\(\int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx\) [413]

Optimal result

Integrand size = 30, antiderivative size = 81 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{7 d (e \sec (c+d x))^{7/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 81, 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 = {3578, 3569} \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {4 i a (a+i a \tan (c+d x))^{3/2}}{21 d e^2 (e \sec (c+d x))^{3/2}}-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{7 d (e \sec (c+d x))^{7/2}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(7/2),x]

[Out]

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

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 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]

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

Mathematica [A] (verified)

Time = 1.40 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {2 a^2 (\cos (2 (c+2 d x))+i \sin (2 (c+2 d x))) (5 i+2 \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{21 d e^2 (e \sec (c+d x))^{3/2} (\cos (d x)+i \sin (d x))^2} \]

[In]

Integrate[(a + I*a*Tan[c + d*x])^(5/2)/(e*Sec[c + d*x])^(7/2),x]

[Out]

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

Maple [A] (verified)

Time = 9.87 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.94

method	result	size
default	\(\frac {2 \left (\tan \left (d x +c \right )-i\right )^{2} a^{2} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (5 i \left (\cos ^{3}\left (d x +c \right )\right )+2 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )\right )}{21 d \sqrt {e \sec \left (d x +c \right )}\, e^{3}}\)	\(76\)
risch	\(-\frac {i a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}+7 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{21 e^{3} \sqrt {\frac {e \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(88\)

[In]

int((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/21/d*(tan(d*x+c)-I)^2*a^2*(a*(1+I*tan(d*x+c)))^(1/2)/(e*sec(d*x+c))^(1/2)/e^3*(5*I*cos(d*x+c)^3+2*cos(d*x+c)
^2*sin(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {{\left (-3 i \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 10 i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} - 7 i \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}}{21 \, d e^{4}} \]

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/21*(-3*I*a^2*e^(5*I*d*x + 5*I*c) - 10*I*a^2*e^(3*I*d*x + 3*I*c) - 7*I*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*
x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c)/(d*e^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(d*x+c))**(5/2)/(e*sec(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.43 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.16 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\frac {{\left (-7 i \, a^{2} \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) - 3 i \, a^{2} \cos \left (\frac {7}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right ) + 7 \, a^{2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 3 \, a^{2} \sin \left (\frac {7}{3} \, \arctan \left (\sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ), \cos \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )\right )\right )\right )} \sqrt {a}}{21 \, d e^{\frac {7}{2}}} \]

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\left (e \sec \left (d x + c\right )\right )^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+I*a*tan(d*x+c))^(5/2)/(e*sec(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)/(e*sec(d*x + c))^(7/2), x)

Mupad [B] (verification not implemented)

Time = 5.99 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.38 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{(e \sec (c+d x))^{7/2}} \, dx=-\frac {a^2\,\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,10{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,3{}\mathrm {i}-10\,\sin \left (2\,c+2\,d\,x\right )-3\,\sin \left (4\,c+4\,d\,x\right )+7{}\mathrm {i}\right )}{42\,d\,e^4} \]

[In]

int((a + a*tan(c + d*x)*1i)^(5/2)/(e/cos(c + d*x))^(7/2),x)

[Out]

-(a^2*(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(2*c + 2*d*x)*10i + cos(4*c + 4*d*x)*3i - 10*sin(2*c + 2*d*x) - 3*sin(4*c + 4*d*x) + 7i))/(42*d*e^4)