\(\int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [682]

Optimal result

Integrand size = 30, antiderivative size = 126 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i (e \cos (c+d x))^{3/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {8 i (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{15 a d} \]

[Out]

2/5*I*(e*cos(d*x+c))^(3/2)/d/(a+I*a*tan(d*x+c))^(1/2)+16/15*I*(e*cos(d*x+c))^(3/2)*sec(d*x+c)^2/d/(a+I*a*tan(d
*x+c))^(1/2)-8/15*I*(e*cos(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^(1/2)/a/d

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3596, 3583, 3578, 3569} \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {8 i \sqrt {a+i a \tan (c+d x)} (e \cos (c+d x))^{3/2}}{15 a d}+\frac {2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i \sec ^2(c+d x) (e \cos (c+d x))^{3/2}}{15 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

(((2*I)/5)*(e*Cos[c + d*x])^(3/2))/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (((16*I)/15)*(e*Cos[c + d*x])^(3/2)*Sec[c
+ d*x]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((8*I)/15)*(e*Cos[c + d*x])^(3/2)*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 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 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 ((e \cos (c+d x))^{3/2} (e \sec (c+d x))^{3/2}\right ) \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {\left (4 (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}{5 a} \\ & = \frac {2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt {a+i a \tan (c+d x)}}-\frac {8 i (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{15 a d}+\frac {\left (8 (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}{15 e^2} \\ & = \frac {2 i (e \cos (c+d x))^{3/2}}{5 d \sqrt {a+i a \tan (c+d x)}}+\frac {16 i (e \cos (c+d x))^{3/2} \sec ^2(c+d x)}{15 d \sqrt {a+i a \tan (c+d x)}}-\frac {8 i (e \cos (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{15 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.60 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.50 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i e^2 (-15+\cos (2 (c+d x))+4 i \sin (2 (c+d x)))}{15 d \sqrt {e \cos (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

((-1/15*I)*e^2*(-15 + Cos[2*(c + d*x)] + (4*I)*Sin[2*(c + d*x)]))/(d*Sqrt[e*Cos[c + d*x]]*Sqrt[a + I*a*Tan[c +
d*x]])

Maple [A] (verified)

Time = 8.20 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.47

method	result	size
default	\(-\frac {2 e \sqrt {e \cos \left (d x +c \right )}\, \left (i \cos \left (d x +c \right )-4 \sin \left (d x +c \right )-8 i \sec \left (d x +c \right )\right )}{15 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(59\)

[In]

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

[Out]

-2/15/d*e*(e*cos(d*x+c))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)*(I*cos(d*x+c)-4*sin(d*x+c)-8*I*sec(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.66 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {\frac {1}{2}} {\left (-5 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 30 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e\right )} \sqrt {e e^{\left (2 i \, d x + 2 i \, c\right )} + e} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{30 \, a d} \]

[In]

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

[Out]

1/30*sqrt(2)*sqrt(1/2)*(-5*I*e*e^(4*I*d*x + 4*I*c) + 30*I*e*e^(2*I*d*x + 2*I*c) + 3*I*e)*sqrt(e*e^(2*I*d*x + 2
*I*c) + e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(-5/2*I*d*x - 5/2*I*c)/(a*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.69 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.08 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (3 i \, e \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) - 5 i \, e \cos \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 i \, e \cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 3 \, e \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, e \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + 30 \, e \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )} \sqrt {e}}{30 \, \sqrt {a} d} \]

[In]

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

[Out]

1/30*(3*I*e*cos(5/2*d*x + 5/2*c) - 5*I*e*cos(3/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*I*e
*cos(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 3*e*sin(5/2*d*x + 5/2*c) + 5*e*sin(3/5*arctan2
(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c))) + 30*e*sin(1/5*arctan2(sin(5/2*d*x + 5/2*c), cos(5/2*d*x + 5/2*c
))))*sqrt(e)/(sqrt(a)*d)

Giac [F]

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

[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 = 1.22 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.79 \[ \int \frac {(e \cos (c+d x))^{3/2}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {e\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (35\,\sin \left (c+d\,x\right )+3\,\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,25{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}\right )}{30\,a\,d} \]

[In]

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

[Out]

(e*(e*cos(c + d*x))^(1/2)*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos
(c + d*x)*25i + 35*sin(c + d*x) + cos(3*c + 3*d*x)*3i + 3*sin(3*c + 3*d*x)))/(30*a*d)