3.176 \(\int \frac {\cos ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx\)

Optimal. Leaf size=106 \[ -\frac {4 \sin ^7(c+d x)}{7 a^3 d}+\frac {9 \sin ^5(c+d x)}{5 a^3 d}-\frac {2 \sin ^3(c+d x)}{a^3 d}+\frac {\sin (c+d x)}{a^3 d}+\frac {4 i \cos ^7(c+d x)}{7 a^3 d}-\frac {i \cos ^5(c+d x)}{5 a^3 d} \]

[Out]

-1/5*I*cos(d*x+c)^5/a^3/d+4/7*I*cos(d*x+c)^7/a^3/d+sin(d*x+c)/a^3/d-2*sin(d*x+c)^3/a^3/d+9/5*sin(d*x+c)^5/a^3/
d-4/7*sin(d*x+c)^7/a^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.23, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3092, 3090, 2633, 2565, 30, 2564, 270, 14} \[ -\frac {4 \sin ^7(c+d x)}{7 a^3 d}+\frac {9 \sin ^5(c+d x)}{5 a^3 d}-\frac {2 \sin ^3(c+d x)}{a^3 d}+\frac {\sin (c+d x)}{a^3 d}+\frac {4 i \cos ^7(c+d x)}{7 a^3 d}-\frac {i \cos ^5(c+d x)}{5 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/5)*Cos[c + d*x]^5)/(a^3*d) + (((4*I)/7)*Cos[c + d*x]^7)/(a^3*d) + Sin[c + d*x]/(a^3*d) - (2*Sin[c + d*x]^
3)/(a^3*d) + (9*Sin[c + d*x]^5)/(5*a^3*d) - (4*Sin[c + d*x]^7)/(7*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 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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 ^4(c+d x)}{(a \cos (c+d x)+i a \sin (c+d x))^3} \, dx &=\frac {i \int \cos ^4(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 ^7(c+d x)-3 a^3 \cos ^6(c+d x) \sin (c+d x)+3 i a^3 \cos ^5(c+d x) \sin ^2(c+d x)+a^3 \cos ^4(c+d x) \sin ^3(c+d x)\right ) \, dx}{a^6}\\ &=\frac {i \int \cos ^4(c+d x) \sin ^3(c+d x) \, dx}{a^3}-\frac {(3 i) \int \cos ^6(c+d x) \sin (c+d x) \, dx}{a^3}+\frac {\int \cos ^7(c+d x) \, dx}{a^3}-\frac {3 \int \cos ^5(c+d x) \sin ^2(c+d x) \, dx}{a^3}\\ &=-\frac {i \operatorname {Subst}\left (\int x^4 \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}+\frac {(3 i) \operatorname {Subst}\left (\int x^6 \, dx,x,\cos (c+d x)\right )}{a^3 d}-\frac {\operatorname {Subst}\left (\int \left (1-3 x^2+3 x^4-x^6\right ) \, dx,x,-\sin (c+d x)\right )}{a^3 d}-\frac {3 \operatorname {Subst}\left (\int x^2 \left (1-x^2\right )^2 \, dx,x,\sin (c+d x)\right )}{a^3 d}\\ &=\frac {3 i \cos ^7(c+d x)}{7 a^3 d}+\frac {\sin (c+d x)}{a^3 d}-\frac {\sin ^3(c+d x)}{a^3 d}+\frac {3 \sin ^5(c+d x)}{5 a^3 d}-\frac {\sin ^7(c+d x)}{7 a^3 d}-\frac {i \operatorname {Subst}\left (\int \left (x^4-x^6\right ) \, dx,x,\cos (c+d x)\right )}{a^3 d}-\frac {3 \operatorname {Subst}\left (\int \left (x^2-2 x^4+x^6\right ) \, dx,x,\sin (c+d x)\right )}{a^3 d}\\ &=-\frac {i \cos ^5(c+d x)}{5 a^3 d}+\frac {4 i \cos ^7(c+d x)}{7 a^3 d}+\frac {\sin (c+d x)}{a^3 d}-\frac {2 \sin ^3(c+d x)}{a^3 d}+\frac {9 \sin ^5(c+d x)}{5 a^3 d}-\frac {4 \sin ^7(c+d x)}{7 a^3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 149, normalized size = 1.41 \[ \frac {5 \sin (c+d x)}{16 a^3 d}+\frac {\sin (3 (c+d x))}{8 a^3 d}+\frac {\sin (5 (c+d x))}{20 a^3 d}+\frac {\sin (7 (c+d x))}{112 a^3 d}+\frac {3 i \cos (c+d x)}{16 a^3 d}+\frac {i \cos (3 (c+d x))}{8 a^3 d}+\frac {i \cos (5 (c+d x))}{20 a^3 d}+\frac {i \cos (7 (c+d x))}{112 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((3*I)/16)*Cos[c + d*x])/(a^3*d) + ((I/8)*Cos[3*(c + d*x)])/(a^3*d) + ((I/20)*Cos[5*(c + d*x)])/(a^3*d) + ((I
/112)*Cos[7*(c + d*x)])/(a^3*d) + (5*Sin[c + d*x])/(16*a^3*d) + Sin[3*(c + d*x)]/(8*a^3*d) + Sin[5*(c + d*x)]/
(20*a^3*d) + Sin[7*(c + d*x)]/(112*a^3*d)

