Integrand size = 19, antiderivative size = 83 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac {\sin (c+d x)}{5 a d (a+a \cos (c+d x))^2}+\frac {\sin (c+d x)}{5 d \left (a^3+a^3 \cos (c+d x)\right )} \] Output:
-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))
Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.53 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\left (1+3 \cos (c+d x)+\cos ^2(c+d x)\right ) \sin (c+d x)}{5 a^3 d (1+\cos (c+d x))^3} \] Input:
Integrate[Cos[c + d*x]/(a + a*Cos[c + d*x])^3,x]
Output:
((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)
Time = 0.37 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 3229, 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.
\(\displaystyle \int \frac {\cos (c+d x)}{(a \cos (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 3229 |
\(\displaystyle \frac {3 \int \frac {1}{(\cos (c+d x) a+a)^2}dx}{5 a}-\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {1}{\left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}-\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\cos (c+d x) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}-\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}-\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3127 |
\(\displaystyle \frac {3 \left (\frac {\sin (c+d x)}{3 a d (a \cos (c+d x)+a)}+\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\right )}{5 a}-\frac {\sin (c+d x)}{5 d (a \cos (c+d x)+a)^3}\) |
Input:
Int[Cos[c + d*x]/(a + a*Cos[c + d*x])^3,x]
Output:
-1/5*Sin[c + d*x]/(d*(a + a*Cos[c + d*x])^3) + (3*(Sin[c + d*x]/(3*d*(a + a*Cos[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]
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] + Simp[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)) I nt[(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)]
Time = 0.77 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.39
method | result | size |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(32\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(32\) |
parallelrisch | \(-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{20 a^{3} d}\) | \(32\) |
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 | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{4 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{20 a d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a^{2}}\) | \(95\) |
Input:
int(cos(d*x+c)/(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+tan(1/2*d*x+1/2*c))
Time = 0.07 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {{\left (\cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{5 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(cos(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="fricas")
Output:
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)
Time = 0.70 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.58 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\begin {cases} - \frac {\tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{20 a^{3} d} + \frac {\tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \cos {\left (c \right )}}{\left (a \cos {\left (c \right )} + a\right )^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)/(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)/(4*a**3*d), Ne(d, 0)), (x*cos(c)/(a*cos(c) + a)**3, True))
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.57 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{20 \, a^{3} d} \] Input:
integrate(cos(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="maxima")
Output:
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)
Time = 0.34 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.37 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{20 \, a^{3} d} \] Input:
integrate(cos(d*x+c)/(a+a*cos(d*x+c))^3,x, algorithm="giac")
Output:
-1/20*(tan(1/2*d*x + 1/2*c)^5 - 5*tan(1/2*d*x + 1/2*c))/(a^3*d)
Time = 41.24 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.36 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-5\right )}{20\,a^3\,d} \] Input:
int(cos(c + d*x)/(a + a*cos(c + d*x))^3,x)
Output:
-(tan(c/2 + (d*x)/2)*(tan(c/2 + (d*x)/2)^4 - 5))/(20*a^3*d)
Time = 0.15 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.39 \[ \int \frac {\cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+5\right )}{20 a^{3} d} \] Input:
int(cos(d*x+c)/(a+a*cos(d*x+c))^3,x)
Output:
(tan((c + d*x)/2)*( - tan((c + d*x)/2)**4 + 5))/(20*a**3*d)