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

Optimal. Leaf size=113 \[ \frac {45 \sin (c+d x)}{512 d (3-5 \cos (c+d x))}-\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d} \]

[Out]

43/2048*ln(cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c))/d-43/2048*ln(cos(1/2*d*x+1/2*c)+2*sin(1/2*d*x+1/2*c))/d-5/
32*sin(d*x+c)/d/(3-5*cos(d*x+c))^2+45/512*sin(d*x+c)/d/(3-5*cos(d*x+c))

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Rubi [A]  time = 0.07, antiderivative size = 113, 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, 207} \[ \frac {45 \sin (c+d x)}{512 d (3-5 \cos (c+d x))}-\frac {5 \sin (c+d x)}{32 d (3-5 \cos (c+d x))^2}+\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d} \]

Antiderivative was successfully verified.

[In]

Int[(3 - 5*Cos[c + d*x])^(-3),x]

[Out]

(43*Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]])/(2048*d) - (43*Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]])/(20
48*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 12

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[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 \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}{-2+8 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{256 d}\\ &=\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \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*}

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Mathematica [A]  time = 0.15, size = 211, normalized size = 1.87 \[ -\frac {45 \sin \left (\frac {1}{2} (c+d x)\right )}{1024 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {45 \sin \left (\frac {1}{2} (c+d x)\right )}{1024 d \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}-\frac {5}{512 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {5}{512 d \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {43 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d}-\frac {43 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2048 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(43*Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]])/(2048*d) - (43*Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]])/(20
48*d) - 5/(512*d*(Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2])^2) - (45*Sin[(c + d*x)/2])/(1024*d*(Cos[(c + d*x)/2]
- 2*Sin[(c + d*x)/2])) + 5/(512*d*(Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2])^2) - (45*Sin[(c + d*x)/2])/(1024*d*(
Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]))

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fricas [A]  time = 0.98, size = 129, normalized size = 1.14 \[ -\frac {43 \, {\left (25 \, \cos \left (d x + c\right )^{2} - 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (-\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 43 \, {\left (25 \, \cos \left (d x + c\right )^{2} - 30 \, \cos \left (d x + c\right ) + 9\right )} \log \left (-\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 40 \, {\left (45 \, \cos \left (d x + c\right ) - 11\right )} \sin \left (d x + c\right )}{4096 \, {\left (25 \, d \cos \left (d x + c\right )^{2} - 30 \, d \cos \left (d x + c\right ) + 9 \, d\right )}} \]

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*c
os(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)
*sin(d*x + c))/(25*d*cos(d*x + c)^2 - 30*d*cos(d*x + c) + 9*d)

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giac [A]  time = 0.40, size = 84, normalized size = 0.74 \[ \frac {\frac {20 \, {\left (28 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 17 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - 43 \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 43 \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{2048 \, d} \]

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

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maple [A]  time = 0.04, size = 120, normalized size = 1.06 \[ \frac {25}{2048 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {35}{2048 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {43 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2048 d}-\frac {25}{2048 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {35}{2048 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {43 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2048 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/2048/d/(2*tan(1/2*d*x+1/2*c)+1)^2+35/2048/d/(2*tan(1/2*d*x+1/2*c)+1)-43/2048/d*ln(2*tan(1/2*d*x+1/2*c)+1)-2
5/2048/d/(2*tan(1/2*d*x+1/2*c)-1)^2+35/2048/d/(2*tan(1/2*d*x+1/2*c)-1)+43/2048/d*ln(2*tan(1/2*d*x+1/2*c)-1)

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maxima [A]  time = 0.67, size = 137, normalized size = 1.21 \[ \frac {\frac {20 \, {\left (\frac {17 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {28 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{\frac {8 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {16 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 1} - 43 \, \log \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + 43 \, \log \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{2048 \, d} \]

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

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mupad [B]  time = 0.67, size = 76, normalized size = 0.67 \[ -\frac {43\,\mathrm {atanh}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{1024\,d}-\frac {\frac {85\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8192}-\frac {35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2048}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}+\frac {1}{16}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(5*cos(c + d*x) - 3)^3,x)

[Out]

- (43*atanh(2*tan(c/2 + (d*x)/2)))/(1024*d) - ((85*tan(c/2 + (d*x)/2))/8192 - (35*tan(c/2 + (d*x)/2)^3)/2048)/
(d*(tan(c/2 + (d*x)/2)^4 - tan(c/2 + (d*x)/2)^2/2 + 1/16))

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sympy [A]  time = 2.48, size = 490, normalized size = 4.34 \[ \begin {cases} \frac {x}{\left (3 - 5 \cos {\left (2 \operatorname {atan}{\left (\frac {1}{2} \right )} \right )}\right )^{3}} & \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 {\relax (c )}\right )^{3}} & \text {for}\: d = 0 \\\frac {688 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {1}{2} \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} - \frac {344 \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 )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} + \frac {43 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} - \frac {1}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} - \frac {688 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + \frac {1}{2} \right )} \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} + \frac {344 \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 )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} - \frac {43 \log {\left (\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )} + \frac {1}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} + \frac {560 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} - \frac {340 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{32768 d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 16384 d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 2048 d} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(3 - 5*cos(2*atan(1/2)))**3, Eq(c, -d*x - 2*atan(1/2)) | Eq(c, -d*x + 2*atan(1/2))), (x/(3 - 5*co
s(c))**3, Eq(d, 0)), (688*log(tan(c/2 + d*x/2) - 1/2)*tan(c/2 + d*x/2)**4/(32768*d*tan(c/2 + d*x/2)**4 - 16384
*d*tan(c/2 + d*x/2)**2 + 2048*d) - 344*log(tan(c/2 + d*x/2) - 1/2)*tan(c/2 + d*x/2)**2/(32768*d*tan(c/2 + d*x/
2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 2048*d) + 43*log(tan(c/2 + d*x/2) - 1/2)/(32768*d*tan(c/2 + d*x/2)**4 -
16384*d*tan(c/2 + d*x/2)**2 + 2048*d) - 688*log(tan(c/2 + d*x/2) + 1/2)*tan(c/2 + d*x/2)**4/(32768*d*tan(c/2 +
d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 2048*d) + 344*log(tan(c/2 + d*x/2) + 1/2)*tan(c/2 + d*x/2)**2/(3276
8*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 2048*d) - 43*log(tan(c/2 + d*x/2) + 1/2)/(32768*d*tan(
c/2 + d*x/2)**4 - 16384*d*tan(c/2 + d*x/2)**2 + 2048*d) + 560*tan(c/2 + d*x/2)**3/(32768*d*tan(c/2 + d*x/2)**4
- 16384*d*tan(c/2 + d*x/2)**2 + 2048*d) - 340*tan(c/2 + d*x/2)/(32768*d*tan(c/2 + d*x/2)**4 - 16384*d*tan(c/2
+ d*x/2)**2 + 2048*d), True))

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