3.234 \(\int e^{c (a+b x)} \tanh ^2(a c+b c x)^{5/2} \, dx\)

Optimal. Leaf size=311 \[ \frac{e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}+\frac{25 e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{4 b c \left (e^{2 c (a+b x)}+1\right )}-\frac{55 e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{6 b c \left (e^{2 c (a+b x)}+1\right )^2}+\frac{26 e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{3 b c \left (e^{2 c (a+b x)}+1\right )^3}-\frac{4 e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^4}-\frac{15 \tan ^{-1}\left (e^{c (a+b x)}\right ) \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{4 b c} \]

[Out]

(E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^2])/(b*c) - (4*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqr
t[Tanh[a*c + b*c*x]^2])/(b*c*(1 + E^(2*c*(a + b*x)))^4) + (26*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c
+ b*c*x]^2])/(3*b*c*(1 + E^(2*c*(a + b*x)))^3) - (55*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^
2])/(6*b*c*(1 + E^(2*c*(a + b*x)))^2) + (25*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^2])/(4*b*
c*(1 + E^(2*c*(a + b*x)))) - (15*ArcTan[E^(c*(a + b*x))]*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^2])/(4*b*c)

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Rubi [A]  time = 0.902018, antiderivative size = 311, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {6720, 2282, 390, 1814, 1157, 385, 203} \[ \frac{e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c}+\frac{25 e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{4 b c \left (e^{2 c (a+b x)}+1\right )}-\frac{55 e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{6 b c \left (e^{2 c (a+b x)}+1\right )^2}+\frac{26 e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{3 b c \left (e^{2 c (a+b x)}+1\right )^3}-\frac{4 e^{c (a+b x)} \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{b c \left (e^{2 c (a+b x)}+1\right )^4}-\frac{15 \tan ^{-1}\left (e^{c (a+b x)}\right ) \sqrt{\tanh ^2(a c+b c x)} \coth (a c+b c x)}{4 b c} \]

Antiderivative was successfully verified.

[In]

Int[E^(c*(a + b*x))*(Tanh[a*c + b*c*x]^2)^(5/2),x]

[Out]

(E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^2])/(b*c) - (4*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqr
t[Tanh[a*c + b*c*x]^2])/(b*c*(1 + E^(2*c*(a + b*x)))^4) + (26*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c
+ b*c*x]^2])/(3*b*c*(1 + E^(2*c*(a + b*x)))^3) - (55*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^
2])/(6*b*c*(1 + E^(2*c*(a + b*x)))^2) + (25*E^(c*(a + b*x))*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^2])/(4*b*
c*(1 + E^(2*c*(a + b*x)))) - (15*ArcTan[E^(c*(a + b*x))]*Coth[a*c + b*c*x]*Sqrt[Tanh[a*c + b*c*x]^2])/(4*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 1814

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x^2, x], f = Coeff[P
olynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[((a
*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*(p + 1)), Int[(a + b*x^2)^(p + 1)*ExpandToS
um[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]

Rule 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + 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, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 385

