Integrand size = 31, antiderivative size = 75 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {3 x}{8 a}+\frac {i \cos ^4(c+d x)}{4 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d} \] Output:
3/8*x/a+1/4*I*cos(d*x+c)^4/a/d+3/8*cos(d*x+c)*sin(d*x+c)/a/d+1/4*cos(d*x+c )^3*sin(d*x+c)/a/d
Time = 0.47 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {12 c+12 d x+4 i \cos (2 (c+d x))+i \cos (4 (c+d x))+8 \sin (2 (c+d x))+\sin (4 (c+d x))}{32 a d} \] Input:
Integrate[Cos[c + d*x]^3/(a*Cos[c + d*x] + I*a*Sin[c + d*x]),x]
Output:
(12*c + 12*d*x + (4*I)*Cos[2*(c + d*x)] + I*Cos[4*(c + d*x)] + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])/(32*a*d)
Time = 0.35 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.04, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3}{a \cos (c+d x)+i a \sin (c+d x)}dx\) |
\(\Big \downarrow \) 3571 |
\(\displaystyle -\frac {i \int \cos ^3(c+d x) (i a \cos (c+d x)+a \sin (c+d x))dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {i \int \cos (c+d x)^3 (i a \cos (c+d x)+a \sin (c+d x))dx}{a^2}\) |
\(\Big \downarrow \) 3569 |
\(\displaystyle -\frac {i \int \left (i a \cos ^4(c+d x)+a \sin (c+d x) \cos ^3(c+d x)\right )dx}{a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {a \cos ^4(c+d x)}{4 d}+\frac {i a \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac {3 i a \sin (c+d x) \cos (c+d x)}{8 d}+\frac {3 i a x}{8}\right )}{a^2}\) |
Input:
Int[Cos[c + d*x]^3/(a*Cos[c + d*x] + I*a*Sin[c + d*x]),x]
Output:
((-I)*(((3*I)/8)*a*x - (a*Cos[c + d*x]^4)/(4*d) + (((3*I)/8)*a*Cos[c + d*x ]*Sin[c + d*x])/d + ((I/4)*a*Cos[c + d*x]^3*Sin[c + d*x])/d))/a^2
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 = 0.77 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.80
method | result | size |
parallelrisch | \(\frac {12 d x -21 i+i \cos \left (4 d x +4 c \right )+4 i \cos \left (2 d x +2 c \right )+\sin \left (4 d x +4 c \right )+8 \sin \left (2 d x +2 c \right )}{32 a d}\) | \(60\) |
risch | \(\frac {3 x}{8 a}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{32 a d}+\frac {i \cos \left (2 d x +2 c \right )}{8 a d}+\frac {\sin \left (2 d x +2 c \right )}{4 a d}\) | \(61\) |
derivativedivides | \(\frac {-\frac {3 i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 \tan \left (d x +c \right )-4 i}+\frac {3 i \ln \left (\tan \left (d x +c \right )+i\right )}{16}+\frac {1}{8 \tan \left (d x +c \right )+8 i}}{d a}\) | \(75\) |
default | \(\frac {-\frac {3 i \ln \left (\tan \left (d x +c \right )-i\right )}{16}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 \tan \left (d x +c \right )-4 i}+\frac {3 i \ln \left (\tan \left (d x +c \right )+i\right )}{16}+\frac {1}{8 \tan \left (d x +c \right )+8 i}}{d a}\) | \(75\) |
orering | \(\text {Expression too large to display}\) | \(923\) |
Input:
int(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/32*(12*d*x-21*I+I*cos(4*d*x+4*c)+4*I*cos(2*d*x+2*c)+sin(4*d*x+4*c)+8*sin (2*d*x+2*c))/a/d
Time = 0.07 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {{\left (12 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a d} \] Input:
integrate(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="fricas" )
Output:
1/32*(12*d*x*e^(4*I*d*x + 4*I*c) - 2*I*e^(6*I*d*x + 6*I*c) + 6*I*e^(2*I*d* x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)/(a*d)
Time = 0.18 (sec) , antiderivative size = 151, normalized size of antiderivative = 2.01 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\begin {cases} \frac {\left (- 512 i a^{2} d^{2} e^{8 i c} e^{2 i d x} + 1536 i a^{2} d^{2} e^{4 i c} e^{- 2 i d x} + 256 i a^{2} d^{2} e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{8192 a^{3} d^{3}} & \text {for}\: a^{3} d^{3} e^{6 i c} \neq 0 \\x \left (\frac {\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 4 i c}}{8 a} - \frac {3}{8 a}\right ) & \text {otherwise} \end {cases} + \frac {3 x}{8 a} \] Input:
integrate(cos(d*x+c)**3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)
Output:
Piecewise(((-512*I*a**2*d**2*exp(8*I*c)*exp(2*I*d*x) + 1536*I*a**2*d**2*ex p(4*I*c)*exp(-2*I*d*x) + 256*I*a**2*d**2*exp(2*I*c)*exp(-4*I*d*x))*exp(-6* I*c)/(8192*a**3*d**3), Ne(a**3*d**3*exp(6*I*c), 0)), (x*((exp(6*I*c) + 3*e xp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-4*I*c)/(8*a) - 3/(8*a)), True)) + 3*x/( 8*a)
Exception generated. \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="maxima" )
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.27 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {-\frac {6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac {6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {2 \, {\left (3 \, \tan \left (d x + c\right ) + 5 i\right )}}{a {\left (-i \, \tan \left (d x + c\right ) + 1\right )}} + \frac {-9 i \, \tan \left (d x + c\right )^{2} - 26 \, \tan \left (d x + c\right ) + 21 i}{a {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{32 \, d} \] Input:
integrate(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="giac")
Output:
-1/32*(-6*I*log(tan(d*x + c) + I)/a + 6*I*log(tan(d*x + c) - I)/a + 2*(3*t an(d*x + c) + 5*I)/(a*(-I*tan(d*x + c) + 1)) + (-9*I*tan(d*x + c)^2 - 26*t an(d*x + c) + 21*I)/(a*(tan(d*x + c) - I)^2))/d
Time = 19.96 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {3\,x}{8\,a}-\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,1{}\mathrm {i}}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,1{}\mathrm {i}}{2}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^2\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^4} \] Input:
int(cos(c + d*x)^3/(a*cos(c + d*x) + a*sin(c + d*x)*1i),x)
Output:
(3*x)/(8*a) - ((5*tan(c/2 + (d*x)/2))/4 + (tan(c/2 + (d*x)/2)^2*1i)/2 - ta n(c/2 + (d*x)/2)^3/2 - (tan(c/2 + (d*x)/2)^4*1i)/2 + (5*tan(c/2 + (d*x)/2) ^5)/4)/(a*d*(tan(c/2 + (d*x)/2) + 1i)^2*(tan(c/2 + (d*x)/2)*1i + 1)^4)
\[ \int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx=\frac {\int \frac {\cos \left (d x +c \right )^{3}}{\cos \left (d x +c \right )+\sin \left (d x +c \right ) i}d x}{a} \] Input:
int(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)
Output:
int(cos(c + d*x)**3/(cos(c + d*x) + sin(c + d*x)*i),x)/a