Integrand size = 31, antiderivative size = 68 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {2 i \cos ^5(c+d x)}{5 a^2 d}+\frac {\sin (c+d x)}{a^2 d}-\frac {\sin ^3(c+d x)}{a^2 d}+\frac {2 \sin ^5(c+d x)}{5 a^2 d} \] Output:
2/5*I*cos(d*x+c)^5/a^2/d+sin(d*x+c)/a^2/d-sin(d*x+c)^3/a^2/d+2/5*sin(d*x+c )^5/a^2/d
Time = 0.57 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.63 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {i \cos (c+d x)}{4 a^2 d}+\frac {i \cos (3 (c+d x))}{8 a^2 d}+\frac {i \cos (5 (c+d x))}{40 a^2 d}+\frac {\sin (c+d x)}{2 a^2 d}+\frac {\sin (3 (c+d x))}{8 a^2 d}+\frac {\sin (5 (c+d x))}{40 a^2 d} \] Input:
Integrate[Cos[c + d*x]^3/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]
Output:
((I/4)*Cos[c + d*x])/(a^2*d) + ((I/8)*Cos[3*(c + d*x)])/(a^2*d) + ((I/40)* Cos[5*(c + d*x)])/(a^2*d) + Sin[c + d*x]/(2*a^2*d) + Sin[3*(c + d*x)]/(8*a ^2*d) + Sin[5*(c + d*x)]/(40*a^2*d)
Time = 0.42 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^3}{(a \cos (c+d x)+i a \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3571 |
\(\displaystyle -\frac {\int \cos ^3(c+d x) (i a \cos (c+d x)+a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {\int \cos (c+d x)^3 (i a \cos (c+d x)+a \sin (c+d x))^2dx}{a^4}\) |
\(\Big \downarrow \) 3569 |
\(\displaystyle -\frac {\int \left (-a^2 \cos ^5(c+d x)+2 i a^2 \sin (c+d x) \cos ^4(c+d x)+a^2 \sin ^2(c+d x) \cos ^3(c+d x)\right )dx}{a^4}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {2 a^2 \sin ^5(c+d x)}{5 d}+\frac {a^2 \sin ^3(c+d x)}{d}-\frac {a^2 \sin (c+d x)}{d}-\frac {2 i a^2 \cos ^5(c+d x)}{5 d}}{a^4}\) |
Input:
Int[Cos[c + d*x]^3/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]
Output:
-(((((-2*I)/5)*a^2*Cos[c + d*x]^5)/d - (a^2*Sin[c + d*x])/d + (a^2*Sin[c + d*x]^3)/d - (2*a^2*Sin[c + d*x]^5)/(5*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 = 1.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.99
method | result | size |
risch | \(\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{8 a^{2} d}+\frac {i {\mathrm e}^{-5 i \left (d x +c \right )}}{40 a^{2} d}+\frac {i \cos \left (d x +c \right )}{4 a^{2} d}+\frac {\sin \left (d x +c \right )}{2 a^{2} d}\) | \(67\) |
derivativedivides | \(\frac {-\frac {2 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {5 i}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}+\frac {2}{8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 i}}{a^{2} d}\) | \(108\) |
default | \(\frac {-\frac {2 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {5 i}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {4}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {3}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {7}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}+\frac {2}{8 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+8 i}}{a^{2} d}\) | \(108\) |
parallelrisch | \(\frac {\frac {4 i}{5}-4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-\frac {6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{5}}{a^{2} d \left (-4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+4 i \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-5 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right )}\) | \(130\) |
orering | \(\text {Expression too large to display}\) | \(901\) |
Input:
int(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/8*I/a^2/d*exp(-3*I*(d*x+c))+1/40*I/a^2/d*exp(-5*I*(d*x+c))+1/4*I/a^2/d*c os(d*x+c)+1/2*sin(d*x+c)/a^2/d
Time = 0.06 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.76 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {{\left (-5 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{2} d} \] Input:
integrate(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/40*(-5*I*e^(6*I*d*x + 6*I*c) + 15*I*e^(4*I*d*x + 4*I*c) + 5*I*e^(2*I*d*x + 2*I*c) + I)*e^(-5*I*d*x - 5*I*c)/(a^2*d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (60) = 120\).
Time = 0.22 (sec) , antiderivative size = 163, normalized size of antiderivative = 2.40 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 2560 i a^{6} d^{3} e^{10 i c} e^{i d x} + 7680 i a^{6} d^{3} e^{8 i c} e^{- i d x} + 2560 i a^{6} d^{3} e^{6 i c} e^{- 3 i d x} + 512 i a^{6} d^{3} e^{4 i c} e^{- 5 i d x}\right ) e^{- 9 i c}}{20480 a^{8} d^{4}} & \text {for}\: a^{8} d^{4} e^{9 i c} \neq 0 \\\frac {x \left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 5 i c}}{8 a^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**3/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)
Output:
Piecewise(((-2560*I*a**6*d**3*exp(10*I*c)*exp(I*d*x) + 7680*I*a**6*d**3*ex p(8*I*c)*exp(-I*d*x) + 2560*I*a**6*d**3*exp(6*I*c)*exp(-3*I*d*x) + 512*I*a **6*d**3*exp(4*I*c)*exp(-5*I*d*x))*exp(-9*I*c)/(20480*a**8*d**4), Ne(a**8* d**4*exp(9*I*c), 0)), (x*(exp(6*I*c) + 3*exp(4*I*c) + 3*exp(2*I*c) + 1)*ex p(-5*I*c)/(8*a**2), True))
Exception generated. \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.15 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {\frac {5}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}} + \frac {35 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 90 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 70 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21}{a^{2} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}}}{20 \, d} \] Input:
integrate(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/20*(5/(a^2*(tan(1/2*d*x + 1/2*c) + I)) + (35*tan(1/2*d*x + 1/2*c)^4 - 90 *I*tan(1/2*d*x + 1/2*c)^3 - 120*tan(1/2*d*x + 1/2*c)^2 + 70*I*tan(1/2*d*x + 1/2*c) + 21)/(a^2*(tan(1/2*d*x + 1/2*c) - I)^5))/d
Time = 17.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.32 \[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {2\,\left (-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,10{}\mathrm {i}+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-2{}\mathrm {i}\right )}{5\,a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-\mathrm {i}\right )}^5\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \] Input:
int(cos(c + d*x)^3/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^2,x)
Output:
-(2*(3*tan(c/2 + (d*x)/2) + 10*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4 *10i - 5*tan(c/2 + (d*x)/2)^5 - 2i))/(5*a^2*d*(tan(c/2 + (d*x)/2) - 1i)^5* (tan(c/2 + (d*x)/2) + 1i))
\[ \int \frac {\cos ^3(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {\int \frac {\cos \left (d x +c \right )^{3}}{\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right ) \sin \left (d x +c \right ) i -\sin \left (d x +c \right )^{2}}d x}{a^{2}} \] Input:
int(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x)
Output:
int(cos(c + d*x)**3/(cos(c + d*x)**2 + 2*cos(c + d*x)*sin(c + d*x)*i - sin (c + d*x)**2),x)/a**2