )}{b c \cosh \left (b c x + a c\right )^{4} + 4 \, b c \cosh \left (b c x + a c\right ) \sinh \left (b c x + a c\right )^{3} + b c \sinh \left (b c x + a c\right )^{4} + 2 \, b c \cosh \left (b c x + a c\right )^{2} + 2 \,{\left (3 \, b c \cosh \left (b c x + a c\right )^{2} + b c\right )} \sinh \left (b c x + a c\right )^{2} + b c + 4 \,{\left (b c \cosh \left (b c x + a c\right )^{3} + b c \cosh \left (b c x + a c\right )\right )} \sinh \left (b c x + a c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(coth(b*c*x+a*c)^2)^(3/2),x, algorithm="fricas")

[Out]

(cosh(b*c*x + a*c)^5 + 5*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^4 + sinh(b*c*x + a*c)^5 + 5*(2*cosh(b*c*x + a*c)^
2 + 1)*sinh(b*c*x + a*c)^3 + 5*cosh(b*c*x + a*c)^3 + 5*(2*cosh(b*c*x + a*c)^3 + 3*cosh(b*c*x + a*c))*sinh(b*c*
x + a*c)^2 - 3*(cosh(b*c*x + a*c)^4 + 4*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^3 + sinh(b*c*x + a*c)^4 + 2*(3*cos
h(b*c*x + a*c)^2 + 1)*sinh(b*c*x + a*c)^2 + 2*cosh(b*c*x + a*c)^2 + 4*(cosh(b*c*x + a*c)^3 + cosh(b*c*x + a*c)
)*sinh(b*c*x + a*c) + 1)*arctan(cosh(b*c*x + a*c) + sinh(b*c*x + a*c)) + (5*cosh(b*c*x + a*c)^4 + 15*cosh(b*c*
x + a*c)^2 + 2)*sinh(b*c*x + a*c) + 2*cosh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^4 + 4*b*c*cosh(b*c*x + a*c)*si
nh(b*c*x + a*c)^3 + b*c*sinh(b*c*x + a*c)^4 + 2*b*c*cosh(b*c*x + a*c)^2 + 2*(3*b*c*cosh(b*c*x + a*c)^2 + b*c)*
sinh(b*c*x + a*c)^2 + b*c + 4*(b*c*cosh(b*c*x + a*c)^3 + b*c*cosh(b*c*x + a*c))*sinh(b*c*x + a*c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(coth(b*c*x+a*c)**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.21802, size = 176, normalized size = 0.91 \begin{align*} -\frac{{\left (3 \, \arctan \left (e^{\left (b c x + a c\right )}\right ) e^{\left (-a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - e^{\left (b c x\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac{3 \, e^{\left (3 \, b c x + 2 \, a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + e^{\left (b c x\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}^{2}}\right )} e^{\left (a c\right )}}{b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))/(coth(b*c*x+a*c)^2)^(3/2),x, algorithm="giac")

[Out]

-(3*arctan(e^(b*c*x + a*c))*e^(-a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) - e^(b*c*x)*sgn(e^(2*b*c*x + 2*a*c) - 1) - (
3*e^(3*b*c*x + 2*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) + e^(b*c*x)*sgn(e^(2*b*c*x + 2*a*c) - 1))/(e^(2*b*c*x + 2*a
*c) + 1)^2)*e^(a*c)/(b*c)