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

Optimal. Leaf size=56 \[ \frac {(2 A+3 B) \sinh (x)}{15 (\cosh (x)+1)}+\frac {(2 A+3 B) \sinh (x)}{15 (\cosh (x)+1)^2}+\frac {(A-B) \sinh (x)}{5 (\cosh (x)+1)^3} \]

[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 = 0.05, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2750, 2650, 2648} \[ \frac {(2 A+3 B) \sinh (x)}{15 (\cosh (x)+1)}+\frac {(2 A+3 B) \sinh (x)}{15 (\cosh (x)+1)^2}+\frac {(A-B) \sinh (x)}{5 (\cosh (x)+1)^3} \]

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])/(15
*(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 = 0.10, size = 42, normalized size = 0.75 \[ \frac {\sinh (x) (6 (2 A+3 B) \cosh (x)+(2 A+3 B) \cosh (2 x)+16 A+9 B)}{30 (\cosh (x)+1)^3} \]

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 = 1.10, size = 127, normalized size = 2.27 \[ -\frac {2 \, {\left (15 \, B \cosh \relax (x)^{2} + 15 \, B \sinh \relax (x)^{2} + 2 \, {\left (11 \, A + 9 \, B\right )} \cosh \relax (x) + 6 \, {\left (5 \, B \cosh \relax (x) + 3 \, A + 2 \, B\right )} \sinh \relax (x) + 10 \, A + 15 \, B\right )}}{15 \, {\left (\cosh \relax (x)^{4} + {\left (4 \, \cosh \relax (x) + 5\right )} \sinh \relax (x)^{3} + \sinh \relax (x)^{4} + 5 \, \cosh \relax (x)^{3} + {\left (6 \, \cosh \relax (x)^{2} + 15 \, \cosh \relax (x) + 10\right )} \sinh \relax (x)^{2} + 10 \, \cosh \relax (x)^{2} + {\left (4 \, \cosh \relax (x)^{3} + 15 \, \cosh \relax (x)^{2} + 20 \, \cosh \relax (x) + 9\right )} \sinh \relax (x) + 11 \, \cosh \relax (x) + 5\right )}} \]

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 = 0.12, size = 46, normalized size = 0.82 \[ -\frac {2 \, {\left (15 \, B e^{\left (3 \, x\right )} + 20 \, A e^{\left (2 \, x\right )} + 15 \, B e^{\left (2 \, x\right )} + 10 \, A e^{x} + 15 \, B e^{x} + 2 \, A + 3 \, B\right )}}{15 \, {\left (e^{x} + 1\right )}^{5}} \]

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 = 0.04, size = 38, normalized size = 0.68 \[ \frac {\left (A -B \right ) \left (\tanh ^{5}\left (\frac {x}{2}\right )\right )}{20}-\frac {A \left (\tanh ^{3}\left (\frac {x}{2}\right )\right )}{6}+\frac {A \tanh \left (\frac {x}{2}\right )}{4}+\frac {B \tanh \left (\frac {x}{2}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.34, size = 263, normalized size = 4.70 \[ \frac {4}{15} \, A {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {10 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {1}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1}\right )} + \frac {2}{5} \, B {\left (\frac {5 \, e^{\left (-x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {5 \, e^{\left (-2 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {5 \, e^{\left (-3 \, x\right )}}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1} + \frac {1}{5 \, e^{\left (-x\right )} + 10 \, e^{\left (-2 \, x\right )} + 10 \, e^{\left (-3 \, x\right )} + 5 \, e^{\left (-4 \, x\right )} + e^{\left (-5 \, x\right )} + 1}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)) + 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) + 10*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))

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mupad [B]  time = 0.92, size = 141, normalized size = 2.52 \[ -\frac {\frac {4\,B\,{\mathrm {e}}^x}{5}+\frac {8\,A\,{\mathrm {e}}^{2\,x}}{5}+\frac {4\,B\,{\mathrm {e}}^{3\,x}}{5}}{10\,{\mathrm {e}}^{2\,x}+10\,{\mathrm {e}}^{3\,x}+5\,{\mathrm {e}}^{4\,x}+{\mathrm {e}}^{5\,x}+5\,{\mathrm {e}}^x+1}-\frac {\frac {B}{5}+\frac {4\,A\,{\mathrm {e}}^x}{5}+\frac {3\,B\,{\mathrm {e}}^{2\,x}}{5}}{6\,{\mathrm {e}}^{2\,x}+4\,{\mathrm {e}}^{3\,x}+{\mathrm {e}}^{4\,x}+4\,{\mathrm {e}}^x+1}-\frac {\frac {4\,A}{15}+\frac {2\,B\,{\mathrm {e}}^x}{5}}{3\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^{3\,x}+3\,{\mathrm {e}}^x+1}-\frac {B}{5\,\left ({\mathrm {e}}^{2\,x}+2\,{\mathrm {e}}^x+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((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 + (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) - B/(5*(exp(2*x) + 2*exp(x) + 1))

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sympy [A]  time = 1.20, size = 46, normalized size = 0.82 \[ \frac {A \tanh ^{5}{\left (\frac {x}{2} \right )}}{20} - \frac {A \tanh ^{3}{\left (\frac {x}{2} \right )}}{6} + \frac {A \tanh {\left (\frac {x}{2} \right )}}{4} - \frac {B \tanh ^{5}{\left (\frac {x}{2} \right )}}{20} + \frac {B \tanh {\left (\frac {x}{2} \right )}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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