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]

5/16*x/a+1/6*I*cos(d*x+c)^6/a/d+5/16*cos(d*x+c)*sin(d*x+c)/a/d+5/24*cos(d*x+c)^3*sin(d*x+c)/a/d+1/6*cos(d*x+c)
^5*sin(d*x+c)/a/d

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Rubi [A]  time = 0.15, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 8

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

Rule 30

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

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 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 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 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]

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*}

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Mathematica [A]  time = 0.14, 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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fricas [A]  time = 0.59, size = 76, normalized size = 0.77 \[ \frac {{\left (120 \, d x e^{\left (6 i \, d x + 6 i \, c\right )} - 3 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 30 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 60 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 15 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{384 \, a d} \]

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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giac [A]  time = 0.24, size = 116, normalized size = 1.17 \[ -\frac {-\frac {30 i \, \log \left (\tan \left (d x + c\right ) + i\right )}{a} + \frac {30 i \, \log \left (\tan \left (d x + c\right ) - i\right )}{a} + \frac {3 \, {\left (-15 i \, \tan \left (d x + c\right )^{2} + 38 \, \tan \left (d x + c\right ) + 25 i\right )}}{a {\left (-i \, \tan \left (d x + c\right ) + 1\right )}^{2}} - \frac {55 i \, \tan \left (d x + c\right )^{3} + 201 \, \tan \left (d x + c\right )^{2} - 255 i \, \tan \left (d x + c\right ) - 117}{a {\left (\tan \left (d x + c\right ) - i\right )}^{3}}}{192 \, d} \]

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

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maple [A]  time = 0.23, size = 137, normalized size = 1.38 \[ \frac {i}{32 a d \left (\tan \left (d x +c \right )+i\right )^{2}}+\frac {5 i \ln \left (\tan \left (d x +c \right )+i\right )}{32 a d}+\frac {1}{8 a d \left (\tan \left (d x +c \right )+i\right )}-\frac {5 i \ln \left (\tan \left (d x +c \right )-i\right )}{32 a d}-\frac {3 i}{32 a d \left (\tan \left (d x +c \right )-i\right )^{2}}-\frac {1}{24 a d \left (\tan \left (d x +c \right )-i\right )^{3}}+\frac {3}{16 a d \left (\tan \left (d x +c \right )-i\right )} \]

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]

1/32*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)-5/32*I/a/d*ln(tan(d*x+c)-I)-3/3
2*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)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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 >> ECL says: Error executing code in Maxima: expt: undefined: 0 to a negative e
xponent.

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mupad [B]  time = 5.14, size = 164, normalized size = 1.66 \[ \frac {5\,x}{16\,a}+\frac {\frac {11\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{8}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,3{}\mathrm {i}}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,1{}\mathrm {i}}{12}+\frac {13\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,1{}\mathrm {i}}{12}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}}{4}+\frac {11\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^4\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5/(a*cos(c + d*x) + a*sin(c + d*x)*1i),x)

[Out]

(5*x)/(16*a) + ((11*tan(c/2 + (d*x)/2))/8 + (tan(c/2 + (d*x)/2)^2*3i)/4 - tan(c/2 + (d*x)/2)^3/3 + (tan(c/2 +
(d*x)/2)^4*1i)/12 + (13*tan(c/2 + (d*x)/2)^5)/4 - (tan(c/2 + (d*x)/2)^6*1i)/12 - tan(c/2 + (d*x)/2)^7/3 - (tan
(c/2 + (d*x)/2)^8*3i)/4 + (11*tan(c/2 + (d*x)/2)^9)/8)/(a*d*(tan(c/2 + (d*x)/2) + 1i)^4*(tan(c/2 + (d*x)/2)*1i
+ 1)^6)

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sympy [A]  time = 0.41, size = 223, normalized size = 2.25 \[ \begin {cases} - \frac {\left (50331648 i a^{4} d^{4} e^{16 i c} e^{4 i d x} + 503316480 i a^{4} d^{4} e^{14 i c} e^{2 i d x} - 1006632960 i a^{4} d^{4} e^{10 i c} e^{- 2 i d x} - 251658240 i a^{4} d^{4} e^{8 i c} e^{- 4 i d x} - 33554432 i a^{4} d^{4} e^{6 i c} e^{- 6 i d x}\right ) e^{- 12 i c}}{6442450944 a^{5} d^{5}} & \text {for}\: 6442450944 a^{5} d^{5} e^{12 i c} \neq 0 \\x \left (\frac {\left (e^{10 i c} + 5 e^{8 i c} + 10 e^{6 i c} + 10 e^{4 i c} + 5 e^{2 i c} + 1\right ) e^{- 6 i c}}{32 a} - \frac {5}{16 a}\right ) & \text {otherwise} \end {cases} + \frac {5 x}{16 a} \]

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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