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

Optimal. Leaf size=99 \[ \frac{i \cos ^6(c+d x)}{6 a d}+\frac{\sin (c+d x) \cos ^5(c+d x)}{6 a d}+\frac{5 \sin (c+d x) \cos ^3(c+d x)}{24 a d}+\frac{5 \sin (c+d x) \cos (c+d x)}{16 a d}+\frac{5 x}{16 a} \]

[Out]

(5*x)/(16*a) + ((I/6)*Cos[c + d*x]^6)/(a*d) + (5*Cos[c + d*x]*Sin[c + d*x])/(16*a*d) + (5*Cos[c + d*x]^3*Sin[c
+ d*x])/(24*a*d) + (Cos[c + d*x]^5*Sin[c + d*x])/(6*a*d)

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Rubi [A]  time = 0.150734, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {3092, 3090, 2635, 8, 2565, 30} \[ \frac{i \cos ^6(c+d x)}{6 a d}+\frac{\sin (c+d x) \cos ^5(c+d x)}{6 a d}+\frac{5 \sin (c+d x) \cos ^3(c+d x)}{24 a d}+\frac{5 \sin (c+d x) \cos (c+d x)}{16 a d}+\frac{5 x}{16 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*x)/(16*a) + ((I/6)*Cos[c + d*x]^6)/(a*d) + (5*Cos[c + d*x]*Sin[c + d*x])/(16*a*d) + (5*Cos[c + d*x]^3*Sin[c
+ d*x])/(24*a*d) + (Cos[c + d*x]^5*Sin[c + d*x])/(6*a*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 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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]

Rubi steps

\begin{align*} \int \frac{\cos ^5(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx &=-\frac{i \int \cos ^5(c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2}\\ &=-\frac{i \int \left (i a \cos ^6(c+d x)+a \cos ^5(c+d x) \sin (c+d x)\right ) \, dx}{a^2}\\ &=-\frac{i \int \cos ^5(c+d x) \sin (c+d x) \, dx}{a}+\frac{\int \cos ^6(c+d x) \, dx}{a}\\ &=\frac{\cos ^5(c+d x) \sin (c+d x)}{6 a d}+\frac{5 \int \cos ^4(c+d x) \, dx}{6 a}+\frac{i \operatorname{Subst}\left (\int x^5 \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac{i \cos ^6(c+d x)}{6 a d}+\frac{5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}+\frac{\cos ^5(c+d x) \sin (c+d x)}{6 a d}+\frac{5 \int \cos ^2(c+d x) \, dx}{8 a}\\ &=\frac{i \cos ^6(c+d x)}{6 a d}+\frac{5 \cos (c+d x) \sin (c+d x)}{16 a d}+\frac{5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}+\frac{\cos ^5(c+d x) \sin (c+d x)}{6 a d}+\frac{5 \int 1 \, dx}{16 a}\\ &=\frac{5 x}{16 a}+\frac{i \cos ^6(c+d x)}{6 a d}+\frac{5 \cos (c+d x) \sin (c+d x)}{16 a d}+\frac{5 \cos ^3(c+d x) \sin (c+d x)}{24 a d}+\frac{\cos ^5(c+d x) \sin (c+d x)}{6 a d}\\ \end{align*}

Mathematica [A]  time = 0.135858, size = 82, normalized size = 0.83 \[ \frac{45 \sin (2 (c+d x))+9 \sin (4 (c+d x))+\sin (6 (c+d x))+15 i \cos (2 (c+d x))+6 i \cos (4 (c+d x))+i \cos (6 (c+d x))+60 c+60 d x}{192 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*c + 60*d*x + (15*I)*Cos[2*(c + d*x)] + (6*I)*Cos[4*(c + d*x)] + I*Cos[6*(c + d*x)] + 45*Sin[2*(c + d*x)] +
9*Sin[4*(c + d*x)] + Sin[6*(c + d*x)])/(192*a*d)

