Integrand size = 29, antiderivative size = 52 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {2 i \cos ^3(c+d x)}{3 a^2 d}+\frac {\sin (c+d x)}{a^2 d}-\frac {2 \sin ^3(c+d x)}{3 a^2 d} \] Output:
2/3*I*cos(d*x+c)^3/a^2/d+sin(d*x+c)/a^2/d-2/3*sin(d*x+c)^3/a^2/d
Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.40 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {i \cos (c+d x)}{2 a^2 d}+\frac {i \cos (3 (c+d x))}{6 a^2 d}+\frac {\sin (c+d x)}{2 a^2 d}+\frac {\sin (3 (c+d x))}{6 a^2 d} \] Input:
Integrate[Cos[c + d*x]/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]
Output:
((I/2)*Cos[c + d*x])/(a^2*d) + ((I/6)*Cos[3*(c + d*x)])/(a^2*d) + Sin[c + d*x]/(2*a^2*d) + Sin[3*(c + d*x)]/(6*a^2*d)
Time = 0.35 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.12, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, 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.
\(\displaystyle \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3571 |
\(\displaystyle -\frac {\int \cos (c+d x) (i a \cos (c+d x)+a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \cos (c+d x) (i a \cos (c+d x)+a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3569 |
\(\displaystyle -\frac {\int \left (-a^2 \cos ^3(c+d x)+2 i a^2 \sin (c+d x) \cos ^2(c+d x)+a^2 \sin ^2(c+d x) \cos (c+d x)\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {2 a^2 \sin ^3(c+d x)}{3 d}-\frac {a^2 \sin (c+d x)}{d}-\frac {2 i a^2 \cos ^3(c+d x)}{3 d}}{a^4}\) |
Input:
Int[Cos[c + d*x]/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]
Output:
-(((((-2*I)/3)*a^2*Cos[c + d*x]^3)/d - (a^2*Sin[c + d*x])/d + (2*a^2*Sin[c + d*x]^3)/(3*d))/a^4)
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 = 0.46 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.73
method | result | size |
risch | \(\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}+\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{6 a^{2} d}\) | \(38\) |
derivativedivides | \(\frac {\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}+\frac {2 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}}{a^{2} d}\) | \(57\) |
default | \(\frac {\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}+\frac {2 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {4}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}}{a^{2} d}\) | \(57\) |
norman | \(\frac {\frac {4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a d}+\frac {4 i}{3 a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d a}}{a \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}\) | \(105\) |
orering | \(\frac {4 i \cos \left (d x +c \right )}{3 d \left (a \cos \left (d x +c \right )+i a \sin \left (d x +c \right )\right )^{2}}+\frac {-\frac {d \sin \left (d x +c \right )}{\left (a \cos \left (d x +c \right )+i a \sin \left (d x +c \right )\right )^{2}}-\frac {2 \cos \left (d x +c \right ) \left (-a d \sin \left (d x +c \right )+i a d \cos \left (d x +c \right )\right )}{\left (a \cos \left (d x +c \right )+i a \sin \left (d x +c \right )\right )^{3}}}{3 d^{2}}\) | \(122\) |
Input:
int(cos(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/2*I/a^2/d*exp(-I*(d*x+c))+1/6*I/a^2/d*exp(-3*I*(d*x+c))
Time = 0.07 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.58 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {{\left (3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{6 \, a^{2} d} \] Input:
integrate(cos(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas" )
Output:
1/6*(3*I*e^(2*I*d*x + 2*I*c) + I)*e^(-3*I*d*x - 3*I*c)/(a^2*d)
Time = 0.15 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.77 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\begin {cases} \frac {\left (6 i a^{2} d e^{3 i c} e^{- i d x} + 2 i a^{2} d e^{i c} e^{- 3 i d x}\right ) e^{- 4 i c}}{12 a^{4} d^{2}} & \text {for}\: a^{4} d^{2} e^{4 i c} \neq 0 \\\frac {x \left (e^{2 i c} + 1\right ) e^{- 3 i c}}{2 a^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)
Output:
Piecewise(((6*I*a**2*d*exp(3*I*c)*exp(-I*d*x) + 2*I*a**2*d*exp(I*c)*exp(-3 *I*d*x))*exp(-4*I*c)/(12*a**4*d**2), Ne(a**4*d**2*exp(4*I*c), 0)), (x*(exp (2*I*c) + 1)*exp(-3*I*c)/(2*a**2), True))
Time = 0.04 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.87 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {i \, \cos \left (3 \, d x + 3 \, c\right ) + 3 i \, \cos \left (d x + c\right ) + \sin \left (3 \, d x + 3 \, c\right ) + 3 \, \sin \left (d x + c\right )}{6 \, a^{2} d} \] Input:
integrate(cos(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima" )
Output:
1/6*(I*cos(3*d*x + 3*c) + 3*I*cos(d*x + c) + sin(3*d*x + 3*c) + 3*sin(d*x + c))/(a^2*d)
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.90 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {2 \, {\left (3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2\right )}}{3 \, a^{2} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{3}} \] Input:
integrate(cos(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")
Output:
2/3*(3*tan(1/2*d*x + 1/2*c)^2 - 3*I*tan(1/2*d*x + 1/2*c) - 2)/(a^2*d*(tan( 1/2*d*x + 1/2*c) - I)^3)
Time = 17.03 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.52 \[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2{}\mathrm {i}\right )}{3\,a^2\,d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,1{}\mathrm {i}-3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,3{}\mathrm {i}+1\right )} \] Input:
int(cos(c + d*x)/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^2,x)
Output:
-(2*(3*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2*3i - 2i))/(3*a^2*d*(tan(c /2 + (d*x)/2)*3i - 3*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*1i + 1))
\[ \int \frac {\cos (c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i -\sin \left (d x +c \right )^{2}}d x}{a^{2}} \] Input:
int(cos(d*x+c)/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x)
Output:
int(cos(c + d*x)/(cos(c + d*x)**2 + 2*cos(c + d*x)*sin(c + d*x)*i - sin(c + d*x)**2),x)/a**2