\(\int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx\) [67]

Optimal result

Integrand size = 19, antiderivative size = 83 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac {\sin (c+d x)}{5 d \left (a^3+a^3 \cos (c+d x)\right )} \]

[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
))

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2829, 2729, 2727} \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\sin (c+d x)}{5 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {\sin (c+d x)}{5 a d (a \cos (c+d x)+a)^2}-\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3} \]

[In]

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

[Out]

-1/5*Sin[c + d*x]/(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 2727

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 2729

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 2829

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*} \text {integral}& = -\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] (verified)

Time = 0.07 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.53 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (1+3 \cos (c+d x)+\cos ^2(c+d x)\right ) \sin (c+d x)}{5 a^3 d (1+\cos (c+d x))^3} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.84 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.37

method	result	size
parallelrisch	\(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )-5\right )}{20 a^{3} d}\)	\(31\)
derivativedivides	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(32\)
default	\(\frac {-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\)	\(32\)
risch	\(\frac {2 i \left (5 \,{\mathrm e}^{3 i \left (d x +c \right )}+5 \,{\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{5 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\)	\(58\)
norman	\(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}+\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d a}-\frac {\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d a}-\frac {\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )}{20 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}\)	\(95\)

[In]

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

[Out]

-1/20*tan(1/2*d*x+1/2*c)*(tan(1/2*d*x+1/2*c)^4-5)/a^3/d

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left (\cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{5 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.81 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.58 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.57 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.57 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{20 \, a^{3} d} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.37 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{20 \, a^{3} d} \]

[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)

Mupad [B] (verification not implemented)

Time = 14.32 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.36 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-5\right )}{20\,a^3\,d} \]

[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)