3.29 \(\int \frac{1}{(-5+3 \cos (c+d x))^4} \, dx\)

Optimal. Leaf size=108 \[ \frac{311 \sin (c+d x)}{8192 d (5-3 \cos (c+d x))}+\frac{25 \sin (c+d x)}{512 d (5-3 \cos (c+d x))^2}+\frac{\sin (c+d x)}{16 d (5-3 \cos (c+d x))^3}+\frac{385 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{16384 d}+\frac{385 x}{32768} \]

[Out]

(385*x)/32768 + (385*ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])])/(16384*d) + Sin[c + d*x]/(16*d*(5 - 3*Cos[c + d*
x])^3) + (25*Sin[c + d*x])/(512*d*(5 - 3*Cos[c + d*x])^2) + (311*Sin[c + d*x])/(8192*d*(5 - 3*Cos[c + d*x]))

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Rubi [A]  time = 0.0949558, antiderivative size = 108, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2664, 2754, 12, 2658} \[ \frac{311 \sin (c+d x)}{8192 d (5-3 \cos (c+d x))}+\frac{25 \sin (c+d x)}{512 d (5-3 \cos (c+d x))^2}+\frac{\sin (c+d x)}{16 d (5-3 \cos (c+d x))^3}+\frac{385 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{16384 d}+\frac{385 x}{32768} \]

Antiderivative was successfully verified.

[In]

Int[(-5 + 3*Cos[c + d*x])^(-4),x]

[Out]

(385*x)/32768 + (385*ArcTan[Sin[c + d*x]/(3 - Cos[c + d*x])])/(16384*d) + Sin[c + d*x]/(16*d*(5 - 3*Cos[c + d*
x])^3) + (25*Sin[c + d*x])/(512*d*(5 - 3*Cos[c + d*x])^2) + (311*Sin[c + d*x])/(8192*d*(5 - 3*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 2658

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

Rubi steps

\begin{align*} \int \frac{1}{(-5+3 \cos (c+d x))^4} \, dx &=\frac{\sin (c+d x)}{16 d (5-3 \cos (c+d x))^3}-\frac{1}{48} \int \frac{15+6 \cos (c+d x)}{(-5+3 \cos (c+d x))^3} \, dx\\ &=\frac{\sin (c+d x)}{16 d (5-3 \cos (c+d x))^3}+\frac{25 \sin (c+d x)}{512 d (5-3 \cos (c+d x))^2}+\frac{\int \frac{186+75 \cos (c+d x)}{(-5+3 \cos (c+d x))^2} \, dx}{1536}\\ &=\frac{\sin (c+d x)}{16 d (5-3 \cos (c+d x))^3}+\frac{25 \sin (c+d x)}{512 d (5-3 \cos (c+d x))^2}+\frac{311 \sin (c+d x)}{8192 d (5-3 \cos (c+d x))}-\frac{\int \frac{1155}{-5+3 \cos (c+d x)} \, dx}{24576}\\ &=\frac{\sin (c+d x)}{16 d (5-3 \cos (c+d x))^3}+\frac{25 \sin (c+d x)}{512 d (5-3 \cos (c+d x))^2}+\frac{311 \sin (c+d x)}{8192 d (5-3 \cos (c+d x))}-\frac{385 \int \frac{1}{-5+3 \cos (c+d x)} \, dx}{8192}\\ &=\frac{385 x}{32768}+\frac{385 \tan ^{-1}\left (\frac{\sin (c+d x)}{3-\cos (c+d x)}\right )}{16384 d}+\frac{\sin (c+d x)}{16 d (5-3 \cos (c+d x))^3}+\frac{25 \sin (c+d x)}{512 d (5-3 \cos (c+d x))^2}+\frac{311 \sin (c+d x)}{8192 d (5-3 \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.025745, size = 66, normalized size = 0.61 \[ \frac{770 \tan ^{-1}\left (2 \tan \left (\frac{1}{2} (c+d x)\right )\right )-\frac{9 (4883 \sin (c+d x)-2340 \sin (2 (c+d x))+311 \sin (3 (c+d x)))}{(3 \cos (c+d x)-5)^3}}{32768 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(-5 + 3*Cos[c + d*x])^(-4),x]

[Out]

(770*ArcTan[2*Tan[(c + d*x)/2]] - (9*(4883*Sin[c + d*x] - 2340*Sin[2*(c + d*x)] + 311*Sin[3*(c + d*x)]))/(-5 +
3*Cos[c + d*x])^3)/(32768*d)

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Maple [A]  time = 0.04, size = 116, normalized size = 1.1 \begin{align*}{\frac{369}{512\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5} \left ( 4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-3}}+{\frac{117}{256\,d} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3} \left ( 4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-3}}+{\frac{639}{8192\,d}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 4\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) ^{-3}}+{\frac{385}{16384\,d}\arctan \left ( 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/(-5+3*cos(d*x+c))^4,x)

[Out]

