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]

1/5*sin(d*x+c)/d/(a+a*cos(d*x+c))^3+2/15*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^2+2/15*sin(d*x+c)/d/(a^3+a^3*cos(d*x+
c))

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Rubi [A]  time = 0.05, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.08, 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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fricas [A]  time = 0.77, size = 75, normalized size = 0.90 \[ \frac {{\left (2 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 7\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

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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giac [A]  time = 0.46, size = 46, normalized size = 0.55 \[ \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \]

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)

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maple [A]  time = 0.04, size = 45, normalized size = 0.54 \[ \frac {\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*cos(d*x+c))^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 = 0.33, size = 67, normalized size = 0.81 \[ \frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{60 \, a^{3} d} \]

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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mupad [B]  time = 0.35, size = 45, normalized size = 0.54 \[ \frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15\right )}{60\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*cos(c + d*x))^3,x)

[Out]

(tan(c/2 + (d*x)/2)*(10*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + 15))/(60*a^3*d)

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sympy [A]  time = 1.63, size = 63, normalized size = 0.76 \[ \begin {cases} \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \cos {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

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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