Integrand size = 31, antiderivative size = 90 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {i \cos ^3(c+d x)}{3 a^3 d}+\frac {4 i \cos ^5(c+d x)}{5 a^3 d}+\frac {\sin (c+d x)}{a^3 d}-\frac {5 \sin ^3(c+d x)}{3 a^3 d}+\frac {4 \sin ^5(c+d x)}{5 a^3 d} \]
-1/3*I*cos(d*x+c)^3/a^3/d+4/5*I*cos(d*x+c)^5/a^3/d+sin(d*x+c)/a^3/d-5/3*si n(d*x+c)^3/a^3/d+4/5*sin(d*x+c)^5/a^3/d
Time = 0.75 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.23 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {i \cos (c+d x)}{4 a^3 d}+\frac {i \cos (3 (c+d x))}{6 a^3 d}+\frac {i \cos (5 (c+d x))}{20 a^3 d}+\frac {\sin (c+d x)}{4 a^3 d}+\frac {\sin (3 (c+d x))}{6 a^3 d}+\frac {\sin (5 (c+d x))}{20 a^3 d} \]
((I/4)*Cos[c + d*x])/(a^3*d) + ((I/6)*Cos[3*(c + d*x)])/(a^3*d) + ((I/20)* Cos[5*(c + d*x)])/(a^3*d) + Sin[c + d*x]/(4*a^3*d) + Sin[3*(c + d*x)]/(6*a ^3*d) + Sin[5*(c + d*x)]/(20*a^3*d)
Time = 0.43 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.11, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3571, 3042, 3569, 2009}
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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^2}{(a \cos (c+d x)+i a \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3571 |
\(\displaystyle \frac {i \int \cos ^2(c+d x) (i a \cos (c+d x)+a \sin (c+d x))^3dx}{a^6}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \int \cos (c+d x)^2 (i a \cos (c+d x)+a \sin (c+d x))^3dx}{a^6}\) |
\(\Big \downarrow \) 3569 |
\(\displaystyle \frac {i \int \left (-i a^3 \cos ^5(c+d x)-3 a^3 \sin (c+d x) \cos ^4(c+d x)+3 i a^3 \sin ^2(c+d x) \cos ^3(c+d x)+a^3 \sin ^3(c+d x) \cos ^2(c+d x)\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i \left (-\frac {4 i a^3 \sin ^5(c+d x)}{5 d}+\frac {5 i a^3 \sin ^3(c+d x)}{3 d}-\frac {i a^3 \sin (c+d x)}{d}+\frac {4 a^3 \cos ^5(c+d x)}{5 d}-\frac {a^3 \cos ^3(c+d x)}{3 d}\right )}{a^6}\) |
(I*(-1/3*(a^3*Cos[c + d*x]^3)/d + (4*a^3*Cos[c + d*x]^5)/(5*d) - (I*a^3*Si n[c + d*x])/d + (((5*I)/3)*a^3*Sin[c + d*x]^3)/d - (((4*I)/5)*a^3*Sin[c + d*x]^5)/d))/a^6
3.2.78.3.1 Defintions of rubi rules used
Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si n[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandTrig[cos[c + d*x]^m*(a *cos[c + d*x] + b*sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d}, x] && Inte gerQ[m] && IGtQ[n, 0]
Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*si n[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a^n*b^n Int[Cos[c + d*x]^m /(b*Cos[c + d*x] + a*Sin[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, m}, x] & & EqQ[a^2 + b^2, 0] && ILtQ[n, 0]
Time = 0.66 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.62
method | result | size |
risch | \(\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{4 d \,a^{3}}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{6 d \,a^{3}}+\frac {i {\mathrm e}^{-5 i \left (d x +c \right )}}{20 d \,a^{3}}\) | \(56\) |
derivativedivides | \(\frac {\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}-\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}}{d \,a^{3}}\) | \(90\) |
default | \(\frac {\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}-\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}}{d \,a^{3}}\) | \(90\) |
norman | \(\frac {\frac {6 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {164 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 a d}-\frac {16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}+\frac {14 i}{15 a d}-\frac {4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3 a d}-\frac {4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a d}+\frac {40 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{3 a d}}{a^{2} \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}\) | \(203\) |
Time = 0.25 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.46 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {{\left (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{60 \, a^{3} d} \]
1/60*(15*I*e^(4*I*d*x + 4*I*c) + 10*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5*I*d *x - 5*I*c)/(a^3*d)
Time = 0.17 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.46 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\begin {cases} \frac {\left (120 i a^{6} d^{2} e^{8 i c} e^{- i d x} + 80 i a^{6} d^{2} e^{6 i c} e^{- 3 i d x} + 24 i a^{6} d^{2} e^{4 i c} e^{- 5 i d x}\right ) e^{- 9 i c}}{480 a^{9} d^{3}} & \text {for}\: a^{9} d^{3} e^{9 i c} \neq 0 \\\frac {x \left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 5 i c}}{4 a^{3}} & \text {otherwise} \end {cases} \]
Piecewise(((120*I*a**6*d**2*exp(8*I*c)*exp(-I*d*x) + 80*I*a**6*d**2*exp(6* I*c)*exp(-3*I*d*x) + 24*I*a**6*d**2*exp(4*I*c)*exp(-5*I*d*x))*exp(-9*I*c)/ (480*a**9*d**3), Ne(a**9*d**3*exp(9*I*c), 0)), (x*(exp(4*I*c) + 2*exp(2*I* c) + 1)*exp(-5*I*c)/(4*a**3), True))
Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 10 i \, \cos \left (3 \, d x + 3 \, c\right ) + 15 i \, \cos \left (d x + c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 10 \, \sin \left (3 \, d x + 3 \, c\right ) + 15 \, \sin \left (d x + c\right )}{60 \, a^{3} d} \]
1/60*(3*I*cos(5*d*x + 5*c) + 10*I*cos(3*d*x + 3*c) + 15*I*cos(d*x + c) + 3 *sin(5*d*x + 5*c) + 10*sin(3*d*x + 3*c) + 15*sin(d*x + c))/(a^3*d)
Time = 0.35 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7\right )}}{15 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}} \]
2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 30*I*tan(1/2*d*x + 1/2*c)^3 - 40*tan(1/2 *d*x + 1/2*c)^2 + 20*I*tan(1/2*d*x + 1/2*c) + 7)/(a^3*d*(tan(1/2*d*x + 1/2 *c) - I)^5)
Time = 23.00 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,15{}\mathrm {i}+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,40{}\mathrm {i}-20\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+7{}\mathrm {i}\right )}{15\,a^3\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5{}\mathrm {i}+1\right )} \]