\(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [33]

Optimal result

Integrand size = 24, antiderivative size = 69 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {3 a^2 \sin (c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x)}{5 d}-\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3577, 2713} \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {a^2 \sin ^3(c+d x)}{5 d}+\frac {3 a^2 \sin (c+d x)}{5 d}-\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \]

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

(3*a^2*Sin[c + d*x])/(5*d) - (a^2*Sin[c + d*x]^3)/(5*d) - (((2*I)/5)*Cos[c + d*x]^5*(a^2 + I*a^2*Tan[c + d*x])
)/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 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*} \text {integral}& = -\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}+\frac {1}{5} \left (3 a^2\right ) \int \cos ^3(c+d x) \, dx \\ & = -\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d}-\frac {\left (3 a^2\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d} \\ & = \frac {3 a^2 \sin (c+d x)}{5 d}-\frac {a^2 \sin ^3(c+d x)}{5 d}-\frac {2 i \cos ^5(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.99 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2 i a^2 \cos ^5(c+d x)}{5 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {a^2 \sin ^3(c+d x)}{d}+\frac {2 a^2 \sin ^5(c+d x)}{5 d} \]

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 14.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.97

method	result	size
risch	\(-\frac {i a^{2} {\mathrm e}^{5 i \left (d x +c \right )}}{40 d}-\frac {i a^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{8 d}-\frac {i a^{2} \cos \left (d x +c \right )}{4 d}+\frac {a^{2} \sin \left (d x +c \right )}{2 d}\)	\(67\)
derivativedivides	\(\frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {2 i a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(91\)
default	\(\frac {-a^{2} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{15}\right )-\frac {2 i a^{2} \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a^{2} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(91\)

[In]

int(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/40*I/d*a^2*exp(5*I*(d*x+c))-1/8*I/d*a^2*exp(3*I*(d*x+c))-1/4*I/d*a^2*cos(d*x+c)+1/2*a^2*sin(d*x+c)/d

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.90 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {{\left (-i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 5 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 15 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a^{2}\right )} e^{\left (-i \, d x - i \, c\right )}}{40 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/40*(-I*a^2*e^(6*I*d*x + 6*I*c) - 5*I*a^2*e^(4*I*d*x + 4*I*c) - 15*I*a^2*e^(2*I*d*x + 2*I*c) + 5*I*a^2)*e^(-I
*d*x - I*c)/d

Sympy [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 153 vs. \(2 (60) = 120\).

Time = 0.23 (sec) , antiderivative size = 153, normalized size of antiderivative = 2.22 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=\begin {cases} \frac {\left (- 512 i a^{2} d^{3} e^{6 i c} e^{5 i d x} - 2560 i a^{2} d^{3} e^{4 i c} e^{3 i d x} - 7680 i a^{2} d^{3} e^{2 i c} e^{i d x} + 2560 i a^{2} d^{3} e^{- i d x}\right ) e^{- i c}}{20480 d^{4}} & \text {for}\: d^{4} e^{i c} \neq 0 \\\frac {x \left (a^{2} e^{6 i c} + 3 a^{2} e^{4 i c} + 3 a^{2} e^{2 i c} + a^{2}\right ) e^{- i c}}{8} & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**5*(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((-512*I*a**2*d**3*exp(6*I*c)*exp(5*I*d*x) - 2560*I*a**2*d**3*exp(4*I*c)*exp(3*I*d*x) - 7680*I*a**2*
d**3*exp(2*I*c)*exp(I*d*x) + 2560*I*a**2*d**3*exp(-I*d*x))*exp(-I*c)/(20480*d**4), Ne(d**4*exp(I*c), 0)), (x*(
a**2*exp(6*I*c) + 3*a**2*exp(4*I*c) + 3*a**2*exp(2*I*c) + a**2)*exp(-I*c)/8, True))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {6 i \, a^{2} \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 5 \, \sin \left (d x + c\right )^{3}\right )} a^{2} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{2}}{15 \, d} \]

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/15*(6*I*a^2*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 5*sin(d*x + c)^3)*a^2 - (3*sin(d*x + c)^5 - 10*sin(d*x + c
)^3 + 15*sin(d*x + c))*a^2)/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. 613 vs. \(2 (59) = 118\).

Time = 0.60 (sec) , antiderivative size = 613, normalized size of antiderivative = 8.88 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {45 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 90 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 45 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 40 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 80 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 40 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 45 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 90 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 45 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 40 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 80 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 40 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 5 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 10 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 5 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 5 \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 10 \, a^{2} e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 5 \, a^{2} e^{\left (i \, d x - i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 4 i \, a^{2} e^{\left (10 i \, d x + 8 i \, c\right )} + 28 i \, a^{2} e^{\left (8 i \, d x + 6 i \, c\right )} + 104 i \, a^{2} e^{\left (6 i \, d x + 4 i \, c\right )} + 120 i \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} + 20 i \, a^{2} e^{\left (2 i \, d x\right )} - 20 i \, a^{2} e^{\left (-2 i \, c\right )}}{160 \, {\left (d e^{\left (5 i \, d x + 3 i \, c\right )} + 2 \, d e^{\left (3 i \, d x + i \, c\right )} + d e^{\left (i \, d x - i \, c\right )}\right )}} \]

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/160*(45*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) + 1) + 90*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c)
+ 1) + 45*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x + I*c) + 1) + 40*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x + I*c) -
1) + 80*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1) + 40*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x + I*c) - 1)
- 45*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 90*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) + 1
) - 45*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x + I*c) + 1) - 40*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x + I*c) - 1
) - 80*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) - 40*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x + I*c) - 1)
- 5*a^2*e^(5*I*d*x + 3*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 10*a^2*e^(3*I*d*x + I*c)*log(I*e^(I*d*x) + e^(-I*c))
- 5*a^2*e^(I*d*x - I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 5*a^2*e^(5*I*d*x + 3*I*c)*log(-I*e^(I*d*x) + e^(-I*c))
+ 10*a^2*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 5*a^2*e^(I*d*x - I*c)*log(-I*e^(I*d*x) + e^(-I*c)) +
4*I*a^2*e^(10*I*d*x + 8*I*c) + 28*I*a^2*e^(8*I*d*x + 6*I*c) + 104*I*a^2*e^(6*I*d*x + 4*I*c) + 120*I*a^2*e^(4*
I*d*x + 2*I*c) + 20*I*a^2*e^(2*I*d*x) - 20*I*a^2*e^(-2*I*c))/(d*e^(5*I*d*x + 3*I*c) + 2*d*e^(3*I*d*x + I*c) +
d*e^(I*d*x - I*c))

Mupad [B] (verification not implemented)

Time = 5.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.03 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {2\,a^2\,\left (\frac {5\,\sin \left (3\,c+3\,d\,x\right )}{16}-\frac {\cos \left (5\,c+5\,d\,x\right )\,1{}\mathrm {i}}{16}-\frac {\cos \left (3\,c+3\,d\,x\right )\,5{}\mathrm {i}}{16}+\frac {\sin \left (5\,c+5\,d\,x\right )}{16}+\frac {5\,\sqrt {3}\,\sin \left (c+d\,x-\frac {\ln \left (3\right )\,1{}\mathrm {i}}{2}\right )}{8}\right )}{5\,d} \]

[In]

int(cos(c + d*x)^5*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

(2*a^2*((5*sin(3*c + 3*d*x))/16 - (cos(5*c + 5*d*x)*1i)/16 - (cos(3*c + 3*d*x)*5i)/16 + sin(5*c + 5*d*x)/16 +
(5*3^(1/2)*sin(c - (log(3)*1i)/2 + d*x))/8))/(5*d)