\(\int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\) [164]

Optimal result

Integrand size = 31, antiderivative size = 101 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {x}{4 a^2}-\frac {1}{16 a^2 d (i-\cot (c+d x))}-\frac {1}{12 a^2 d (i+\cot (c+d x))^3}-\frac {3 i}{8 a^2 d (i+\cot (c+d x))^2}+\frac {11}{16 a^2 d (i+\cot (c+d x))} \]

[Out]

1/4*x/a^2-1/16/a^2/d/(I-cot(d*x+c))-1/12/a^2/d/(I+cot(d*x+c))^3-3/8*I/a^2/d/(I+cot(d*x+c))^2+11/16/a^2/d/(I+co
t(d*x+c))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {3167, 862, 90, 209} \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {1}{16 a^2 d (-\cot (c+d x)+i)}+\frac {11}{16 a^2 d (\cot (c+d x)+i)}-\frac {3 i}{8 a^2 d (\cot (c+d x)+i)^2}-\frac {1}{12 a^2 d (\cot (c+d x)+i)^3}+\frac {x}{4 a^2} \]

[In]

Int[Cos[c + d*x]^4/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

x/(4*a^2) - 1/(16*a^2*d*(I - Cot[c + d*x])) - 1/(12*a^2*d*(I + Cot[c + d*x])^3) - ((3*I)/8)/(a^2*d*(I + Cot[c
+ d*x])^2) + 11/(16*a^2*d*(I + Cot[c + d*x]))

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 862

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c/e)*x)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 3167

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[-d^(-1), Subst[Int[x^m*((b + a*x)^n/(1 + x^2)^((m + n + 2)/2)), x], x, Cot[c + d*x]], x] /; FreeQ[
{a, b, c, d}, x] && IntegerQ[n] && IntegerQ[(m + n)/2] && NeQ[n, -1] &&  !(GtQ[n, 0] && GtQ[m, 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \frac {x^4}{(i a+a x)^2 \left (1+x^2\right )^2} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \frac {x^4}{\left (-\frac {i}{a}+\frac {x}{a}\right )^2 (i a+a x)^4} \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {\text {Subst}\left (\int \left (\frac {1}{16 a^2 (-i+x)^2}-\frac {1}{4 a^2 (i+x)^4}-\frac {3 i}{4 a^2 (i+x)^3}+\frac {11}{16 a^2 (i+x)^2}+\frac {1}{4 a^2 \left (1+x^2\right )}\right ) \, dx,x,\cot (c+d x)\right )}{d} \\ & = -\frac {1}{16 a^2 d (i-\cot (c+d x))}-\frac {1}{12 a^2 d (i+\cot (c+d x))^3}-\frac {3 i}{8 a^2 d (i+\cot (c+d x))^2}+\frac {11}{16 a^2 d (i+\cot (c+d x))}-\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\cot (c+d x)\right )}{4 a^2 d} \\ & = \frac {x}{4 a^2}-\frac {1}{16 a^2 d (i-\cot (c+d x))}-\frac {1}{12 a^2 d (i+\cot (c+d x))^3}-\frac {3 i}{8 a^2 d (i+\cot (c+d x))^2}+\frac {11}{16 a^2 d (i+\cot (c+d x))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.77 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {24 c+24 d x+15 i \cos (2 (c+d x))+6 i \cos (4 (c+d x))+i \cos (6 (c+d x))+21 \sin (2 (c+d x))+6 \sin (4 (c+d x))+\sin (6 (c+d x))}{96 a^2 d} \]

[In]

Integrate[Cos[c + d*x]^4/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

(24*c + 24*d*x + (15*I)*Cos[2*(c + d*x)] + (6*I)*Cos[4*(c + d*x)] + I*Cos[6*(c + d*x)] + 21*Sin[2*(c + d*x)] +
6*Sin[4*(c + d*x)] + Sin[6*(c + d*x)])/(96*a^2*d)

Maple [A] (verified)

Time = 0.82 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.78

method	result	size
risch	\(\frac {x}{4 a^{2}}+\frac {i {\mathrm e}^{-4 i \left (d x +c \right )}}{16 a^{2} d}+\frac {i {\mathrm e}^{-6 i \left (d x +c \right )}}{96 a^{2} d}+\frac {5 i \cos \left (2 d x +2 c \right )}{32 a^{2} d}+\frac {7 \sin \left (2 d x +2 c \right )}{32 a^{2} d}\)	\(79\)
derivativedivides	\(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {3}{16 \left (\tan \left (d x +c \right )-i\right )}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8}+\frac {1}{16 \tan \left (d x +c \right )+16 i}}{d \,a^{2}}\)	\(88\)
default	\(\frac {-\frac {i \ln \left (\tan \left (d x +c \right )-i\right )}{8}-\frac {i}{8 \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{12 \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {3}{16 \left (\tan \left (d x +c \right )-i\right )}+\frac {i \ln \left (\tan \left (d x +c \right )+i\right )}{8}+\frac {1}{16 \tan \left (d x +c \right )+16 i}}{d \,a^{2}}\)	\(88\)

