3.178 \(\int \frac{\cos ^2(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=90 \[ \frac{4 \sin ^5(c+d x)}{5 a^3 d}-\frac{5 \sin ^3(c+d x)}{3 a^3 d}+\frac{\sin (c+d x)}{a^3 d}+\frac{4 i \cos ^5(c+d x)}{5 a^3 d}-\frac{i \cos ^3(c+d x)}{3 a^3 d} \]

[Out]

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

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Rubi [A]  time = 0.220501, antiderivative size = 90, normalized size of antiderivative = 1., number of steps used = 13, 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, 14} \[ \frac{4 \sin ^5(c+d x)}{5 a^3 d}-\frac{5 \sin ^3(c+d x)}{3 a^3 d}+\frac{\sin (c+d x)}{a^3 d}+\frac{4 i \cos ^5(c+d x)}{5 a^3 d}-\frac{i \cos ^3(c+d x)}{3 a^3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

((-I/3)*Cos[c + d*x]^3)/(a^3*d) + (((4*I)/5)*Cos[c + d*x]^5)/(a^3*d) + Sin[c + d*x]/(a^3*d) - (5*Sin[c + d*x]^
3)/(3*a^3*d) + (4*Sin[c + d*x]^5)/(5*a^3*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0787693, size = 111, normalized size = 1.23 \[ \frac{\sin (c+d x)}{4 a^3 d}+\frac{\sin (3 (c+d x))}{6 a^3 d}+\frac{\sin (5 (c+d x))}{20 a^3 d}+\frac{i \cos (c+d x)}{4 a^3 d}+\frac{i \cos (3 (c+d x))}{6 a^3 d}+\frac{i \cos (5 (c+d x))}{20 a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/(a*Cos[c + d*x] + I*a*Sin[c + d*x])^3,x]

[Out]

((I/4)*Cos[c + d*x])/(a^3*d) + ((I/6)*Cos[3*(c + d*x)])/(a^3*d) + ((I/20)*Cos[5*(c + d*x)])/(a^3*d) + Sin[c +
d*x]/(4*a^3*d) + Sin[3*(c + d*x)]/(6*a^3*d) + Sin[5*(c + d*x)]/(20*a^3*d)

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Maple [A]  time = 0.137, size = 90, normalized size = 1. \begin{align*} 2\,{\frac{1}{d{a}^{3}} \left ({\frac{-2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{4}}}+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-1}+4/5\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-5}+{\frac{2\,i}{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{2}}}-8/3\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -i \right ) ^{-3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x)

[Out]

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

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Maxima [A]  time = 1.13629, size = 93, normalized size = 1.03 \begin{align*} \frac{3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 10 i \, \cos \left (3 \, d x + 3 \, c\right ) + 15 i \, \cos \left (d x + c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 10 \, \sin \left (3 \, d x + 3 \, c\right ) + 15 \, \sin \left (d x + c\right )}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/60*(3*I*cos(5*d*x + 5*c) + 10*I*cos(3*d*x + 3*c) + 15*I*cos(d*x + c) + 3*sin(5*d*x + 5*c) + 10*sin(3*d*x + 3
*c) + 15*sin(d*x + c))/(a^3*d)

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Fricas [A]  time = 0.461456, size = 128, normalized size = 1.42 \begin{align*} \frac{{\left (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{60 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/60*(15*I*e^(4*I*d*x + 4*I*c) + 10*I*e^(2*I*d*x + 2*I*c) + 3*I)*e^(-5*I*d*x - 5*I*c)/(a^3*d)

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Sympy [A]  time = 0.64325, size = 133, normalized size = 1.48 \begin{align*} \begin{cases} \frac{\left (120 i a^{6} d^{2} e^{8 i c} e^{- i d x} + 80 i a^{6} d^{2} e^{6 i c} e^{- 3 i d x} + 24 i a^{6} d^{2} e^{4 i c} e^{- 5 i d x}\right ) e^{- 9 i c}}{480 a^{9} d^{3}} & \text{for}\: 480 a^{9} d^{3} e^{9 i c} \neq 0 \\\frac{x \left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 5 i c}}{4 a^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2/(a*cos(d*x+c)+I*a*sin(d*x+c))**3,x)

[Out]

Piecewise(((120*I*a**6*d**2*exp(8*I*c)*exp(-I*d*x) + 80*I*a**6*d**2*exp(6*I*c)*exp(-3*I*d*x) + 24*I*a**6*d**2*
exp(4*I*c)*exp(-5*I*d*x))*exp(-9*I*c)/(480*a**9*d**3), Ne(480*a**9*d**3*exp(9*I*c), 0)), (x*(exp(4*I*c) + 2*ex
p(2*I*c) + 1)*exp(-5*I*c)/(4*a**3), True))

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Giac [A]  time = 1.19289, size = 99, normalized size = 1.1 \begin{align*} \frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 30 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 20 i \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7\right )}}{15 \, a^{3} d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - i\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

2/15*(15*tan(1/2*d*x + 1/2*c)^4 - 30*I*tan(1/2*d*x + 1/2*c)^3 - 40*tan(1/2*d*x + 1/2*c)^2 + 20*I*tan(1/2*d*x +
1/2*c) + 7)/(a^3*d*(tan(1/2*d*x + 1/2*c) - I)^5)