3.72 \(\int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx\)

Optimal. Leaf size=48 \[ \frac {5 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}-\frac {3 \tan ^{-1}\left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{32 d} \]

[Out]

-3/32*arctan(1/2*tanh(1/2*d*x+1/2*c))/d+5/16*sinh(d*x+c)/d/(3+5*cosh(d*x+c))

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Rubi [A]  time = 0.03, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2664, 12, 2659, 206} \[ \frac {5 \sinh (c+d x)}{16 d (5 \cosh (c+d x)+3)}-\frac {3 \tan ^{-1}\left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Cosh[c + d*x])^(-2),x]

[Out]

(-3*ArcTan[Tanh[(c + d*x)/2]/2])/(32*d) + (5*Sinh[c + d*x])/(16*d*(3 + 5*Cosh[c + d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 2664

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

Rubi steps

\begin {align*} \int \frac {1}{(3+5 \cosh (c+d x))^2} \, dx &=\frac {5 \sinh (c+d x)}{16 d (3+5 \cosh (c+d x))}+\frac {1}{16} \int -\frac {3}{3+5 \cosh (c+d x)} \, dx\\ &=\frac {5 \sinh (c+d x)}{16 d (3+5 \cosh (c+d x))}-\frac {3}{16} \int \frac {1}{3+5 \cosh (c+d x)} \, dx\\ &=\frac {5 \sinh (c+d x)}{16 d (3+5 \cosh (c+d x))}+\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{8-2 x^2} \, dx,x,\tan \left (\frac {1}{2} (i c+i d x)\right )\right )}{8 d}\\ &=-\frac {3 \tan ^{-1}\left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{32 d}+\frac {5 \sinh (c+d x)}{16 d (3+5 \cosh (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 45, normalized size = 0.94 \[ \frac {\frac {10 \sinh (c+d x)}{5 \cosh (c+d x)+3}-3 \tan ^{-1}\left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )}{32 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(3 + 5*Cosh[c + d*x])^(-2),x]

[Out]

(-3*ArcTan[Tanh[(c + d*x)/2]/2] + (10*Sinh[c + d*x])/(3 + 5*Cosh[c + d*x]))/(32*d)

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fricas [B]  time = 0.60, size = 147, normalized size = 3.06 \[ -\frac {3 \, {\left (5 \, \cosh \left (d x + c\right )^{2} + 2 \, {\left (5 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right ) + 5 \, \sinh \left (d x + c\right )^{2} + 6 \, \cosh \left (d x + c\right ) + 5\right )} \arctan \left (\frac {5}{4} \, \cosh \left (d x + c\right ) + \frac {5}{4} \, \sinh \left (d x + c\right ) + \frac {3}{4}\right ) + 12 \, \cosh \left (d x + c\right ) + 12 \, \sinh \left (d x + c\right ) + 20}{32 \, {\left (5 \, d \cosh \left (d x + c\right )^{2} + 5 \, d \sinh \left (d x + c\right )^{2} + 6 \, d \cosh \left (d x + c\right ) + 2 \, {\left (5 \, d \cosh \left (d x + c\right ) + 3 \, d\right )} \sinh \left (d x + c\right ) + 5 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*cosh(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/32*(3*(5*cosh(d*x + c)^2 + 2*(5*cosh(d*x + c) + 3)*sinh(d*x + c) + 5*sinh(d*x + c)^2 + 6*cosh(d*x + c) + 5)
*arctan(5/4*cosh(d*x + c) + 5/4*sinh(d*x + c) + 3/4) + 12*cosh(d*x + c) + 12*sinh(d*x + c) + 20)/(5*d*cosh(d*x
+ c)^2 + 5*d*sinh(d*x + c)^2 + 6*d*cosh(d*x + c) + 2*(5*d*cosh(d*x + c) + 3*d)*sinh(d*x + c) + 5*d)

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giac [A]  time = 0.12, size = 54, normalized size = 1.12 \[ -\frac {\frac {4 \, {\left (3 \, e^{\left (d x + c\right )} + 5\right )}}{5 \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} + 5} + 3 \, \arctan \left (\frac {5}{4} \, e^{\left (d x + c\right )} + \frac {3}{4}\right )}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*cosh(d*x+c))^2,x, algorithm="giac")

[Out]

-1/32*(4*(3*e^(d*x + c) + 5)/(5*e^(2*d*x + 2*c) + 6*e^(d*x + c) + 5) + 3*arctan(5/4*e^(d*x + c) + 3/4))/d

