3.2.46 \(\int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [146]

Optimal. Leaf size=121 \[ \frac {10 \sin (c+d x)}{21 a^3 d}-\frac {20 \sin ^3(c+d x)}{63 a^3 d}+\frac {2 \sin ^5(c+d x)}{21 a^3 d}+\frac {i \cos ^3(c+d x)}{9 d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{21 d \left (a^3+i a^3 \tan (c+d x)\right )} \]

[Out]

10/21*sin(d*x+c)/a^3/d-20/63*sin(d*x+c)^3/a^3/d+2/21*sin(d*x+c)^5/a^3/d+1/9*I*cos(d*x+c)^3/d/(a+I*a*tan(d*x+c)
)^3+4/21*I*cos(d*x+c)^5/d/(a^3+I*a^3*tan(d*x+c))

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Rubi [A]
time = 0.08, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3583, 3581, 2713} \begin {gather*} \frac {2 \sin ^5(c+d x)}{21 a^3 d}-\frac {20 \sin ^3(c+d x)}{63 a^3 d}+\frac {10 \sin (c+d x)}{21 a^3 d}+\frac {4 i \cos ^5(c+d x)}{21 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {i \cos ^3(c+d x)}{9 d (a+i a \tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sin[c + d*x])/(21*a^3*d) - (20*Sin[c + d*x]^3)/(63*a^3*d) + (2*Sin[c + d*x]^5)/(21*a^3*d) + ((I/9)*Cos[c +
d*x]^3)/(d*(a + I*a*Tan[c + d*x])^3) + (((4*I)/21)*Cos[c + d*x]^5)/(d*(a^3 + I*a^3*Tan[c + d*x]))

Rule 2713

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 3581

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*d^2*
(d*Sec[e + f*x])^(m - 2)*((a + b*Tan[e + f*x])^(n + 1)/(b*f*(m + 2*n))), x] - Dist[d^2*((m - 2)/(b^2*(m + 2*n)
)), Int[(d*Sec[e + f*x])^(m - 2)*(a + b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a
^2 + b^2, 0] && LtQ[n, -1] && ((ILtQ[n/2, 0] && IGtQ[m - 1/2, 0]) || EqQ[n, -2] || IGtQ[m + n, 0] || (Integers
Q[n, m + 1/2] && GtQ[2*m + n + 1, 0])) && IntegerQ[2*m]

Rule 3583

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(b*f*(m + 2*n))), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=\frac {i \cos ^3(c+d x)}{9 d (a+i a \tan (c+d x))^3}+\frac {2 \int \frac {\cos ^3(c+d x)}{(a+i a \tan (c+d x))^2} \, dx}{3 a}\\ &=\frac {i \cos ^3(c+d x)}{9 d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{21 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac {10 \int \cos ^5(c+d x) \, dx}{21 a^3}\\ &=\frac {i \cos ^3(c+d x)}{9 d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{21 d \left (a^3+i a^3 \tan (c+d x)\right )}-\frac {10 \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{21 a^3 d}\\ &=\frac {10 \sin (c+d x)}{21 a^3 d}-\frac {20 \sin ^3(c+d x)}{63 a^3 d}+\frac {2 \sin ^5(c+d x)}{21 a^3 d}+\frac {i \cos ^3(c+d x)}{9 d (a+i a \tan (c+d x))^3}+\frac {4 i \cos ^5(c+d x)}{21 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 98, normalized size = 0.81 \begin {gather*} \frac {\sec ^3(c+d x) (-210-567 \cos (2 (c+d x))+162 \cos (4 (c+d x))+7 \cos (6 (c+d x))-378 i \sin (2 (c+d x))+216 i \sin (4 (c+d x))+14 i \sin (6 (c+d x)))}{2016 a^3 d (-i+\tan (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^3*(-210 - 567*Cos[2*(c + d*x)] + 162*Cos[4*(c + d*x)] + 7*Cos[6*(c + d*x)] - (378*I)*Sin[2*(c +
d*x)] + (216*I)*Sin[4*(c + d*x)] + (14*I)*Sin[6*(c + d*x)]))/(2016*a^3*d*(-I + Tan[c + d*x])^3)

