\(\int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx\) [675]

Optimal result

Integrand size = 30, antiderivative size = 85 \[ \int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {4 i a e \sqrt {e \cos (c+d x)} \sec (c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.01, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3596, 3578, 3569} \[ \int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {4 i a \sec ^2(c+d x) (e \cos (c+d x))^{3/2}}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2}}{3 d} \]

[In]

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

[Out]

(((4*I)/3)*a*(e*Cos[c + d*x])^(3/2)*Sec[c + d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((2*I)/3)*(e*Cos[c + d*x
])^(3/2)*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 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[e + f*x])^m*((a + b*Tan[e + f*x])^n/(a*f*m)), x] + Dist[a*((m + n)/(m*d^2)), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && 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 ((e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx \\ & = -\frac {2 i (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\left (2 a (e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 e^2} \\ & = \frac {4 i a (e \cos (c+d x))^{3/2} \sec ^2(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}}-\frac {2 i (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.66 \[ \int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 e \sqrt {e \cos (c+d x)} (i \cos (c+d x)+2 \sin (c+d x)) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 8.58 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.59

method	result	size
default	\(\frac {2 \left (i \cos \left (d x +c \right )+2 \sin \left (d x +c \right )\right ) \sqrt {e \cos \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, e}{3 d}\)	\(50\)
risch	\(-\frac {i e \sqrt {2}\, \sqrt {e \cos \left (d x +c \right )}\, \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (-2 \cos \left (d x +c \right )+4 i \sin \left (d x +c \right )\right )}{3 d}\)	\(65\)

[In]

int((e*cos(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.80 \[ \int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} {\left (-i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {1}{2} i \, d x - \frac {1}{2} i \, c\right )}}{3 \, d} \]

[In]

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

Timed out. \[ \int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.73 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {{\left (-i \, e \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 i \, e \cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + e \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, e \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a} \sqrt {e}}{3 \, d} \]

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx=\int { \left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}} \sqrt {i \, a \tan \left (d x + c\right ) + a} \,d x } \]

[In]

integrate((e*cos(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((e*cos(d*x + c))^(3/2)*sqrt(I*a*tan(d*x + c) + a), x)

Mupad [B] (verification not implemented)

Time = 0.58 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.04 \[ \int (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2\,e\,\sqrt {e\,\left (2\,{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}\,\left ({\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,2{}\mathrm {i}+2\,\sin \left (c+d\,x\right )-\mathrm {i}\right )\,\sqrt {\frac {a\,\left (2\,{\cos \left (c+d\,x\right )}^2+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{2\,{\cos \left (c+d\,x\right )}^2}}}{3\,d} \]

[In]

int((e*cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

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