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

Optimal. Leaf size=73 \[ \frac{43 \tan ^{-1}\left (\frac{1}{2} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{1024 d}-\frac{45 \sinh (c+d x)}{512 d (5 \cosh (c+d x)+3)}+\frac{5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2} \]

[Out]

(43*ArcTan[Tanh[(c + d*x)/2]/2])/(1024*d) + (5*Sinh[c + d*x])/(32*d*(3 + 5*Cosh[c + d*x])^2) - (45*Sinh[c + d*
x])/(512*d*(3 + 5*Cosh[c + d*x]))

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Rubi [A]  time = 0.0646426, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {2664, 2754, 12, 2659, 206} \[ \frac{43 \tan ^{-1}\left (\frac{1}{2} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{1024 d}-\frac{45 \sinh (c+d x)}{512 d (5 \cosh (c+d x)+3)}+\frac{5 \sinh (c+d x)}{32 d (5 \cosh (c+d x)+3)^2} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Cosh[c + d*x])^(-3),x]

[Out]

(43*ArcTan[Tanh[(c + d*x)/2]/2])/(1024*d) + (5*Sinh[c + d*x])/(32*d*(3 + 5*Cosh[c + d*x])^2) - (45*Sinh[c + d*
x])/(512*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 2754

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 + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

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))^3} \, dx &=\frac{5 \sinh (c+d x)}{32 d (3+5 \cosh (c+d x))^2}+\frac{1}{32} \int \frac{-6+5 \cosh (c+d x)}{(3+5 \cosh (c+d x))^2} \, dx\\ &=\frac{5 \sinh (c+d x)}{32 d (3+5 \cosh (c+d x))^2}-\frac{45 \sinh (c+d x)}{512 d (3+5 \cosh (c+d x))}+\frac{1}{512} \int \frac{43}{3+5 \cosh (c+d x)} \, dx\\ &=\frac{5 \sinh (c+d x)}{32 d (3+5 \cosh (c+d x))^2}-\frac{45 \sinh (c+d x)}{512 d (3+5 \cosh (c+d x))}+\frac{43}{512} \int \frac{1}{3+5 \cosh (c+d x)} \, dx\\ &=\frac{5 \sinh (c+d x)}{32 d (3+5 \cosh (c+d x))^2}-\frac{45 \sinh (c+d x)}{512 d (3+5 \cosh (c+d x))}-\frac{(43 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 )}{256 d}\\ &=\frac{43 \tan ^{-1}\left (\frac{1}{2} \tanh \left (\frac{1}{2} (c+d x)\right )\right )}{1024 d}+\frac{5 \sinh (c+d x)}{32 d (3+5 \cosh (c+d x))^2}-\frac{45 \sinh (c+d x)}{512 d (3+5 \cosh (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.152716, size = 55, normalized size = 0.75 \[ \frac{43 \tan ^{-1}\left (\frac{1}{2} \tanh \left (\frac{1}{2} (c+d x)\right )\right )-\frac{10 \sinh (c+d x) (45 \cosh (c+d x)+11)}{(5 \cosh (c+d x)+3)^2}}{1024 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(43*ArcTan[Tanh[(c + d*x)/2]/2] - (10*(11 + 45*Cosh[c + d*x])*Sinh[c + d*x])/(3 + 5*Cosh[c + d*x])^2)/(1024*d)

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Maple [A]  time = 0.016, size = 79, normalized size = 1.1 \begin{align*} -{\frac{85}{512\,d} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}+4 \right ) ^{-2}}-{\frac{35}{128\,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 ) ^{-2}}+{\frac{43}{1024\,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))^3,x)

[Out]

-85/512/d/(tanh(1/2*d*x+1/2*c)^2+4)^2*tanh(1/2*d*x+1/2*c)^3-35/128/d/(tanh(1/2*d*x+1/2*c)^2+4)^2*tanh(1/2*d*x+
1/2*c)+43/1024*arctan(1/2*tanh(1/2*d*x+1/2*c))/d

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Maxima [A]  time = 1.55, size = 146, normalized size = 2. \begin{align*} -\frac{43 \, \arctan \left (\frac{5}{4} \, e^{\left (-d x - c\right )} + \frac{3}{4}\right )}{1024 \, d} - \frac{325 \, e^{\left (-d x - c\right )} + 387 \, e^{\left (-2 \, d x - 2 \, c\right )} + 215 \, e^{\left (-3 \, d x - 3 \, c\right )} + 225}{256 \, d{\left (60 \, e^{\left (-d x - c\right )} + 86 \, e^{\left (-2 \, d x - 2 \, c\right )} + 60 \, e^{\left (-3 \, d x - 3 \, c\right )} + 25 \, e^{\left (-4 \, d x - 4 \, c\right )} + 25\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-43/1024*arctan(5/4*e^(-d*x - c) + 3/4)/d - 1/256*(325*e^(-d*x - c) + 387*e^(-2*d*x - 2*c) + 215*e^(-3*d*x - 3
*c) + 225)/(d*(60*e^(-d*x - c) + 86*e^(-2*d*x - 2*c) + 60*e^(-3*d*x - 3*c) + 25*e^(-4*d*x - 4*c) + 25))

