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

Optimal. Leaf size=115 \[ -\frac{45 \sin (c+d x)}{512 d (5 \cos (c+d x)+3)}+\frac{5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}-\frac{43 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}+\frac{43 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d} \]

[Out]

(-43*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2048*d) + (43*Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2
048*d) + (5*Sin[c + d*x])/(32*d*(3 + 5*Cos[c + d*x])^2) - (45*Sin[c + d*x])/(512*d*(3 + 5*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.0750301, antiderivative size = 115, 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} \[ -\frac{45 \sin (c+d x)}{512 d (5 \cos (c+d x)+3)}+\frac{5 \sin (c+d x)}{32 d (5 \cos (c+d x)+3)^2}-\frac{43 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}+\frac{43 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*Cos[c + d*x])^(-3),x]

[Out]

(-43*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2048*d) + (43*Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2
048*d) + (5*Sin[c + d*x])/(32*d*(3 + 5*Cos[c + d*x])^2) - (45*Sin[c + d*x])/(512*d*(3 + 5*Cos[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 \cos (c+d x))^3} \, dx &=\frac{5 \sin (c+d x)}{32 d (3+5 \cos (c+d x))^2}+\frac{1}{32} \int \frac{-6+5 \cos (c+d x)}{(3+5 \cos (c+d x))^2} \, dx\\ &=\frac{5 \sin (c+d x)}{32 d (3+5 \cos (c+d x))^2}-\frac{45 \sin (c+d x)}{512 d (3+5 \cos (c+d x))}+\frac{1}{512} \int \frac{43}{3+5 \cos (c+d x)} \, dx\\ &=\frac{5 \sin (c+d x)}{32 d (3+5 \cos (c+d x))^2}-\frac{45 \sin (c+d x)}{512 d (3+5 \cos (c+d x))}+\frac{43}{512} \int \frac{1}{3+5 \cos (c+d x)} \, dx\\ &=\frac{5 \sin (c+d x)}{32 d (3+5 \cos (c+d x))^2}-\frac{45 \sin (c+d x)}{512 d (3+5 \cos (c+d x))}+\frac{43 \operatorname{Subst}\left (\int \frac{1}{8-2 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{256 d}\\ &=-\frac{43 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}+\frac{43 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )+\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}+\frac{5 \sin (c+d x)}{32 d (3+5 \cos (c+d x))^2}-\frac{45 \sin (c+d x)}{512 d (3+5 \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.193965, size = 217, normalized size = 1.89 \[ -\frac{45 \sin \left (\frac{1}{2} (c+d x)\right )}{2048 d \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{45 \sin \left (\frac{1}{2} (c+d x)\right )}{2048 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}+\frac{5}{512 d \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{5}{512 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{43 \log \left (2 \cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d}+\frac{43 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+2 \cos \left (\frac{1}{2} (c+d x)\right )\right )}{2048 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-43*Log[2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2]])/(2048*d) + (43*Log[2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]])/(2
048*d) + 5/(512*d*(2*Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) - (45*Sin[(c + d*x)/2])/(2048*d*(2*Cos[(c + d*x)/
2] - Sin[(c + d*x)/2])) - 5/(512*d*(2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) - (45*Sin[(c + d*x)/2])/(2048*d*
(2*Cos[(c + d*x)/2] + Sin[(c + d*x)/2]))

________________________________________________________________________________________

Maple [A]  time = 0.042, size = 108, normalized size = 0.9 \begin{align*} -{\frac{25}{512\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +2 \right ) ^{-2}}+{\frac{85}{1024\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +2 \right ) ^{-1}}+{\frac{43}{2048\,d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +2 \right ) }+{\frac{25}{512\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -2 \right ) ^{-2}}+{\frac{85}{1024\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -2 \right ) ^{-1}}-{\frac{43}{2048\,d}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -2 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/512/d/(tan(1/2*d*x+1/2*c)+2)^2+85/1024/d/(tan(1/2*d*x+1/2*c)+2)+43/2048/d*ln(tan(1/2*d*x+1/2*c)+2)+25/512/
d/(tan(1/2*d*x+1/2*c)-2)^2+85/1024/d/(tan(1/2*d*x+1/2*c)-2)-43/2048/d*ln(tan(1/2*d*x+1/2*c)-2)

