\(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx\) [73]

Optimal result

Integrand size = 24, antiderivative size = 32 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^5}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {3569} \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^5}{5 d} \]

[In]

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

[Out]

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

Rule 3569

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

Rubi steps \begin{align*} \text {integral}& = -\frac {i \cos ^5(c+d x) (a+i a \tan (c+d x))^5}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.42 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i a^5 (\cos (c+d x)+i \sin (c+d x))^5}{5 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 81.45 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59

method	result	size
risch	\(-\frac {i a^{5} {\mathrm e}^{5 i \left (d x +c \right )}}{5 d}\)	\(19\)
derivativedivides	\(\frac {-\frac {i a^{5} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}+a^{5} \left (\sin ^{5}\left (d x +c \right )\right )-10 i a^{5} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-i a^{5} \left (\cos ^{5}\left (d x +c \right )\right )+\frac {a^{5} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(170\)
default	\(\frac {-\frac {i a^{5} \left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )}{5}+a^{5} \left (\sin ^{5}\left (d x +c \right )\right )-10 i a^{5} \left (-\frac {\left (\cos ^{3}\left (d x +c \right )\right ) \left (\sin ^{2}\left (d x +c \right )\right )}{5}-\frac {2 \left (\cos ^{3}\left (d x +c \right )\right )}{15}\right )-10 a^{5} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-i a^{5} \left (\cos ^{5}\left (d x +c \right )\right )+\frac {a^{5} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(170\)

[In]

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

[Out]

-1/5*I/d*a^5*exp(5*I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.53 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {i \, a^{5} e^{\left (5 i \, d x + 5 i \, c\right )}}{5 \, d} \]

[In]

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

[Out]

-1/5*I*a^5*e^(5*I*d*x + 5*I*c)/d

Sympy [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.12 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx=\begin {cases} - \frac {i a^{5} e^{5 i c} e^{5 i d x}}{5 d} & \text {for}\: d \neq 0 \\a^{5} x e^{5 i c} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-I*a**5*exp(5*I*c)*exp(5*I*d*x)/(5*d), Ne(d, 0)), (a**5*x*exp(5*I*c), True))

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 152 vs. \(2 (26) = 52\).

Time = 0.63 (sec) , antiderivative size = 152, normalized size of antiderivative = 4.75 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx=-\frac {15 i \, a^{5} \cos \left (d x + c\right )^{5} - 15 \, a^{5} \sin \left (d x + c\right )^{5} + 10 i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3}\right )} a^{5} + i \, {\left (3 \, \cos \left (d x + c\right )^{5} - 10 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )\right )} a^{5} - 10 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{5}}{15 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1669 vs. \(2 (26) = 52\).

Time = 0.89 (sec) , antiderivative size = 1669, normalized size of antiderivative = 52.16 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/40960*(11375*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 91000*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(
I*d*x + I*c) + 1) + 318500*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 637000*a^5*e^(10*I*d*x + 2*I*
c)*log(I*e^(I*d*x + I*c) + 1) + 637000*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 318500*a^5*e^(4*I*
d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 91000*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 796250*a^
5*e^(8*I*d*x)*log(I*e^(I*d*x + I*c) + 1) + 11375*a^5*e^(-8*I*c)*log(I*e^(I*d*x + I*c) + 1) + 11590*a^5*e^(16*I
*d*x + 8*I*c)*log(I*e^(I*d*x + I*c) - 1) + 92720*a^5*e^(14*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 324520*
a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 649040*a^5*e^(10*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) -
1) + 649040*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 324520*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(I*d*x
+ I*c) - 1) + 92720*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 811300*a^5*e^(8*I*d*x)*log(I*e^(I*d*
x + I*c) - 1) + 11590*a^5*e^(-8*I*c)*log(I*e^(I*d*x + I*c) - 1) - 11375*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d
*x + I*c) + 1) - 91000*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 318500*a^5*e^(12*I*d*x + 4*I*c)*
log(-I*e^(I*d*x + I*c) + 1) - 637000*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 637000*a^5*e^(6*I*
d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 318500*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 91000*
a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 796250*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x + I*c) + 1) - 113
75*a^5*e^(-8*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 11590*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 9
2720*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 324520*a^5*e^(12*I*d*x + 4*I*c)*log(-I*e^(I*d*x +
I*c) - 1) - 649040*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 649040*a^5*e^(6*I*d*x - 2*I*c)*log(-
I*e^(I*d*x + I*c) - 1) - 324520*a^5*e^(4*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 92720*a^5*e^(2*I*d*x - 6
*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 811300*a^5*e^(8*I*d*x)*log(-I*e^(I*d*x + I*c) - 1) - 11590*a^5*e^(-8*I*c)*
log(-I*e^(I*d*x + I*c) - 1) + 215*a^5*e^(16*I*d*x + 8*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 1720*a^5*e^(14*I*d*x
+ 6*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 6020*a^5*e^(12*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 12040*a^5*e
^(10*I*d*x + 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 12040*a^5*e^(6*I*d*x - 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) +
6020*a^5*e^(4*I*d*x - 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 1720*a^5*e^(2*I*d*x - 6*I*c)*log(I*e^(I*d*x) + e^(-
I*c)) + 15050*a^5*e^(8*I*d*x)*log(I*e^(I*d*x) + e^(-I*c)) + 215*a^5*e^(-8*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 2
15*a^5*e^(16*I*d*x + 8*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 1720*a^5*e^(14*I*d*x + 6*I*c)*log(-I*e^(I*d*x) + e^
(-I*c)) - 6020*a^5*e^(12*I*d*x + 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 12040*a^5*e^(10*I*d*x + 2*I*c)*log(-I*e
^(I*d*x) + e^(-I*c)) - 12040*a^5*e^(6*I*d*x - 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 6020*a^5*e^(4*I*d*x - 4*I*
c)*log(-I*e^(I*d*x) + e^(-I*c)) - 1720*a^5*e^(2*I*d*x - 6*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 15050*a^5*e^(8*I
*d*x)*log(-I*e^(I*d*x) + e^(-I*c)) - 215*a^5*e^(-8*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 8192*I*a^5*e^(21*I*d*x
+ 13*I*c) + 65536*I*a^5*e^(19*I*d*x + 11*I*c) + 229376*I*a^5*e^(17*I*d*x + 9*I*c) + 458752*I*a^5*e^(15*I*d*x +
7*I*c) + 573440*I*a^5*e^(13*I*d*x + 5*I*c) + 458752*I*a^5*e^(11*I*d*x + 3*I*c) + 229376*I*a^5*e^(9*I*d*x + I*
c) + 65536*I*a^5*e^(7*I*d*x - I*c) + 8192*I*a^5*e^(5*I*d*x - 3*I*c))/(d*e^(16*I*d*x + 8*I*c) + 8*d*e^(14*I*d*x
+ 6*I*c) + 28*d*e^(12*I*d*x + 4*I*c) + 56*d*e^(10*I*d*x + 2*I*c) + 56*d*e^(6*I*d*x - 2*I*c) + 28*d*e^(4*I*d*x
- 4*I*c) + 8*d*e^(2*I*d*x - 6*I*c) + 70*d*e^(8*I*d*x) + d*e^(-8*I*c))

Mupad [B] (verification not implemented)

Time = 4.31 (sec) , antiderivative size = 104, normalized size of antiderivative = 3.25 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^5 \, dx=\frac {2\,a^5\,\left (5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{5\,d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]

[In]

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

[Out]

(2*a^5*(5*tan(c/2 + (d*x)/2)^4 - 10*tan(c/2 + (d*x)/2)^2 + 1))/(5*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))