3.99 \(\int \frac{A+B \cosh (x)}{(1-\cosh (x))^3} \, dx\)

Optimal. Leaf size=60 \[ -\frac{(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))}-\frac{(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))^2}-\frac{(A+B) \sinh (x)}{5 (1-\cosh (x))^3} \]

[Out]

-((A + B)*Sinh[x])/(5*(1 - Cosh[x])^3) - ((2*A - 3*B)*Sinh[x])/(15*(1 - Cosh[x])^2) - ((2*A - 3*B)*Sinh[x])/(1
5*(1 - Cosh[x]))

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Rubi [A]  time = 0.0550135, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2750, 2650, 2648} \[ -\frac{(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))}-\frac{(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))^2}-\frac{(A+B) \sinh (x)}{5 (1-\cosh (x))^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Cosh[x])/(1 - Cosh[x])^3,x]

[Out]

-((A + B)*Sinh[x])/(5*(1 - Cosh[x])^3) - ((2*A - 3*B)*Sinh[x])/(15*(1 - Cosh[x])^2) - ((2*A - 3*B)*Sinh[x])/(1
5*(1 - Cosh[x]))

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{A+B \cosh (x)}{(1-\cosh (x))^3} \, dx &=-\frac{(A+B) \sinh (x)}{5 (1-\cosh (x))^3}+\frac{1}{5} (2 A-3 B) \int \frac{1}{(1-\cosh (x))^2} \, dx\\ &=-\frac{(A+B) \sinh (x)}{5 (1-\cosh (x))^3}-\frac{(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))^2}+\frac{1}{15} (2 A-3 B) \int \frac{1}{1-\cosh (x)} \, dx\\ &=-\frac{(A+B) \sinh (x)}{5 (1-\cosh (x))^3}-\frac{(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))^2}-\frac{(2 A-3 B) \sinh (x)}{15 (1-\cosh (x))}\\ \end{align*}

Mathematica [A]  time = 0.0727499, size = 42, normalized size = 0.7 \[ \frac{\sinh (x) (-6 (2 A-3 B) \cosh (x)+(2 A-3 B) \cosh (2 x)+16 A-9 B)}{30 (\cosh (x)-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cosh[x])/(1 - Cosh[x])^3,x]

[Out]

((16*A - 9*B - 6*(2*A - 3*B)*Cosh[x] + (2*A - 3*B)*Cosh[2*x])*Sinh[x])/(30*(-1 + Cosh[x])^3)

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Maple [A]  time = 0.013, size = 39, normalized size = 0.7 \begin{align*} -{\frac{-A+B}{4} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{A}{6} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{-A-B}{20} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{-5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cosh(x))/(1-cosh(x))^3,x)

[Out]

-1/4*(-A+B)/tanh(1/2*x)-1/6*A/tanh(1/2*x)^3-1/20*(-A-B)/tanh(1/2*x)^5

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Maxima [B]  time = 1.0524, size = 360, normalized size = 6. \begin{align*} -\frac{2}{5} \, B{\left (\frac{5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac{5 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} + \frac{5 \, e^{\left (-3 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac{1}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1}\right )} + \frac{4}{15} \, A{\left (\frac{5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac{10 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac{1}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(1-cosh(x))^3,x, algorithm="maxima")

[Out]

-2/5*B*(5*e^(-x)/(5*e^(-x) - 10*e^(-2*x) + 10*e^(-3*x) - 5*e^(-4*x) + e^(-5*x) - 1) - 5*e^(-2*x)/(5*e^(-x) - 1
0*e^(-2*x) + 10*e^(-3*x) - 5*e^(-4*x) + e^(-5*x) - 1) + 5*e^(-3*x)/(5*e^(-x) - 10*e^(-2*x) + 10*e^(-3*x) - 5*e
^(-4*x) + e^(-5*x) - 1) - 1/(5*e^(-x) - 10*e^(-2*x) + 10*e^(-3*x) - 5*e^(-4*x) + e^(-5*x) - 1)) + 4/15*A*(5*e^
(-x)/(5*e^(-x) - 10*e^(-2*x) + 10*e^(-3*x) - 5*e^(-4*x) + e^(-5*x) - 1) - 10*e^(-2*x)/(5*e^(-x) - 10*e^(-2*x)
+ 10*e^(-3*x) - 5*e^(-4*x) + e^(-5*x) - 1) - 1/(5*e^(-x) - 10*e^(-2*x) + 10*e^(-3*x) - 5*e^(-4*x) + e^(-5*x) -
1))

