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

Optimal. Leaf size=197 \[ -\frac{3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x) \sqrt{\coth ^2(a c+b c x)}}{b c}+\frac{e^{c (a+b x)} \tanh (a c+b c x) \sqrt{\coth ^2(a c+b c x)}}{b c}+\frac{3 e^{c (a+b x)} \tanh (a c+b c x) \sqrt{\coth ^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )}-\frac{2 e^{c (a+b x)} \tanh (a c+b c x) \sqrt{\coth ^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.2836, antiderivative size = 197, 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, 207} \[ -\frac{3 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x) \sqrt{\coth ^2(a c+b c x)}}{b c}+\frac{e^{c (a+b x)} \tanh (a c+b c x) \sqrt{\coth ^2(a c+b c x)}}{b c}+\frac{3 e^{c (a+b x)} \tanh (a c+b c x) \sqrt{\coth ^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )}-\frac{2 e^{c (a+b x)} \tanh (a c+b c x) \sqrt{\coth ^2(a c+b c x)}}{b c \left (1-e^{2 c (a+b x)}\right )^2} \]

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

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 207

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

Rubi steps

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

Mathematica [C]  time = 3.77009, size = 334, normalized size = 1.7 \[ -\frac{e^{-5 c (a+b x)} \tanh ^3(c (a+b x)) \coth ^2(c (a+b x))^{3/2} \left (256 e^{8 c (a+b x)} \left (e^{2 c (a+b x)}+1\right )^3 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2,2\right \},\left \{1,1,1,1,\frac{11}{2}\right \},e^{2 c (a+b x)}\right )+384 e^{8 c (a+b x)} \left (5 e^{2 c (a+b x)}+7\right ) \left (e^{2 c (a+b x)}+1\right )^2 \text{HypergeometricPFQ}\left (\left \{\frac{3}{2},2,2,2,2\right \},\left \{1,1,1,\frac{11}{2}\right \},e^{2 c (a+b x)}\right )-21 \left (507305 e^{2 c (a+b x)}+173916 e^{4 c (a+b x)}-154296 e^{6 c (a+b x)}-73885 e^{8 c (a+b x)}+4887 e^{10 c (a+b x)}+252105\right )-\frac{315 \left (-28218 e^{2 c (a+b x)}+1173 e^{4 c (a+b x)}+17748 e^{6 c (a+b x)}+4299 e^{8 c (a+b x)}-1434 e^{10 c (a+b x)}+7 e^{12 c (a+b x)}-16807\right ) \tanh ^{-1}\left (\sqrt{e^{2 c (a+b x)}}\right )}{\sqrt{e^{2 c (a+b x)}}}\right )}{60480 b c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-((Coth[c*(a + b*x)]^2)^(3/2)*(-21*(252105 + 507305*E^(2*c*(a + b*x)) + 173916*E^(4*c*(a + b*x)) - 154296*E^(6
*c*(a + b*x)) - 73885*E^(8*c*(a + b*x)) + 4887*E^(10*c*(a + b*x))) - (315*(-16807 - 28218*E^(2*c*(a + b*x)) +
1173*E^(4*c*(a + b*x)) + 17748*E^(6*c*(a + b*x)) + 4299*E^(8*c*(a + b*x)) - 1434*E^(10*c*(a + b*x)) + 7*E^(12*
c*(a + b*x)))*ArcTanh[Sqrt[E^(2*c*(a + b*x))]])/Sqrt[E^(2*c*(a + b*x))] + 384*E^(8*c*(a + b*x))*(1 + E^(2*c*(a
+ b*x)))^2*(7 + 5*E^(2*c*(a + b*x)))*HypergeometricPFQ[{3/2, 2, 2, 2, 2}, {1, 1, 1, 11/2}, E^(2*c*(a + b*x))]
+ 256*E^(8*c*(a + b*x))*(1 + E^(2*c*(a + b*x)))^3*HypergeometricPFQ[{3/2, 2, 2, 2, 2, 2}, {1, 1, 1, 1, 11/2},
E^(2*c*(a + b*x))])*Tanh[c*(a + b*x)]^3)/(60480*b*c*E^(5*c*(a + b*x)))

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Maple [A]  time = 0.199, size = 298, normalized size = 1.5 \begin{align*}{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ){{\rm e}^{c \left ( bx+a \right ) }}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) cb}\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}\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{ \left ( 3\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-3 \right ) \ln \left ({{\rm e}^{c \left ( bx+a \right ) }}+1 \right ) }{ \left ( 2+2\,{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) cb}\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{ \left ( 3\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-3 \right ) \ln \left ({{\rm e}^{c \left ( bx+a \right ) }}-1 \right ) }{ \left ( 2+2\,{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) cb}\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)))*(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*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/(1+exp(2*c*(b*x+a)))*(exp(2*c*(b*x+a))-1)*((1+exp(2*c*(b*x+a)))^2/(ex
p(2*c*(b*x+a))-1)^2)^(1/2)/c/b*ln(exp(c*(b*x+a))+1)+3/2/(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))-1)

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Maxima [A]  time = 1.82596, size = 151, normalized size = 0.77 \begin{align*} -\frac{3 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{2 \, b c} + \frac{3 \, \log \left (e^{\left (b c x + a c\right )} - 1\right )}{2 \, 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]

-3/2*log(e^(b*c*x + a*c) + 1)/(b*c) + 3/2*log(e^(b*c*x + a*c) - 1)/(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 = 2.42228, size = 1574, normalized size = 7.99 \begin{align*} \text{result too large to display} \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]

1/2*(2*cosh(b*c*x + a*c)^5 + 10*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^4 + 2*sinh(b*c*x + a*c)^5 + 10*(2*cosh(b*c
*x + a*c)^2 - 1)*sinh(b*c*x + a*c)^3 - 10*cosh(b*c*x + a*c)^3 + 10*(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*cosh(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)*log(cosh(b*c*x + a*c) + sinh(b*c*x + a*c) + 1) + 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*cosh(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)*log(cosh(b*c
*x + a*c) + sinh(b*c*x + a*c) - 1) + 2*(5*cosh(b*c*x + a*c)^4 - 15*cosh(b*c*x + a*c)^2 + 2)*sinh(b*c*x + a*c)
+ 4*cosh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^4 + 4*b*c*cosh(b*c*x + a*c)*sinh(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 = 1.27028, size = 209, normalized size = 1.06 \begin{align*} \frac{\frac{2 \, e^{\left (b c x + a c\right )}}{\mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} - \frac{3 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{\mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} + \frac{3 \, \log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right )}{\mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )} - \frac{2 \,{\left (3 \, e^{\left (3 \, b c x + 3 \, a c\right )} - e^{\left (b c x + a c\right )}\right )}}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{2} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}}{2 \, 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]

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