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

Optimal. Leaf size=101 \[ -\frac{1}{16 a^2 d (-\cot (c+d x)+i)}+\frac{11}{16 a^2 d (\cot (c+d x)+i)}-\frac{3 i}{8 a^2 d (\cot (c+d x)+i)^2}-\frac{1}{12 a^2 d (\cot (c+d x)+i)^3}+\frac{x}{4 a^2} \]

[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]))

________________________________________________________________________________________

Rubi [A]  time = 0.100526, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {3088, 848, 88, 203} \[ -\frac{1}{16 a^2 d (-\cot (c+d x)+i)}+\frac{11}{16 a^2 d (\cot (c+d x)+i)}-\frac{3 i}{8 a^2 d (\cot (c+d x)+i)^2}-\frac{1}{12 a^2 d (\cot (c+d x)+i)^3}+\frac{x}{4 a^2} \]

Antiderivative was successfully verified.

[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 3088

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])

Rule 848

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*x)/e)^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 88

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 203

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

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx &=-\frac{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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]  time = 0.127354, size = 82, normalized size = 0.81 \[ \frac{21 \sin (2 (c+d x))+6 \sin (4 (c+d x))+\sin (6 (c+d x))+15 i \cos (2 (c+d x))+6 i \cos (4 (c+d x))+i \cos (6 (c+d x))+24 c+24 d x}{96 a^2 d} \]

Antiderivative was successfully verified.

[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]  time = 0.134, size = 117, normalized size = 1.2 \begin{align*}{\frac{-{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{d{a}^{2}}}-{\frac{{\frac{i}{8}}}{d{a}^{2} \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{1}{12\,d{a}^{2} \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\frac{3}{16\,d{a}^{2} \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{i}{8}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{d{a}^{2}}}+{\frac{1}{16\,d{a}^{2} \left ( \tan \left ( dx+c \right ) +i \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*I/d/a^2*ln(tan(d*x+c)-I)-1/8*I/d/a^2/(tan(d*x+c)-I)^2-1/12/d/a^2/(tan(d*x+c)-I)^3+3/16/d/a^2/(tan(d*x+c)-
I)+1/8*I/d/a^2*ln(tan(d*x+c)+I)+1/16/d/a^2/(tan(d*x+c)+I)

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

Fricas [A]  time = 0.470143, size = 198, normalized size = 1.96 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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]  time = 0.58235, size = 190, normalized size = 1.88 \begin{align*} \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}\: 786432 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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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(786432*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)

________________________________________________________________________________________

Giac [A]  time = 1.1448, size = 139, normalized size = 1.38 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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