3.1.23 \(\int \frac {1}{(5-3 \cos (c+d x))^2} \, dx\) [23]

Optimal. Leaf size=58 \[ \frac {5 x}{64}+\frac {5 \text {ArcTan}\left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{32 d}+\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))} \]

[Out]

5/64*x+5/32*arctan(sin(d*x+c)/(3-cos(d*x+c)))/d+3/16*sin(d*x+c)/d/(5-3*cos(d*x+c))

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Rubi [A]
time = 0.02, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2743, 12, 2736} \begin {gather*} \frac {5 \text {ArcTan}\left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{32 d}+\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}+\frac {5 x}{64} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(5 - 3*Cos[c + d*x])^(-2),x]

[Out]

(5*x)/64 + (5*ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])])/(32*d) + (3*Sin[c + d*x])/(16*d*(5 - 3*Cos[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 2736

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{q = Rt[a^2 - b^2, 2]}, Simp[x/q, x] + Simp
[(2/(d*q))*ArcTan[b*(Cos[c + d*x]/(a + q + b*Sin[c + d*x]))], x]] /; FreeQ[{a, b, c, d}, x] && GtQ[a^2 - b^2,
0] && PosQ[a]

Rule 2743

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] && Integ
erQ[2*n]

Rubi steps

\begin {align*} \int \frac {1}{(5-3 \cos (c+d x))^2} \, dx &=\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}-\frac {1}{16} \int -\frac {5}{5-3 \cos (c+d x)} \, dx\\ &=\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}+\frac {5}{16} \int \frac {1}{5-3 \cos (c+d x)} \, dx\\ &=\frac {5 x}{64}+\frac {5 \tan ^{-1}\left (\frac {\sin (c+d x)}{3-\cos (c+d x)}\right )}{32 d}+\frac {3 \sin (c+d x)}{16 d (5-3 \cos (c+d x))}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 43, normalized size = 0.74 \begin {gather*} \frac {5 \text {ArcTan}\left (2 \tan \left (\frac {1}{2} (c+d x)\right )\right )-\frac {6 \sin (c+d x)}{-5+3 \cos (c+d x)}}{32 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(5*ArcTan[2*Tan[(c + d*x)/2]] - (6*Sin[c + d*x])/(-5 + 3*Cos[c + d*x]))/(32*d)

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Maple [A]
time = 0.07, size = 46, normalized size = 0.79

method	result	size
derivativedivides	\(\frac {\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{4}\right )}+\frac {5 \arctan \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32}}{d}\)	\(46\)
default	\(\frac {\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{64 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )+\frac {1}{4}\right )}+\frac {5 \arctan \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{32}}{d}\)	\(46\)
risch	\(\frac {i \left (5 \,{\mathrm e}^{i \left (d x +c \right )}-3\right )}{8 d \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-10 \,{\mathrm e}^{i \left (d x +c \right )}+3\right )}+\frac {5 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-3\right )}{64 d}-\frac {5 i \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {1}{3}\right )}{64 d}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(3/64*tan(1/2*d*x+1/2*c)/(tan(1/2*d*x+1/2*c)^2+1/4)+5/32*arctan(2*tan(1/2*d*x+1/2*c)))

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Maxima [A]
time = 0.50, size = 69, normalized size = 1.19 \begin {gather*} \frac {\frac {6 \, \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 )}} + 5 \, \arctan \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{32 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(6*sin(d*x + c)/((4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 1)*(cos(d*x + c) + 1)) + 5*arctan(2*sin(d*x + c
)/(cos(d*x + c) + 1)))/d

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Fricas [A]
time = 0.37, size = 59, normalized size = 1.02 \begin {gather*} -\frac {5 \, {\left (3 \, \cos \left (d x + c\right ) - 5\right )} \arctan \left (\frac {5 \, \cos \left (d x + c\right ) - 3}{4 \, \sin \left (d x + c\right )}\right ) + 12 \, \sin \left (d x + c\right )}{64 \, {\left (3 \, d \cos \left (d x + c\right ) - 5 \, d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(5*(3*cos(d*x + c) - 5)*arctan(1/4*(5*cos(d*x + c) - 3)/sin(d*x + c)) + 12*sin(d*x + c))/(3*d*cos(d*x +
c) - 5*d)

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Sympy [C] Result contains complex when optimal does not.
time = 0.71, size = 192, normalized size = 3.31 \begin {gather*} \begin {cases} \frac {x}{\left (5 - 3 \cosh {\left (2 \operatorname {atanh}{\left (\frac {1}{2} \right )} \right )}\right )^{2}} & \text {for}\: c = - d x - 2 i \operatorname {atanh}{\left (\frac {1}{2} \right )} \vee c = - d x + 2 i \operatorname {atanh}{\left (\frac {1}{2} \right )} \\\frac {x}{\left (5 - 3 \cos {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\\frac {20 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right ) \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{128 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32 d} + \frac {5 \left (\operatorname {atan}{\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} \right )} + \pi \left \lfloor {\frac {\frac {c}{2} + \frac {d x}{2} - \frac {\pi }{2}}{\pi }}\right \rfloor \right )}{128 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32 d} + \frac {6 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{128 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 32 d} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(5 - 3*cosh(2*atanh(1/2)))**2, Eq(c, -d*x - 2*I*atanh(1/2)) | Eq(c, -d*x + 2*I*atanh(1/2))), (x/(
5 - 3*cos(c))**2, Eq(d, 0)), (20*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/
2)**2/(128*d*tan(c/2 + d*x/2)**2 + 32*d) + 5*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))/(1
28*d*tan(c/2 + d*x/2)**2 + 32*d) + 6*tan(c/2 + d*x/2)/(128*d*tan(c/2 + d*x/2)**2 + 32*d), True))

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Giac [A]
time = 0.44, size = 61, normalized size = 1.05 \begin {gather*} \frac {5 \, d x + 5 \, c + \frac {12 \, \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} - 10 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) - 3}\right )}{64 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(5*d*x + 5*c + 12*tan(1/2*d*x + 1/2*c)/(4*tan(1/2*d*x + 1/2*c)^2 + 1) - 10*arctan(sin(d*x + c)/(cos(d*x +
c) - 3)))/d

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Mupad [B]
time = 0.31, size = 67, normalized size = 1.16 \begin {gather*} \frac {5\,\mathrm {atan}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{32\,d}-\frac {5\,\left (\mathrm {atan}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )-\frac {d\,x}{2}\right )}{32\,d}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{64\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*cos(c + d*x) - 5)^2,x)

[Out]

(5*atan(2*tan(c/2 + (d*x)/2)))/(32*d) - (5*(atan(tan(c/2 + (d*x)/2)) - (d*x)/2))/(32*d) + (3*tan(c/2 + (d*x)/2
))/(64*d*(tan(c/2 + (d*x)/2)^2 + 1/4))

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