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Maple [A]
time = 0.29, size = 207, normalized size = 1.71

method	result	size
risch	\(\frac {3 i {\mathrm e}^{-5 i \left (d x +c \right )}}{64 a^{3} d}+\frac {3 i {\mathrm e}^{-7 i \left (d x +c \right )}}{224 a^{3} d}+\frac {i {\mathrm e}^{-9 i \left (d x +c \right )}}{576 a^{3} d}+\frac {9 i \cos \left (d x +c \right )}{64 a^{3} d}+\frac {21 \sin \left (d x +c \right )}{64 a^{3} d}+\frac {19 i \cos \left (3 d x +3 c \right )}{192 a^{3} d}+\frac {7 \sin \left (3 d x +3 c \right )}{64 a^{3} d}\)	\(120\)
derivativedivides	\(\frac {-\frac {i}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {7}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}+\frac {46 i}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {9 i}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {59 i}{4 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {8}{9 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {68}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {35}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {19}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {57}{32 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{a^{3} d}\)	\(207\)
default	\(\frac {-\frac {i}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{2}}-\frac {1}{24 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )^{3}}+\frac {7}{32 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+i\right )}+\frac {46 i}{3 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}-\frac {4 i}{\left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{8}}+\frac {9 i}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {59 i}{4 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {8}{9 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{9}}-\frac {68}{7 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7}}+\frac {35}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{5}}-\frac {19}{2 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {57}{32 \left (-i+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}{a^{3} d}\)	\(207\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

2/d/a^3*(-1/32*I/(tan(1/2*d*x+1/2*c)+I)^2-1/48/(tan(1/2*d*x+1/2*c)+I)^3+7/64/(tan(1/2*d*x+1/2*c)+I)+23/3*I/(-I
+tan(1/2*d*x+1/2*c))^6-2*I/(-I+tan(1/2*d*x+1/2*c))^8+9/4*I/(-I+tan(1/2*d*x+1/2*c))^2-59/8*I/(-I+tan(1/2*d*x+1/
2*c))^4+4/9/(-I+tan(1/2*d*x+1/2*c))^9-34/7/(-I+tan(1/2*d*x+1/2*c))^7+35/4/(-I+tan(1/2*d*x+1/2*c))^5-19/4/(-I+t
an(1/2*d*x+1/2*c))^3+57/64/(-I+tan(1/2*d*x+1/2*c)))

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3/(a+I*a*tan(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.

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Fricas [A]
time = 0.37, size = 85, normalized size = 0.70 \begin {gather*} \frac {{\left (-21 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 378 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 945 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 420 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 189 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 54 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 7 i\right )} e^{\left (-9 i \, d x - 9 i \, c\right )}}{4032 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4032*(-21*I*e^(12*I*d*x + 12*I*c) - 378*I*e^(10*I*d*x + 10*I*c) + 945*I*e^(8*I*d*x + 8*I*c) + 420*I*e^(6*I*d
*x + 6*I*c) + 189*I*e^(4*I*d*x + 4*I*c) + 54*I*e^(2*I*d*x + 2*I*c) + 7*I)*e^(-9*I*d*x - 9*I*c)/(a^3*d)

