3.1.49 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [49]

Optimal. Leaf size=88 \[ -\frac {i a^3 \cos ^3(c+d x)}{15 d}+\frac {a^3 \sin (c+d x)}{5 d}-\frac {a^3 \sin ^3(c+d x)}{15 d}-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d} \]

[Out]

-1/15*I*a^3*cos(d*x+c)^3/d+1/5*a^3*sin(d*x+c)/d-1/15*a^3*sin(d*x+c)^3/d-2/5*I*a*cos(d*x+c)^5*(a+I*a*tan(d*x+c)
)^2/d

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Rubi [A]
time = 0.06, antiderivative size = 88, 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 = {3577, 3567, 2713} \begin {gather*} -\frac {a^3 \sin ^3(c+d x)}{15 d}+\frac {a^3 \sin (c+d x)}{5 d}-\frac {i a^3 \cos ^3(c+d x)}{15 d}-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^3,x]

[Out]

((-1/15*I)*a^3*Cos[c + d*x]^3)/d + (a^3*Sin[c + d*x])/(5*d) - (a^3*Sin[c + d*x]^3)/(15*d) - (((2*I)/5)*a*Cos[c
+ d*x]^5*(a + I*a*Tan[c + d*x])^2)/d

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 3567

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

Rule 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \cos ^5(c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {1}{5} a^2 \int \cos ^3(c+d x) (a+i a \tan (c+d x)) \, dx\\ &=-\frac {i a^3 \cos ^3(c+d x)}{15 d}-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}+\frac {1}{5} a^3 \int \cos ^3(c+d x) \, dx\\ &=-\frac {i a^3 \cos ^3(c+d x)}{15 d}-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}-\frac {a^3 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=-\frac {i a^3 \cos ^3(c+d x)}{15 d}+\frac {a^3 \sin (c+d x)}{5 d}-\frac {a^3 \sin ^3(c+d x)}{15 d}-\frac {2 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^2}{5 d}\\ \end {align*}

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Mathematica [A]
time = 0.33, size = 55, normalized size = 0.62 \begin {gather*} \frac {a^3 (5+9 \cos (2 (c+d x))-6 i \sin (2 (c+d x))) (-i \cos (3 (c+d x))+\sin (3 (c+d x)))}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*(5 + 9*Cos[2*(c + d*x)] - (6*I)*Sin[2*(c + d*x)])*((-I)*Cos[3*(c + d*x)] + Sin[3*(c + d*x)]))/(30*d)

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Maple [A]
time = 0.21, size = 126, normalized size = 1.43

method	result	size
risch	\(-\frac {i a^{3} {\mathrm e}^{5 i \left (d x +c \right )}}{20 d}-\frac {i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{6 d}-\frac {i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{4 d}\)	\(56\)
derivativedivides	\(\frac {-i a^{3} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 i a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(126\)
default	\(\frac {-i a^{3} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-3 a^{3} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{15}\right )-\frac {3 i a^{3} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(126\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-I*a^3*(-1/5*cos(d*x+c)^3*sin(d*x+c)^2-2/15*cos(d*x+c)^3)-3*a^3*(-1/5*sin(d*x+c)*cos(d*x+c)^4+1/15*(cos(d
*x+c)^2+2)*sin(d*x+c))-3/5*I*a^3*cos(d*x+c)^5+1/5*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c))

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Maxima [A]
time = 0.28, size = 105, normalized size = 1.19 \begin {gather*} -\frac {9 i \, a^{3} \cos \left (d x + c\right )^{5} + i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{3} - 3 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{3} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/15*(9*I*a^3*cos(d*x + c)^5 + I*(3*cos(d*x + c)^5 - 5*cos(d*x + c)^3)*a^3 - 3*(3*sin(d*x + c)^5 - 5*sin(d*x
+ c)^3)*a^3 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3)/d

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Fricas [A]
time = 0.36, size = 48, normalized size = 0.55 \begin {gather*} \frac {-3 i \, a^{3} e^{\left (5 i \, d x + 5 i \, c\right )} - 10 i \, a^{3} e^{\left (3 i \, d x + 3 i \, c\right )} - 15 i \, a^{3} e^{\left (i \, d x + i \, c\right )}}{60 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 0.23, size = 116, normalized size = 1.32 \begin {gather*} \begin {cases} \frac {- 24 i a^{3} d^{2} e^{5 i c} e^{5 i d x} - 80 i a^{3} d^{2} e^{3 i c} e^{3 i d x} - 120 i a^{3} d^{2} e^{i c} e^{i d x}}{480 d^{3}} & \text {for}\: d^{3} \neq 0 \\x \left (\frac {a^{3} e^{5 i c}}{4} + \frac {a^{3} e^{3 i c}}{2} + \frac {a^{3} e^{i c}}{4}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**3,x)

[Out]

