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

Optimal. Leaf size=319 \[ \frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}-\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt {\tanh ^2(a c+b c x)}} \]

[Out]

exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(tanh(b*c*x+a*c)^2)^(1/2)-4*exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(
b*x+a)))^4/(tanh(b*c*x+a*c)^2)^(1/2)+26/3*exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))^3/(tanh(b*c*
x+a*c)^2)^(1/2)-55/6*exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))^2/(tanh(b*c*x+a*c)^2)^(1/2)+25/4*
exp(c*(b*x+a))*tanh(b*c*x+a*c)/b/c/(1-exp(2*c*(b*x+a)))/(tanh(b*c*x+a*c)^2)^(1/2)-15/4*arctanh(exp(c*(b*x+a)))
*tanh(b*c*x+a*c)/b/c/(tanh(b*c*x+a*c)^2)^(1/2)

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Rubi [A]  time = 1.77, antiderivative size = 319, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6720, 2282, 390, 1814, 1157, 385, 207} \[ \frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}} \]

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

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])

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 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 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 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 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 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])

Rubi steps

\begin {align*} \int \frac {e^{c (a+b x)}}{\tanh ^2(a c+b c x)^{5/2}} \, dx &=\frac {\tanh (a c+b c x) \int e^{c (a+b x)} \coth ^5(a c+b c x) \, dx}{\sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {\tanh (a c+b c x) \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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {\tanh (a c+b c x) \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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}+\frac {(2 \tanh (a c+b c x)) \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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {\tanh (a c+b c x) \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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}+\frac {\tanh (a c+b c x) \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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {\tanh (a c+b c x) \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 \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}+\frac {(15 \tanh (a c+b c x)) \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,e^{c (a+b x)}\right )}{4 b c \sqrt {\tanh ^2(a c+b c x)}}\\ &=\frac {e^{c (a+b x)} \tanh (a c+b c x)}{b c \sqrt {\tanh ^2(a c+b c x)}}-\frac {4 e^{c (a+b x)} \tanh (a c+b c x)}{b c \left (1-e^{2 c (a+b x)}\right )^4 \sqrt {\tanh ^2(a c+b c x)}}+\frac {26 e^{c (a+b x)} \tanh (a c+b c x)}{3 b c \left (1-e^{2 c (a+b x)}\right )^3 \sqrt {\tanh ^2(a c+b c x)}}-\frac {55 e^{c (a+b x)} \tanh (a c+b c x)}{6 b c \left (1-e^{2 c (a+b x)}\right )^2 \sqrt {\tanh ^2(a c+b c x)}}+\frac {25 e^{c (a+b x)} \tanh (a c+b c x)}{4 b c \left (1-e^{2 c (a+b x)}\right ) \sqrt {\tanh ^2(a c+b c x)}}-\frac {15 \tanh ^{-1}\left (e^{c (a+b x)}\right ) \tanh (a c+b c x)}{4 b c \sqrt {\tanh ^2(a c+b c x)}}\\ \end {align*}

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Mathematica [A]  time = 10.69, size = 164, normalized size = 0.51 \[ \frac {\left (66 e^{c (a+b x)}-314 e^{3 c (a+b x)}+374 e^{5 c (a+b x)}-246 e^{7 c (a+b x)}+24 e^{9 c (a+b x)}+45 \left (e^{2 c (a+b x)}-1\right )^4 \log \left (1-e^{c (a+b x)}\right )-45 \left (e^{2 c (a+b x)}-1\right )^4 \log \left (e^{c (a+b x)}+1\right )\right ) \tanh (c (a+b x))}{24 b c \left (e^{2 c (a+b x)}-1\right )^4 \sqrt {\tanh ^2(c (a+b x))}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((66*E^(c*(a + b*x)) - 314*E^(3*c*(a + b*x)) + 374*E^(5*c*(a + b*x)) - 246*E^(7*c*(a + b*x)) + 24*E^(9*c*(a +
b*x)) + 45*(-1 + E^(2*c*(a + b*x)))^4*Log[1 - E^(c*(a + b*x))] - 45*(-1 + E^(2*c*(a + b*x)))^4*Log[1 + E^(c*(a
+ b*x))])*Tanh[c*(a + b*x)])/(24*b*c*(-1 + E^(2*c*(a + b*x)))^4*Sqrt[Tanh[c*(a + b*x)]^2])

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fricas [B]  time = 0.93, size = 1617, normalized size = 5.07 \[ \text {result too large to display} \]

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]

