3.13 \(\int \cosh ^m(a+b x) \sinh ^3(a+b x) \, dx\)

Optimal. Leaf size=40 \[ \frac {\cosh ^{m+3}(a+b x)}{b (m+3)}-\frac {\cosh ^{m+1}(a+b x)}{b (m+1)} \]

[Out]

-cosh(b*x+a)^(1+m)/b/(1+m)+cosh(b*x+a)^(3+m)/b/(3+m)

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Rubi [A]  time = 0.05, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {2565, 14} \[ \frac {\cosh ^{m+3}(a+b x)}{b (m+3)}-\frac {\cosh ^{m+1}(a+b x)}{b (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[Cosh[a + b*x]^m*Sinh[a + b*x]^3,x]

[Out]

-(Cosh[a + b*x]^(1 + m)/(b*(1 + m))) + Cosh[a + b*x]^(3 + m)/(b*(3 + m))

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rubi steps

\begin {align*} \int \cosh ^m(a+b x) \sinh ^3(a+b x) \, dx &=-\frac {\operatorname {Subst}\left (\int x^m \left (1-x^2\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int \left (x^m-x^{2+m}\right ) \, dx,x,\cosh (a+b x)\right )}{b}\\ &=-\frac {\cosh ^{1+m}(a+b x)}{b (1+m)}+\frac {\cosh ^{3+m}(a+b x)}{b (3+m)}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 44, normalized size = 1.10 \[ \frac {\cosh ^{m+1}(a+b x) ((m+1) \cosh (2 (a+b x))-m-5)}{2 b (m+1) (m+3)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cosh[a + b*x]^m*Sinh[a + b*x]^3,x]

[Out]

(Cosh[a + b*x]^(1 + m)*(-5 - m + (1 + m)*Cosh[2*(a + b*x)]))/(2*b*(1 + m)*(3 + m))

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fricas [B]  time = 0.43, size = 189, normalized size = 4.72 \[ \frac {{\left ({\left (m + 1\right )} \cosh \left (b x + a\right )^{3} + 3 \, {\left (m + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - {\left (m + 9\right )} \cosh \left (b x + a\right )\right )} \cosh \left (m \log \left (\cosh \left (b x + a\right )\right )\right ) + {\left ({\left (m + 1\right )} \cosh \left (b x + a\right )^{3} + 3 \, {\left (m + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} - {\left (m + 9\right )} \cosh \left (b x + a\right )\right )} \sinh \left (m \log \left (\cosh \left (b x + a\right )\right )\right )}{4 \, {\left ({\left (b m^{2} + 4 \, b m + 3 \, b\right )} \cosh \left (b x + a\right )^{4} - 2 \, {\left (b m^{2} + 4 \, b m + 3 \, b\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + {\left (b m^{2} + 4 \, b m + 3 \, b\right )} \sinh \left (b x + a\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(((m + 1)*cosh(b*x + a)^3 + 3*(m + 1)*cosh(b*x + a)*sinh(b*x + a)^2 - (m + 9)*cosh(b*x + a))*cosh(m*log(co
sh(b*x + a))) + ((m + 1)*cosh(b*x + a)^3 + 3*(m + 1)*cosh(b*x + a)*sinh(b*x + a)^2 - (m + 9)*cosh(b*x + a))*si
nh(m*log(cosh(b*x + a))))/((b*m^2 + 4*b*m + 3*b)*cosh(b*x + a)^4 - 2*(b*m^2 + 4*b*m + 3*b)*cosh(b*x + a)^2*sin
h(b*x + a)^2 + (b*m^2 + 4*b*m + 3*b)*sinh(b*x + a)^4)

