3.86 \(\int \frac{\cos ^3(c+d x)}{(a+a \cos (c+d x))^5} \, dx\)

Optimal. Leaf size=147 \[ \frac{5 \sin (c+d x)}{63 d \left (a^5 \cos (c+d x)+a^5\right )}+\frac{5 \sin (c+d x)}{63 a^3 d (a \cos (c+d x)+a)^2}-\frac{17 \sin (c+d x)}{63 a^2 d (a \cos (c+d x)+a)^3}-\frac{\sin (c+d x) \cos ^2(c+d x)}{9 d (a \cos (c+d x)+a)^5}+\frac{\sin (c+d x)}{7 a d (a \cos (c+d x)+a)^4} \]

[Out]

-(Cos[c + d*x]^2*Sin[c + d*x])/(9*d*(a + a*Cos[c + d*x])^5) + Sin[c + d*x]/(7*a*d*(a + a*Cos[c + d*x])^4) - (1
7*Sin[c + d*x])/(63*a^2*d*(a + a*Cos[c + d*x])^3) + (5*Sin[c + d*x])/(63*a^3*d*(a + a*Cos[c + d*x])^2) + (5*Si
n[c + d*x])/(63*d*(a^5 + a^5*Cos[c + d*x]))

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Rubi [A]  time = 0.228439, antiderivative size = 147, normalized size of antiderivative = 1., 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, 3019, 2750, 2650, 2648} \[ \frac{5 \sin (c+d x)}{63 d \left (a^5 \cos (c+d x)+a^5\right )}+\frac{5 \sin (c+d x)}{63 a^3 d (a \cos (c+d x)+a)^2}-\frac{17 \sin (c+d x)}{63 a^2 d (a \cos (c+d x)+a)^3}-\frac{\sin (c+d x) \cos ^2(c+d x)}{9 d (a \cos (c+d x)+a)^5}+\frac{\sin (c+d x)}{7 a d (a \cos (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a + a*Cos[c + d*x])^5,x]

[Out]

-(Cos[c + d*x]^2*Sin[c + d*x])/(9*d*(a + a*Cos[c + d*x])^5) + Sin[c + d*x]/(7*a*d*(a + a*Cos[c + d*x])^4) - (1
7*Sin[c + d*x])/(63*a^2*d*(a + a*Cos[c + d*x])^3) + (5*Sin[c + d*x])/(63*a^3*d*(a + a*Cos[c + d*x])^2) + (5*Si
n[c + d*x])/(63*d*(a^5 + a^5*Cos[c + d*x]))

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]

