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

Optimal. Leaf size=138 \[ -\frac {995 \sin (c+d x)}{24576 d (3-5 \cos (c+d x))}+\frac {25 \sin (c+d x)}{512 d (3-5 \cos (c+d x))^2}-\frac {5 \sin (c+d x)}{48 d (3-5 \cos (c+d x))^3}-\frac {279 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{32768 d}+\frac {279 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{32768 d} \]

[Out]

-279/32768*ln(cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c))/d+279/32768*ln(cos(1/2*d*x+1/2*c)+2*sin(1/2*d*x+1/2*c))
/d-5/48*sin(d*x+c)/d/(3-5*cos(d*x+c))^3+25/512*sin(d*x+c)/d/(3-5*cos(d*x+c))^2-995/24576*sin(d*x+c)/d/(3-5*cos
(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {2664, 2754, 12, 2659, 206} \[ -\frac {995 \sin (c+d x)}{24576 d (3-5 \cos (c+d x))}+\frac {25 \sin (c+d x)}{512 d (3-5 \cos (c+d x))^2}-\frac {5 \sin (c+d x)}{48 d (3-5 \cos (c+d x))^3}-\frac {279 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{32768 d}+\frac {279 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{32768 d} \]

Antiderivative was successfully verified.

[In]

Int[(-3 + 5*Cos[c + d*x])^(-4),x]

[Out]

(-279*Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]])/(32768*d) + (279*Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]])
/(32768*d) - (5*Sin[c + d*x])/(48*d*(3 - 5*Cos[c + d*x])^3) + (25*Sin[c + d*x])/(512*d*(3 - 5*Cos[c + d*x])^2)
- (995*Sin[c + d*x])/(24576*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 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])

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))^4} \, dx &=-\frac {5 \sin (c+d x)}{48 d (3-5 \cos (c+d x))^3}+\frac {1}{48} \int \frac {9+10 \cos (c+d x)}{(-3+5 \cos (c+d x))^3} \, dx\\ &=-\frac {5 \sin (c+d x)}{48 d (3-5 \cos (c+d x))^3}+\frac {25 \sin (c+d x)}{512 d (3-5 \cos (c+d x))^2}+\frac {\int \frac {154+75 \cos (c+d x)}{(-3+5 \cos (c+d x))^2} \, dx}{1536}\\ &=-\frac {5 \sin (c+d x)}{48 d (3-5 \cos (c+d x))^3}+\frac {25 \sin (c+d x)}{512 d (3-5 \cos (c+d x))^2}-\frac {995 \sin (c+d x)}{24576 d (3-5 \cos (c+d x))}+\frac {\int \frac {837}{-3+5 \cos (c+d x)} \, dx}{24576}\\ &=-\frac {5 \sin (c+d x)}{48 d (3-5 \cos (c+d x))^3}+\frac {25 \sin (c+d x)}{512 d (3-5 \cos (c+d x))^2}-\frac {995 \sin (c+d x)}{24576 d (3-5 \cos (c+d x))}+\frac {279 \int \frac {1}{-3+5 \cos (c+d x)} \, dx}{8192}\\ &=-\frac {5 \sin (c+d x)}{48 d (3-5 \cos (c+d x))^3}+\frac {25 \sin (c+d x)}{512 d (3-5 \cos (c+d x))^2}-\frac {995 \sin (c+d x)}{24576 d (3-5 \cos (c+d x))}+\frac {279 \operatorname {Subst}\left (\int \frac {1}{2-8 x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{4096 d}\\ &=-\frac {279 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{32768 d}+\frac {279 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{32768 d}-\frac {5 \sin (c+d x)}{48 d (3-5 \cos (c+d x))^3}+\frac {25 \sin (c+d x)}{512 d (3-5 \cos (c+d x))^2}-\frac {995 \sin (c+d x)}{24576 d (3-5 \cos (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 0.02, size = 288, normalized size = 2.09 \[ \frac {226140 \sin (c+d x)-190800 \sin (2 (c+d x))+99500 \sin (3 (c+d x))-104625 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )+467046 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )-765855 \cos (c+d x) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+376650 \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+104625 \cos (3 (c+d x)) \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )-467046 \log \left (2 \sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{393216 d (5 \cos (c+d x)-3)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(467046*Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]] - 104625*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2] - 2*Sin[(c +
d*x)/2]] - 765855*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]]) + 376650*Cos[2*(c + d*x)]*(Log[Cos[(c + d*x)/2] - 2*Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + 2*S
in[(c + d*x)/2]]) - 467046*Log[Cos[(c + d*x)/2] + 2*Sin[(c + d*x)/2]] + 104625*Cos[3*(c + d*x)]*Log[Cos[(c + d
*x)/2] + 2*Sin[(c + d*x)/2]] + 226140*Sin[c + d*x] - 190800*Sin[2*(c + d*x)] + 99500*Sin[3*(c + d*x)])/(393216
*d*(-3 + 5*Cos[c + d*x])^3)

