3.65 \(\int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=96 \[ -\frac {29 \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {x}{a^3}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {7 \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

[Out]

x/a^3-1/5*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+7/15*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2-29/15*sin(d*x+c)
/d/(a^3+a^3*cos(d*x+c))

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Rubi [A]  time = 0.18, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2765, 2968, 3019, 2735, 2648} \[ -\frac {29 \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {x}{a^3}-\frac {\sin (c+d x) \cos ^2(c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {7 \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^3,x]

[Out]

x/a^3 - (Cos[c + d*x]^2*Sin[c + d*x])/(5*d*(a + a*Cos[c + d*x])^3) + (7*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d
*x])^2) - (29*Sin[c + d*x])/(15*d*(a^3 + a^3*Cos[c + d*x]))

Rule 2648

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

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3019

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_
.)*(x_)]^2), x_Symbol] :> Simp[((A*b - a*B + b*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + D
ist[1/(a^2*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[a*A*(m + 1) + m*(b*B - a*C) + b*C*(2*m + 1)*Sin[e
+ f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {\int \frac {\cos (c+d x) (2 a-5 a \cos (c+d x))}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}-\frac {\int \frac {2 a \cos (c+d x)-5 a \cos ^2(c+d x)}{(a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {\int \frac {-14 a^2+15 a^2 \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{15 a^4}\\ &=\frac {x}{a^3}-\frac {\cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {29 \int \frac {1}{a+a \cos (c+d x)} \, dx}{15 a^2}\\ &=\frac {x}{a^3}-\frac {\cos ^2(c+d x) \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {7 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac {29 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 154, normalized size = 1.60 \[ \frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (60 d x \cos ^5\left (\frac {1}{2} (c+d x)\right )+26 \tan \left (\frac {c}{2}\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )-3 \tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )-3 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-128 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )+26 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{15 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^3,x]

[Out]

(2*Cos[(c + d*x)/2]*(60*d*x*Cos[(c + d*x)/2]^5 - 3*Sec[c/2]*Sin[(d*x)/2] + 26*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[
(d*x)/2] - 128*Cos[(c + d*x)/2]^4*Sec[c/2]*Sin[(d*x)/2] - 3*Cos[(c + d*x)/2]*Tan[c/2] + 26*Cos[(c + d*x)/2]^3*
Tan[c/2]))/(15*a^3*d*(1 + Cos[c + d*x])^3)

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fricas [A]  time = 1.75, size = 116, normalized size = 1.21 \[ \frac {15 \, d x \cos \left (d x + c\right )^{3} + 45 \, d x \cos \left (d x + c\right )^{2} + 45 \, d x \cos \left (d x + c\right ) + 15 \, d x - {\left (32 \, \cos \left (d x + c\right )^{2} + 51 \, \cos \left (d x + c\right ) + 22\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(15*d*x*cos(d*x + c)^3 + 45*d*x*cos(d*x + c)^2 + 45*d*x*cos(d*x + c) + 15*d*x - (32*cos(d*x + c)^2 + 51*c
os(d*x + c) + 22)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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giac [A]  time = 0.47, size = 68, normalized size = 0.71 \[ \frac {\frac {60 \, {\left (d x + c\right )}}{a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 20 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(60*(d*x + c)/a^3 - (3*a^12*tan(1/2*d*x + 1/2*c)^5 - 20*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d

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maple [A]  time = 0.06, size = 75, normalized size = 0.78 \[ -\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d \,a^{3}}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{3 d \,a^{3}}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x)

[Out]

-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5+1/3/d/a^3*tan(1/2*d*x+1/2*c)^3-7/4/d/a^3*tan(1/2*d*x+1/2*c)+2/d/a^3*arctan(ta
n(1/2*d*x+1/2*c))

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maxima [A]  time = 0.69, size = 92, normalized size = 0.96 \[ -\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3)/d

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mupad [B]  time = 0.42, size = 81, normalized size = 0.84 \[ \frac {x}{a^3}-\frac {\frac {32\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{15}-\frac {13\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{30}+\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{20}}{a^3\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + a*cos(c + d*x))^3,x)

[Out]

x/a^3 - (sin(c/2 + (d*x)/2)/20 - (13*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2))/30 + (32*cos(c/2 + (d*x)/2)^4*si
n(c/2 + (d*x)/2))/15)/(a^3*d*cos(c/2 + (d*x)/2)^5)

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sympy [A]  time = 5.15, size = 75, normalized size = 0.78 \[ \begin {cases} \frac {x}{a^{3}} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{3 a^{3} d} - \frac {7 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((x/a**3 - tan(c/2 + d*x/2)**5/(20*a**3*d) + tan(c/2 + d*x/2)**3/(3*a**3*d) - 7*tan(c/2 + d*x/2)/(4*a
**3*d), Ne(d, 0)), (x*cos(c)**3/(a*cos(c) + a)**3, True))

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