1/24*(24*cosh(b*c*x + a*c)^9 + 216*cosh(b*c*x + a*c)*sinh(b*c*x + a*c)^8 + 24*sinh(b*c*x + a*c)^9 + 6*(144*cos
h(b*c*x + a*c)^2 - 41)*sinh(b*c*x + a*c)^7 - 246*cosh(b*c*x + a*c)^7 + 42*(48*cosh(b*c*x + a*c)^3 - 41*cosh(b*
c*x + a*c))*sinh(b*c*x + a*c)^6 + 2*(1512*cosh(b*c*x + a*c)^4 - 2583*cosh(b*c*x + a*c)^2 + 187)*sinh(b*c*x + a
*c)^5 + 374*cosh(b*c*x + a*c)^5 + 2*(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 + 2*(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)*sinh(b*c*x + a*c)^3 - 314*cosh(b*c*x + a*c)^3 + 2*(432*cosh(b*c*x + a*c)^7 - 2583*cosh(b*c*x + a*c)^5 + 1
870*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)*sinh(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*c
osh(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)*log(cosh(b*c*x + a*c) + si
nh(b*c*x + a*c) + 1) + 45*(cosh(b*c*x + a*c)^8 + 8*cosh(b*c*x + a*c)*sinh(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*co
sh(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)*log(cosh(b*c*x + a*c) + sinh(b*c*x + a*c) - 1) + 2*(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) + 66*c
osh(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)*sinh(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))*sinh(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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giac [A]  time = 0.36, size = 215, normalized size = 0.67 \[ \frac {24 \, e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 45 \, \log \left (e^{\left (b c x + a c\right )} + 1\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 45 \, \log \left ({\left | e^{\left (b c x + a c\right )} - 1 \right |}\right ) \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - \frac {2 \, {\left (75 \, e^{\left (7 \, b c x + 7 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 115 \, e^{\left (5 \, b c x + 5 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) + 109 \, e^{\left (3 \, b c x + 3 \, a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right ) - 21 \, e^{\left (b c x + a c\right )} \mathrm {sgn}\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )\right )}}{{\left (e^{\left (2 \, b c x + 2 \, a c\right )} - 1\right )}^{4}}}{24 \, b c} \]

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]

1/24*(24*e^(b*c*x + a*c)*sgn(e^(2*b*c*x + 2*a*c) - 1) - 45*log(e^(b*c*x + a*c) + 1)*sgn(e^(2*b*c*x + 2*a*c) -
1) + 45*log(abs(e^(b*c*x + a*c) - 1))*sgn(e^(2*b*c*x + 2*a*c) - 1) - 2*(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)

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maple [A]  time = 0.99, size = 320, normalized size = 1.00 \[ \frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) {\mathrm e}^{c \left (b x +a \right )}}{\sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {{\mathrm e}^{c \left (b x +a \right )} \left (75 \,{\mathrm e}^{6 c \left (b x +a \right )}-115 \,{\mathrm e}^{4 c \left (b x +a \right )}+109 \,{\mathrm e}^{2 c \left (b x +a \right )}-21\right )}{12 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{3} \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, c b}+\frac {15 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left ({\mathrm e}^{c \left (b x +a \right )}-1\right )}{8 \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b}-\frac {15 \left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right ) \ln \left (1+{\mathrm e}^{c \left (b x +a \right )}\right )}{8 \sqrt {\frac {\left ({\mathrm e}^{2 c \left (b x +a \right )}-1\right )^{2}}{\left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right )^{2}}}\, \left (1+{\mathrm e}^{2 c \left (b x +a \right )}\right ) c b} \]

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)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)/(1+exp(2*c*(b*x+a)))*(exp(2*c*(b*x+a))-1)*exp(c*(b*x+a
))/c/b-1/12/(exp(2*c*(b*x+a))-1)^3/(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))*(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/((exp(2*c*(b*x+a))-
1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)/(1+exp(2*c*(b*x+a)))*(exp(2*c*(b*x+a))-1)/c/b*ln(exp(c*(b*x+a))-1)-15/8/((e
xp(2*c*(b*x+a))-1)^2/(1+exp(2*c*(b*x+a)))^2)^(1/2)/(1+exp(2*c*(b*x+a)))*(exp(2*c*(b*x+a))-1)/c/b*ln(1+exp(c*(b
*x+a)))

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maxima [A]  time = 0.49, size = 167, normalized size = 0.52 \[ -\frac {15 \, \log \left (e^{\left (b c x + a c\right )} + 1\right )}{8 \, b c} + \frac {15 \, \log \left (e^{\left (b c x + a c\right )} - 1\right )}{8 \, b c} + \frac {12 \, e^{\left (9 \, b c x + 9 \, a c\right )} - 123 \, e^{\left (7 \, b c x + 7 \, a c\right )} + 187 \, e^{\left (5 \, b c x + 5 \, a c\right )} - 157 \, e^{\left (3 \, b c x + 3 \, a c\right )} + 33 \, e^{\left (b c x + a c\right )}}{12 \, b c {\left (e^{\left (8 \, b c x + 8 \, a c\right )} - 4 \, e^{\left (6 \, b c x + 6 \, a c\right )} + 6 \, e^{\left (4 \, b c x + 4 \, a c\right )} - 4 \, e^{\left (2 \, b c x + 2 \, a c\right )} + 1\right )}} \]

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]

-15/8*log(e^(b*c*x + a*c) + 1)/(b*c) + 15/8*log(e^(b*c*x + a*c) - 1)/(b*c) + 1/12*(12*e^(9*b*c*x + 9*a*c) - 12
3*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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {e}}^{c\,\left (a+b\,x\right )}}{{\left ({\mathrm {tanh}\left (a\,c+b\,c\,x\right )}^2\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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