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]

43/1024*arctan(1/2*tanh(1/2*d*x+1/2*c))/d+5/32*sinh(d*x+c)/d/(3+5*cosh(d*x+c))^2-45/512*sinh(d*x+c)/d/(3+5*cos
h(d*x+c))

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Rubi [A]  time = 0.06, antiderivative size = 73, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.16, 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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fricas [B]  time = 0.46, size = 408, normalized size = 5.59 \[ \frac {860 \, \cosh \left (d x + c\right )^{3} + 516 \, {\left (5 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right )^{2} + 860 \, \sinh \left (d x + c\right )^{3} + 43 \, {\left (25 \, \cosh \left (d x + c\right )^{4} + 20 \, {\left (5 \, \cosh \left (d x + c\right ) + 3\right )} \sinh \left (d x + c\right )^{3} + 25 \, \sinh \left (d x + c\right )^{4} + 60 \, \cosh \left (d x + c\right )^{3} + 2 \, {\left (75 \, \cosh \left (d x + c\right )^{2} + 90 \, \cosh \left (d x + c\right ) + 43\right )} \sinh \left (d x + c\right )^{2} + 86 \, \cosh \left (d x + c\right )^{2} + 4 \, {\left (25 \, \cosh \left (d x + c\right )^{3} + 45 \, \cosh \left (d x + c\right )^{2} + 43 \, \cosh \left (d x + c\right ) + 15\right )} \sinh \left (d x + c\right ) + 60 \, \cosh \left (d x + c\right ) + 25\right )} \arctan \left (\frac {5}{4} \, \cosh \left (d x + c\right ) + \frac {5}{4} \, \sinh \left (d x + c\right ) + \frac {3}{4}\right ) + 1548 \, \cosh \left (d x + c\right )^{2} + 4 \, {\left (645 \, \cosh \left (d x + c\right )^{2} + 774 \, \cosh \left (d x + c\right ) + 325\right )} \sinh \left (d x + c\right ) + 1300 \, \cosh \left (d x + c\right ) + 900}{1024 \, {\left (25 \, d \cosh \left (d x + c\right )^{4} + 25 \, d \sinh \left (d x + c\right )^{4} + 60 \, d \cosh \left (d x + c\right )^{3} + 20 \, {\left (5 \, d \cosh \left (d x + c\right ) + 3 \, d\right )} \sinh \left (d x + c\right )^{3} + 86 \, d \cosh \left (d x + c\right )^{2} + 2 \, {\left (75 \, d \cosh \left (d x + c\right )^{2} + 90 \, d \cosh \left (d x + c\right ) + 43 \, d\right )} \sinh \left (d x + c\right )^{2} + 60 \, d \cosh \left (d x + c\right ) + 4 \, {\left (25 \, d \cosh \left (d x + c\right )^{3} + 45 \, d \cosh \left (d x + c\right )^{2} + 43 \, d \cosh \left (d x + c\right ) + 15 \, d\right )} \sinh \left (d x + c\right ) + 25 \, d\right )}} \]

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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giac [A]  time = 0.12, size = 76, normalized size = 1.04 \[ \frac {\frac {4 \, {\left (215 \, e^{\left (3 \, d x + 3 \, c\right )} + 387 \, e^{\left (2 \, d x + 2 \, c\right )} + 325 \, e^{\left (d x + c\right )} + 225\right )}}{{\left (5 \, e^{\left (2 \, d x + 2 \, c\right )} + 6 \, e^{\left (d x + c\right )} + 5\right )}^{2}} + 43 \, \arctan \left (\frac {5}{4} \, e^{\left (d x + c\right )} + \frac {3}{4}\right )}{1024 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 79, normalized size = 1.08 \[ -\frac {85 \left (\tanh ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{512 d \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )^{2}}-\frac {35 \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{128 d \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+4\right )^{2}}+\frac {43 \arctan \left (\frac {\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}\right )}{1024 d} \]

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 = 0.45, size = 108, normalized size = 1.48 \[ -\frac {43 \, \arctan \left (\frac {5}{4} \, e^{\left (-d x - c\right )} + \frac {3}{4}\right )}{1024 \, d} - \frac {325 \, e^{\left (-d x - c\right )} + 387 \, e^{\left (-2 \, d x - 2 \, c\right )} + 215 \, e^{\left (-3 \, d x - 3 \, c\right )} + 225}{256 \, d {\left (60 \, e^{\left (-d x - c\right )} + 86 \, e^{\left (-2 \, d x - 2 \, c\right )} + 60 \, e^{\left (-3 \, d x - 3 \, c\right )} + 25 \, e^{\left (-4 \, d x - 4 \, c\right )} + 25\right )}} \]

