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

Optimal. Leaf size=56 \[ -\frac {3 \sinh (c+d x)}{16 d (3 \cosh (c+d x)+5)}-\frac {5 \tanh ^{-1}\left (\frac {\sinh (c+d x)}{\cosh (c+d x)+3}\right )}{32 d}+\frac {5 x}{64} \]

[Out]

5/64*x-5/32*arctanh(sinh(d*x+c)/(3+cosh(d*x+c)))/d-3/16*sinh(d*x+c)/d/(5+3*cosh(d*x+c))

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Rubi [A]  time = 0.04, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2664, 12, 2657} \[ -\frac {3 \sinh (c+d x)}{16 d (3 \cosh (c+d x)+5)}-\frac {5 \tanh ^{-1}\left (\frac {\sinh (c+d x)}{\cosh (c+d x)+3}\right )}{32 d}+\frac {5 x}{64} \]

Antiderivative was successfully verified.

[In]

Int[(5 + 3*Cosh[c + d*x])^(-2),x]

[Out]

(5*x)/64 - (5*ArcTanh[Sinh[c + d*x]/(3 + Cosh[c + d*x])])/(32*d) - (3*Sinh[c + d*x])/(16*d*(5 + 3*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 2657

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

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}{(5+3 \cosh (c+d x))^2} \, dx &=-\frac {3 \sinh (c+d x)}{16 d (5+3 \cosh (c+d x))}-\frac {1}{16} \int -\frac {5}{5+3 \cosh (c+d x)} \, dx\\ &=-\frac {3 \sinh (c+d x)}{16 d (5+3 \cosh (c+d x))}+\frac {5}{16} \int \frac {1}{5+3 \cosh (c+d x)} \, dx\\ &=\frac {5 x}{64}-\frac {5 \tanh ^{-1}\left (\frac {\sinh (c+d x)}{3+\cosh (c+d x)}\right )}{32 d}-\frac {3 \sinh (c+d x)}{16 d (5+3 \cosh (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 45, normalized size = 0.80 \[ \frac {5 \tanh ^{-1}\left (\frac {1}{2} \tanh \left (\frac {1}{2} (c+d x)\right )\right )-\frac {6 \sinh (c+d x)}{3 \cosh (c+d x)+5}}{32 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTanh[Tanh[(c + d*x)/2]/2] - (6*Sinh[c + d*x])/(5 + 3*Cosh[c + d*x]))/(32*d)

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fricas [B]  time = 1.03, size = 212, normalized size = 3.79 \[ \frac {5 \, {\left (3 \, \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, \cosh \left (d x + c\right ) + 5\right )} \sinh \left (d x + c\right ) + 3 \, \sinh \left (d x + c\right )^{2} + 10 \, \cosh \left (d x + c\right ) + 3\right )} \log \left (3 \, \cosh \left (d x + c\right ) + 3 \, \sinh \left (d x + c\right ) + 1\right ) - 5 \, {\left (3 \, \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, \cosh \left (d x + c\right ) + 5\right )} \sinh \left (d x + c\right ) + 3 \, \sinh \left (d x + c\right )^{2} + 10 \, \cosh \left (d x + c\right ) + 3\right )} \log \left (\cosh \left (d x + c\right ) + \sinh \left (d x + c\right ) + 3\right ) + 40 \, \cosh \left (d x + c\right ) + 40 \, \sinh \left (d x + c\right ) + 24}{64 \, {\left (3 \, d \cosh \left (d x + c\right )^{2} + 3 \, d \sinh \left (d x + c\right )^{2} + 10 \, d \cosh \left (d x + c\right ) + 2 \, {\left (3 \, d \cosh \left (d x + c\right ) + 5 \, d\right )} \sinh \left (d x + c\right ) + 3 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(5*(3*cosh(d*x + c)^2 + 2*(3*cosh(d*x + c) + 5)*sinh(d*x + c) + 3*sinh(d*x + c)^2 + 10*cosh(d*x + c) + 3)
*log(3*cosh(d*x + c) + 3*sinh(d*x + c) + 1) - 5*(3*cosh(d*x + c)^2 + 2*(3*cosh(d*x + c) + 5)*sinh(d*x + c) + 3
*sinh(d*x + c)^2 + 10*cosh(d*x + c) + 3)*log(cosh(d*x + c) + sinh(d*x + c) + 3) + 40*cosh(d*x + c) + 40*sinh(d
*x + c) + 24)/(3*d*cosh(d*x + c)^2 + 3*d*sinh(d*x + c)^2 + 10*d*cosh(d*x + c) + 2*(3*d*cosh(d*x + c) + 5*d)*si
nh(d*x + c) + 3*d)

