\(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx\) [32]

Optimal result

Integrand size = 24, antiderivative size = 51 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \sin (c+d x)}{3 d}-\frac {2 i \cos ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3577, 2717} \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {a^2 \sin (c+d x)}{3 d}-\frac {2 i \cos ^3(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )}{3 d} \]

[In]

Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Rule 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2 i a^2 \cos ^3(c+d x)}{3 d}+\frac {a^2 \sin (c+d x)}{d}-\frac {2 a^2 \sin ^3(c+d x)}{3 d} \]

[In]

Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^2,x]

[Out]

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

Maple [A] (verified)

Time = 3.13 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.75

method	result	size
risch	\(-\frac {i a^{2} {\mathrm e}^{3 i \left (d x +c \right )}}{6 d}-\frac {i a^{2} {\mathrm e}^{i \left (d x +c \right )}}{2 d}\)	\(38\)
derivativedivides	\(\frac {-\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {2 i a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(54\)
default	\(\frac {-\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3}-\frac {2 i a^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{3}+\frac {a^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(54\)

[In]

int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

-1/6*I/d*a^2*exp(3*I*(d*x+c))-1/2*I/d*a^2*exp(I*(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.67 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\frac {-i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 3 i \, a^{2} e^{\left (i \, d x + i \, c\right )}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

1/6*(-I*a^2*e^(3*I*d*x + 3*I*c) - 3*I*a^2*e^(I*d*x + I*c))/d

Sympy [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.47 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=\begin {cases} \frac {- 2 i a^{2} d e^{3 i c} e^{3 i d x} - 6 i a^{2} d e^{i c} e^{i d x}}{12 d^{2}} & \text {for}\: d^{2} \neq 0 \\x \left (\frac {a^{2} e^{3 i c}}{2} + \frac {a^{2} e^{i c}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**2,x)

[Out]

Piecewise(((-2*I*a**2*d*exp(3*I*c)*exp(3*I*d*x) - 6*I*a**2*d*exp(I*c)*exp(I*d*x))/(12*d**2), Ne(d**2, 0)), (x*
(a**2*exp(3*I*c)/2 + a**2*exp(I*c)/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.65 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2 i \, a^{2} \cos \left (d x + c\right )^{3} + a^{2} \sin \left (d x + c\right )^{3} + {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{2}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/3*(2*I*a^2*cos(d*x + c)^3 + a^2*sin(d*x + c)^3 + (sin(d*x + c)^3 - 3*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. 531 vs. \(2 (43) = 86\).

Time = 0.52 (sec) , antiderivative size = 531, normalized size of antiderivative = 10.41 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {24 \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 48 \, a^{2} e^{\left (2 i \, d x\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 24 \, a^{2} e^{\left (-2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 27 \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 54 \, a^{2} e^{\left (2 i \, d x\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 27 \, a^{2} e^{\left (-2 i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 24 \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 48 \, a^{2} e^{\left (2 i \, d x\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 24 \, a^{2} e^{\left (-2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 27 \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 54 \, a^{2} e^{\left (2 i \, d x\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 27 \, a^{2} e^{\left (-2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) + 3 \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 6 \, a^{2} e^{\left (2 i \, d x\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 3 \, a^{2} e^{\left (-2 i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 3 \, a^{2} e^{\left (4 i \, d x + 2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 6 \, a^{2} e^{\left (2 i \, d x\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) - 3 \, a^{2} e^{\left (-2 i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 16 i \, a^{2} e^{\left (7 i \, d x + 5 i \, c\right )} + 80 i \, a^{2} e^{\left (5 i \, d x + 3 i \, c\right )} + 112 i \, a^{2} e^{\left (3 i \, d x + i \, c\right )} + 48 i \, a^{2} e^{\left (i \, d x - i \, c\right )}}{96 \, {\left (d e^{\left (4 i \, d x + 2 i \, c\right )} + 2 \, d e^{\left (2 i \, d x\right )} + d e^{\left (-2 i \, c\right )}\right )}} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/96*(24*a^2*e^(4*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 48*a^2*e^(2*I*d*x)*log(I*e^(I*d*x + I*c) + 1) +
24*a^2*e^(-2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 27*a^2*e^(4*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 54*a^2
*e^(2*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 27*a^2*e^(-2*I*c)*log(I*e^(I*d*x + I*c) - 1) - 24*a^2*e^(4*I*d*x + 2
*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 48*a^2*e^(2*I*d*x)*log(-I*e^(I*d*x + I*c) + 1) - 24*a^2*e^(-2*I*c)*log(-I*
e^(I*d*x + I*c) + 1) - 27*a^2*e^(4*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 54*a^2*e^(2*I*d*x)*log(-I*e^(I
*d*x + I*c) - 1) - 27*a^2*e^(-2*I*c)*log(-I*e^(I*d*x + I*c) - 1) + 3*a^2*e^(4*I*d*x + 2*I*c)*log(I*e^(I*d*x) +
e^(-I*c)) + 6*a^2*e^(2*I*d*x)*log(I*e^(I*d*x) + e^(-I*c)) + 3*a^2*e^(-2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 3*
a^2*e^(4*I*d*x + 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 6*a^2*e^(2*I*d*x)*log(-I*e^(I*d*x) + e^(-I*c)) - 3*a^2*
e^(-2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 16*I*a^2*e^(7*I*d*x + 5*I*c) + 80*I*a^2*e^(5*I*d*x + 3*I*c) + 112*I*
a^2*e^(3*I*d*x + I*c) + 48*I*a^2*e^(I*d*x - I*c))/(d*e^(4*I*d*x + 2*I*c) + 2*d*e^(2*I*d*x) + d*e^(-2*I*c))

Mupad [B] (verification not implemented)

Time = 3.71 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.53 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^2 \, dx=-\frac {2\,a^2\,\left (3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,3{}\mathrm {i}-2\right )}{3\,d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,3{}\mathrm {i}+3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^2,x)

[Out]

-(2*a^2*(tan(c/2 + (d*x)/2)*3i + 3*tan(c/2 + (d*x)/2)^2 - 2))/(3*d*(3*tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)/2)^
2*3i - tan(c/2 + (d*x)/2)^3 + 1i))