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]

((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]))

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Rubi [A]  time = 0.0482171, antiderivative size = 56, normalized size of antiderivative = 1., 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 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 = 0.0758902, 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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Maple [A]  time = 0.009, size = 38, normalized size = 0.7 \begin{align*}{\frac{A-B}{20} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{5}}-{\frac{A}{6} \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{3}}+{\frac{A}{4}\tanh \left ({\frac{x}{2}} \right ) }+{\frac{B}{4}\tanh \left ({\frac{x}{2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(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 = 1.04365, size = 355, normalized size = 6.34 \begin{align*} \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 )} \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]

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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Fricas [B]  time = 2.10774, size = 427, normalized size = 7.62 \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 = 1.81946, size = 46, normalized size = 0.82 \begin{align*} \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} \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*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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Giac [A]  time = 1.17739, size = 62, normalized size = 1.11 \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