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fricas [A]  time = 1.60, size = 170, normalized size = 1.23 \[ \frac {837 \, {\left (125 \, \cos \left (d x + c\right )^{3} - 225 \, \cos \left (d x + c\right )^{2} + 135 \, \cos \left (d x + c\right ) - 27\right )} \log \left (-\frac {3}{2} \, \cos \left (d x + c\right ) + 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) - 837 \, {\left (125 \, \cos \left (d x + c\right )^{3} - 225 \, \cos \left (d x + c\right )^{2} + 135 \, \cos \left (d x + c\right ) - 27\right )} \log \left (-\frac {3}{2} \, \cos \left (d x + c\right ) - 2 \, \sin \left (d x + c\right ) + \frac {5}{2}\right ) + 40 \, {\left (4975 \, \cos \left (d x + c\right )^{2} - 4770 \, \cos \left (d x + c\right ) + 1583\right )} \sin \left (d x + c\right )}{196608 \, {\left (125 \, d \cos \left (d x + c\right )^{3} - 225 \, d \cos \left (d x + c\right )^{2} + 135 \, d \cos \left (d x + c\right ) - 27 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/196608*(837*(125*cos(d*x + c)^3 - 225*cos(d*x + c)^2 + 135*cos(d*x + c) - 27)*log(-3/2*cos(d*x + c) + 2*sin(
d*x + c) + 5/2) - 837*(125*cos(d*x + c)^3 - 225*cos(d*x + c)^2 + 135*cos(d*x + c) - 27)*log(-3/2*cos(d*x + c)
- 2*sin(d*x + c) + 5/2) + 40*(4975*cos(d*x + c)^2 - 4770*cos(d*x + c) + 1583)*sin(d*x + c))/(125*d*cos(d*x + c
)^3 - 225*d*cos(d*x + c)^2 + 135*d*cos(d*x + c) - 27*d)

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giac [A]  time = 0.85, size = 97, normalized size = 0.70 \[ -\frac {\frac {20 \, {\left (2832 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1696 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 447 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (4 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}} - 837 \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 837 \, \log \left ({\left | 2 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{98304 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/98304*(20*(2832*tan(1/2*d*x + 1/2*c)^5 - 1696*tan(1/2*d*x + 1/2*c)^3 + 447*tan(1/2*d*x + 1/2*c))/(4*tan(1/2
*d*x + 1/2*c)^2 - 1)^3 - 837*log(abs(2*tan(1/2*d*x + 1/2*c) + 1)) + 837*log(abs(2*tan(1/2*d*x + 1/2*c) - 1)))/
d

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maple [A]  time = 0.04, size = 160, normalized size = 1.16 \[ -\frac {125}{49152 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {25}{8192 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {295}{32768 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {279 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32768 d}-\frac {125}{49152 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {25}{8192 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {295}{32768 d \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {279 \ln \left (2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32768 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/49152/d/(2*tan(1/2*d*x+1/2*c)+1)^3-25/8192/d/(2*tan(1/2*d*x+1/2*c)+1)^2-295/32768/d/(2*tan(1/2*d*x+1/2*c)
+1)+279/32768/d*ln(2*tan(1/2*d*x+1/2*c)+1)-125/49152/d/(2*tan(1/2*d*x+1/2*c)-1)^3+25/8192/d/(2*tan(1/2*d*x+1/2
*c)-1)^2-295/32768/d/(2*tan(1/2*d*x+1/2*c)-1)-279/32768/d*ln(2*tan(1/2*d*x+1/2*c)-1)

