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

Optimal. Leaf size=143 \[ \frac{2 \sin (c+d x)}{63 d \left (a^5 \cos (c+d x)+a^5\right )}+\frac{2 \sin (c+d x)}{63 a d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac{\sin (c+d x)}{21 a^2 d (a \cos (c+d x)+a)^3}+\frac{5 \sin (c+d x)}{63 a d (a \cos (c+d x)+a)^4}-\frac{\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5} \]

[Out]

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

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Rubi [A]  time = 0.105471, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2750, 2650, 2648} \[ \frac{2 \sin (c+d x)}{63 d \left (a^5 \cos (c+d x)+a^5\right )}+\frac{2 \sin (c+d x)}{63 a d \left (a^2 \cos (c+d x)+a^2\right )^2}+\frac{\sin (c+d x)}{21 a^2 d (a \cos (c+d x)+a)^3}+\frac{5 \sin (c+d x)}{63 a d (a \cos (c+d x)+a)^4}-\frac{\sin (c+d x)}{9 d (a \cos (c+d x)+a)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.17237, size = 97, normalized size = 0.68 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-63 \sin \left (c+\frac{d x}{2}\right )+84 \sin \left (c+\frac{3 d x}{2}\right )+36 \sin \left (2 c+\frac{5 d x}{2}\right )+9 \sin \left (3 c+\frac{7 d x}{2}\right )+\sin \left (4 c+\frac{9 d x}{2}\right )+63 \sin \left (\frac{d x}{2}\right )\right ) \sec ^9\left (\frac{1}{2} (c+d x)\right )}{8064 a^5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^9*(63*Sin[(d*x)/2] - 63*Sin[c + (d*x)/2] + 84*Sin[c + (3*d*x)/2] + 36*Sin[2*c + (5*
d*x)/2] + 9*Sin[3*c + (7*d*x)/2] + Sin[4*c + (9*d*x)/2]))/(8064*a^5*d)

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Maple [A]  time = 0.037, 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)/(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 = 1.16703, size = 117, normalized size = 0.82 \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)/(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 = 1.56325, size = 312, normalized size = 2.18 \begin{align*} \frac{{\left (2 \, \cos \left (d x + c\right )^{4} + 10 \, \cos \left (d x + c\right )^{3} + 21 \, \cos \left (d x + c\right )^{2} + 25 \, \cos \left (d x + c\right ) + 5\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)/(a+a*cos(d*x+c))^5,x, algorithm="fricas")

[Out]

1/63*(2*cos(d*x + c)^4 + 10*cos(d*x + c)^3 + 21*cos(d*x + c)^2 + 25*cos(d*x + c) + 5)*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 [A]  time = 19.2459, size = 85, normalized size = 0.59 \begin{align*} \begin{cases} - \frac{\tan ^{9}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{144 a^{5} d} - \frac{\tan ^{7}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{56 a^{5} d} + \frac{\tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{24 a^{5} d} + \frac{\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{16 a^{5} d} & \text{for}\: d \neq 0 \\\frac{x \cos{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{5}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-tan(c/2 + d*x/2)**9/(144*a**5*d) - tan(c/2 + d*x/2)**7/(56*a**5*d) + tan(c/2 + d*x/2)**3/(24*a**5*
d) + tan(c/2 + d*x/2)/(16*a**5*d), Ne(d, 0)), (x*cos(c)/(a*cos(c) + a)**5, True))

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Giac [A]  time = 1.37065, size = 80, normalized size = 0.56 \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)/(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)