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2650

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

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]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x)}{(a+a \cos (c+d x))^5} \, dx &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{\int \frac{\cos (c+d x) (2 a-7 a \cos (c+d x))}{(a+a \cos (c+d x))^4} \, dx}{9 a^2}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}-\frac{\int \frac{2 a \cos (c+d x)-7 a \cos ^2(c+d x)}{(a+a \cos (c+d x))^4} \, dx}{9 a^2}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac{\sin (c+d x)}{7 a d (a+a \cos (c+d x))^4}+\frac{\int \frac{-36 a^2+49 a^2 \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx}{63 a^4}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac{\sin (c+d x)}{7 a d (a+a \cos (c+d x))^4}-\frac{17 \sin (c+d x)}{63 a^2 d (a+a \cos (c+d x))^3}+\frac{5 \int \frac{1}{(a+a \cos (c+d x))^2} \, dx}{21 a^3}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac{\sin (c+d x)}{7 a d (a+a \cos (c+d x))^4}-\frac{17 \sin (c+d x)}{63 a^2 d (a+a \cos (c+d x))^3}+\frac{5 \sin (c+d x)}{63 a^3 d (a+a \cos (c+d x))^2}+\frac{5 \int \frac{1}{a+a \cos (c+d x)} \, dx}{63 a^4}\\ &=-\frac{\cos ^2(c+d x) \sin (c+d x)}{9 d (a+a \cos (c+d x))^5}+\frac{\sin (c+d x)}{7 a d (a+a \cos (c+d x))^4}-\frac{17 \sin (c+d x)}{63 a^2 d (a+a \cos (c+d x))^3}+\frac{5 \sin (c+d x)}{63 a^3 d (a+a \cos (c+d x))^2}+\frac{5 \sin (c+d x)}{63 d \left (a^5+a^5 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.229102, size = 125, normalized size = 0.85 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-315 \sin \left (c+\frac{d x}{2}\right )+273 \sin \left (c+\frac{3 d x}{2}\right )-147 \sin \left (2 c+\frac{3 d x}{2}\right )+117 \sin \left (2 c+\frac{5 d x}{2}\right )-63 \sin \left (3 c+\frac{5 d x}{2}\right )+45 \sin \left (3 c+\frac{7 d x}{2}\right )+5 \sin \left (4 c+\frac{9 d x}{2}\right )+315 \sin \left (\frac{d x}{2}\right )\right ) \sec ^9\left (\frac{1}{2} (c+d x)\right )}{16128 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^9*(315*Sin[(d*x)/2] - 315*Sin[c + (d*x)/2] + 273*Sin[c + (3*d*x)/2] - 147*Sin[2*c +
(3*d*x)/2] + 117*Sin[2*c + (5*d*x)/2] - 63*Sin[3*c + (5*d*x)/2] + 45*Sin[3*c + (7*d*x)/2] + 5*Sin[4*c + (9*d*
x)/2]))/(16128*a^5*d)

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Maple [A]  time = 0.042, size = 58, normalized size = 0.4 \begin{align*}{\frac{1}{16\,d{a}^{5}} \left ( -{\frac{1}{9} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{9}}+{\frac{2}{7} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{7}}-{\frac{2}{3} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}+\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d/a^5*(-1/9*tan(1/2*d*x+1/2*c)^9+2/7*tan(1/2*d*x+1/2*c)^7-2/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1/2*c))

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Maxima [A]  time = 1.13758, size = 117, normalized size = 0.8 \begin{align*} \frac{\frac{63 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{42 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{18 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{7 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}}}{1008 \, a^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1008*(63*sin(d*x + c)/(cos(d*x + c) + 1) - 42*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 18*sin(d*x + c)^7/(cos(d
*x + c) + 1)^7 - 7*sin(d*x + c)^9/(cos(d*x + c) + 1)^9)/(a^5*d)

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Fricas [A]  time = 1.52687, size = 312, normalized size = 2.12 \begin{align*} \frac{{\left (5 \, \cos \left (d x + c\right )^{4} + 25 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 10 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{63 \,{\left (a^{5} d \cos \left (d x + c\right )^{5} + 5 \, a^{5} d \cos \left (d x + c\right )^{4} + 10 \, a^{5} d \cos \left (d x + c\right )^{3} + 10 \, a^{5} d \cos \left (d x + c\right )^{2} + 5 \, a^{5} d \cos \left (d x + c\right ) + a^{5} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/63*(5*cos(d*x + c)^4 + 25*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 10*cos(d*x + c) + 2)*sin(d*x + c)/(a^5*d*cos(
d*x + c)^5 + 5*a^5*d*cos(d*x + c)^4 + 10*a^5*d*cos(d*x + c)^3 + 10*a^5*d*cos(d*x + c)^2 + 5*a^5*d*cos(d*x + c)
+ a^5*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.32374, size = 80, normalized size = 0.54 \begin{align*} -\frac{7 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 18 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 42 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 63 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{1008 \, a^{5} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1008*(7*tan(1/2*d*x + 1/2*c)^9 - 18*tan(1/2*d*x + 1/2*c)^7 + 42*tan(1/2*d*x + 1/2*c)^3 - 63*tan(1/2*d*x + 1
/2*c))/(a^5*d)