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giac [B]  time = 0.35, size = 325, normalized size = 8.12 \[ \frac {m e^{\left (7 \, b x + m \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-b x - a\right )}\right ) + 7 \, a\right )} - m e^{\left (5 \, b x + m \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-b x - a\right )}\right ) + 5 \, a\right )} - m e^{\left (3 \, b x + m \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-b x - a\right )}\right ) + 3 \, a\right )} + m e^{\left (b x + m \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-b x - a\right )}\right ) + a\right )} + e^{\left (7 \, b x + m \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-b x - a\right )}\right ) + 7 \, a\right )} - 9 \, e^{\left (5 \, b x + m \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-b x - a\right )}\right ) + 5 \, a\right )} - 9 \, e^{\left (3 \, b x + m \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-b x - a\right )}\right ) + 3 \, a\right )} + e^{\left (b x + m \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} + 1\right )} e^{\left (-b x - a\right )}\right ) + a\right )}}{8 \, {\left (b m^{2} e^{\left (4 \, b x + 4 \, a\right )} + 4 \, b m e^{\left (4 \, b x + 4 \, a\right )} + 3 \, b e^{\left (4 \, b x + 4 \, a\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(m*e^(7*b*x + m*log(1/2*(e^(2*b*x + 2*a) + 1)*e^(-b*x - a)) + 7*a) - m*e^(5*b*x + m*log(1/2*(e^(2*b*x + 2*
a) + 1)*e^(-b*x - a)) + 5*a) - m*e^(3*b*x + m*log(1/2*(e^(2*b*x + 2*a) + 1)*e^(-b*x - a)) + 3*a) + m*e^(b*x +
m*log(1/2*(e^(2*b*x + 2*a) + 1)*e^(-b*x - a)) + a) + e^(7*b*x + m*log(1/2*(e^(2*b*x + 2*a) + 1)*e^(-b*x - a))
+ 7*a) - 9*e^(5*b*x + m*log(1/2*(e^(2*b*x + 2*a) + 1)*e^(-b*x - a)) + 5*a) - 9*e^(3*b*x + m*log(1/2*(e^(2*b*x
+ 2*a) + 1)*e^(-b*x - a)) + 3*a) + e^(b*x + m*log(1/2*(e^(2*b*x + 2*a) + 1)*e^(-b*x - a)) + a))/(b*m^2*e^(4*b*
x + 4*a) + 4*b*m*e^(4*b*x + 4*a) + 3*b*e^(4*b*x + 4*a))

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maple [C]  time = 1.02, size = 923, normalized size = 23.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(b*x+a)^m*sinh(b*x+a)^3,x)

[Out]

1/8/b/(3+m)*(1+exp(2*b*x+2*a))^m*(1/2)^m*exp(b*x+a)^(-m)*exp(-3*b*x-3*a)*exp(-1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2
*a)))*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a))))*exp(1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*cs
gn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(1/2*I*Pi*m*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a
)))^2)*exp(-1/2*I*Pi*m*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^3)+1/8*exp(b*x+a)^(-m)*(1/2)^m*(1+exp(2*b*x+2*a)
)^m/b/(3+m)*exp(3*b*x+3*a)*exp(-1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(
1+exp(2*b*x+2*a))))*exp(1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(1/
2*I*Pi*m*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(-1/2*I*Pi*m*csgn(I*exp(-b*x-a)*(1+e
xp(2*b*x+2*a)))^3)-1/8*exp(b*x+a)^(-m)*(1/2)^m*(1+exp(2*b*x+2*a))^m*(m+9)/b/(m^2+4*m+3)*exp(-b*x-a)*exp(-1/2*I
*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a))))*exp(1/2*I*Pi*m*cs
gn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(1/2*I*Pi*m*csgn(I*exp(-b*x-a))*csgn(I*e
xp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(-1/2*I*Pi*m*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^3)-1/8*exp(b*x+a)^(-m
)*(1/2)^m*(1+exp(2*b*x+2*a))^m*(m+9)/b/(m^2+4*m+3)*exp(b*x+a)*exp(-1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(
I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a))))*exp(1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b
*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(1/2*I*Pi*m*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(
-1/2*I*Pi*m*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^3)

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maxima [B]  time = 0.58, size = 293, normalized size = 7.32 \[ \frac {m e^{\left ({\left (b x + a\right )} m + 3 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + 3 \, a\right )}}{8 \, {\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} - \frac {{\left (m + 9\right )} e^{\left ({\left (b x + a\right )} m + b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + a\right )}}{8 \, {\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} - \frac {{\left (m + 9\right )} e^{\left ({\left (b x + a\right )} m - b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - a\right )}}{8 \, {\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} + \frac {{\left (m + 1\right )} e^{\left ({\left (b x + a\right )} m - 3 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - 3 \, a\right )}}{8 \, {\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} + \frac {e^{\left ({\left (b x + a\right )} m + 3 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + 3 \, a\right )}}{8 \, {\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*m*e^((b*x + a)*m + 3*b*x + m*log(e^(-2*b*x - 2*a) + 1) + 3*a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b) - 1/8*(m
+ 9)*e^((b*x + a)*m + b*x + m*log(e^(-2*b*x - 2*a) + 1) + a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b) - 1/8*(m + 9
)*e^((b*x + a)*m - b*x + m*log(e^(-2*b*x - 2*a) + 1) - a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b) + 1/8*(m + 1)*e^
((b*x + a)*m - 3*b*x + m*log(e^(-2*b*x - 2*a) + 1) - 3*a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b) + 1/8*e^((b*x +
a)*m + 3*b*x + m*log(e^(-2*b*x - 2*a) + 1) + 3*a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b)