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

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 929 vs. \(2 (74) = 148\).
time = 1.00, size = 929, normalized size = 10.56 \begin {gather*} \frac {1785 \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 7140 \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 7140 \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 10710 \, a^{3} e^{\left (4 i \, d x\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 1785 \, a^{3} e^{\left (-4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 1530 \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 6120 \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 6120 \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 9180 \, a^{3} e^{\left (4 i \, d x\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 1530 \, a^{3} e^{\left (-4 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 1785 \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 7140 \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 7140 \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 10710 \, a^{3} e^{\left (4 i \, d x\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 1785 \, a^{3} e^{\left (-4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 1530 \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 6120 \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 6120 \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 9180 \, a^{3} e^{\left (4 i \, d x\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 1530 \, a^{3} e^{\left (-4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 255 \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 1020 \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 1020 \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 1530 \, a^{3} e^{\left (4 i \, d x\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 255 \, a^{3} e^{\left (-4 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 255 \, a^{3} e^{\left (8 i \, d x + 4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 1020 \, a^{3} e^{\left (6 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 1020 \, a^{3} e^{\left (2 i \, d x - 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 1530 \, a^{3} e^{\left (4 i \, d x\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 255 \, a^{3} e^{\left (-4 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 384 i \, a^{3} e^{\left (13 i \, d x + 9 i \, c\right )} - 2816 i \, a^{3} e^{\left (11 i \, d x + 7 i \, c\right )} - 9344 i \, a^{3} e^{\left (9 i \, d x + 5 i \, c\right )} - 16896 i \, a^{3} e^{\left (7 i \, d x + 3 i \, c\right )} - 17024 i \, a^{3} e^{\left (5 i \, d x + i \, c\right )} - 8960 i \, a^{3} e^{\left (3 i \, d x - i \, c\right )} - 1920 i \, a^{3} e^{\left (i \, d x - 3 i \, c\right )}}{7680 \, {\left (d e^{\left (8 i \, d x + 4 i \, c\right )} + 4 \, d e^{\left (6 i \, d x + 2 i \, c\right )} + 4 \, d e^{\left (2 i \, d x - 2 i \, c\right )} + 6 \, d e^{\left (4 i \, d x\right )} + d e^{\left (-4 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

1/7680*(1785*a^3*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 7140*a^3*e^(6*I*d*x + 2*I*c)*log(I*e^(I*d*x
+ I*c) + 1) + 7140*a^3*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 10710*a^3*e^(4*I*d*x)*log(I*e^(I*d*x +
I*c) + 1) + 1785*a^3*e^(-4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1530*a^3*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*
c) - 1) + 6120*a^3*e^(6*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 6120*a^3*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*
x + I*c) - 1) + 9180*a^3*e^(4*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 1530*a^3*e^(-4*I*c)*log(I*e^(I*d*x + I*c) -
1) - 1785*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 7140*a^3*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x +
I*c) + 1) - 7140*a^3*e^(2*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 10710*a^3*e^(4*I*d*x)*log(-I*e^(I*d*x
+ I*c) + 1) - 1785*a^3*e^(-4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1530*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*x +
I*c) - 1) - 6120*a^3*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 6120*a^3*e^(2*I*d*x - 2*I*c)*log(-I*e^
(I*d*x + I*c) - 1) - 9180*a^3*e^(4*I*d*x)*log(-I*e^(I*d*x + I*c) - 1) - 1530*a^3*e^(-4*I*c)*log(-I*e^(I*d*x +
I*c) - 1) - 255*a^3*e^(8*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 1020*a^3*e^(6*I*d*x + 2*I*c)*log(I*e^(I*
d*x) + e^(-I*c)) - 1020*a^3*e^(2*I*d*x - 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 1530*a^3*e^(4*I*d*x)*log(I*e^(I*
d*x) + e^(-I*c)) - 255*a^3*e^(-4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 255*a^3*e^(8*I*d*x + 4*I*c)*log(-I*e^(I*d*
x) + e^(-I*c)) + 1020*a^3*e^(6*I*d*x + 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 1020*a^3*e^(2*I*d*x - 2*I*c)*log(
-I*e^(I*d*x) + e^(-I*c)) + 1530*a^3*e^(4*I*d*x)*log(-I*e^(I*d*x) + e^(-I*c)) + 255*a^3*e^(-4*I*c)*log(-I*e^(I*
d*x) + e^(-I*c)) - 384*I*a^3*e^(13*I*d*x + 9*I*c) - 2816*I*a^3*e^(11*I*d*x + 7*I*c) - 9344*I*a^3*e^(9*I*d*x +
5*I*c) - 16896*I*a^3*e^(7*I*d*x + 3*I*c) - 17024*I*a^3*e^(5*I*d*x + I*c) - 8960*I*a^3*e^(3*I*d*x - I*c) - 1920
*I*a^3*e^(I*d*x - 3*I*c))/(d*e^(8*I*d*x + 4*I*c) + 4*d*e^(6*I*d*x + 2*I*c) + 4*d*e^(2*I*d*x - 2*I*c) + 6*d*e^(
4*I*d*x) + d*e^(-4*I*c))

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Mupad [B]
time = 3.57, size = 130, normalized size = 1.48 \begin {gather*} \frac {2\,a^3\,\left (15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,30{}\mathrm {i}-40\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,20{}\mathrm {i}+7\right )}{15\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

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

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