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Fricas [B]  time = 2.03179, size = 425, normalized size = 7.08 \begin{align*} \frac{2 \,{\left (15 \, B \cosh \left (x\right )^{2} + 15 \, B \sinh \left (x\right )^{2} + 2 \,{\left (11 \, A - 9 \, B\right )} \cosh \left (x\right ) + 6 \,{\left (5 \, B \cosh \left (x\right ) + 3 \, A - 2 \, B\right )} \sinh \left (x\right ) - 10 \, A + 15 \, B\right )}}{15 \,{\left (\cosh \left (x\right )^{4} +{\left (4 \, \cosh \left (x\right ) - 5\right )} \sinh \left (x\right )^{3} + \sinh \left (x\right )^{4} - 5 \, \cosh \left (x\right )^{3} +{\left (6 \, \cosh \left (x\right )^{2} - 15 \, \cosh \left (x\right ) + 10\right )} \sinh \left (x\right )^{2} + 10 \, \cosh \left (x\right )^{2} +{\left (4 \, \cosh \left (x\right )^{3} - 15 \, \cosh \left (x\right )^{2} + 20 \, \cosh \left (x\right ) - 9\right )} \sinh \left (x\right ) - 11 \, \cosh \left (x\right ) + 5\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(1-cosh(x))^3,x, algorithm="fricas")

[Out]

2/15*(15*B*cosh(x)^2 + 15*B*sinh(x)^2 + 2*(11*A - 9*B)*cosh(x) + 6*(5*B*cosh(x) + 3*A - 2*B)*sinh(x) - 10*A +
15*B)/(cosh(x)^4 + (4*cosh(x) - 5)*sinh(x)^3 + sinh(x)^4 - 5*cosh(x)^3 + (6*cosh(x)^2 - 15*cosh(x) + 10)*sinh(
x)^2 + 10*cosh(x)^2 + (4*cosh(x)^3 - 15*cosh(x)^2 + 20*cosh(x) - 9)*sinh(x) - 11*cosh(x) + 5)

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Sympy [A]  time = 2.33447, size = 46, normalized size = 0.77 \begin{align*} \frac{A}{4 \tanh{\left (\frac{x}{2} \right )}} - \frac{A}{6 \tanh ^{3}{\left (\frac{x}{2} \right )}} + \frac{A}{20 \tanh ^{5}{\left (\frac{x}{2} \right )}} - \frac{B}{4 \tanh{\left (\frac{x}{2} \right )}} + \frac{B}{20 \tanh ^{5}{\left (\frac{x}{2} \right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(1-cosh(x))**3,x)

[Out]

A/(4*tanh(x/2)) - A/(6*tanh(x/2)**3) + A/(20*tanh(x/2)**5) - B/(4*tanh(x/2)) + B/(20*tanh(x/2)**5)

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Giac [A]  time = 1.179, size = 62, normalized size = 1.03 \begin{align*} \frac{2 \,{\left (15 \, B e^{\left (3 \, x\right )} + 20 \, A e^{\left (2 \, x\right )} - 15 \, B e^{\left (2 \, x\right )} - 10 \, A e^{x} + 15 \, B e^{x} + 2 \, A - 3 \, B\right )}}{15 \,{\left (e^{x} - 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cosh(x))/(1-cosh(x))^3,x, algorithm="giac")

[Out]

2/15*(15*B*e^(3*x) + 20*A*e^(2*x) - 15*B*e^(2*x) - 10*A*e^x + 15*B*e^x + 2*A - 3*B)/(e^x - 1)^5