[In]

int(cos(d*x+c)^4/(cos(d*x+c)*a+I*a*sin(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/4*x/a^2+1/16*I/a^2/d*exp(-4*I*(d*x+c))+1/96*I/a^2/d*exp(-6*I*(d*x+c))+5/32*I/a^2/d*cos(2*d*x+2*c)+7/32/a^2/d
*sin(2*d*x+2*c)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {{\left (24 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 18 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{96 \, a^{2} d} \]

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/96*(24*d*x*e^(6*I*d*x + 6*I*c) - 3*I*e^(8*I*d*x + 8*I*c) + 18*I*e^(4*I*d*x + 4*I*c) + 6*I*e^(2*I*d*x + 2*I*c
) + I)*e^(-6*I*d*x - 6*I*c)/(a^2*d)

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.87 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\begin {cases} \frac {\left (- 24576 i a^{6} d^{3} e^{14 i c} e^{2 i d x} + 147456 i a^{6} d^{3} e^{10 i c} e^{- 2 i d x} + 49152 i a^{6} d^{3} e^{8 i c} e^{- 4 i d x} + 8192 i a^{6} d^{3} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{786432 a^{8} d^{4}} & \text {for}\: a^{8} d^{4} e^{12 i c} \neq 0 \\x \left (\frac {\left (e^{8 i c} + 4 e^{6 i c} + 6 e^{4 i c} + 4 e^{2 i c} + 1\right ) e^{- 6 i c}}{16 a^{2}} - \frac {1}{4 a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {x}{4 a^{2}} \]

[In]

integrate(cos(d*x+c)**4/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)

[Out]

Piecewise(((-24576*I*a**6*d**3*exp(14*I*c)*exp(2*I*d*x) + 147456*I*a**6*d**3*exp(10*I*c)*exp(-2*I*d*x) + 49152
*I*a**6*d**3*exp(8*I*c)*exp(-4*I*d*x) + 8192*I*a**6*d**3*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(786432*a**8*d
**4), Ne(a**8*d**4*exp(12*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(-6*
I*c)/(16*a**2) - 1/(4*a**2)), True)) + x/(4*a**2)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negative exponent.

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=-\frac {-\frac {6 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a^{2}} + \frac {6 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a^{2}} + \frac {3 \, {\left (2 i \, \tan \left (d x + c\right ) - 3\right )}}{a^{2} {\left (\tan \left (d x + c\right ) + i\right )}} + \frac {-11 i \, \tan \left (d x + c\right )^{3} - 42 \, \tan \left (d x + c\right )^{2} + 57 i \, \tan \left (d x + c\right ) + 30}{a^{2} {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{48 \, d} \]

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

-1/48*(-6*I*log(tan(d*x + c) + I)/a^2 + 6*I*log(tan(d*x + c) - I)/a^2 + 3*(2*I*tan(d*x + c) - 3)/(a^2*(tan(d*x
+ c) + I)) + (-11*I*tan(d*x + c)^3 - 42*tan(d*x + c)^2 + 57*I*tan(d*x + c) + 30)/(a^2*(tan(d*x + c) - I)^3))/
d

Mupad [B] (verification not implemented)

Time = 27.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.37 \[ \int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx=\frac {x}{4\,a^2}-\frac {-\frac {3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{2}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,2{}\mathrm {i}+\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{6}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,4{}\mathrm {i}}{3}-\frac {7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,2{}\mathrm {i}+\frac {3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{a^2\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^2\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^6} \]

[In]

int(cos(c + d*x)^4/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^2,x)

[Out]

x/(4*a^2) - ((3*tan(c/2 + (d*x)/2))/2 + tan(c/2 + (d*x)/2)^2*2i - (7*tan(c/2 + (d*x)/2)^3)/6 + (tan(c/2 + (d*x
)/2)^4*4i)/3 + (7*tan(c/2 + (d*x)/2)^5)/6 + tan(c/2 + (d*x)/2)^6*2i - (3*tan(c/2 + (d*x)/2)^7)/2)/(a^2*d*(tan(
c/2 + (d*x)/2) + 1i)^2*(tan(c/2 + (d*x)/2)*1i + 1)^6)