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

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Rubi [A]  time = 0.0316992, antiderivative size = 48, normalized size of antiderivative = 1., 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 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]

Rule 12

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

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 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])

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*}

Mathematica [A]  time = 0.100258, 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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Maple [A]  time = 0.015, size = 48, normalized size = 1. \begin{align*}{\frac{5}{16\,d}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+4 \right ) ^{-1}}-{\frac{3}{32\,d}\arctan \left ({\frac{1}{2}\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) } \right ) } \end{align*}

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 = 1.53041, size = 86, normalized size = 1.79 \begin{align*} \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 )}} \end{align*}

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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Fricas [B]  time = 2.09781, size = 435, normalized size = 9.06 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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Giac [A]  time = 1.15567, size = 74, normalized size = 1.54 \begin{align*} -\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 (5 \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} + 5\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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