369/512/d/(4*tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)^5+117/256/d/(4*tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x
+1/2*c)^3+639/8192/d/(4*tan(1/2*d*x+1/2*c)^2+1)^3*tan(1/2*d*x+1/2*c)+385/16384/d*arctan(2*tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 2.08888, size = 205, normalized size = 1.9 \begin{align*} \frac{\frac{18 \,{\left (\frac{71 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{416 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{656 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{\frac{12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{48 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{64 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + 1} + 385 \, \arctan \left (\frac{2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{16384 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*cos(d*x+c))^4,x, algorithm="maxima")

[Out]

1/16384*(18*(71*sin(d*x + c)/(cos(d*x + c) + 1) + 416*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 656*sin(d*x + c)^5
/(cos(d*x + c) + 1)^5)/(12*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 48*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 64*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 1) + 385*arctan(2*sin(d*x + c)/(cos(d*x + c) + 1)))/d

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Fricas [A]  time = 1.65929, size = 362, normalized size = 3.35 \begin{align*} -\frac{385 \,{\left (27 \, \cos \left (d x + c\right )^{3} - 135 \, \cos \left (d x + c\right )^{2} + 225 \, \cos \left (d x + c\right ) - 125\right )} \arctan \left (\frac{5 \, \cos \left (d x + c\right ) - 3}{4 \, \sin \left (d x + c\right )}\right ) + 36 \,{\left (311 \, \cos \left (d x + c\right )^{2} - 1170 \, \cos \left (d x + c\right ) + 1143\right )} \sin \left (d x + c\right )}{32768 \,{\left (27 \, d \cos \left (d x + c\right )^{3} - 135 \, d \cos \left (d x + c\right )^{2} + 225 \, d \cos \left (d x + c\right ) - 125 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*cos(d*x+c))^4,x, algorithm="fricas")

[Out]

-1/32768*(385*(27*cos(d*x + c)^3 - 135*cos(d*x + c)^2 + 225*cos(d*x + c) - 125)*arctan(1/4*(5*cos(d*x + c) - 3
)/sin(d*x + c)) + 36*(311*cos(d*x + c)^2 - 1170*cos(d*x + c) + 1143)*sin(d*x + c))/(27*d*cos(d*x + c)^3 - 135*
d*cos(d*x + c)^2 + 225*d*cos(d*x + c) - 125*d)

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Sympy [A]  time = 14.6756, size = 597, normalized size = 5.53 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*cos(d*x+c))**4,x)

[Out]

Piecewise((x/(-5 + 3*cosh(2*atanh(1/2)))**4, Eq(c, -d*x - 2*I*atanh(1/2)) | Eq(c, -d*x + 2*I*atanh(1/2))), (x/
(3*cos(c) - 5)**4, Eq(d, 0)), (24640*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 +
d*x/2)**6/(1048576*d*tan(c/2 + d*x/2)**6 + 786432*d*tan(c/2 + d*x/2)**4 + 196608*d*tan(c/2 + d*x/2)**2 + 16384
*d) + 18480*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2 + d*x/2 - pi/2)/pi))*tan(c/2 + d*x/2)**4/(1048576*d*tan(
c/2 + d*x/2)**6 + 786432*d*tan(c/2 + d*x/2)**4 + 196608*d*tan(c/2 + d*x/2)**2 + 16384*d) + 4620*(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/(1048576*d*tan(c/2 + d*x/2)**6 + 786432*d
*tan(c/2 + d*x/2)**4 + 196608*d*tan(c/2 + d*x/2)**2 + 16384*d) + 385*(atan(2*tan(c/2 + d*x/2)) + pi*floor((c/2
+ d*x/2 - pi/2)/pi))/(1048576*d*tan(c/2 + d*x/2)**6 + 786432*d*tan(c/2 + d*x/2)**4 + 196608*d*tan(c/2 + d*x/2
)**2 + 16384*d) + 11808*tan(c/2 + d*x/2)**5/(1048576*d*tan(c/2 + d*x/2)**6 + 786432*d*tan(c/2 + d*x/2)**4 + 19
6608*d*tan(c/2 + d*x/2)**2 + 16384*d) + 7488*tan(c/2 + d*x/2)**3/(1048576*d*tan(c/2 + d*x/2)**6 + 786432*d*tan
(c/2 + d*x/2)**4 + 196608*d*tan(c/2 + d*x/2)**2 + 16384*d) + 1278*tan(c/2 + d*x/2)/(1048576*d*tan(c/2 + d*x/2)
**6 + 786432*d*tan(c/2 + d*x/2)**4 + 196608*d*tan(c/2 + d*x/2)**2 + 16384*d), True))

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Giac [A]  time = 1.15696, size = 122, normalized size = 1.13 \begin{align*} \frac{385 \, d x + 385 \, c + \frac{36 \,{\left (656 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 416 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 71 \, \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 )}^{3}} - 770 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) - 3}\right )}{32768 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-5+3*cos(d*x+c))^4,x, algorithm="giac")

[Out]

1/32768*(385*d*x + 385*c + 36*(656*tan(1/2*d*x + 1/2*c)^5 + 416*tan(1/2*d*x + 1/2*c)^3 + 71*tan(1/2*d*x + 1/2*
c))/(4*tan(1/2*d*x + 1/2*c)^2 + 1)^3 - 770*arctan(sin(d*x + c)/(cos(d*x + c) - 3)))/d