3.163 \(\int \frac{\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=85 \[ -\frac{2 \sin ^7(c+d x)}{7 a^2 d}+\frac{\sin ^5(c+d x)}{a^2 d}-\frac{4 \sin ^3(c+d x)}{3 a^2 d}+\frac{\sin (c+d x)}{a^2 d}+\frac{2 i \cos ^7(c+d x)}{7 a^2 d} \]

[Out]

(((2*I)/7)*Cos[c + d*x]^7)/(a^2*d) + Sin[c + d*x]/(a^2*d) - (4*Sin[c + d*x]^3)/(3*a^2*d) + Sin[c + d*x]^5/(a^2
*d) - (2*Sin[c + d*x]^7)/(7*a^2*d)

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Rubi [A]  time = 0.185731, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3092, 3090, 2633, 2565, 30, 2564, 270} \[ -\frac{2 \sin ^7(c+d x)}{7 a^2 d}+\frac{\sin ^5(c+d x)}{a^2 d}-\frac{4 \sin ^3(c+d x)}{3 a^2 d}+\frac{\sin (c+d x)}{a^2 d}+\frac{2 i \cos ^7(c+d x)}{7 a^2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

(((2*I)/7)*Cos[c + d*x]^7)/(a^2*d) + Sin[c + d*x]/(a^2*d) - (4*Sin[c + d*x]^3)/(3*a^2*d) + Sin[c + d*x]^5/(a^2
*d) - (2*Sin[c + d*x]^7)/(7*a^2*d)

Rule 3092

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symb
ol] :> Dist[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]

Rule 3090

Int[cos[(c_.) + (d_.)*(x_)]^(m_.)*(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_.), x_Sym
bol] :> 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] &&
IntegerQ[m] && IGtQ[n, 0]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^2} \, dx &=-\frac{\int \cos ^5(c+d x) (i a \cos (c+d x)+a \sin (c+d x))^2 \, dx}{a^4}\\ &=-\frac{\int \left (-a^2 \cos ^7(c+d x)+2 i a^2 \cos ^6(c+d x) \sin (c+d x)+a^2 \cos ^5(c+d x) \sin ^2(c+d x)\right ) \, dx}{a^4}\\ &=-\frac{(2 i) \int \cos ^6(c+d x) \sin (c+d x) \, dx}{a^2}+\frac{\int \cos ^7(c+d x) \, dx}{a^2}-\frac{\int \cos ^5(c+d x) \sin ^2(c+d x) \, dx}{a^2}\\ &=\frac{(2 i) \operatorname{Subst}\left (\int x^6 \, dx,x,\cos (c+d x)\right )}{a^2 d}-\frac{\operatorname{Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (c+d x)\right )}{a^2 d}-\frac{\operatorname{Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=\frac{2 i \cos ^7(c+d x)}{7 a^2 d}+\frac{\sin (c+d x)}{a^2 d}-\frac{\sin ^3(c+d x)}{a^2 d}+\frac{3 \sin ^5(c+d x)}{5 a^2 d}-\frac{\sin ^7(c+d x)}{7 a^2 d}-\frac{\operatorname{Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (c+d x)\right )}{a^2 d}\\ &=\frac{2 i \cos ^7(c+d x)}{7 a^2 d}+\frac{\sin (c+d x)}{a^2 d}-\frac{4 \sin ^3(c+d x)}{3 a^2 d}+\frac{\sin ^5(c+d x)}{a^2 d}-\frac{2 \sin ^7(c+d x)}{7 a^2 d}\\ \end{align*}

Mathematica [A]  time = 0.103847, size = 149, normalized size = 1.75 \[ \frac{15 \sin (c+d x)}{32 a^2 d}+\frac{11 \sin (3 (c+d x))}{96 a^2 d}+\frac{\sin (5 (c+d x))}{32 a^2 d}+\frac{\sin (7 (c+d x))}{224 a^2 d}+\frac{5 i \cos (c+d x)}{32 a^2 d}+\frac{3 i \cos (3 (c+d x))}{32 a^2 d}+\frac{i \cos (5 (c+d x))}{32 a^2 d}+\frac{i \cos (7 (c+d x))}{224 a^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^2,x]

[Out]

(((5*I)/32)*Cos[c + d*x])/(a^2*d) + (((3*I)/32)*Cos[3*(c + d*x)])/(a^2*d) + ((I/32)*Cos[5*(c + d*x)])/(a^2*d)
+ ((I/224)*Cos[7*(c + d*x)])/(a^2*d) + (15*Sin[c + d*x])/(32*a^2*d) + (11*Sin[3*(c + d*x)])/(96*a^2*d) + Sin[5
*(c + d*x)]/(32*a^2*d) + Sin[7*(c + d*x)]/(224*a^2*d)

