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]

-1/3*I*cos(d*x+c)^3/a^3/d+4/5*I*cos(d*x+c)^5/a^3/d+sin(d*x+c)/a^3/d-5/3*sin(d*x+c)^3/a^3/d+4/5*sin(d*x+c)^5/a^
3/d

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Rubi [A]  time = 0.22, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 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]]

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 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 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 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 ^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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.91, size = 41, normalized size = 0.46 \[ \frac {{\left (15 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{60 \, a^{3} d} \]

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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giac [A]  time = 1.50, size = 73, normalized size = 0.81 \[ \frac {2 \, {\left (15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 30 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 20 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7\right )}}{15 \, a^{3} d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{5}} \]

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)

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maple [A]  time = 0.19, size = 90, normalized size = 1.00 \[ \frac {-\frac {16}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}-\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}+\frac {2}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i}+\frac {8}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}}{d \,a^{3}} \]

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

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maxima [A]  time = 0.53, size = 69, normalized size = 0.77 \[ \frac {3 i \, \cos \left (5 \, d x + 5 \, c\right ) + 10 i \, \cos \left (3 \, d x + 3 \, c\right ) + 15 i \, \cos \left (d x + c\right ) + 3 \, \sin \left (5 \, d x + 5 \, c\right ) + 10 \, \sin \left (3 \, d x + 3 \, c\right ) + 15 \, \sin \left (d x + c\right )}{60 \, a^{3} d} \]

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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mupad [B]  time = 0.87, size = 133, normalized size = 1.48 \[ \frac {2\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,15{}\mathrm {i}+30\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,40{}\mathrm {i}-20\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+7{}\mathrm {i}\right )}{15\,a^3\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,1{}\mathrm {i}+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,10{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,5{}\mathrm {i}+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^3,x)

[Out]

(2*(30*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^2*40i - 20*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^4*15i + 7i
))/(15*a^3*d*(tan(c/2 + (d*x)/2)*5i - 10*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*10i + 5*tan(c/2 + (d*x)/2
)^4 + tan(c/2 + (d*x)/2)^5*1i + 1))

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sympy [A]  time = 0.35, size = 136, normalized size = 1.51 \[ \begin {cases} - \frac {\left (- 120 i a^{6} d^{2} e^{8 i c} e^{- i d x} - 80 i a^{6} d^{2} e^{6 i c} e^{- 3 i d x} - 24 i a^{6} d^{2} e^{4 i c} e^{- 5 i d x}\right ) e^{- 9 i c}}{480 a^{9} d^{3}} & \text {for}\: 480 a^{9} d^{3} e^{9 i c} \neq 0 \\\frac {x \left (e^{4 i c} + 2 e^{2 i c} + 1\right ) e^{- 5 i c}}{4 a^{3}} & \text {otherwise} \end {cases} \]

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*
exp(2*I*c) + 1)*exp(-5*I*c)/(4*a**3), True))

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