\(\int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [434]

Optimal result

Integrand size = 30, antiderivative size = 121 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i \sqrt {e \sec (c+d x)}}{45 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

16/45*I*(e*sec(d*x+c))^(1/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+2/9*I*(e*sec(d*x+c))^(1/2)/d/(a+I*a*tan(d*x+c))^(5
/2)+8/45*I*(e*sec(d*x+c))^(1/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3583, 3569} \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {16 i \sqrt {e \sec (c+d x)}}{45 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

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

[Out]

(((2*I)/9)*Sqrt[e*Sec[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((8*I)/45)*Sqrt[e*Sec[c + d*x]])/(a*d*(a
+ I*a*Tan[c + d*x])^(3/2)) + (((16*I)/45)*Sqrt[e*Sec[c + d*x]])/(a^2*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)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {4 \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx}{9 a} \\ & = \frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {8 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{45 a^2} \\ & = \frac {2 i \sqrt {e \sec (c+d x)}}{9 d (a+i a \tan (c+d x))^{5/2}}+\frac {8 i \sqrt {e \sec (c+d x)}}{45 a d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i \sqrt {e \sec (c+d x)}}{45 a^2 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.35 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \sec ^2(c+d x) \sqrt {e \sec (c+d x)} (9+25 \cos (2 (c+d x))+20 i \sin (2 (c+d x)))}{45 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

((-1/45*I)*Sec[c + d*x]^2*Sqrt[e*Sec[c + d*x]]*(9 + 25*Cos[2*(c + d*x)] + (20*I)*Sin[2*(c + d*x)]))/(a^2*d*(-I
+ Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 13.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.57

method	result	size
default	\(\frac {2 i \sqrt {e \sec \left (d x +c \right )}\, \left (20 i \tan \left (d x +c \right )+25-8 \left (\sec ^{2}\left (d x +c \right )\right )\right )}{45 d \left (1+i \tan \left (d x +c \right )\right )^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}\)	\(69\)

[In]

int((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/45*I/d*(e*sec(d*x+c))^(1/2)/(1+I*tan(d*x+c))^2/(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(20*I*tan(d*x+c)+25-8*sec(d*x+
c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (45 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 63 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 23 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 5 i\right )} e^{\left (-\frac {9}{2} i \, d x - \frac {9}{2} i \, c\right )}}{90 \, a^{3} d} \]

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/90*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(45*I*e^(6*I*d*x + 6*I*c) + 63*I*e^(4
*I*d*x + 4*I*c) + 23*I*e^(2*I*d*x + 2*I*c) + 5*I)*e^(-9/2*I*d*x - 9/2*I*c)/(a^3*d)

Sympy [F]

\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {e \sec {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral(sqrt(e*sec(c + d*x))/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.39 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {e} {\left (5 i \, \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 18 i \, \cos \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 45 i \, \cos \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 5 \, \sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ) + 18 \, \sin \left (\frac {5}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right ) + 45 \, \sin \left (\frac {1}{9} \, \arctan \left (\sin \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right ), \cos \left (\frac {9}{2} \, d x + \frac {9}{2} \, c\right )\right )\right )\right )}}{90 \, a^{\frac {5}{2}} d} \]

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/90*sqrt(e)*(5*I*cos(9/2*d*x + 9/2*c) + 18*I*cos(5/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 4
5*I*cos(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 5*sin(9/2*d*x + 9/2*c) + 18*sin(5/9*arctan2
(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))) + 45*sin(1/9*arctan2(sin(9/2*d*x + 9/2*c), cos(9/2*d*x + 9/2*c))
))/(a^(5/2)*d)

Giac [F]

\[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\sqrt {e \sec \left (d x + c\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(1/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(e*sec(d*x + c))/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 5.34 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.90 \[ \int \frac {\sqrt {e \sec (c+d x)}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,18{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,5{}\mathrm {i}+18\,\sin \left (2\,c+2\,d\,x\right )+5\,\sin \left (4\,c+4\,d\,x\right )+45{}\mathrm {i}\right )}{90\,a^2\,d\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]

[In]

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

[Out]

((e/cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*18i + cos(4*c + 4*d*x)*5i + 18*sin(2*c + 2*d*x) + 5*sin(4*c + 4*d*x)
+ 45i))/(90*a^2*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))