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giac [A]  time = 0.14, size = 65, normalized size = 1.16 \[ \frac {\frac {8 \, {\left (5 \, e^{\left (d x + c\right )} + 3\right )}}{3 \, e^{\left (2 \, d x + 2 \, c\right )} + 10 \, e^{\left (d x + c\right )} + 3} + 5 \, \log \left (3 \, e^{\left (d x + c\right )} + 1\right ) - 5 \, \log \left (e^{\left (d x + c\right )} + 3\right )}{64 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(8*(5*e^(d*x + c) + 3)/(3*e^(2*d*x + 2*c) + 10*e^(d*x + c) + 3) + 5*log(3*e^(d*x + c) + 1) - 5*log(e^(d*x
+ c) + 3))/d

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maple [A]  time = 0.07, size = 72, normalized size = 1.29 \[ \frac {3}{32 d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}+\frac {5 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+2\right )}{64 d}+\frac {3}{32 d \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}-\frac {5 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-2\right )}{64 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/32/d/(tanh(1/2*d*x+1/2*c)+2)+5/64/d*ln(tanh(1/2*d*x+1/2*c)+2)+3/32/d/(tanh(1/2*d*x+1/2*c)-2)-5/64/d*ln(tanh(
1/2*d*x+1/2*c)-2)

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maxima [A]  time = 0.31, size = 81, normalized size = 1.45 \[ -\frac {5 \, \log \left (3 \, e^{\left (-d x - c\right )} + 1\right )}{64 \, d} + \frac {5 \, \log \left (e^{\left (-d x - c\right )} + 3\right )}{64 \, d} - \frac {5 \, e^{\left (-d x - c\right )} + 3}{8 \, d {\left (10 \, e^{\left (-d x - c\right )} + 3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/64*log(3*e^(-d*x - c) + 1)/d + 5/64*log(e^(-d*x - c) + 3)/d - 1/8*(5*e^(-d*x - c) + 3)/(d*(10*e^(-d*x - c)
+ 3*e^(-2*d*x - 2*c) + 3))

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mupad [B]  time = 0.94, size = 77, normalized size = 1.38 \[ \frac {\frac {5\,{\mathrm {e}}^{c+d\,x}}{8\,d}+\frac {3}{8\,d}}{10\,{\mathrm {e}}^{c+d\,x}+3\,{\mathrm {e}}^{2\,c+2\,d\,x}+3}-\frac {5\,\mathrm {atan}\left (\left (\frac {5}{4\,d}+\frac {3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4\,d}\right )\,\sqrt {-d^2}\right )}{32\,\sqrt {-d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*cosh(c + d*x) + 5)^2,x)

[Out]

((5*exp(c + d*x))/(8*d) + 3/(8*d))/(10*exp(c + d*x) + 3*exp(2*c + 2*d*x) + 3) - (5*atan((5/(4*d) + (3*exp(d*x)
*exp(c))/(4*d))*(-d^2)^(1/2)))/(32*(-d^2)^(1/2))

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sympy [A]  time = 1.65, size = 199, normalized size = 3.55 \[ \begin {cases} - \frac {5 \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )} \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} + \frac {20 \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 2 \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} + \frac {5 \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )} \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} - \frac {20 \log {\left (\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2 \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} + \frac {12 \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{64 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 256 d} & \text {for}\: d \neq 0 \\\frac {x}{\left (3 \cosh {\relax (c )} + 5\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-5*log(tanh(c/2 + d*x/2) - 2)*tanh(c/2 + d*x/2)**2/(64*d*tanh(c/2 + d*x/2)**2 - 256*d) + 20*log(tan
h(c/2 + d*x/2) - 2)/(64*d*tanh(c/2 + d*x/2)**2 - 256*d) + 5*log(tanh(c/2 + d*x/2) + 2)*tanh(c/2 + d*x/2)**2/(6
4*d*tanh(c/2 + d*x/2)**2 - 256*d) - 20*log(tanh(c/2 + d*x/2) + 2)/(64*d*tanh(c/2 + d*x/2)**2 - 256*d) + 12*tan
h(c/2 + d*x/2)/(64*d*tanh(c/2 + d*x/2)**2 - 256*d), Ne(d, 0)), (x/(3*cosh(c) + 5)**2, True))

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