\(\int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [683]

Optimal result

Integrand size = 30, antiderivative size = 80 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i \sqrt {e \cos (c+d x)}}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {4 i \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d} \]

[Out]

2/3*I*(e*cos(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(1/2)-4/3*I*(e*cos(d*x+c))^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/a/d

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3596, 3583, 3569} \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i \sqrt {e \cos (c+d x)}}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {4 i \sqrt {a+i a \tan (c+d x)} \sqrt {e \cos (c+d x)}}{3 a d} \]

[In]

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

[Out]

(((2*I)/3)*Sqrt[e*Cos[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((4*I)/3)*Sqrt[e*Cos[c + d*x]]*Sqrt[a + I*a
*Tan[c + d*x]])/(a*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 3583

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(d*
Sec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(b*f*(m + 2*n))), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

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 {1}{\sqrt {e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {2 i \sqrt {e \cos (c+d x)}}{3 d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (2 \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}{3 a} \\ & = \frac {2 i \sqrt {e \cos (c+d x)}}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {4 i \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}}{3 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 \sqrt {e \cos (c+d x)} (-i+2 \tan (c+d x))}{3 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 7.82 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.52

method	result	size
default	\(-\frac {2 \sqrt {e \cos \left (d x +c \right )}\, \left (i-2 \tan \left (d x +c \right )\right )}{3 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(42\)
risch	\(-\frac {i \sqrt {2}\, \sqrt {e \cos \left (d x +c \right )}\, \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{3 \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(72\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-3 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac {3}{2} i \, d x - \frac {3}{2} i \, c\right )}}{3 \, a d} \]

[In]

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

[Out]

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

Sympy [F]

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

[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 [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e} {\left (i \, \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )}}{3 \, \sqrt {a} d} \]

[In]

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

[Out]

1/3*sqrt(e)*(I*cos(3/2*d*x + 3/2*c) - 3*I*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sin(3
/2*d*x + 3/2*c) + 3*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))/(sqrt(a)*d)

Giac [F]

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

[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 [B] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.02 \[ \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )-3{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}}{3\,a\,d} \]

[In]

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

[Out]

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