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

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 e^{c (a+b x)} \tanh ^2(a c+b c x)^{5/2} \, dx &=\left (\coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}\right ) \int e^{c (a+b x)} \tanh ^5(a c+b c x) \, dx\\ &=\frac{\left (\coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}\right ) \operatorname{Subst}\left (\int \frac{\left (-1+x^2\right )^5}{\left (1+x^2\right )^5} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac{\left (\coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}\right ) \operatorname{Subst}\left (\int \left (1-\frac{2 \left (1+10 x^4+5 x^8\right )}{\left (1+x^2\right )^5}\right ) \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c}-\frac{\left (2 \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}\right ) \operatorname{Subst}\left (\int \frac{1+10 x^4+5 x^8}{\left (1+x^2\right )^5} \, dx,x,e^{c (a+b x)}\right )}{b c}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c}-\frac{4 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac{\left (\coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}\right ) \operatorname{Subst}\left (\int \frac{8-120 x^2+40 x^4-40 x^6}{\left (1+x^2\right )^4} \, dx,x,e^{c (a+b x)}\right )}{4 b c}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c}-\frac{4 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac{26 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{3 b c \left (1+e^{2 c (a+b x)}\right )^3}-\frac{\left (\coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}\right ) \operatorname{Subst}\left (\int \frac{160-480 x^2+240 x^4}{\left (1+x^2\right )^3} \, dx,x,e^{c (a+b x)}\right )}{24 b c}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c}-\frac{4 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac{26 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{3 b c \left (1+e^{2 c (a+b x)}\right )^3}-\frac{55 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{6 b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac{\left (\coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}\right ) \operatorname{Subst}\left (\int \frac{240-960 x^2}{\left (1+x^2\right )^2} \, dx,x,e^{c (a+b x)}\right )}{96 b c}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c}-\frac{4 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac{26 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{3 b c \left (1+e^{2 c (a+b x)}\right )^3}-\frac{55 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{6 b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac{25 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{4 b c \left (1+e^{2 c (a+b x)}\right )}-\frac{\left (15 \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,e^{c (a+b x)}\right )}{4 b c}\\ &=\frac{e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c}-\frac{4 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{b c \left (1+e^{2 c (a+b x)}\right )^4}+\frac{26 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{3 b c \left (1+e^{2 c (a+b x)}\right )^3}-\frac{55 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{6 b c \left (1+e^{2 c (a+b x)}\right )^2}+\frac{25 e^{c (a+b x)} \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{4 b c \left (1+e^{2 c (a+b x)}\right )}-\frac{15 \tan ^{-1}\left (e^{c (a+b x)}\right ) \coth (a c+b c x) \sqrt{\tanh ^2(a c+b c x)}}{4 b c}\\ \end{align*}

Mathematica [A]  time = 0.217265, size = 133, normalized size = 0.43 \[ \frac{\left (e^{c (a+b x)} \left (157 e^{2 c (a+b x)}+187 e^{4 c (a+b x)}+123 e^{6 c (a+b x)}+12 e^{8 c (a+b x)}+33\right )-45 \left (e^{2 c (a+b x)}+1\right )^4 \tan ^{-1}\left (e^{c (a+b x)}\right )\right ) \sqrt{\tanh ^2(c (a+b x))} \coth (c (a+b x))}{12 b c \left (e^{2 c (a+b x)}+1\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(c*(a + b*x))*(Tanh[a*c + b*c*x]^2)^(5/2),x]

[Out]

((E^(c*(a + b*x))*(33 + 157*E^(2*c*(a + b*x)) + 187*E^(4*c*(a + b*x)) + 123*E^(6*c*(a + b*x)) + 12*E^(8*c*(a +
b*x))) - 45*(1 + E^(2*c*(a + b*x)))^4*ArcTan[E^(c*(a + b*x))])*Coth[c*(a + b*x)]*Sqrt[Tanh[c*(a + b*x)]^2])/(
12*b*c*(1 + E^(2*c*(a + b*x)))^4)

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Maple [C]  time = 0.277, size = 324, normalized size = 1. \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}\sqrt{{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}}+{\frac{{{\rm e}^{c \left ( bx+a \right ) }} \left ( 75\,{{\rm e}^{6\,c \left ( bx+a \right ) }}+115\,{{\rm e}^{4\,c \left ( bx+a \right ) }}+109\,{{\rm e}^{2\,c \left ( bx+a \right ) }}+21 \right ) }{ \left ( 12\,{{\rm e}^{2\,c \left ( bx+a \right ) }}-12 \right ) \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{3}cb}\sqrt{{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}}+{\frac{{\frac{15\,i}{8}} \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}\sqrt{{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}}-{\frac{{\frac{15\,i}{8}} \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}\sqrt{{\frac{ \left ({{\rm e}^{2\,c \left ( bx+a \right ) }}-1 \right ) ^{2}}{ \left ( 1+{{\rm e}^{2\,c \left ( bx+a \right ) }} \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(5/2),x)

[Out]

