3.152 \(\int \frac {\cos ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx\)

Optimal. Leaf size=75 \[ \frac {i \cos ^4(c+d x)}{4 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac {3 \sin (c+d x) \cos (c+d x)}{8 a d}+\frac {3 x}{8 a} \]

[Out]

3/8*x/a+1/4*I*cos(d*x+c)^4/a/d+3/8*cos(d*x+c)*sin(d*x+c)/a/d+1/4*cos(d*x+c)^3*sin(d*x+c)/a/d

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Rubi [A]  time = 0.13, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ \frac {i \cos ^4(c+d x)}{4 a d}+\frac {\sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac {3 \sin (c+d x) \cos (c+d x)}{8 a d}+\frac {3 x}{8 a} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3/(a*Cos[c + d*x] + I*a*Sin[c + d*x]),x]

[Out]

(3*x)/(8*a) + ((I/4)*Cos[c + d*x]^4)/(a*d) + (3*Cos[c + d*x]*Sin[c + d*x])/(8*a*d) + (Cos[c + d*x]^3*Sin[c + d
*x])/(4*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 ^3(c+d x)}{a \cos (c+d x)+i a \sin (c+d x)} \, dx &=-\frac {i \int \cos ^3(c+d x) (i a \cos (c+d x)+a \sin (c+d x)) \, dx}{a^2}\\ &=-\frac {i \int \left (i a \cos ^4(c+d x)+a \cos ^3(c+d x) \sin (c+d x)\right ) \, dx}{a^2}\\ &=-\frac {i \int \cos ^3(c+d x) \sin (c+d x) \, dx}{a}+\frac {\int \cos ^4(c+d x) \, dx}{a}\\ &=\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac {3 \int \cos ^2(c+d x) \, dx}{4 a}+\frac {i \operatorname {Subst}\left (\int x^3 \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac {i \cos ^4(c+d x)}{4 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d}+\frac {3 \int 1 \, dx}{8 a}\\ &=\frac {3 x}{8 a}+\frac {i \cos ^4(c+d x)}{4 a d}+\frac {3 \cos (c+d x) \sin (c+d x)}{8 a d}+\frac {\cos ^3(c+d x) \sin (c+d x)}{4 a d}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 60, normalized size = 0.80 \[ \frac {8 \sin (2 (c+d x))+\sin (4 (c+d x))+4 i \cos (2 (c+d x))+i \cos (4 (c+d x))+12 c+12 d x}{32 a d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3/(a*Cos[c + d*x] + I*a*Sin[c + d*x]),x]

[Out]

(12*c + 12*d*x + (4*I)*Cos[2*(c + d*x)] + I*Cos[4*(c + d*x)] + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)])/(32*a*d)

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fricas [A]  time = 0.56, size = 54, normalized size = 0.72 \[ \frac {{\left (12 \, d x e^{\left (4 i \, d x + 4 i \, c\right )} - 2 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{32 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/32*(12*d*x*e^(4*I*d*x + 4*I*c) - 2*I*e^(6*I*d*x + 6*I*c) + 6*I*e^(2*I*d*x + 2*I*c) + I)*e^(-4*I*d*x - 4*I*c)
/(a*d)

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giac [A]  time = 0.19, size = 99, normalized size = 1.32 \[ -\frac {\frac {6 i \, \log \left (i \, \tan \left (d x + c\right ) + 1\right )}{a} - \frac {6 i \, \log \left (i \, \tan \left (d x + c\right ) - 1\right )}{a} + \frac {2 \, {\left (3 \, \tan \left (d x + c\right ) + 5 i\right )}}{a {\left (-i \, \tan \left (d x + c\right ) + 1\right )}} + \frac {-9 i \, \tan \left (d x + c\right )^{2} - 26 \, \tan \left (d x + c\right ) + 21 i}{a {\left (\tan \left (d x + c\right ) - i\right )}^{2}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/32*(6*I*log(I*tan(d*x + c) + 1)/a - 6*I*log(I*tan(d*x + c) - 1)/a + 2*(3*tan(d*x + c) + 5*I)/(a*(-I*tan(d*x
+ c) + 1)) + (-9*I*tan(d*x + c)^2 - 26*tan(d*x + c) + 21*I)/(a*(tan(d*x + c) - I)^2))/d

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maple [A]  time = 0.16, size = 98, normalized size = 1.31 \[ \frac {3 i \ln \left (\tan \left (d x +c \right )+i\right )}{16 a d}+\frac {1}{8 a d \left (\tan \left (d x +c \right )+i\right )}-\frac {3 i \ln \left (\tan \left (d x +c \right )-i\right )}{16 a d}-\frac {i}{8 a d \left (\tan \left (d x +c \right )-i\right )^{2}}+\frac {1}{4 a d \left (\tan \left (d x +c \right )-i\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)

[Out]

3/16*I/a/d*ln(tan(d*x+c)+I)+1/8/a/d/(tan(d*x+c)+I)-3/16*I/a/d*ln(tan(d*x+c)-I)-1/8*I/a/d/(tan(d*x+c)-I)^2+1/4/
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)^3/(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 = 3.43, size = 111, normalized size = 1.48 \[ \frac {3\,x}{8\,a}-\frac {\frac {5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,1{}\mathrm {i}}{2}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{2}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,1{}\mathrm {i}}{2}+\frac {5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^2\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3/(a*cos(c + d*x) + a*sin(c + d*x)*1i),x)

[Out]

(3*x)/(8*a) - ((5*tan(c/2 + (d*x)/2))/4 + (tan(c/2 + (d*x)/2)^2*1i)/2 - tan(c/2 + (d*x)/2)^3/2 - (tan(c/2 + (d
*x)/2)^4*1i)/2 + (5*tan(c/2 + (d*x)/2)^5)/4)/(a*d*(tan(c/2 + (d*x)/2) + 1i)^2*(tan(c/2 + (d*x)/2)*1i + 1)^4)

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sympy [A]  time = 0.30, size = 155, normalized size = 2.07 \[ \begin {cases} - \frac {\left (512 i a^{2} d^{2} e^{8 i c} e^{2 i d x} - 1536 i a^{2} d^{2} e^{4 i c} e^{- 2 i d x} - 256 i a^{2} d^{2} e^{2 i c} e^{- 4 i d x}\right ) e^{- 6 i c}}{8192 a^{3} d^{3}} & \text {for}\: 8192 a^{3} d^{3} e^{6 i c} \neq 0 \\x \left (\frac {\left (e^{6 i c} + 3 e^{4 i c} + 3 e^{2 i c} + 1\right ) e^{- 4 i c}}{8 a} - \frac {3}{8 a}\right ) & \text {otherwise} \end {cases} + \frac {3 x}{8 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3/(a*cos(d*x+c)+I*a*sin(d*x+c)),x)

[Out]

Piecewise((-(512*I*a**2*d**2*exp(8*I*c)*exp(2*I*d*x) - 1536*I*a**2*d**2*exp(4*I*c)*exp(-2*I*d*x) - 256*I*a**2*
d**2*exp(2*I*c)*exp(-4*I*d*x))*exp(-6*I*c)/(8192*a**3*d**3), Ne(8192*a**3*d**3*exp(6*I*c), 0)), (x*((exp(6*I*c
) + 3*exp(4*I*c) + 3*exp(2*I*c) + 1)*exp(-4*I*c)/(8*a) - 3/(8*a)), True)) + 3*x/(8*a)

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