Integrand size = 12, antiderivative size = 83 \[ \int \frac {1}{(a+a \cos (c+d x))^3} \, dx=\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {2 \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac {2 \sin (c+d x)}{15 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
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))
Time = 0.09 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78 \[ \int \frac {1}{(a+a \cos (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (10 \sin \left (\frac {1}{2} (c+d x)\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )+\sin \left (\frac {5}{2} (c+d x)\right )\right )}{15 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[(a + a*Cos[c + d*x])^(-3),x]
Output:
(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)
Time = 0.35 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 3129, 3042, 3129, 3042, 3127}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {1}{(a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {2 \int \frac {1}{(\cos (c+d x) a+a)^2}dx}{5 a}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{\cos (c+d x) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}+\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}+\frac {2 \left (\frac {\sin (c+d x)}{3 a d (a \cos (c+d x)+a)}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}\) |
Input:
Int[(a + a*Cos[c + d*x])^(-3),x]
Output:
Sin[c + d*x]/(5*d*(a + a*Cos[c + d*x])^3) + (2*(Sin[c + d*x]/(3*d*(a + a*C os[c + d*x])^2) + Sin[c + d*x]/(3*a*d*(a + a*Cos[c + d*x]))))/(5*a)
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]
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] + Simp[(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]
Time = 0.59 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(45\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(45\) |
risch | \(\frac {4 i \left (10 \,{\mathrm e}^{2 i \left (d x +c \right )}+5 \,{\mathrm e}^{i \left (d x +c \right )}+1\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(47\) |
parallelrisch | \(\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 a^{3} d}\) | \(47\) |
norman | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 a d}}{a^{2}}\) | \(61\) |
Input:
int(1/(a+a*cos(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
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))
Time = 0.08 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left (2 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 7\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(1/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
Output:
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)
Time = 0.60 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.76 \[ \int \frac {1}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {\tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{6 a^{3} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(1/(a+a*cos(d*x+c))**3,x)
Output:
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))
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int \frac {1}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{60 \, a^{3} d} \] Input:
integrate(1/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
Output:
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)
Time = 0.35 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.55 \[ \int \frac {1}{(a+a \cos (c+d x))^3} \, dx=\frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \] Input:
integrate(1/(a+a*cos(d*x+c))^3,x, algorithm="giac")
Output:
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)
Time = 41.12 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(a+a \cos (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15\right )}{60\,a^3\,d} \] Input:
int(1/(a + a*cos(c + d*x))^3,x)
Output:
(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)
Time = 0.15 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(a+a \cos (c+d x))^3} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+15\right )}{60 a^{3} d} \] Input:
int(1/(a+a*cos(d*x+c))^3,x)
Output:
(tan((c + d*x)/2)*(3*tan((c + d*x)/2)**4 + 10*tan((c + d*x)/2)**2 + 15))/( 60*a**3*d)