\(\int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx\) [248]

Optimal result

Integrand size = 22, antiderivative size = 32 \[ \int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i (a \cos (c+d x)+i a \sin (c+d x))^n}{d n} \]

[Out]

-I*(a*cos(d*x+c)+I*a*sin(d*x+c))^n/d/n

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {3150} \[ \int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i (a \cos (c+d x)+i a \sin (c+d x))^n}{d n} \]

[In]

Int[(a*Cos[c + d*x] + I*a*Sin[c + d*x])^n,x]

[Out]

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

Rule 3150

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a*Cos[c + d*x]
+ b*Sin[c + d*x])^n/(b*d*n)), x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[a^2 + b^2, 0]

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97 \[ \int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i (a (\cos (c+d x)+i \sin (c+d x)))^n}{d n} \]

[In]

Integrate[(a*Cos[c + d*x] + I*a*Sin[c + d*x])^n,x]

[Out]

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

Maple [A] (verified)

Time = 0.64 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(-\frac {i \left (\cos \left (d x +c \right ) a +i a \sin \left (d x +c \right )\right )^{n}}{d n}\)	\(31\)
default	\(-\frac {i \left (\cos \left (d x +c \right ) a +i a \sin \left (d x +c \right )\right )^{n}}{d n}\)	\(31\)
norman	\(-\frac {i {\mathrm e}^{n \ln \left (\frac {\left (1-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) a}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {2 i a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\right )}}{n d}\)	\(75\)
risch	\(-\frac {i a^{n} \left ({\mathrm e}^{i \left (d x +c \right )}\right )^{n} {\mathrm e}^{-\frac {i \operatorname {csgn}\left (i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) \pi n \left (-\operatorname {csgn}\left (i a \,{\mathrm e}^{i \left (d x +c \right )}\right )+\operatorname {csgn}\left (i {\mathrm e}^{i \left (d x +c \right )}\right )\right ) \left (-\operatorname {csgn}\left (i a \,{\mathrm e}^{i \left (d x +c \right )}\right )+\operatorname {csgn}\left (i a \right )\right )}{2}}}{n d}\)	\(96\)

[In]

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

[Out]

-I*(cos(d*x+c)*a+I*a*sin(d*x+c))^n/d/n

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i \, e^{\left (i \, d n x + i \, c n + n \log \left (a\right )\right )}}{d n} \]

[In]

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

[Out]

-I*e^(I*d*n*x + I*c*n + n*log(a))/(d*n)

Sympy [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.31 \[ \int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\begin {cases} x & \text {for}\: d = 0 \wedge n = 0 \\x \left (i a \sin {\left (c \right )} + a \cos {\left (c \right )}\right )^{n} & \text {for}\: d = 0 \\x & \text {for}\: n = 0 \\- \frac {i \left (i a \sin {\left (c + d x \right )} + a \cos {\left (c + d x \right )}\right )^{n}}{d n} & \text {otherwise} \end {cases} \]

[In]

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))**n,x)

[Out]

Piecewise((x, Eq(d, 0) & Eq(n, 0)), (x*(I*a*sin(c) + a*cos(c))**n, Eq(d, 0)), (x, Eq(n, 0)), (-I*(I*a*sin(c +
d*x) + a*cos(c + d*x))**n/(d*n), 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. 59 vs. \(2 (28) = 56\).

Time = 0.33 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.84 \[ \int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i \, a^{n} e^{\left (-n \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i\right ) + n \log \left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i\right )\right )}}{d n} \]

[In]

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

[Out]

-I*a^n*e^(-n*log(sin(d*x + c)/(cos(d*x + c) + 1) + I) + n*log(-sin(d*x + c)/(cos(d*x + c) + 1) + I))/(d*n)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=-\frac {i \, e^{\left (i \, d n x + i \, c n + n \log \left (a\right )\right )}}{d n} \]

[In]

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

[Out]

-I*e^(I*d*n*x + I*c*n + n*log(a))/(d*n)

Mupad [F(-1)]

Timed out. \[ \int (a \cos (c+d x)+i a \sin (c+d x))^n \, dx=\int {\left (a\,\cos \left (c+d\,x\right )+a\,\sin \left (c+d\,x\right )\,1{}\mathrm {i}\right )}^n \,d x \]

[In]

int((a*cos(c + d*x) + a*sin(c + d*x)*1i)^n,x)

[Out]

int((a*cos(c + d*x) + a*sin(c + d*x)*1i)^n, x)