\(\int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx\) [17]

Optimal result

Integrand size = 22, antiderivative size = 62 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {i a \cos ^5(c+d x)}{5 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3567, 2713} \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {a \sin ^5(c+d x)}{5 d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin (c+d x)}{d}-\frac {i a \cos ^5(c+d x)}{5 d} \]

[In]

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

[Out]

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

Rubi steps \begin{align*} \text {integral}& = -\frac {i a \cos ^5(c+d x)}{5 d}+a \int \cos ^5(c+d x) \, dx \\ & = -\frac {i a \cos ^5(c+d x)}{5 d}-\frac {a \text {Subst}\left (\int \left (1-2 x^2+x^4\right ) \, dx,x,-\sin (c+d x)\right )}{d} \\ & = -\frac {i a \cos ^5(c+d x)}{5 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {i a \cos ^5(c+d x)}{5 d}+\frac {a \sin (c+d x)}{d}-\frac {2 a \sin ^3(c+d x)}{3 d}+\frac {a \sin ^5(c+d x)}{5 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 6.79 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {-\frac {i a \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(47\)
default	\(\frac {-\frac {i a \left (\cos ^{5}\left (d x +c \right )\right )}{5}+\frac {a \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}}{d}\)	\(47\)
risch	\(-\frac {i a \,{\mathrm e}^{5 i \left (d x +c \right )}}{80 d}-\frac {i a \cos \left (d x +c \right )}{8 d}+\frac {5 a \sin \left (d x +c \right )}{8 d}-\frac {i a \cos \left (3 d x +3 c \right )}{16 d}+\frac {5 a \sin \left (3 d x +3 c \right )}{48 d}\)	\(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.06 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\frac {{\left (-3 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 20 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 90 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 60 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i \, a\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{240 \, d} \]

[In]

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

[Out]

1/240*(-3*I*a*e^(8*I*d*x + 8*I*c) - 20*I*a*e^(6*I*d*x + 6*I*c) - 90*I*a*e^(4*I*d*x + 4*I*c) + 60*I*a*e^(2*I*d*
x + 2*I*c) + 5*I*a)*e^(-3*I*d*x - 3*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. 184 vs. \(2 (53) = 106\).

Time = 0.24 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.97 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx=\begin {cases} \frac {\left (- 18432 i a d^{4} e^{9 i c} e^{5 i d x} - 122880 i a d^{4} e^{7 i c} e^{3 i d x} - 552960 i a d^{4} e^{5 i c} e^{i d x} + 368640 i a d^{4} e^{3 i c} e^{- i d x} + 30720 i a d^{4} e^{i c} e^{- 3 i d x}\right ) e^{- 4 i c}}{1474560 d^{5}} & \text {for}\: d^{5} e^{4 i c} \neq 0 \\\frac {x \left (a e^{8 i c} + 4 a e^{6 i c} + 6 a e^{4 i c} + 4 a e^{2 i c} + a\right ) e^{- 3 i c}}{16} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise(((-18432*I*a*d**4*exp(9*I*c)*exp(5*I*d*x) - 122880*I*a*d**4*exp(7*I*c)*exp(3*I*d*x) - 552960*I*a*d**
4*exp(5*I*c)*exp(I*d*x) + 368640*I*a*d**4*exp(3*I*c)*exp(-I*d*x) + 30720*I*a*d**4*exp(I*c)*exp(-3*I*d*x))*exp(
-4*I*c)/(1474560*d**5), Ne(d**5*exp(4*I*c), 0)), (x*(a*exp(8*I*c) + 4*a*exp(6*I*c) + 6*a*exp(4*I*c) + 4*a*exp(
2*I*c) + a)*exp(-3*I*c)/16, True))

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.79 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {3 i \, a \cos \left (d x + c\right )^{5} - {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a}{15 \, d} \]

[In]

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

[Out]

-1/15*(3*I*a*cos(d*x + c)^5 - (3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a)/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. 220 vs. \(2 (54) = 108\).

Time = 0.53 (sec) , antiderivative size = 220, normalized size of antiderivative = 3.55 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {{\left (135 \, a e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) + 90 \, a e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 135 \, a e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} + 1\right ) - 90 \, a e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x + i \, c\right )} - 1\right ) - 45 \, a e^{\left (3 i \, d x + i \, c\right )} \log \left (i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 45 \, a e^{\left (3 i \, d x + i \, c\right )} \log \left (-i \, e^{\left (i \, d x\right )} + e^{\left (-i \, c\right )}\right ) + 12 i \, a e^{\left (8 i \, d x + 6 i \, c\right )} + 80 i \, a e^{\left (6 i \, d x + 4 i \, c\right )} + 360 i \, a e^{\left (4 i \, d x + 2 i \, c\right )} - 240 i \, a e^{\left (2 i \, d x\right )} - 20 i \, a e^{\left (-2 i \, c\right )}\right )} e^{\left (-3 i \, d x - i \, c\right )}}{960 \, d} \]

[In]

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

[Out]

-1/960*(135*a*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c) + 1) + 90*a*e^(3*I*d*x + I*c)*log(I*e^(I*d*x + I*c) - 1)
- 135*a*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) + 1) - 90*a*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x + I*c) - 1) -
45*a*e^(3*I*d*x + I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 45*a*e^(3*I*d*x + I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 12*
I*a*e^(8*I*d*x + 6*I*c) + 80*I*a*e^(6*I*d*x + 4*I*c) + 360*I*a*e^(4*I*d*x + 2*I*c) - 240*I*a*e^(2*I*d*x) - 20*
I*a*e^(-2*I*c))*e^(-3*I*d*x - I*c)/d

Mupad [B] (verification not implemented)

Time = 6.00 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.13 \[ \int \cos ^5(c+d x) (a+i a \tan (c+d x)) \, dx=-\frac {2\,a\,\left (-\frac {75\,\sin \left (c+d\,x\right )}{16}-\frac {25\,\sin \left (3\,c+3\,d\,x\right )}{32}-\frac {3\,\sin \left (5\,c+5\,d\,x\right )}{32}+\frac {\cos \left (c+d\,x\right )\,15{}\mathrm {i}}{16}+\frac {\cos \left (3\,c+3\,d\,x\right )\,15{}\mathrm {i}}{32}+\frac {\cos \left (5\,c+5\,d\,x\right )\,3{}\mathrm {i}}{32}\right )}{15\,d} \]

[In]

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

[Out]

-(2*a*((cos(c + d*x)*15i)/16 - (75*sin(c + d*x))/16 + (cos(3*c + 3*d*x)*15i)/32 + (cos(5*c + 5*d*x)*3i)/32 - (
25*sin(3*c + 3*d*x))/32 - (3*sin(5*c + 5*d*x))/32))/(15*d)