\(\int \sqrt {a \cos (c+d x)+i a \sin (c+d x)} \, dx\) [259]

Optimal result

Integrand size = 24, antiderivative size = 31 \[ \int \sqrt {a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {2 i \sqrt {a \cos (c+d x)+i a \sin (c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, 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 = {3150} \[ \int \sqrt {a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {2 i \sqrt {a \cos (c+d x)+i a \sin (c+d x)}}{d} \]

[In]

Int[Sqrt[a*Cos[c + d*x] + I*a*Sin[c + d*x]],x]

[Out]

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

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 {2 i \sqrt {a \cos (c+d x)+i a \sin (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int \sqrt {a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {2 i \sqrt {a (\cos (c+d x)+i \sin (c+d x))}}{d} \]

[In]

Integrate[Sqrt[a*Cos[c + d*x] + I*a*Sin[c + d*x]],x]

[Out]

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

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.65

method	result	size
risch	\(-\frac {2 i \sqrt {a \,{\mathrm e}^{i \left (d x +c \right )}}}{d}\)	\(20\)
derivativedivides	\(-\frac {2 i \sqrt {\cos \left (d x +c \right ) a +i a \sin \left (d x +c \right )}}{d}\)	\(28\)
default	\(-\frac {2 i \sqrt {\cos \left (d x +c \right ) a +i a \sin \left (d x +c \right )}}{d}\)	\(28\)

[In]

int((cos(d*x+c)*a+I*a*sin(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.55 \[ \int \sqrt {a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {2 i \, \sqrt {a} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{d} \]

[In]

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sqrt {a \cos (c+d x)+i a \sin (c+d x)} \, dx=\int \sqrt {i a \sin {\left (c + d x \right )} + a \cos {\left (c + d x \right )}}\, dx \]

[In]

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(I*a*sin(c + d*x) + a*cos(c + d*x)), x)

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. 51 vs. \(2 (25) = 50\).

Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.65 \[ \int \sqrt {a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {2 i \, \sqrt {a} \sqrt {-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i}}{d \sqrt {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + i}} \]

[In]

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2*I*sqrt(a)*sqrt(-sin(d*x + c)/(cos(d*x + c) + 1) + I)/(d*sqrt(sin(d*x + c)/(cos(d*x + c) + 1) + I))

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.81 \[ \int \sqrt {a \cos (c+d x)+i a \sin (c+d x)} \, dx=-\frac {2 i \, \sqrt {a \cos \left (d x + c\right ) + i \, a \sin \left (d x + c\right )}}{d} \]

[In]

integrate((a*cos(d*x+c)+I*a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-2*I*sqrt(a*cos(d*x + c) + I*a*sin(d*x + c))/d

Mupad [F(-1)]

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

[In]

int((a*cos(c + d*x) + a*sin(c + d*x)*1i)^(1/2),x)

[Out]

int((a*cos(c + d*x) + a*sin(c + d*x)*1i)^(1/2), x)