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]

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

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Rubi [A]  time = 0.06, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 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]

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)]

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*}

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Mathematica [A]  time = 0.14, 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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fricas [A]  time = 0.76, size = 73, normalized size = 0.88 \[ \frac {{\left (\cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{5 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

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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giac [A]  time = 0.41, size = 31, normalized size = 0.37 \[ -\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{20 \, a^{3} d} \]

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)

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maple [A]  time = 0.04, size = 32, normalized size = 0.39 \[ \frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)/(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+tan(1/2*d*x+1/2*c))

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maxima [A]  time = 0.48, size = 47, normalized size = 0.57 \[ \frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{20 \, a^{3} d} \]

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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mupad [B]  time = 0.34, size = 30, normalized size = 0.36 \[ -\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-5\right )}{20\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(tan(c/2 + (d*x)/2)*(tan(c/2 + (d*x)/2)^4 - 5))/(20*a^3*d)

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sympy [A]  time = 2.28, size = 48, normalized size = 0.58 \[ \begin {cases} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos {\relax (c )}}{\left (a \cos {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

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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