1/(exp(2*c*(b*x+a))-1)*(1+exp(2*c*(b*x+a)))*((exp(2*c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)*exp(c*(b*x+a
))/c/b+1/12/(exp(2*c*(b*x+a))-1)/(1+exp(2*c*(b*x+a)))^3*((exp(2*c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)*
exp(c*(b*x+a))*(75*exp(6*c*(b*x+a))+115*exp(4*c*(b*x+a))+109*exp(2*c*(b*x+a))+21)/c/b+15/8*I/(exp(2*c*(b*x+a))
-1)*(1+exp(2*c*(b*x+a)))*((exp(2*c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)/c/b*ln(exp(c*(b*x+a))-I)-15/8*I
/(exp(2*c*(b*x+a))-1)*(1+exp(2*c*(b*x+a)))*((exp(2*c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)/c/b*ln(exp(c*
(b*x+a))+I)

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Maxima [A]  time = 1.70385, size = 196, normalized size = 0.63 \begin{align*} -\frac{15 \, \arctan \left (e^{\left (b c x + a c\right )}\right )}{4 \, b c} + \frac{12 \, e^{\left (9 \, b c x + 9 \, a c\right )} + 123 \, e^{\left (7 \, b c x + 7 \, a c\right )} + 187 \, e^{\left (5 \, b c x + 5 \, a c\right )} + 157 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 33 \, e^{\left (b c x + a c\right )}}{12 \, b c{\left (e^{\left (8 \, b c x + 8 \, a c\right )} + 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} + 4 \, 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))*(tanh(b*c*x+a*c)^2)^(5/2),x, algorithm="maxima")

[Out]

-15/4*arctan(e^(b*c*x + a*c))/(b*c) + 1/12*(12*e^(9*b*c*x + 9*a*c) + 123*e^(7*b*c*x + 7*a*c) + 187*e^(5*b*c*x
+ 5*a*c) + 157*e^(3*b*c*x + 3*a*c) + 33*e^(b*c*x + a*c))/(b*c*(e^(8*b*c*x + 8*a*c) + 4*e^(6*b*c*x + 6*a*c) + 6
*e^(4*b*c*x + 4*a*c) + 4*e^(2*b*c*x + 2*a*c) + 1))

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Fricas [B]  time = 2.1185, size = 3186, normalized size = 10.24 \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))*(tanh(b*c*x+a*c)^2)^(5/2),x, algorithm="fricas")

[Out]

