3.68 \(\int \frac{1}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=83 \[ \frac{2 \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{2 \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac{\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.0463566, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2650, 2648} \[ \frac{2 \sin (c+d x)}{15 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{2 \sin (c+d x)}{15 a d (a \cos (c+d x)+a)^2}+\frac{\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])^(-3),x]

[Out]

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

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

Mathematica [A]  time = 0.0771736, size = 65, normalized size = 0.78 \[ \frac{\left (10 \sin \left (\frac{1}{2} (c+d x)\right )+5 \sin \left (\frac{3}{2} (c+d x)\right )+\sin \left (\frac{5}{2} (c+d x)\right )\right ) \cos \left (\frac{1}{2} (c+d x)\right )}{15 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(10*Sin[(c + d*x)/2] + 5*Sin[(3*(c + d*x))/2] + Sin[(5*(c + d*x))/2]))/(15*a^3*d*(1 + Cos[c
+ d*x])^3)

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Maple [A]  time = 0.034, size = 45, normalized size = 0.5 \begin{align*}{\frac{1}{4\,d{a}^{3}} \left ({\frac{1}{5} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\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(1/(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+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 = 1.16851, size = 90, normalized size = 1.08 \begin{align*} \frac{\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \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}}}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*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*d)

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Fricas [A]  time = 1.51177, size = 186, normalized size = 2.24 \begin{align*} \frac{{\left (2 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 7\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

1/15*(2*cos(d*x + c)^2 + 6*cos(d*x + c) + 7)*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 = 2.38436, size = 63, normalized size = 0.76 \begin{align*} \begin{cases} \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 )}}{6 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}{\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(1/(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)**3/(6*a**3*d) + tan(c/2 + d*x/2)/(4*a**3*d), Ne(
d, 0)), (x/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.14434, size = 62, normalized size = 0.75 \begin{align*} \frac{3 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 10 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(3*tan(1/2*d*x + 1/2*c)^5 + 10*tan(1/2*d*x + 1/2*c)^3 + 15*tan(1/2*d*x + 1/2*c))/(a^3*d)