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Sympy [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (105) = 210\).
time = 0.38, size = 265, normalized size = 2.19 \begin {gather*} \begin {cases} \frac {\left (- 811748818944 i a^{18} d^{6} e^{28 i c} e^{3 i d x} - 14611478740992 i a^{18} d^{6} e^{26 i c} e^{i d x} + 36528696852480 i a^{18} d^{6} e^{24 i c} e^{- i d x} + 16234976378880 i a^{18} d^{6} e^{22 i c} e^{- 3 i d x} + 7305739370496 i a^{18} d^{6} e^{20 i c} e^{- 5 i d x} + 2087354105856 i a^{18} d^{6} e^{18 i c} e^{- 7 i d x} + 270582939648 i a^{18} d^{6} e^{16 i c} e^{- 9 i d x}\right ) e^{- 25 i c}}{155855773237248 a^{21} d^{7}} & \text {for}\: a^{21} d^{7} e^{25 i c} \neq 0 \\\frac {x \left (e^{12 i c} + 6 e^{10 i c} + 15 e^{8 i c} + 20 e^{6 i c} + 15 e^{4 i c} + 6 e^{2 i c} + 1\right ) e^{- 9 i c}}{64 a^{3}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((-811748818944*I*a**18*d**6*exp(28*I*c)*exp(3*I*d*x) - 14611478740992*I*a**18*d**6*exp(26*I*c)*exp(
I*d*x) + 36528696852480*I*a**18*d**6*exp(24*I*c)*exp(-I*d*x) + 16234976378880*I*a**18*d**6*exp(22*I*c)*exp(-3*
I*d*x) + 7305739370496*I*a**18*d**6*exp(20*I*c)*exp(-5*I*d*x) + 2087354105856*I*a**18*d**6*exp(18*I*c)*exp(-7*
I*d*x) + 270582939648*I*a**18*d**6*exp(16*I*c)*exp(-9*I*d*x))*exp(-25*I*c)/(155855773237248*a**21*d**7), Ne(a*
*21*d**7*exp(25*I*c), 0)), (x*(exp(12*I*c) + 6*exp(10*I*c) + 15*exp(8*I*c) + 20*exp(6*I*c) + 15*exp(4*I*c) + 6
*exp(2*I*c) + 1)*exp(-9*I*c)/(64*a**3), True))

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Giac [A]
time = 0.75, size = 171, normalized size = 1.41 \begin {gather*} \frac {\frac {21 \, {\left (21 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 36 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 19\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + i\right )}^{3}} + \frac {3591 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} - 19656 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 56196 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 95760 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 107730 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 79464 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 38484 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10944 i \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1615}{a^{3} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - i\right )}^{9}}}{2016 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2016*(21*(21*tan(1/2*d*x + 1/2*c)^2 + 36*I*tan(1/2*d*x + 1/2*c) - 19)/(a^3*(tan(1/2*d*x + 1/2*c) + I)^3) + (
3591*tan(1/2*d*x + 1/2*c)^8 - 19656*I*tan(1/2*d*x + 1/2*c)^7 - 56196*tan(1/2*d*x + 1/2*c)^6 + 95760*I*tan(1/2*
d*x + 1/2*c)^5 + 107730*tan(1/2*d*x + 1/2*c)^4 - 79464*I*tan(1/2*d*x + 1/2*c)^3 - 38484*tan(1/2*d*x + 1/2*c)^2
+ 10944*I*tan(1/2*d*x + 1/2*c) + 1615)/(a^3*(tan(1/2*d*x + 1/2*c) - I)^9))/d

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Mupad [B]
time = 6.61, size = 188, normalized size = 1.55 \begin {gather*} \frac {\left (63\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,189{}\mathrm {i}-273\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,63{}\mathrm {i}-378\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,294{}\mathrm {i}-306\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,450{}\mathrm {i}+235\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,39{}\mathrm {i}+51\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-19{}\mathrm {i}\right )\,2{}\mathrm {i}}{63\,a^3\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )}^3\,{\left (1+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}\right )}^9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((51*tan(c/2 + (d*x)/2) + tan(c/2 + (d*x)/2)^2*39i + 235*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4*450i - 30
6*tan(c/2 + (d*x)/2)^5 + tan(c/2 + (d*x)/2)^6*294i - 378*tan(c/2 + (d*x)/2)^7 - tan(c/2 + (d*x)/2)^8*63i - 273
*tan(c/2 + (d*x)/2)^9 - tan(c/2 + (d*x)/2)^10*189i + 63*tan(c/2 + (d*x)/2)^11 - 19i)*2i)/(63*a^3*d*(tan(c/2 +
(d*x)/2) + 1i)^3*(tan(c/2 + (d*x)/2)*1i + 1)^9)

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