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Fricas [B]  time = 2.21371, size = 1214, normalized size = 16.63 \begin{align*} \frac{860 \, \cosh \left (d x + c\right )^{3} + 516 \,{\left (5 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right )^{2} + 860 \, \sinh \left (d x + c\right )^{3} + 43 \,{\left (25 \, \cosh \left (d x + c\right )^{4} + 20 \,{\left (5 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right )^{3} + 25 \, \sinh \left (d x + c\right )^{4} + 60 \, \cosh \left (d x + c\right )^{3} + 2 \,{\left (75 \, \cosh \left (d x + c\right )^{2} + 90 \, \cosh \left (d x + c\right ) + 43\right )} \sinh \left (d x + c\right )^{2} + 86 \, \cosh \left (d x + c\right )^{2} + 4 \,{\left (25 \, \cosh \left (d x + c\right )^{3} + 45 \, \cosh \left (d x + c\right )^{2} + 43 \, \cosh \left (d x + c\right ) + 15\right )} \sinh \left (d x + c\right ) + 60 \, \cosh \left (d x + c\right ) + 25\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 ) + 1548 \, \cosh \left (d x + c\right )^{2} + 4 \,{\left (645 \, \cosh \left (d x + c\right )^{2} + 774 \, \cosh \left (d x + c\right ) + 325\right )} \sinh \left (d x + c\right ) + 1300 \, \cosh \left (d x + c\right ) + 900}{1024 \,{\left (25 \, d \cosh \left (d x + c\right )^{4} + 25 \, d \sinh \left (d x + c\right )^{4} + 60 \, d \cosh \left (d x + c\right )^{3} + 20 \,{\left (5 \, d \cosh \left (d x + c\right ) + 3 \, d\right )} \sinh \left (d x + c\right )^{3} + 86 \, d \cosh \left (d x + c\right )^{2} + 2 \,{\left (75 \, d \cosh \left (d x + c\right )^{2} + 90 \, d \cosh \left (d x + c\right ) + 43 \, d\right )} \sinh \left (d x + c\right )^{2} + 60 \, d \cosh \left (d x + c\right ) + 4 \,{\left (25 \, d \cosh \left (d x + c\right )^{3} + 45 \, d \cosh \left (d x + c\right )^{2} + 43 \, d \cosh \left (d x + c\right ) + 15 \, d\right )} \sinh \left (d x + c\right ) + 25 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1024*(860*cosh(d*x + c)^3 + 516*(5*cosh(d*x + c) + 3)*sinh(d*x + c)^2 + 860*sinh(d*x + c)^3 + 43*(25*cosh(d*
x + c)^4 + 20*(5*cosh(d*x + c) + 3)*sinh(d*x + c)^3 + 25*sinh(d*x + c)^4 + 60*cosh(d*x + c)^3 + 2*(75*cosh(d*x
+ c)^2 + 90*cosh(d*x + c) + 43)*sinh(d*x + c)^2 + 86*cosh(d*x + c)^2 + 4*(25*cosh(d*x + c)^3 + 45*cosh(d*x +
c)^2 + 43*cosh(d*x + c) + 15)*sinh(d*x + c) + 60*cosh(d*x + c) + 25)*arctan(5/4*cosh(d*x + c) + 5/4*sinh(d*x +
c) + 3/4) + 1548*cosh(d*x + c)^2 + 4*(645*cosh(d*x + c)^2 + 774*cosh(d*x + c) + 325)*sinh(d*x + c) + 1300*cos
h(d*x + c) + 900)/(25*d*cosh(d*x + c)^4 + 25*d*sinh(d*x + c)^4 + 60*d*cosh(d*x + c)^3 + 20*(5*d*cosh(d*x + c)
+ 3*d)*sinh(d*x + c)^3 + 86*d*cosh(d*x + c)^2 + 2*(75*d*cosh(d*x + c)^2 + 90*d*cosh(d*x + c) + 43*d)*sinh(d*x
+ c)^2 + 60*d*cosh(d*x + c) + 4*(25*d*cosh(d*x + c)^3 + 45*d*cosh(d*x + c)^2 + 43*d*cosh(d*x + c) + 15*d)*sinh
(d*x + c) + 25*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (5 \cosh{\left (c + d x \right )} + 3\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((5*cosh(c + d*x) + 3)**(-3), x)

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Giac [A]  time = 1.13102, size = 104, normalized size = 1.42 \begin{align*} \frac{43 \, \arctan \left (\frac{5}{4} \, e^{\left (d x + c\right )} + \frac{3}{4}\right )}{1024 \, d} + \frac{215 \, e^{\left (3 \, d x + 3 \, c\right )} + 387 \, e^{\left (2 \, d x + 2 \, c\right )} + 325 \, e^{\left (d x + c\right )} + 225}{256 \, d{\left (5 \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} + 5\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

43/1024*arctan(5/4*e^(d*x + c) + 3/4)/d + 1/256*(215*e^(3*d*x + 3*c) + 387*e^(2*d*x + 2*c) + 325*e^(d*x + c) +
225)/(d*(5*e^(2*d*x + 2*c) + 6*e^(d*x + c) + 5)^2)