1/12*(12*cosh(b*c*x + a*c)^9 + 108*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^8 + 12*sinh(b*c*x + a*c)^9 + 3*(144*cos
h(b*c*x + a*c)^2 + 41)*sinh(b*c*x + a*c)^7 + 123*cosh(b*c*x + a*c)^7 + 21*(48*cosh(b*c*x + a*c)^3 + 41*cosh(b*
c*x + a*c))*sinh(b*c*x + a*c)^6 + (1512*cosh(b*c*x + a*c)^4 + 2583*cosh(b*c*x + a*c)^2 + 187)*sinh(b*c*x + a*c
)^5 + 187*cosh(b*c*x + a*c)^5 + (1512*cosh(b*c*x + a*c)^5 + 4305*cosh(b*c*x + a*c)^3 + 935*cosh(b*c*x + a*c))*
sinh(b*c*x + a*c)^4 + (1008*cosh(b*c*x + a*c)^6 + 4305*cosh(b*c*x + a*c)^4 + 1870*cosh(b*c*x + a*c)^2 + 157)*s
inh(b*c*x + a*c)^3 + 157*cosh(b*c*x + a*c)^3 + (432*cosh(b*c*x + a*c)^7 + 2583*cosh(b*c*x + a*c)^5 + 1870*cosh
(b*c*x + a*c)^3 + 471*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^2 - 45*(cosh(b*c*x + a*c)^8 + 8*cosh(b*c*x + a*c)*s
inh(b*c*x + a*c)^7 + sinh(b*c*x + a*c)^8 + 4*(7*cosh(b*c*x + a*c)^2 + 1)*sinh(b*c*x + a*c)^6 + 4*cosh(b*c*x +
a*c)^6 + 8*(7*cosh(b*c*x + a*c)^3 + 3*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^5 + 2*(35*cosh(b*c*x + a*c)^4 + 30*
cosh(b*c*x + a*c)^2 + 3)*sinh(b*c*x + a*c)^4 + 6*cosh(b*c*x + a*c)^4 + 8*(7*cosh(b*c*x + a*c)^5 + 10*cosh(b*c*
x + a*c)^3 + 3*cosh(b*c*x + a*c))*sinh(b*c*x + a*c)^3 + 4*(7*cosh(b*c*x + a*c)^6 + 15*cosh(b*c*x + a*c)^4 + 9*
cosh(b*c*x + a*c)^2 + 1)*sinh(b*c*x + a*c)^2 + 4*cosh(b*c*x + a*c)^2 + 8*(cosh(b*c*x + a*c)^7 + 3*cosh(b*c*x +
a*c)^5 + 3*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)) + (108*cosh(b*c*x + a*c)^8 + 861*cosh(b*c*x + a*c)^6 + 935*cosh(b*c*x + a*c)^4 + 471*cosh(b*c*x +
a*c)^2 + 33)*sinh(b*c*x + a*c) + 33*cosh(b*c*x + a*c))/(b*c*cosh(b*c*x + a*c)^8 + 8*b*c*cosh(b*c*x + a*c)*sinh
(b*c*x + a*c)^7 + b*c*sinh(b*c*x + a*c)^8 + 4*b*c*cosh(b*c*x + a*c)^6 + 4*(7*b*c*cosh(b*c*x + a*c)^2 + b*c)*si
nh(b*c*x + a*c)^6 + 6*b*c*cosh(b*c*x + a*c)^4 + 8*(7*b*c*cosh(b*c*x + a*c)^3 + 3*b*c*cosh(b*c*x + a*c))*sinh(b
*c*x + a*c)^5 + 2*(35*b*c*cosh(b*c*x + a*c)^4 + 30*b*c*cosh(b*c*x + a*c)^2 + 3*b*c)*sinh(b*c*x + a*c)^4 + 4*b*
c*cosh(b*c*x + a*c)^2 + 8*(7*b*c*cosh(b*c*x + a*c)^5 + 10*b*c*cosh(b*c*x + a*c)^3 + 3*b*c*cosh(b*c*x + a*c))*s
inh(b*c*x + a*c)^3 + 4*(7*b*c*cosh(b*c*x + a*c)^6 + 15*b*c*cosh(b*c*x + a*c)^4 + 9*b*c*cosh(b*c*x + a*c)^2 + b
*c)*sinh(b*c*x + a*c)^2 + b*c + 8*(b*c*cosh(b*c*x + a*c)^7 + 3*b*c*cosh(b*c*x + a*c)^5 + 3*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))*(tanh(b*c*x+a*c)**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.33277, size = 248, normalized size = 0.8 \begin{align*} -\frac{45 \, \arctan \left (e^{\left (b c x + a c\right )}\right ) \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 12 \, e^{\left (b c x + a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac{75 \, e^{\left (7 \, b c x + 7 \, a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 115 \, e^{\left (5 \, b c x + 5 \, a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 109 \, e^{\left (3 \, b c x + 3 \, a c\right )} \mathrm{sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 21 \, e^{\left (b c x + a c\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 )}^{4}}}{12 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(c*(b*x+a))*(tanh(b*c*x+a*c)^2)^(5/2),x, algorithm="giac")

[Out]

-1/12*(45*arctan(e^(b*c*x + a*c))*sgn(e^(2*b*c*x + 2*a*c) - 1) - 12*e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) -
1) - (75*e^(7*b*c*x + 7*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) + 115*e^(5*b*c*x + 5*a*c)*sgn(e^(2*b*c*x + 2*a*c) -
1) + 109*e^(3*b*c*x + 3*a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) + 21*e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1))/(
e^(2*b*c*x + 2*a*c) + 1)^4)/(b*c)