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Maple [B]  time = 0.138, size = 174, normalized size = 2.1 \begin{align*} 2\,{\frac{1}{d{a}^{2}} \left ({\frac{i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{6}}}-{\frac{5/2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{4}}}+{\frac{{\frac{23\,i}{16}}}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}-2/7\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-7}+2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-5}-{\frac{55}{24\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{3}}}+{\frac{13}{16\,\tan \left ( 1/2\,dx+c/2 \right ) -16\,i}}-{\frac{i/16}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) +i \right ) ^{2}}}-1/24\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +i \right ) ^{-3}+3/16\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +i \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x)

[Out]

2/d/a^2*(I/(tan(1/2*d*x+1/2*c)-I)^6-5/2*I/(tan(1/2*d*x+1/2*c)-I)^4+23/16*I/(tan(1/2*d*x+1/2*c)-I)^2-2/7/(tan(1
/2*d*x+1/2*c)-I)^7+2/(tan(1/2*d*x+1/2*c)-I)^5-55/24/(tan(1/2*d*x+1/2*c)-I)^3+13/16/(tan(1/2*d*x+1/2*c)-I)-1/16
*I/(tan(1/2*d*x+1/2*c)+I)^2-1/24/(tan(1/2*d*x+1/2*c)+I)^3+3/16/(tan(1/2*d*x+1/2*c)+I))

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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)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 0.469126, size = 244, normalized size = 2.87 \begin{align*} \frac{{\left (-7 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 105 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 210 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 21 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{672 \, a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/672*(-7*I*e^(10*I*d*x + 10*I*c) - 105*I*e^(8*I*d*x + 8*I*c) + 210*I*e^(6*I*d*x + 6*I*c) + 70*I*e^(4*I*d*x +
4*I*c) + 21*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-7*I*d*x - 7*I*c)/(a^2*d)

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Sympy [A]  time = 1.81458, size = 233, normalized size = 2.74 \begin{align*} \begin{cases} \frac{\left (- 176160768 i a^{10} d^{5} e^{19 i c} e^{3 i d x} - 2642411520 i a^{10} d^{5} e^{17 i c} e^{i d x} + 5284823040 i a^{10} d^{5} e^{15 i c} e^{- i d x} + 1761607680 i a^{10} d^{5} e^{13 i c} e^{- 3 i d x} + 528482304 i a^{10} d^{5} e^{11 i c} e^{- 5 i d x} + 75497472 i a^{10} d^{5} e^{9 i c} e^{- 7 i d x}\right ) e^{- 16 i c}}{16911433728 a^{12} d^{6}} & \text{for}\: 16911433728 a^{12} d^{6} e^{16 i c} \neq 0 \\\frac{x \left (e^{10 i c} + 5 e^{8 i c} + 10 e^{6 i c} + 10 e^{4 i c} + 5 e^{2 i c} + 1\right ) e^{- 7 i c}}{32 a^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5/(a*cos(d*x+c)+I*a*sin(d*x+c))**2,x)

[Out]

Piecewise(((-176160768*I*a**10*d**5*exp(19*I*c)*exp(3*I*d*x) - 2642411520*I*a**10*d**5*exp(17*I*c)*exp(I*d*x)
+ 5284823040*I*a**10*d**5*exp(15*I*c)*exp(-I*d*x) + 1761607680*I*a**10*d**5*exp(13*I*c)*exp(-3*I*d*x) + 528482
304*I*a**10*d**5*exp(11*I*c)*exp(-5*I*d*x) + 75497472*I*a**10*d**5*exp(9*I*c)*exp(-7*I*d*x))*exp(-16*I*c)/(169
11433728*a**12*d**6), Ne(16911433728*a**12*d**6*exp(16*I*c), 0)), (x*(exp(10*I*c) + 5*exp(8*I*c) + 10*exp(6*I*
c) + 10*exp(4*I*c) + 5*exp(2*I*c) + 1)*exp(-7*I*c)/(32*a**2), True))

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Giac [A]  time = 1.11403, size = 196, normalized size = 2.31 \begin{align*} \frac{\frac{7 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 15 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + i\right )}^{3}} + \frac{273 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{6} - 1155 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 2450 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2870 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2037 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 791 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 152}{a^{2}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{7}}}{168 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5/(a*cos(d*x+c)+I*a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/168*(7*(9*tan(1/2*d*x + 1/2*c)^2 + 15*I*tan(1/2*d*x + 1/2*c) - 8)/(a^2*(tan(1/2*d*x + 1/2*c) + I)^3) + (273*
tan(1/2*d*x + 1/2*c)^6 - 1155*I*tan(1/2*d*x + 1/2*c)^5 - 2450*tan(1/2*d*x + 1/2*c)^4 + 2870*I*tan(1/2*d*x + 1/
2*c)^3 + 2037*tan(1/2*d*x + 1/2*c)^2 - 791*I*tan(1/2*d*x + 1/2*c) - 152)/(a^2*(tan(1/2*d*x + 1/2*c) - I)^7))/d