________________________________________________________________________________________

fricas [A]  time = 2.12, size = 63, normalized size = 0.59 \[ \frac {{\left (-35 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 140 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 70 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 28 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-7 i \, d x - 7 i \, c\right )}}{560 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/560*(-35*I*e^(8*I*d*x + 8*I*c) + 140*I*e^(6*I*d*x + 6*I*c) + 70*I*e^(4*I*d*x + 4*I*c) + 28*I*e^(2*I*d*x + 2*
I*c) + 5*I)*e^(-7*I*d*x - 7*I*c)/(a^3*d)

________________________________________________________________________________________

giac [A]  time = 0.26, size = 119, normalized size = 1.12 \[ \frac {\frac {35}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}} + \frac {525 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 1960 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4025 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 4480 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3143 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1176 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 243}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{7}}}{280 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/280*(35/(a^3*(tan(1/2*d*x + 1/2*c) + I)) + (525*tan(1/2*d*x + 1/2*c)^6 - 1960*I*tan(1/2*d*x + 1/2*c)^5 - 402
5*tan(1/2*d*x + 1/2*c)^4 + 4480*I*tan(1/2*d*x + 1/2*c)^3 + 3143*tan(1/2*d*x + 1/2*c)^2 - 1176*I*tan(1/2*d*x +
1/2*c) - 243)/(a^3*(tan(1/2*d*x + 1/2*c) - I)^7))/d

________________________________________________________________________________________

maple [A]  time = 0.19, size = 141, normalized size = 1.33 \[ \frac {\frac {2}{16 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+16 i}+\frac {4 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{6}}-\frac {9 i}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{4}}+\frac {17 i}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{2}}-\frac {8}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{7}}+\frac {38}{5 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{5}}-\frac {15}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )^{3}}+\frac {15}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-i\right )}}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x)

[Out]

2/d/a^3*(1/16/(tan(1/2*d*x+1/2*c)+I)+2*I/(tan(1/2*d*x+1/2*c)-I)^6-9/2*I/(tan(1/2*d*x+1/2*c)-I)^4+17/8*I/(tan(1
/2*d*x+1/2*c)-I)^2-4/7/(tan(1/2*d*x+1/2*c)-I)^7+19/5/(tan(1/2*d*x+1/2*c)-I)^5-15/4/(tan(1/2*d*x+1/2*c)-I)^3+15
/16/(tan(1/2*d*x+1/2*c)-I))

________________________________________________________________________________________

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)^4/(a*cos(d*x+c)+I*a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 3.17, size = 134, normalized size = 1.26 \[ -\frac {\left (35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,105{}\mathrm {i}-175\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,105{}\mathrm {i}-7\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,77{}\mathrm {i}+43\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-13{}\mathrm {i}\right )\,2{}\mathrm {i}}{35\,a^3\,d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4/(a*cos(c + d*x) + a*sin(c + d*x)*1i)^3,x)

[Out]

-((43*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2*77i - 7*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4*105i - 175
*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x)/2)^6*105i + 35*tan(c/2 + (d*x)/2)^7 - 13i)*2i)/(35*a^3*d*(tan(c/2 + (d
*x)/2) + 1i)*(tan(c/2 + (d*x)/2)*1i + 1)^7)

________________________________________________________________________________________

sympy [A]  time = 0.48, size = 201, normalized size = 1.90 \[ \begin {cases} - \frac {\left (71680 i a^{12} d^{4} e^{17 i c} e^{i d x} - 286720 i a^{12} d^{4} e^{15 i c} e^{- i d x} - 143360 i a^{12} d^{4} e^{13 i c} e^{- 3 i d x} - 57344 i a^{12} d^{4} e^{11 i c} e^{- 5 i d x} - 10240 i a^{12} d^{4} e^{9 i c} e^{- 7 i d x}\right ) e^{- 16 i c}}{1146880 a^{15} d^{5}} & \text {for}\: 1146880 a^{15} d^{5} e^{16 i c} \neq 0 \\\frac {x \left (e^{8 i c} + 4 e^{6 i c} + 6 e^{4 i c} + 4 e^{2 i c} + 1\right ) e^{- 7 i c}}{16 a^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4/(a*cos(d*x+c)+I*a*sin(d*x+c))**3,x)

[Out]

Piecewise((-(71680*I*a**12*d**4*exp(17*I*c)*exp(I*d*x) - 286720*I*a**12*d**4*exp(15*I*c)*exp(-I*d*x) - 143360*
I*a**12*d**4*exp(13*I*c)*exp(-3*I*d*x) - 57344*I*a**12*d**4*exp(11*I*c)*exp(-5*I*d*x) - 10240*I*a**12*d**4*exp
(9*I*c)*exp(-7*I*d*x))*exp(-16*I*c)/(1146880*a**15*d**5), Ne(1146880*a**15*d**5*exp(16*I*c), 0)), (x*(exp(8*I*
c) + 4*exp(6*I*c) + 6*exp(4*I*c) + 4*exp(2*I*c) + 1)*exp(-7*I*c)/(16*a**3), True))

________________________________________________________________________________________