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Maple [A]  time = 0.119, size = 137, normalized size = 1.4 \begin{align*}{\frac{-{\frac{5\,i}{32}}\ln \left ( \tan \left ( dx+c \right ) -i \right ) }{ad}}-{\frac{{\frac{3\,i}{32}}}{ad \left ( \tan \left ( dx+c \right ) -i \right ) ^{2}}}-{\frac{1}{24\,ad \left ( \tan \left ( dx+c \right ) -i \right ) ^{3}}}+{\frac{3}{16\,ad \left ( \tan \left ( dx+c \right ) -i \right ) }}+{\frac{{\frac{i}{32}}}{ad \left ( \tan \left ( dx+c \right ) +i \right ) ^{2}}}+{\frac{{\frac{5\,i}{32}}\ln \left ( \tan \left ( dx+c \right ) +i \right ) }{ad}}+{\frac{1}{8\,ad \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)^5/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)

[Out]

-5/32*I/a/d*ln(tan(d*x+c)-I)-3/32*I/a/d/(tan(d*x+c)-I)^2-1/24/a/d/(tan(d*x+c)-I)^3+3/16/a/d/(tan(d*x+c)-I)+1/3
2*I/a/d/(tan(d*x+c)+I)^2+5/32*I/a/d*ln(tan(d*x+c)+I)+1/8/a/d/(tan(d*x+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)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 0.472815, size = 242, normalized size = 2.44 \begin{align*} \frac{{\left (120 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 30 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 60 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{384 \, a 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)),x, algorithm="fricas")

[Out]

1/384*(120*d*x*e^(6*I*d*x + 6*I*c) - 3*I*e^(10*I*d*x + 10*I*c) - 30*I*e^(8*I*d*x + 8*I*c) + 60*I*e^(4*I*d*x +
4*I*c) + 15*I*e^(2*I*d*x + 2*I*c) + 2*I)*e^(-6*I*d*x - 6*I*c)/(a*d)

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Sympy [A]  time = 0.861229, size = 221, normalized size = 2.23 \begin{align*} \begin{cases} \frac{\left (- 50331648 i a^{4} d^{4} e^{16 i c} e^{4 i d x} - 503316480 i a^{4} d^{4} e^{14 i c} e^{2 i d x} + 1006632960 i a^{4} d^{4} e^{10 i c} e^{- 2 i d x} + 251658240 i a^{4} d^{4} e^{8 i c} e^{- 4 i d x} + 33554432 i a^{4} d^{4} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{6442450944 a^{5} d^{5}} & \text{for}\: 6442450944 a^{5} d^{5} e^{12 i c} \neq 0 \\x \left (\frac{\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^{- 6 i c}}{32 a} - \frac{5}{16 a}\right ) & \text{otherwise} \end{cases} + \frac{5 x}{16 a} \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)),x)

[Out]

Piecewise(((-50331648*I*a**4*d**4*exp(16*I*c)*exp(4*I*d*x) - 503316480*I*a**4*d**4*exp(14*I*c)*exp(2*I*d*x) +
1006632960*I*a**4*d**4*exp(10*I*c)*exp(-2*I*d*x) + 251658240*I*a**4*d**4*exp(8*I*c)*exp(-4*I*d*x) + 33554432*I
*a**4*d**4*exp(6*I*c)*exp(-6*I*d*x))*exp(-12*I*c)/(6442450944*a**5*d**5), Ne(6442450944*a**5*d**5*exp(12*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(-6*I*c)/(32*a) -
5/(16*a)), True)) + 5*x/(16*a)

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Giac [A]  time = 1.13472, size = 157, normalized size = 1.59 \begin{align*} -\frac{-\frac{30 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac{30 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac{3 \,{\left (-15 i \, \tan \left (d x + c\right )^{2} + 38 \, \tan \left (d x + c\right ) + 25 i\right )}}{a{\left (-i \, \tan \left (d x + c\right ) + 1\right )}^{2}} - \frac{55 i \, \tan \left (d x + c\right )^{3} + 201 \, \tan \left (d x + c\right )^{2} - 255 i \, \tan \left (d x + c\right ) - 117}{a{\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{192 \, 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)),x, algorithm="giac")

[Out]

-1/192*(-30*I*log(tan(d*x + c) + I)/a + 30*I*log(tan(d*x + c) - I)/a + 3*(-15*I*tan(d*x + c)^2 + 38*tan(d*x +
c) + 25*I)/(a*(-I*tan(d*x + c) + 1)^2) - (55*I*tan(d*x + c)^3 + 201*tan(d*x + c)^2 - 255*I*tan(d*x + c) - 117)
/(a*(tan(d*x + c) - I)^3))/d