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mupad [B]  time = 0.18, size = 132, normalized size = 3.30 \[ {\left (\frac {1}{2}\right )}^m\,{\mathrm {e}}^{-3\,a-3\,b\,x}\,{\left ({\mathrm {e}}^{a+b\,x}+{\mathrm {e}}^{-a-b\,x}\right )}^m\,\left (\frac {\frac {m}{8}+\frac {1}{8}}{b\,\left (m^2+4\,m+3\right )}-\frac {{\mathrm {e}}^{2\,a+2\,b\,x}\,\left (m+9\right )}{8\,b\,\left (m^2+4\,m+3\right )}+\frac {{\mathrm {e}}^{6\,a+6\,b\,x}\,\left (m+1\right )}{8\,b\,\left (m^2+4\,m+3\right )}-\frac {{\mathrm {e}}^{4\,a+4\,b\,x}\,\left (m+9\right )}{8\,b\,\left (m^2+4\,m+3\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cosh(a + b*x)^m*sinh(a + b*x)^3,x)

[Out]

(1/2)^m*exp(- 3*a - 3*b*x)*(exp(a + b*x) + exp(- a - b*x))^m*((m/8 + 1/8)/(b*(4*m + m^2 + 3)) - (exp(2*a + 2*b
*x)*(m + 9))/(8*b*(4*m + m^2 + 3)) + (exp(6*a + 6*b*x)*(m + 1))/(8*b*(4*m + m^2 + 3)) - (exp(4*a + 4*b*x)*(m +
9))/(8*b*(4*m + m^2 + 3)))

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sympy [A]  time = 8.55, size = 648, normalized size = 16.20 \[ \begin {cases} x \sinh ^{3}{\relax (a )} \cosh ^{m}{\relax (a )} & \text {for}\: b = 0 \\\frac {\log {\left (\cosh {\left (a + b x \right )} \right )}}{b} - \frac {\sinh ^{2}{\left (a + b x \right )}}{2 b \cosh ^{2}{\left (a + b x \right )}} & \text {for}\: m = -3 \\- \frac {b x \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {2 b x \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {b x}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {2 \log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )} \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {4 \log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )} \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {2 \log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {\log {\left (\tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )} \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {2 \log {\left (\tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )} \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {\log {\left (\tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {2 \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} & \text {for}\: m = -1 \\\frac {m \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )} \cosh ^{m}{\left (a + b x \right )}}{b m^{2} + 4 b m + 3 b} + \frac {3 \sinh ^{2}{\left (a + b x \right )} \cosh {\left (a + b x \right )} \cosh ^{m}{\left (a + b x \right )}}{b m^{2} + 4 b m + 3 b} - \frac {2 \cosh ^{3}{\left (a + b x \right )} \cosh ^{m}{\left (a + b x \right )}}{b m^{2} + 4 b m + 3 b} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cosh(b*x+a)**m*sinh(b*x+a)**3,x)

[Out]

Piecewise((x*sinh(a)**3*cosh(a)**m, Eq(b, 0)), (log(cosh(a + b*x))/b - sinh(a + b*x)**2/(2*b*cosh(a + b*x)**2)
, Eq(m, -3)), (-b*x*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + 2*b*x*tanh(
a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - b*x/(b*tanh(a/2 + b*x/2)**4 - 2*b*ta
nh(a/2 + b*x/2)**2 + b) + 2*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh
(a/2 + b*x/2)**2 + b) - 4*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a
/2 + b*x/2)**2 + b) + 2*log(tanh(a/2 + b*x/2) + 1)/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - l
og(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + 2*
log(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - l
og(tanh(a/2 + b*x/2)**2 + 1)/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + 2*tanh(a/2 + b*x/2)**2/
(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b), Eq(m, -1)), (m*sinh(a + b*x)**2*cosh(a + b*x)*cosh(a
+ b*x)**m/(b*m**2 + 4*b*m + 3*b) + 3*sinh(a + b*x)**2*cosh(a + b*x)*cosh(a + b*x)**m/(b*m**2 + 4*b*m + 3*b) -
2*cosh(a + b*x)**3*cosh(a + b*x)**m/(b*m**2 + 4*b*m + 3*b), True))

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