3.43 \(\int \frac{1}{(-3+5 \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=88 \[ -\frac{5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}-\frac{3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-2 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{3 \log \left (2 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d} \]

[Out]

(-3*Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]])/(64*d) + (3*Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]])/(64*d)
- (5*Sin[c + d*x])/(16*d*(3 - 5*Cos[c + d*x]))

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Rubi [A]  time = 0.039762, antiderivative size = 88, 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} \[ -\frac{5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}-\frac{3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-2 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{3 \log \left (2 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d} \]

Antiderivative was successfully verified.

[In]

Int[(-3 + 5*Cos[c + d*x])^(-2),x]

[Out]

(-3*Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]])/(64*d) + (3*Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]])/(64*d)
- (5*Sin[c + d*x])/(16*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 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))^2} \, dx &=-\frac{5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}+\frac{1}{16} \int \frac{3}{-3+5 \cos (c+d x)} \, dx\\ &=-\frac{5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}+\frac{3}{16} \int \frac{1}{-3+5 \cos (c+d x)} \, dx\\ &=-\frac{5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{2-8 x^2} \, dx,x,\tan \left (\frac{1}{2} (c+d x)\right )\right )}{8 d}\\ &=-\frac{3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-2 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}+\frac{3 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )+2 \sin \left (\frac{1}{2} (c+d x)\right )\right )}{64 d}-\frac{5 \sin (c+d x)}{16 d (3-5 \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.020595, size = 139, normalized size = 1.58 \[ \frac{20 \sin (c+d x)+9 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-2 \sin \left (\frac{1}{2} (c+d x)\right )\right )-15 \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-2 \sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (2 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-9 \log \left (2 \sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{64 d (5 \cos (c+d x)-3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(9*Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]] - 15*Cos[c + d*x]*(Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]] -
Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]]) - 9*Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]] + 20*Sin[c + d*x])/
(64*d*(-3 + 5*Cos[c + d*x]))

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Maple [A]  time = 0.039, size = 80, normalized size = 0.9 \begin{align*} -{\frac{5}{64\,d} \left ( 2\,\tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{-1}}-{\frac{3}{64\,d}\ln \left ( 2\,\tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) }-{\frac{5}{64\,d} \left ( 1+2\,\tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{-1}}+{\frac{3}{64\,d}\ln \left ( 1+2\,\tan \left ( 1/2\,dx+c/2 \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.54867, size = 127, normalized size = 1.44 \begin{align*} -\frac{\frac{20 \, \sin \left (d x + c\right )}{{\left (\frac{4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}{\left (\cos \left (d x + c\right ) + 1\right )}} - 3 \, \log \left (\frac{2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + 3 \, \log \left (\frac{2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(20*sin(d*x + c)/((4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)*(cos(d*x + c) + 1)) - 3*log(2*sin(d*x + c)
/(cos(d*x + c) + 1) + 1) + 3*log(2*sin(d*x + c)/(cos(d*x + c) + 1) - 1))/d

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Fricas [A]  time = 1.65209, size = 259, normalized size = 2.94 \begin{align*} \frac{3 \,{\left (5 \, \cos \left (d x + c\right ) - 3\right )} \log \left (-\frac{3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac{5}{2}\right ) - 3 \,{\left (5 \, \cos \left (d x + c\right ) - 3\right )} \log \left (-\frac{3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac{5}{2}\right ) + 40 \, \sin \left (d x + c\right )}{128 \,{\left (5 \, d \cos \left (d x + c\right ) - 3 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/128*(3*(5*cos(d*x + c) - 3)*log(-3/2*cos(d*x + c) + 2*sin(d*x + c) + 5/2) - 3*(5*cos(d*x + c) - 3)*log(-3/2*
cos(d*x + c) - 2*sin(d*x + c) + 5/2) + 40*sin(d*x + c))/(5*d*cos(d*x + c) - 3*d)

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Sympy [A]  time = 2.30547, size = 240, normalized size = 2.73 \begin{align*} \begin{cases} \frac{x}{\left (-3 + 5 \cos{\left (2 \operatorname{atan}{\left (\frac{1}{2} \right )} \right )}\right )^{2}} & \text{for}\: c = - d x - 2 \operatorname{atan}{\left (\frac{1}{2} \right )} \vee c = - d x + 2 \operatorname{atan}{\left (\frac{1}{2} \right )} \\\frac{x}{\left (5 \cos{\left (c \right )} - 3\right )^{2}} & \text{for}\: d = 0 \\- \frac{12 \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} - \frac{1}{2} \right )} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{256 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 64 d} + \frac{3 \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} - \frac{1}{2} \right )}}{256 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 64 d} + \frac{12 \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + \frac{1}{2} \right )} \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{256 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 64 d} - \frac{3 \log{\left (\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )} + \frac{1}{2} \right )}}{256 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 64 d} - \frac{20 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{256 d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 64 d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(-3 + 5*cos(2*atan(1/2)))**2, Eq(c, -d*x - 2*atan(1/2)) | Eq(c, -d*x + 2*atan(1/2))), (x/(5*cos(c
) - 3)**2, Eq(d, 0)), (-12*log(tan(c/2 + d*x/2) - 1/2)*tan(c/2 + d*x/2)**2/(256*d*tan(c/2 + d*x/2)**2 - 64*d)
+ 3*log(tan(c/2 + d*x/2) - 1/2)/(256*d*tan(c/2 + d*x/2)**2 - 64*d) + 12*log(tan(c/2 + d*x/2) + 1/2)*tan(c/2 +
d*x/2)**2/(256*d*tan(c/2 + d*x/2)**2 - 64*d) - 3*log(tan(c/2 + d*x/2) + 1/2)/(256*d*tan(c/2 + d*x/2)**2 - 64*d
) - 20*tan(c/2 + d*x/2)/(256*d*tan(c/2 + d*x/2)**2 - 64*d), True))

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Giac [A]  time = 1.16221, size = 92, normalized size = 1.05 \begin{align*} -\frac{\frac{20 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{4 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} - 3 \, \log \left ({\left | 2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) + 3 \, \log \left ({\left | 2 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right )}{64 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(20*tan(1/2*d*x + 1/2*c)/(4*tan(1/2*d*x + 1/2*c)^2 - 1) - 3*log(abs(2*tan(1/2*d*x + 1/2*c) + 1)) + 3*log
(abs(2*tan(1/2*d*x + 1/2*c) - 1)))/d