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maple [A]  time = 0.07, size = 48, normalized size = 1.00 \[ \frac {5 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )}-\frac {3 \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{32 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3+5*cosh(d*x+c))^2,x)

[Out]

5/16/d*tanh(1/2*d*x+1/2*c)/(tanh(1/2*d*x+1/2*c)^2+4)-3/32*arctan(1/2*tanh(1/2*d*x+1/2*c))/d

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maxima [A]  time = 0.40, size = 64, normalized size = 1.33 \[ \frac {3 \, \arctan \left (\frac {5}{4} \, e^{\left (-d x - c\right )} + \frac {3}{4}\right )}{32 \, d} + \frac {3 \, e^{\left (-d x - c\right )} + 5}{8 \, d {\left (6 \, e^{\left (-d x - c\right )} + 5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 5\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*cosh(d*x+c))^2,x, algorithm="maxima")

[Out]

3/32*arctan(5/4*e^(-d*x - c) + 3/4)/d + 1/8*(3*e^(-d*x - c) + 5)/(d*(6*e^(-d*x - c) + 5*e^(-2*d*x - 2*c) + 5))

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mupad [B]  time = 0.94, size = 74, normalized size = 1.54 \[ -\frac {\frac {3\,{\mathrm {e}}^{c+d\,x}}{8\,d}+\frac {5}{8\,d}}{6\,{\mathrm {e}}^{c+d\,x}+5\,{\mathrm {e}}^{2\,c+2\,d\,x}+5}-\frac {3\,\mathrm {atan}\left (\left (\frac {3}{4\,d}+\frac {5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4\,d}\right )\,\sqrt {d^2}\right )}{32\,\sqrt {d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*cosh(c + d*x) + 3)^2,x)

[Out]

- ((3*exp(c + d*x))/(8*d) + 5/(8*d))/(6*exp(c + d*x) + 5*exp(2*c + 2*d*x) + 5) - (3*atan((3/(4*d) + (5*exp(d*x
)*exp(c))/(4*d))*(d^2)^(1/2)))/(32*(d^2)^(1/2))

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sympy [A]  time = 2.52, size = 316, normalized size = 6.58 \[ \begin {cases} \frac {x}{\left (5 \cosh {\left (\log {\left (- \frac {3}{5} - \frac {4 i}{5} \right )} \right )} + 3\right )^{2}} & \text {for}\: c = \log {\left (- \frac {3}{5} - \frac {4 i}{5} \right )} \wedge d = 0 \\- \frac {\log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )}}{25 d \cosh ^{2}{\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 30 d \cosh {\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 9 d} + \frac {\log {\relax (5 )}}{25 d \cosh ^{2}{\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 30 d \cosh {\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 9 d} & \text {for}\: c = \log {\left (\frac {\left (-3 - 4 i\right ) e^{- d x}}{5} \right )} \\\frac {x}{\left (5 \cosh {\relax (c )} + 3\right )^{2}} & \text {for}\: d = 0 \\- \frac {3 \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} \operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{32 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 128 d} + \frac {10 \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 128 d} - \frac {12 \operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{32 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 128 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(3+5*cosh(d*x+c))**2,x)

[Out]

Piecewise((x/(5*cosh(log(-3/5 - 4*I/5)) + 3)**2, Eq(d, 0) & Eq(c, log(-3/5 - 4*I/5))), (-log(-3*exp(-d*x) - 4*
I*exp(-d*x))/(25*d*cosh(d*x + log(-3*exp(-d*x) - 4*I*exp(-d*x)) - log(5))**2 + 30*d*cosh(d*x + log(-3*exp(-d*x
) - 4*I*exp(-d*x)) - log(5)) + 9*d) + log(5)/(25*d*cosh(d*x + log(-3*exp(-d*x) - 4*I*exp(-d*x)) - log(5))**2 +
30*d*cosh(d*x + log(-3*exp(-d*x) - 4*I*exp(-d*x)) - log(5)) + 9*d), Eq(c, log((-3 - 4*I)*exp(-d*x)/5))), (x/(
5*cosh(c) + 3)**2, Eq(d, 0)), (-3*tanh(c/2 + d*x/2)**2*atan(tanh(c/2 + d*x/2)/2)/(32*d*tanh(c/2 + d*x/2)**2 +
128*d) + 10*tanh(c/2 + d*x/2)/(32*d*tanh(c/2 + d*x/2)**2 + 128*d) - 12*atan(tanh(c/2 + d*x/2)/2)/(32*d*tanh(c/
2 + d*x/2)**2 + 128*d), True))

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