Integrand size = 31, antiderivative size = 106 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {i \cos ^5(c+d x)}{5 a^3 d}+\frac {4 i \cos ^7(c+d x)}{7 a^3 d}+\frac {\sin (c+d x)}{a^3 d}-\frac {2 \sin ^3(c+d x)}{a^3 d}+\frac {9 \sin ^5(c+d x)}{5 a^3 d}-\frac {4 \sin ^7(c+d x)}{7 a^3 d} \] Output:
-1/5*I*cos(d*x+c)^5/a^3/d+4/7*I*cos(d*x+c)^7/a^3/d+sin(d*x+c)/a^3/d-2*sin( d*x+c)^3/a^3/d+9/5*sin(d*x+c)^5/a^3/d-4/7*sin(d*x+c)^7/a^3/d
Time = 0.61 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.41 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {3 i \cos (c+d x)}{16 a^3 d}+\frac {i \cos (3 (c+d x))}{8 a^3 d}+\frac {i \cos (5 (c+d x))}{20 a^3 d}+\frac {i \cos (7 (c+d x))}{112 a^3 d}+\frac {5 \sin (c+d x)}{16 a^3 d}+\frac {\sin (3 (c+d x))}{8 a^3 d}+\frac {\sin (5 (c+d x))}{20 a^3 d}+\frac {\sin (7 (c+d x))}{112 a^3 d} \] Input:
Integrate[Cos[c + d*x]^4/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]
Output:
(((3*I)/16)*Cos[c + d*x])/(a^3*d) + ((I/8)*Cos[3*(c + d*x)])/(a^3*d) + ((I /20)*Cos[5*(c + d*x)])/(a^3*d) + ((I/112)*Cos[7*(c + d*x)])/(a^3*d) + (5*S in[c + d*x])/(16*a^3*d) + Sin[3*(c + d*x)]/(8*a^3*d) + Sin[5*(c + d*x)]/(2 0*a^3*d) + Sin[7*(c + d*x)]/(112*a^3*d)
Time = 0.48 (sec) , antiderivative size = 118, 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.
\(\displaystyle \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{(a \cos (c+d x)+i a \sin (c+d x))^3}dx\) |
\(\Big \downarrow \) 3571 |
\(\displaystyle \frac {i \int \cos ^4(c+d x) (i a \cos (c+d x)+a \sin (c+d x))^3dx}{a^6}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {i \int \cos (c+d x)^4 (i a \cos (c+d x)+a \sin (c+d x))^3dx}{a^6}\) |
\(\Big \downarrow \) 3569 |
\(\displaystyle \frac {i \int \left (-i a^3 \cos ^7(c+d x)-3 a^3 \sin (c+d x) \cos ^6(c+d x)+3 i a^3 \sin ^2(c+d x) \cos ^5(c+d x)+a^3 \sin ^3(c+d x) \cos ^4(c+d x)\right )dx}{a^6}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i \left (\frac {4 i a^3 \sin ^7(c+d x)}{7 d}-\frac {9 i a^3 \sin ^5(c+d x)}{5 d}+\frac {2 i a^3 \sin ^3(c+d x)}{d}-\frac {i a^3 \sin (c+d x)}{d}+\frac {4 a^3 \cos ^7(c+d x)}{7 d}-\frac {a^3 \cos ^5(c+d x)}{5 d}\right )}{a^6}\) |
Input:
Int[Cos[c + d*x]^4/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]
Output:
(I*(-1/5*(a^3*Cos[c + d*x]^5)/d + (4*a^3*Cos[c + d*x]^7)/(7*d) - (I*a^3*Si n[c + d*x])/d + ((2*I)*a^3*Sin[c + d*x]^3)/d - (((9*I)/5)*a^3*Sin[c + d*x] ^5)/d + (((4*I)/7)*a^3*Sin[c + d*x]^7)/d))/a^6
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 = 1.43 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.80
method | result | size |
risch | \(\frac {i {\mathrm e}^{-3 i \left (d x +c \right )}}{8 a^{3} d}+\frac {i {\mathrm e}^{-5 i \left (d x +c \right )}}{20 a^{3} d}+\frac {i {\mathrm e}^{-7 i \left (d x +c \right )}}{112 a^{3} d}+\frac {3 i \cos \left (d x +c \right )}{16 a^{3} d}+\frac {5 \sin \left (d x +c \right )}{16 a^{3} d}\) | \(85\) |
derivativedivides | \(\frac {\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{6}}-\frac {9 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {17 i}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{7}}+\frac {38}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {15}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}+\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}}{a^{3} d}\) | \(141\) |
default | \(\frac {\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{6}}-\frac {9 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {17 i}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{7}}+\frac {38}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {15}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}+\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}}{a^{3} d}\) | \(141\) |
orering | \(\text {Expression too large to display}\) | \(1944\) |
Input:
int(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/8*I/a^3/d*exp(-3*I*(d*x+c))+1/20*I/a^3/d*exp(-5*I*(d*x+c))+1/112*I/a^3/d *exp(-7*I*(d*x+c))+3/16*I/a^3/d*cos(d*x+c)+5/16*sin(d*x+c)/a^3/d
Time = 0.06 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.59 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {{\left (-35 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 140 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 28 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{560 \, a^{3} d} \] Input:
integrate(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="frica s")
Output:
1/560*(-35*I*e^(8*I*d*x + 8*I*c) + 140*I*e^(6*I*d*x + 6*I*c) + 70*I*e^(4*I *d*x + 4*I*c) + 28*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-7*I*d*x - 7*I*c)/(a^3* d)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 197 vs. \(2 (95) = 190\).
Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.86 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\begin {cases} \frac {\left (- 71680 i a^{12} d^{4} e^{17 i c} e^{i d x} + 286720 i a^{12} d^{4} e^{15 i c} e^{- i d x} + 143360 i a^{12} d^{4} e^{13 i c} e^{- 3 i d x} + 57344 i a^{12} d^{4} e^{11 i c} e^{- 5 i d x} + 10240 i a^{12} d^{4} e^{9 i c} e^{- 7 i d x}\right ) e^{- 16 i c}}{1146880 a^{15} d^{5}} & \text {for}\: a^{15} d^{5} e^{16 i c} \neq 0 \\\frac {x \left (e^{8 i c} + 4 e^{6 i c} + 6 e^{4 i c} + 4 e^{2 i c} + 1\right ) e^{- 7 i c}}{16 a^{3}} & \text {otherwise} \end {cases} \] Input:
integrate(cos(d*x+c)**4/(a*cos(d*x+c)+I*a*sin(d*x+c))**3,x)
Output:
Piecewise(((-71680*I*a**12*d**4*exp(17*I*c)*exp(I*d*x) + 286720*I*a**12*d* *4*exp(15*I*c)*exp(-I*d*x) + 143360*I*a**12*d**4*exp(13*I*c)*exp(-3*I*d*x) + 57344*I*a**12*d**4*exp(11*I*c)*exp(-5*I*d*x) + 10240*I*a**12*d**4*exp(9 *I*c)*exp(-7*I*d*x))*exp(-16*I*c)/(1146880*a**15*d**5), Ne(a**15*d**5*exp( 16*I*c), 0)), (x*(exp(8*I*c) + 4*exp(6*I*c) + 6*exp(4*I*c) + 4*exp(2*I*c) + 1)*exp(-7*I*c)/(16*a**3), True))
Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="maxim a")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.16 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {\frac {35}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}} + \frac {525 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1960 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4025 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4480 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3143 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1176 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 243}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{7}}}{280 \, d} \] Input:
integrate(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="giac" )
Output:
1/280*(35/(a^3*(tan(1/2*d*x + 1/2*c) + I)) + (525*tan(1/2*d*x + 1/2*c)^6 - 1960*I*tan(1/2*d*x + 1/2*c)^5 - 4025*tan(1/2*d*x + 1/2*c)^4 + 4480*I*tan( 1/2*d*x + 1/2*c)^3 + 3143*tan(1/2*d*x + 1/2*c)^2 - 1176*I*tan(1/2*d*x + 1/ 2*c) - 243)/(a^3*(tan(1/2*d*x + 1/2*c) - I)^7))/d
Time = 20.77 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=-\frac {\left (35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,105{}\mathrm {i}-175\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,105{}\mathrm {i}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,77{}\mathrm {i}+43\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-13{}\mathrm {i}\right )\,2{}\mathrm {i}}{35\,a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^7} \] Input:
int(cos(c + d*x)^4/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^3,x)
Output:
-((43*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2*77i - 7*tan(c/2 + (d*x)/2) ^3 + tan(c/2 + (d*x)/2)^4*105i - 175*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x )/2)^6*105i + 35*tan(c/2 + (d*x)/2)^7 - 13i)*2i)/(35*a^3*d*(tan(c/2 + (d*x )/2) + 1i)*(tan(c/2 + (d*x)/2)*1i + 1)^7)
\[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx=\frac {\int \frac {\cos \left (d x +c \right )^{4}}{\cos \left (d x +c \right )^{3}+3 \cos \left (d x +c \right )^{2} \sin \left (d x +c \right ) i -3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2}-\sin \left (d x +c \right )^{3} i}d x}{a^{3}} \] Input:
int(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x)
Output:
int(cos(c + d*x)**4/(cos(c + d*x)**3 + 3*cos(c + d*x)**2*sin(c + d*x)*i - 3*cos(c + d*x)*sin(c + d*x)**2 - sin(c + d*x)**3*i),x)/a**3