\(\int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [426]

Optimal result

Integrand size = 30, antiderivative size = 80 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}+\frac {4 i \sqrt {e \sec (c+d x)}}{5 a d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

4/5*I*(e*sec(d*x+c))^(1/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)+2/5*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 80, 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 = {3583, 3569} \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {4 i \sqrt {e \sec (c+d x)}}{5 a d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

(((2*I)/5)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((4*I)/5)*Sqrt[e*Sec[c + d*x]])/(a*d*Sqrt
[a + I*a*Tan[c + d*x]])

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}+\frac {2 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{5 a} \\ & = \frac {2 i \sqrt {e \sec (c+d x)}}{5 d (a+i a \tan (c+d x))^{3/2}}+\frac {4 i \sqrt {e \sec (c+d x)}}{5 a d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {2 \sqrt {e \sec (c+d x)} (3+2 i \tan (c+d x))}{5 a d (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

(2*Sqrt[e*Sec[c + d*x]]*(3 + (2*I)*Tan[c + d*x]))/(5*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 14.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.74

method	result	size
default	\(\frac {2 i \sqrt {e \sec \left (d x +c \right )}\, \left (2 i \tan \left (d x +c \right )+3\right )}{5 d \left (1+i \tan \left (d x +c \right )\right ) a \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(59\)

[In]

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

[Out]

2/5*I/d*(e*sec(d*x+c))^(1/2)/(1+I*tan(d*x+c))/a/(a*(1+I*tan(d*x+c)))^(1/2)*(2*I*tan(d*x+c)+3)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} e^{\left (-\frac {5}{2} i \, d x - \frac {5}{2} i \, c\right )}}{5 \, a^{2} d} \]

[In]

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

[Out]

1/5*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(5*I*e^(4*I*d*x + 4*I*c) + 6*I*e^(2*I*
d*x + 2*I*c) + I)*e^(-5/2*I*d*x - 5/2*I*c)/(a^2*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.00 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {e} {\left (i \, \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 i \, \cos \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right ) + \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sin \left (\frac {1}{5} \, \arctan \left (\sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ), \cos \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )\right )\right )\right )}}{5 \, a^{\frac {3}{2}} d} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.84 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.05 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+5{}\mathrm {i}\right )}{5\,a\,d\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]

[In]

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

[Out]

((e/cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*1i + sin(2*c + 2*d*x) + 5i))/(5*a*d*((a*(cos(2*c + 2*d*x) + sin(2*c
+ 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2))