\(\int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [433]

Optimal result

Integrand size = 30, antiderivative size = 80 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}}+\frac {4 i (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

2/7*I*(e*sec(d*x+c))^(3/2)/d/(a+I*a*tan(d*x+c))^(5/2)+4/21*I*(e*sec(d*x+c))^(3/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.18 (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} \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {4 i (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}}+\frac {2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

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

[Out]

(((2*I)/7)*(e*Sec[c + d*x])^(3/2))/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((4*I)/21)*(e*Sec[c + d*x])^(3/2))/(a*d
*(a + I*a*Tan[c + d*x])^(3/2))

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 (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}}+\frac {2 \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{3/2}} \, dx}{7 a} \\ & = \frac {2 i (e \sec (c+d x))^{3/2}}{7 d (a+i a \tan (c+d x))^{5/2}}+\frac {4 i (e \sec (c+d x))^{3/2}}{21 a d (a+i a \tan (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.22 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.79 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 (e \sec (c+d x))^{3/2} (-5 i+2 \tan (c+d x))}{21 a^2 d (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

(2*(e*Sec[c + d*x])^(3/2)*(-5*I + 2*Tan[c + d*x]))/(21*a^2*d*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 14.33 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.91

method	result	size
default	\(\frac {2 i \sqrt {e \sec \left (d x +c \right )}\, e \left (2 i \tan \left (d x +c \right ) \sec \left (d x +c \right )+5 \sec \left (d x +c \right )\right )}{21 d \left (1+i \tan \left (d x +c \right )\right )^{2} a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(73\)

[In]

int((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/21*I/d*(e*sec(d*x+c))^(1/2)*e/(1+I*tan(d*x+c))^2/a^2/(a*(1+I*tan(d*x+c)))^(1/2)*(2*I*sec(d*x+c)*tan(d*x+c)+5
*sec(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (7 i \, e e^{\left (4 i \, d x + 4 i \, c\right )} + 10 i \, e e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, e\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (-\frac {7}{2} i \, d x - \frac {7}{2} i \, c\right )}}{21 \, a^{3} d} \]

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/21*(7*I*e*e^(4*I*d*x + 4*I*c) + 10*I*e*e^(2*I*d*x + 2*I*c) + 3*I*e)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e
/(e^(2*I*d*x + 2*I*c) + 1))*e^(-7/2*I*d*x - 7/2*I*c)/(a^3*d)

Sympy [F]

\[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

Integral((e*sec(c + d*x))**(3/2)/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.08 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (3 i \, e \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 i \, e \cos \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right ) + 3 \, e \sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ) + 7 \, e \sin \left (\frac {3}{7} \, \arctan \left (\sin \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right ), \cos \left (\frac {7}{2} \, d x + \frac {7}{2} \, c\right )\right )\right )\right )} \sqrt {e}}{21 \, a^{\frac {5}{2}} d} \]

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/21*(3*I*e*cos(7/2*d*x + 7/2*c) + 7*I*e*cos(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))) + 3*e*si
n(7/2*d*x + 7/2*c) + 7*e*sin(3/7*arctan2(sin(7/2*d*x + 7/2*c), cos(7/2*d*x + 7/2*c))))*sqrt(e)/(a^(5/2)*d)

Giac [F]

\[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 4.77 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.28 \[ \int \frac {(e \sec (c+d x))^{3/2}}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {e\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (7\,\sin \left (c+d\,x\right )+3\,\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,7{}\mathrm {i}+\cos \left (3\,c+3\,d\,x\right )\,3{}\mathrm {i}\right )}{21\,a^2\,d\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]

[In]

int((e/cos(c + d*x))^(3/2)/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

(e*(e/cos(c + d*x))^(1/2)*(cos(c + d*x)*7i + 7*sin(c + d*x) + cos(3*c + 3*d*x)*3i + 3*sin(3*c + 3*d*x)))/(21*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))