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]

-1/5*(A+B)*sinh(x)/(1-cosh(x))^3-1/15*(2*A-3*B)*sinh(x)/(1-cosh(x))^2-1/15*(2*A-3*B)*sinh(x)/(1-cosh(x))

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Rubi [A]  time = 0.06, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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]

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 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)]

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*}

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Mathematica [A]  time = 0.09, size = 42, normalized size = 0.70 \[ \frac {\sinh (x) (-6 (2 A-3 B) \cosh (x)+(2 A-3 B) \cosh (2 x)+16 A-9 B)}{30 (\cosh (x)-1)^3} \]

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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fricas [B]  time = 1.49, size = 127, normalized size = 2.12 \[ \frac {2 \, {\left (15 \, B \cosh \relax (x)^{2} + 15 \, B \sinh \relax (x)^{2} + 2 \, {\left (11 \, A - 9 \, B\right )} \cosh \relax (x) + 6 \, {\left (5 \, B \cosh \relax (x) + 3 \, A - 2 \, B\right )} \sinh \relax (x) - 10 \, A + 15 \, B\right )}}{15 \, {\left (\cosh \relax (x)^{4} + {\left (4 \, \cosh \relax (x) - 5\right )} \sinh \relax (x)^{3} + \sinh \relax (x)^{4} - 5 \, \cosh \relax (x)^{3} + {\left (6 \, \cosh \relax (x)^{2} - 15 \, \cosh \relax (x) + 10\right )} \sinh \relax (x)^{2} + 10 \, \cosh \relax (x)^{2} + {\left (4 \, \cosh \relax (x)^{3} - 15 \, \cosh \relax (x)^{2} + 20 \, \cosh \relax (x) - 9\right )} \sinh \relax (x) - 11 \, \cosh \relax (x) + 5\right )}} \]

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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giac [A]  time = 0.11, size = 46, normalized size = 0.77 \[ \frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 20 \, A e^{\left (2 \, x\right )} - 15 \, B e^{\left (2 \, x\right )} - 10 \, A e^{x} + 15 \, B e^{x} + 2 \, A - 3 \, B\right )}}{15 \, {\left (e^{x} - 1\right )}^{5}} \]

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

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maple [A]  time = 0.06, size = 39, normalized size = 0.65 \[ -\frac {A}{6 \tanh \left (\frac {x}{2}\right )^{3}}-\frac {-A +B}{4 \tanh \left (\frac {x}{2}\right )}-\frac {-A -B}{20 \tanh \left (\frac {x}{2}\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.33, size = 267, normalized size = 4.45 \[ -\frac {2}{5} \, B {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac {5 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} + \frac {5 \, e^{\left (-3 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac {1}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1}\right )} + \frac {4}{15} \, A {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac {10 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1} - \frac {1}{5 \, e^{\left (-x\right )} - 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} - 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} - 1}\right )} \]

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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mupad [B]  time = 0.08, size = 143, normalized size = 2.38 \[ \frac {\frac {B}{5}+\frac {4\,A\,{\mathrm {e}}^x}{5}+\frac {3\,B\,{\mathrm {e}}^{2\,x}}{5}}{6\,{\mathrm {e}}^{2\,x}-4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}-4\,{\mathrm {e}}^x+1}-\frac {\frac {4\,A}{15}+\frac {2\,B\,{\mathrm {e}}^x}{5}}{3\,{\mathrm {e}}^{2\,x}-{\mathrm {e}}^{3\,x}-3\,{\mathrm {e}}^x+1}-\frac {\frac {4\,B\,{\mathrm {e}}^x}{5}+\frac {8\,A\,{\mathrm {e}}^{2\,x}}{5}+\frac {4\,B\,{\mathrm {e}}^{3\,x}}{5}}{10\,{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^{3\,x}+5\,{\mathrm {e}}^{4\,x}-{\mathrm {e}}^{5\,x}-5\,{\mathrm {e}}^x+1}+\frac {B}{5\,\left ({\mathrm {e}}^{2\,x}-2\,{\mathrm {e}}^x+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B/5 + (4*A*exp(x))/5 + (3*B*exp(2*x))/5)/(6*exp(2*x) - 4*exp(3*x) + exp(4*x) - 4*exp(x) + 1) - ((4*A)/15 + (2
*B*exp(x))/5)/(3*exp(2*x) - exp(3*x) - 3*exp(x) + 1) - ((4*B*exp(x))/5 + (8*A*exp(2*x))/5 + (4*B*exp(3*x))/5)/
(10*exp(2*x) - 10*exp(3*x) + 5*exp(4*x) - exp(5*x) - 5*exp(x) + 1) + B/(5*(exp(2*x) - 2*exp(x) + 1))

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sympy [A]  time = 1.46, size = 46, normalized size = 0.77 \[ \frac {A}{4 \tanh {\left (\frac {x}{2} \right )}} - \frac {A}{6 \tanh ^{3}{\left (\frac {x}{2} \right )}} + \frac {A}{20 \tanh ^{5}{\left (\frac {x}{2} \right )}} - \frac {B}{4 \tanh {\left (\frac {x}{2} \right )}} + \frac {B}{20 \tanh ^{5}{\left (\frac {x}{2} \right )}} \]

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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