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maxima [A]  time = 1.11, size = 177, normalized size = 1.28 \[ -\frac {\frac {20 \, {\left (\frac {447 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {1696 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {2832 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{\frac {12 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {48 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {64 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - 1} - 837 \, \log \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right ) + 837 \, \log \left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{98304 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/98304*(20*(447*sin(d*x + c)/(cos(d*x + c) + 1) - 1696*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 2832*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*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 1) - 837*log(2*sin(d*x + c)/(cos(d*x + c) + 1) + 1) + 837*log(2*sin(d
*x + c)/(cos(d*x + c) + 1) - 1))/d

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mupad [B]  time = 0.00, size = 102, normalized size = 0.74 \[ \frac {279\,\mathrm {atanh}\left (2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{16384\,d}-\frac {\frac {295\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{32768}-\frac {265\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{49152}+\frac {745\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{524288}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{4}+\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{16}-\frac {1}{64}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(5*cos(c + d*x) - 3)^4,x)

[Out]

(279*atanh(2*tan(c/2 + (d*x)/2)))/(16384*d) - ((745*tan(c/2 + (d*x)/2))/524288 - (265*tan(c/2 + (d*x)/2)^3)/49
152 + (295*tan(c/2 + (d*x)/2)^5)/32768)/(d*((3*tan(c/2 + (d*x)/2)^2)/16 - (3*tan(c/2 + (d*x)/2)^4)/4 + tan(c/2
+ (d*x)/2)^6 - 1/64))

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sympy [A]  time = 5.28, size = 831, normalized size = 6.02 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x/(-3 + 5*cos(2*atan(1/2)))**4, Eq(c, -d*x - 2*atan(1/2)) | Eq(c, -d*x + 2*atan(1/2))), (x/(5*cos(c
) - 3)**4, Eq(d, 0)), (-53568*log(tan(c/2 + d*x/2) - 1/2)*tan(c/2 + d*x/2)**6/(6291456*d*tan(c/2 + d*x/2)**6 -
4718592*d*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 + d*x/2)**2 - 98304*d) + 40176*log(tan(c/2 + d*x/2) - 1/2)*
tan(c/2 + d*x/2)**4/(6291456*d*tan(c/2 + d*x/2)**6 - 4718592*d*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 + d*x/2
)**2 - 98304*d) - 10044*log(tan(c/2 + d*x/2) - 1/2)*tan(c/2 + d*x/2)**2/(6291456*d*tan(c/2 + d*x/2)**6 - 47185
92*d*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 + d*x/2)**2 - 98304*d) + 837*log(tan(c/2 + d*x/2) - 1/2)/(6291456
*d*tan(c/2 + d*x/2)**6 - 4718592*d*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 + d*x/2)**2 - 98304*d) + 53568*log(
tan(c/2 + d*x/2) + 1/2)*tan(c/2 + d*x/2)**6/(6291456*d*tan(c/2 + d*x/2)**6 - 4718592*d*tan(c/2 + d*x/2)**4 + 1
179648*d*tan(c/2 + d*x/2)**2 - 98304*d) - 40176*log(tan(c/2 + d*x/2) + 1/2)*tan(c/2 + d*x/2)**4/(6291456*d*tan
(c/2 + d*x/2)**6 - 4718592*d*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 + d*x/2)**2 - 98304*d) + 10044*log(tan(c/
2 + d*x/2) + 1/2)*tan(c/2 + d*x/2)**2/(6291456*d*tan(c/2 + d*x/2)**6 - 4718592*d*tan(c/2 + d*x/2)**4 + 1179648
*d*tan(c/2 + d*x/2)**2 - 98304*d) - 837*log(tan(c/2 + d*x/2) + 1/2)/(6291456*d*tan(c/2 + d*x/2)**6 - 4718592*d
*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 + d*x/2)**2 - 98304*d) - 56640*tan(c/2 + d*x/2)**5/(6291456*d*tan(c/2
+ d*x/2)**6 - 4718592*d*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 + d*x/2)**2 - 98304*d) + 33920*tan(c/2 + d*x/
2)**3/(6291456*d*tan(c/2 + d*x/2)**6 - 4718592*d*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 + d*x/2)**2 - 98304*d
) - 8940*tan(c/2 + d*x/2)/(6291456*d*tan(c/2 + d*x/2)**6 - 4718592*d*tan(c/2 + d*x/2)**4 + 1179648*d*tan(c/2 +
d*x/2)**2 - 98304*d), True))

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