\(\int \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx\) [676]

Optimal result

Integrand size = 30, antiderivative size = 36 \[ \int \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3596, 3569} \[ \int \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{d} \]

[In]

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

[Out]

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

Rule 3569

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0] &
& EqQ[Simplify[m + n], 0]

Rule 3596

Int[(cos[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(d*Co
s[e + f*x])^m*(d*Sec[e + f*x])^m, Int[(a + b*Tan[e + f*x])^n/(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e,
f, m, n}, x] &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {e \cos (c+d x)} \sqrt {e \sec (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {e \sec (c+d x)}} \, dx \\ & = -\frac {2 i \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 7.31 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {2 i \sqrt {e \cos \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{d}\)	\(32\)
risch	\(-\frac {2 i \sqrt {2}\, \sqrt {e \cos \left (d x +c \right )}\, \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}{d}\)	\(46\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.44 \[ \int \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \, \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{d} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(sqrt(e*cos(c + d*x))*sqrt(I*a*(tan(c + d*x) - I)), 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. 76 vs. \(2 (28) = 56\).

Time = 0.50 (sec) , antiderivative size = 76, normalized size of antiderivative = 2.11 \[ \int \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {2 i \, \sqrt {a} \sqrt {e} \sqrt {-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}}{d \sqrt {-\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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