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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mupad [B]  time = 0.96, size = 137, normalized size = 1.88 \[ \frac {\frac {43\,{\mathrm {e}}^{c+d\,x}}{256\,d}+\frac {129}{1280\,d}}{6\,{\mathrm {e}}^{c+d\,x}+5\,{\mathrm {e}}^{2\,c+2\,d\,x}+5}-\frac {\frac {7\,{\mathrm {e}}^{c+d\,x}}{40\,d}-\frac {3}{8\,d}}{60\,{\mathrm {e}}^{c+d\,x}+86\,{\mathrm {e}}^{2\,c+2\,d\,x}+60\,{\mathrm {e}}^{3\,c+3\,d\,x}+25\,{\mathrm {e}}^{4\,c+4\,d\,x}+25}+\frac {43\,\mathrm {atan}\left (\left (\frac {3}{4\,d}+\frac {5\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c}{4\,d}\right )\,\sqrt {d^2}\right )}{1024\,\sqrt {d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*cosh(c + d*x) + 3)^3,x)

[Out]

((43*exp(c + d*x))/(256*d) + 129/(1280*d))/(6*exp(c + d*x) + 5*exp(2*c + 2*d*x) + 5) - ((7*exp(c + d*x))/(40*d
) - 3/(8*d))/(60*exp(c + d*x) + 86*exp(2*c + 2*d*x) + 60*exp(3*c + 3*d*x) + 25*exp(4*c + 4*d*x) + 25) + (43*at
an((3/(4*d) + (5*exp(d*x)*exp(c))/(4*d))*(d^2)^(1/2)))/(1024*(d^2)^(1/2))

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sympy [A]  time = 6.59, size = 530, normalized size = 7.26 \[ \begin {cases} - \frac {\log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )}}{125 d \cosh ^{3}{\left (d x + \log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 225 d \cosh ^{2}{\left (d x + \log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 135 d \cosh {\left (d x + \log {\left (- 3 e^{- d x} + 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 27 d} & \text {for}\: c = \log {\left (\left (-3 + 4 i\right ) e^{- d x} \right )} - \log {\relax (5 )} \\- \frac {\log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )}}{125 d \cosh ^{3}{\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 225 d \cosh ^{2}{\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 135 d \cosh {\left (d x + \log {\left (- 3 e^{- d x} - 4 i e^{- d x} \right )} - \log {\relax (5 )} \right )} + 27 d} & \text {for}\: c = \log {\left (- \left (3 + 4 i\right ) e^{- d x} \right )} - \log {\relax (5 )} \\\frac {x}{\left (5 \cosh {\relax (c )} + 3\right )^{3}} & \text {for}\: d = 0 \\\frac {43 \tanh ^{4}{\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 )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} - \frac {170 \tanh ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} + \frac {344 \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 )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} - \frac {280 \tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} + \frac {688 \operatorname {atan}{\left (\frac {\tanh {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{2} \right )}}{1024 d \tanh ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 8192 d \tanh ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 16384 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-log(-3*exp(-d*x) + 4*I*exp(-d*x))/(125*d*cosh(d*x + log(-3*exp(-d*x) + 4*I*exp(-d*x)) - log(5))**3
+ 225*d*cosh(d*x + log(-3*exp(-d*x) + 4*I*exp(-d*x)) - log(5))**2 + 135*d*cosh(d*x + log(-3*exp(-d*x) + 4*I*e
xp(-d*x)) - log(5)) + 27*d), Eq(c, log((-3 + 4*I)*exp(-d*x)) - log(5))), (-log(-3*exp(-d*x) - 4*I*exp(-d*x))/(
125*d*cosh(d*x + log(-3*exp(-d*x) - 4*I*exp(-d*x)) - log(5))**3 + 225*d*cosh(d*x + log(-3*exp(-d*x) - 4*I*exp(
-d*x)) - log(5))**2 + 135*d*cosh(d*x + log(-3*exp(-d*x) - 4*I*exp(-d*x)) - log(5)) + 27*d), Eq(c, log(-(3 + 4*
I)*exp(-d*x)) - log(5))), (x/(5*cosh(c) + 3)**3, Eq(d, 0)), (43*tanh(c/2 + d*x/2)**4*atan(tanh(c/2 + d*x/2)/2)
/(1024*d*tanh(c/2 + d*x/2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d) - 170*tanh(c/2 + d*x/2)**3/(1024*d*tanh
(c/2 + d*x/2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d) + 344*tanh(c/2 + d*x/2)**2*atan(tanh(c/2 + d*x/2)/2)
/(1024*d*tanh(c/2 + d*x/2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d) - 280*tanh(c/2 + d*x/2)/(1024*d*tanh(c/
2 + d*x/2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d) + 688*atan(tanh(c/2 + d*x/2)/2)/(1024*d*tanh(c/2 + d*x/
2)**4 + 8192*d*tanh(c/2 + d*x/2)**2 + 16384*d), True))

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