________________________________________________________________________________________

Maxima [A]  time = 1.16337, size = 182, normalized size = 1.58 \begin{align*} \frac{\frac{20 \,{\left (\frac{28 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{17 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac{8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{\sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 16} + 43 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 2\right ) - 43 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 2\right )}{2048 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2048*(20*(28*sin(d*x + c)/(cos(d*x + c) + 1) - 17*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(8*sin(d*x + c)^2/(co
s(d*x + c) + 1)^2 - sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 16) + 43*log(sin(d*x + c)/(cos(d*x + c) + 1) + 2) -
43*log(sin(d*x + c)/(cos(d*x + c) + 1) - 2))/d

________________________________________________________________________________________

Fricas [A]  time = 1.62764, size = 379, normalized size = 3.3 \begin{align*} \frac{43 \,{\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (\frac{3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac{5}{2}\right ) - 43 \,{\left (25 \, \cos \left (d x + c\right )^{2} + 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (\frac{3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac{5}{2}\right ) - 40 \,{\left (45 \, \cos \left (d x + c\right ) + 11\right )} \sin \left (d x + c\right )}{4096 \,{\left (25 \, d \cos \left (d x + c\right )^{2} + 30 \, d \cos \left (d x + c\right ) + 9 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4096*(43*(25*cos(d*x + c)^2 + 30*cos(d*x + c) + 9)*log(3/2*cos(d*x + c) + 2*sin(d*x + c) + 5/2) - 43*(25*cos
(d*x + c)^2 + 30*cos(d*x + c) + 9)*log(3/2*cos(d*x + c) - 2*sin(d*x + c) + 5/2) - 40*(45*cos(d*x + c) + 11)*si
n(d*x + c))/(25*d*cos(d*x + c)^2 + 30*d*cos(d*x + c) + 9*d)

________________________________________________________________________________________

Sympy [A]  time = 4.9541, size = 474, normalized size = 4.12 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(5*cos(2*atan(2)) + 3)**3, Eq(c, -d*x - 2*atan(2)) | Eq(c, -d*x + 2*atan(2))), (x/(5*cos(c) + 3)*
*3, Eq(d, 0)), (-43*log(tan(c/2 + d*x/2) - 2)*tan(c/2 + d*x/2)**4/(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/
2 + d*x/2)**2 + 32768*d) + 344*log(tan(c/2 + d*x/2) - 2)*tan(c/2 + d*x/2)**2/(2048*d*tan(c/2 + d*x/2)**4 - 163
84*d*tan(c/2 + d*x/2)**2 + 32768*d) - 688*log(tan(c/2 + d*x/2) - 2)/(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(
c/2 + d*x/2)**2 + 32768*d) + 43*log(tan(c/2 + d*x/2) + 2)*tan(c/2 + d*x/2)**4/(2048*d*tan(c/2 + d*x/2)**4 - 16
384*d*tan(c/2 + d*x/2)**2 + 32768*d) - 344*log(tan(c/2 + d*x/2) + 2)*tan(c/2 + d*x/2)**2/(2048*d*tan(c/2 + d*x
/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 32768*d) + 688*log(tan(c/2 + d*x/2) + 2)/(2048*d*tan(c/2 + d*x/2)**4 -
16384*d*tan(c/2 + d*x/2)**2 + 32768*d) + 340*tan(c/2 + d*x/2)**3/(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2
+ d*x/2)**2 + 32768*d) - 560*tan(c/2 + d*x/2)/(2048*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 327
68*d), True))

________________________________________________________________________________________

Giac [A]  time = 1.17883, size = 105, normalized size = 0.91 \begin{align*} \frac{\frac{20 \,{\left (17 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 28 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4\right )}^{2}} + 43 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \right |}\right ) - 43 \, \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \right |}\right )}{2048 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2048*(20*(17*tan(1/2*d*x + 1/2*c)^3 - 28*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 4)^2 + 43*log(abs(t
an(1/2*d*x + 1/2*c) + 2)) - 43*log(abs(tan(1/2*d*x + 1/2*c) - 2)))/d