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

Optimal. Leaf size=83 \[ \frac{\sin (c+d x)}{5 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{\sin (c+d x)}{5 a d (a \cos (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.0581915, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2750, 2650, 2648} \[ \frac{\sin (c+d x)}{5 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{\sin (c+d x)}{5 a d (a \cos (c+d x)+a)^2}-\frac{\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-Sin[c + d*x]/(5*d*(a + a*Cos[c + d*x])^3) + Sin[c + d*x]/(5*a*d*(a + a*Cos[c + d*x])^2) + Sin[c + d*x]/(5*d*(
a^3 + a^3*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))^3} \, dx &=-\frac{\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{3 \int \frac{1}{(a+a \cos (c+d x))^2} \, dx}{5 a}\\ &=-\frac{\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{1}{a+a \cos (c+d x)} \, dx}{5 a^2}\\ &=-\frac{\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac{\sin (c+d x)}{5 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.136478, size = 71, normalized size = 0.86 \[ \frac{\sec \left (\frac{c}{2}\right ) \left (-5 \sin \left (c+\frac{d x}{2}\right )+5 \sin \left (c+\frac{3 d x}{2}\right )+\sin \left (2 c+\frac{5 d x}{2}\right )+5 \sin \left (\frac{d x}{2}\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )}{80 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(5*Sin[(d*x)/2] - 5*Sin[c + (d*x)/2] + 5*Sin[c + (3*d*x)/2] + Sin[2*c + (5*d*x)/2
]))/(80*a^3*d)

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Maple [A]  time = 0.036, size = 32, normalized size = 0.4 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ( -{\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+\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)^3,x)

[Out]

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

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Maxima [A]  time = 1.14184, size = 63, normalized size = 0.76 \begin{align*} \frac{\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{20 \, a^{3} 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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.51877, size = 182, normalized size = 2.19 \begin{align*} \frac{{\left (\cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{5 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*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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Sympy [A]  time = 3.41239, size = 48, normalized size = 0.58 \begin{align*} \begin{cases} - \frac{\tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{20 a^{3} d} + \frac{\tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \cos{\left (c \right )}}{\left (a \cos{\left (c \right )} + a\right )^{3}} & \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))**3,x)

[Out]

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

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Giac [A]  time = 1.14257, size = 42, normalized size = 0.51 \begin{align*} -\frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{20 \, a^{3} 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))^3,x, algorithm="giac")

[Out]

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