3.1.39 \(\int \frac {1}{(3-5 \cos (c+d x))^2} \, dx\) [39]

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

[Out]

-3/64*ln(cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c))/d+3/64*ln(cos(1/2*d*x+1/2*c)+2*sin(1/2*d*x+1/2*c))/d-5/16*si
n(d*x+c)/d/(3-5*cos(d*x+c))

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Rubi [A]
time = 0.03, antiderivative size = 88, normalized size of antiderivative = 1.00, 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 = {2743, 12, 2738, 213} \begin {gather*} -\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} \end {gather*}

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 2738

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 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}{(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 \text {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*}

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Mathematica [A]
time = 0.06, size = 139, normalized size = 1.58 \begin {gather*} \frac {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 (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-9 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )+20 \sin (c+d x)}{64 d (-3+5 \cos (c+d x))} \end {gather*}

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 = 0.07, size = 72, normalized size = 0.82

method	result	size
norman	\(-\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{16 d \left (4 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right )}-\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64 d}+\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64 d}\)	\(71\)
derivativedivides	\(\frac {-\frac {5}{64 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}-\frac {5}{64 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64}}{d}\)	\(72\)
default	\(\frac {-\frac {5}{64 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{64}-\frac {5}{64 \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{64}}{d}\)	\(72\)
risch	\(-\frac {i \left (3 \,{\mathrm e}^{i \left (d x +c \right )}-5\right )}{8 d \left (5 \,{\mathrm e}^{2 i \left (d x +c \right )}-6 \,{\mathrm e}^{i \left (d x +c \right )}+5\right )}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {3}{5}-\frac {4 i}{5}\right )}{64 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {3}{5}+\frac {4 i}{5}\right )}{64 d}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.30, size = 94, normalized size = 1.07 \begin {gather*} -\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 {gather*}

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 = 0.37, size = 88, normalized size = 1.00 \begin {gather*} \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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 240 vs. \(2 (78) = 156\).
time = 0.70, size = 240, normalized size = 2.73 \begin {gather*} \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 (3 - 5 \cos {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {12 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \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 (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - 1 \right )}}{256 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 64 d} + \frac {12 \log {\left (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \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 (2 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + 1 \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 {gather*}

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/(3 - 5*co
s(c))**2, Eq(d, 0)), (-12*log(2*tan(c/2 + d*x/2) - 1)*tan(c/2 + d*x/2)**2/(256*d*tan(c/2 + d*x/2)**2 - 64*d) +
3*log(2*tan(c/2 + d*x/2) - 1)/(256*d*tan(c/2 + d*x/2)**2 - 64*d) + 12*log(2*tan(c/2 + d*x/2) + 1)*tan(c/2 + d
*x/2)**2/(256*d*tan(c/2 + d*x/2)**2 - 64*d) - 3*log(2*tan(c/2 + d*x/2) + 1)/(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 = 0.52, size = 68, normalized size = 0.77 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 0.34, size = 47, normalized size = 0.53 \begin {gather*} \frac {3\,\mathrm {atanh}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{32\,d}-\frac {5\,\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/(5*cos(c + d*x) - 3)^2,x)

[Out]

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

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