3.55 \(\int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=80 \[ \frac {4 \sin (c+d x)}{3 a^2 d}+\frac {2 \sin (c+d x)}{a^2 d (\cos (c+d x)+1)}-\frac {2 x}{a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

-2*x/a^2+4/3*sin(d*x+c)/a^2/d+2*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*cos(d*x+c))
^2

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Rubi [A]  time = 0.17, antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2765, 2968, 3023, 12, 2735, 2648} \[ \frac {4 \sin (c+d x)}{3 a^2 d}+\frac {2 \sin (c+d x)}{a^2 d (\cos (c+d x)+1)}-\frac {2 x}{a^2}-\frac {\sin (c+d x) \cos ^2(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^2,x]

[Out]

(-2*x)/a^2 + (4*Sin[c + d*x])/(3*a^2*d) + (2*Sin[c + d*x])/(a^2*d*(1 + Cos[c + d*x])) - (Cos[c + d*x]^2*Sin[c
+ d*x])/(3*d*(a + a*Cos[c + d*x])^2)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 3023

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

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {\int \frac {\cos (c+d x) (2 a-4 a \cos (c+d x))}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {\int \frac {2 a \cos (c+d x)-4 a \cos ^2(c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=\frac {4 \sin (c+d x)}{3 a^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {\int \frac {6 a^2 \cos (c+d x)}{a+a \cos (c+d x)} \, dx}{3 a^3}\\ &=\frac {4 \sin (c+d x)}{3 a^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac {2 \int \frac {\cos (c+d x)}{a+a \cos (c+d x)} \, dx}{a}\\ &=-\frac {2 x}{a^2}+\frac {4 \sin (c+d x)}{3 a^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {2 \int \frac {1}{a+a \cos (c+d x)} \, dx}{a}\\ &=-\frac {2 x}{a^2}+\frac {4 \sin (c+d x)}{3 a^2 d}-\frac {\cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac {2 \sin (c+d x)}{d \left (a^2+a^2 \cos (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 114, normalized size = 1.42 \[ -\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (-6 (\sin (c+d x)-2 d x) \cos ^3\left (\frac {1}{2} (c+d x)\right )+\tan \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )+\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )-16 \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^2,x]

[Out]

(-2*Cos[(c + d*x)/2]*(Sec[c/2]*Sin[(d*x)/2] - 16*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] - 6*Cos[(c + d*x)/2]
^3*(-2*d*x + Sin[c + d*x]) + Cos[(c + d*x)/2]*Tan[c/2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 1.34, size = 90, normalized size = 1.12 \[ -\frac {6 \, d x \cos \left (d x + c\right )^{2} + 12 \, d x \cos \left (d x + c\right ) + 6 \, d x - {\left (3 \, \cos \left (d x + c\right )^{2} + 14 \, \cos \left (d x + c\right ) + 10\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(6*d*x*cos(d*x + c)^2 + 12*d*x*cos(d*x + c) + 6*d*x - (3*cos(d*x + c)^2 + 14*cos(d*x + c) + 10)*sin(d*x +
c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [A]  time = 0.46, size = 79, normalized size = 0.99 \[ -\frac {\frac {12 \, {\left (d x + c\right )}}{a^{2}} - \frac {12 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} a^{2}} + \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

-1/6*(12*(d*x + c)/a^2 - 12*tan(1/2*d*x + 1/2*c)/((tan(1/2*d*x + 1/2*c)^2 + 1)*a^2) + (a^4*tan(1/2*d*x + 1/2*c
)^3 - 15*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [A]  time = 0.06, size = 88, normalized size = 1.10 \[ -\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d \,a^{2}}+\frac {5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+a*cos(d*x+c))^2,x)

[Out]

-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3+5/2/d/a^2*tan(1/2*d*x+1/2*c)+2/d/a^2*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1/2*c)^
2)-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))

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maxima [A]  time = 1.50, size = 118, normalized size = 1.48 \[ \frac {\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

1/6*((15*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(
cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)))
/d

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mupad [B]  time = 0.42, size = 91, normalized size = 1.14 \[ -\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-16\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+12\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (c+d\,x\right )}{6\,a^2\,d\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a + a*cos(c + d*x))^2,x)

[Out]

-(sin(c/2 + (d*x)/2) - 16*cos(c/2 + (d*x)/2)^2*sin(c/2 + (d*x)/2) - 12*cos(c/2 + (d*x)/2)^4*sin(c/2 + (d*x)/2)
+ 12*cos(c/2 + (d*x)/2)^3*(c + d*x))/(6*a^2*d*cos(c/2 + (d*x)/2)^3)

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sympy [A]  time = 3.64, size = 201, normalized size = 2.51 \[ \begin {cases} - \frac {12 d x \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {12 d x}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {14 \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} + \frac {27 \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{2} d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 6 a^{2} d} & \text {for}\: d \neq 0 \\\frac {x \cos ^{3}{\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{2}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a+a*cos(d*x+c))**2,x)

[Out]

Piecewise((-12*d*x*tan(c/2 + d*x/2)**2/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) - 12*d*x/(6*a**2*d*tan(c/2 +
d*x/2)**2 + 6*a**2*d) - tan(c/2 + d*x/2)**5/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 14*tan(c/2 + d*x/2)**3
/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d) + 27*tan(c/2 + d*x/2)/(6*a**2*d*tan(c/2 + d*x/2)**2 + 6*a**2*d), Ne
(d, 0)), (x*cos(c)**3/(a*cos(c) + a)**2, True))

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