\(\int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx\) [436]

Optimal result

Integrand size = 30, antiderivative size = 206 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {2 i}{13 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}}+\frac {16 i}{117 a d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {32 i}{195 a^2 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {256 i \sqrt {e \sec (c+d x)}}{585 a^2 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{585 a^3 d (e \sec (c+d x))^{3/2}} \]

[Out]

32/195*I/a^2/d/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(1/2)+256/585*I*(e*sec(d*x+c))^(1/2)/a^2/d/e^2/(a+I*a*t
an(d*x+c))^(1/2)-128/585*I*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/(e*sec(d*x+c))^(3/2)+2/13*I/d/(e*sec(d*x+c))^(3/2)/(
a+I*a*tan(d*x+c))^(5/2)+16/117*I/a/d/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3583, 3578, 3569} \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{585 a^3 d (e \sec (c+d x))^{3/2}}+\frac {256 i \sqrt {e \sec (c+d x)}}{585 a^2 d e^2 \sqrt {a+i a \tan (c+d x)}}+\frac {32 i}{195 a^2 d \sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{3/2}}+\frac {16 i}{117 a d (a+i a \tan (c+d x))^{3/2} (e \sec (c+d x))^{3/2}}+\frac {2 i}{13 d (a+i a \tan (c+d x))^{5/2} (e \sec (c+d x))^{3/2}} \]

[In]

Int[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((2*I)/13)/(d*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + ((16*I)/117)/(a*d*(e*Sec[c + d*x])^(3/2)*
(a + I*a*Tan[c + d*x])^(3/2)) + ((32*I)/195)/(a^2*d*(e*Sec[c + d*x])^(3/2)*Sqrt[a + I*a*Tan[c + d*x]]) + (((25
6*I)/585)*Sqrt[e*Sec[c + d*x]])/(a^2*d*e^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((128*I)/585)*Sqrt[a + I*a*Tan[c + d
*x]])/(a^3*d*(e*Sec[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 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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 i}{13 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}}+\frac {8 \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}} \, dx}{13 a} \\ & = \frac {2 i}{13 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}}+\frac {16 i}{117 a d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {16 \int \frac {1}{(e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}} \, dx}{39 a^2} \\ & = \frac {2 i}{13 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}}+\frac {16 i}{117 a d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {32 i}{195 a^2 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {64 \int \frac {\sqrt {a+i a \tan (c+d x)}}{(e \sec (c+d x))^{3/2}} \, dx}{195 a^3} \\ & = \frac {2 i}{13 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}}+\frac {16 i}{117 a d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {32 i}{195 a^2 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{585 a^3 d (e \sec (c+d x))^{3/2}}+\frac {128 \int \frac {\sqrt {e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{585 a^2 e^2} \\ & = \frac {2 i}{13 d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}}+\frac {16 i}{117 a d (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{3/2}}+\frac {32 i}{195 a^2 d (e \sec (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}+\frac {256 i \sqrt {e \sec (c+d x)}}{585 a^2 d e^2 \sqrt {a+i a \tan (c+d x)}}-\frac {128 i \sqrt {a+i a \tan (c+d x)}}{585 a^3 d (e \sec (c+d x))^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.57 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.52 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^4(c+d x) (-351 i-1300 i \cos (2 (c+d x))+75 i \cos (4 (c+d x))+1040 \sin (2 (c+d x))-120 \sin (4 (c+d x)))}{2340 a^2 d (e \sec (c+d x))^{3/2} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[1/((e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

(Sec[c + d*x]^4*(-351*I - (1300*I)*Cos[2*(c + d*x)] + (75*I)*Cos[4*(c + d*x)] + 1040*Sin[2*(c + d*x)] - 120*Si
n[4*(c + d*x)]))/(2340*a^2*d*(e*Sec[c + d*x])^(3/2)*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 8.52 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.50

method	result	size
default	\(-\frac {2 \left (75 i \cos \left (d x +c \right )-120 \sin \left (d x +c \right )-400 i \sec \left (d x +c \right )+320 \sec \left (d x +c \right ) \tan \left (d x +c \right )+128 i \left (\sec ^{3}\left (d x +c \right )\right )\right )}{585 d \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (1+i \tan \left (d x +c \right )\right )^{2} \sqrt {e \sec \left (d x +c \right )}\, a^{2} e}\)	\(102\)

[In]

int(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/585/d/(a*(1+I*tan(d*x+c)))^(1/2)/(1+I*tan(d*x+c))^2/(e*sec(d*x+c))^(1/2)/a^2/e*(75*I*cos(d*x+c)-120*sin(d*x
+c)-400*I*sec(d*x+c)+320*sec(d*x+c)*tan(d*x+c)+128*I*sec(d*x+c)^3)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.54 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-195 i \, e^{\left (10 i \, d x + 10 i \, c\right )} + 2145 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 3042 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 962 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 305 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 45 i\right )} e^{\left (-\frac {13}{2} i \, d x - \frac {13}{2} i \, c\right )}}{4680 \, a^{3} d e^{2}} \]

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/4680*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(e/(e^(2*I*d*x + 2*I*c) + 1))*(-195*I*e^(10*I*d*x + 10*I*c) + 214
5*I*e^(8*I*d*x + 8*I*c) + 3042*I*e^(6*I*d*x + 6*I*c) + 962*I*e^(4*I*d*x + 4*I*c) + 305*I*e^(2*I*d*x + 2*I*c) +
45*I)*e^(-13/2*I*d*x - 13/2*I*c)/(a^3*d*e^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.64 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.10 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {45 i \, \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ) + 260 i \, \cos \left (\frac {9}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 702 i \, \cos \left (\frac {5}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) - 195 i \, \cos \left (\frac {3}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 2340 i \, \cos \left (\frac {1}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 45 \, \sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ) + 260 \, \sin \left (\frac {9}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 702 \, \sin \left (\frac {5}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 195 \, \sin \left (\frac {3}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right ) + 2340 \, \sin \left (\frac {1}{13} \, \arctan \left (\sin \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right ), \cos \left (\frac {13}{2} \, d x + \frac {13}{2} \, c\right )\right )\right )}{4680 \, a^{\frac {5}{2}} d e^{\frac {3}{2}}} \]

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/4680*(45*I*cos(13/2*d*x + 13/2*c) + 260*I*cos(9/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)))
+ 702*I*cos(5/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) - 195*I*cos(3/13*arctan2(sin(13/2*d*
x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 2340*I*cos(1/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c)
)) + 45*sin(13/2*d*x + 13/2*c) + 260*sin(9/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 702*s
in(5/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))) + 195*sin(3/13*arctan2(sin(13/2*d*x + 13/2*c)
, cos(13/2*d*x + 13/2*c))) + 2340*sin(1/13*arctan2(sin(13/2*d*x + 13/2*c), cos(13/2*d*x + 13/2*c))))/(a^(5/2)*
d*e^(3/2))

Giac [F]

\[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(e*sec(d*x+c))^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.95 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.66 \[ \int \frac {1}{(e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\frac {e}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,507{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,260{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,45{}\mathrm {i}+897\,\sin \left (2\,c+2\,d\,x\right )+260\,\sin \left (4\,c+4\,d\,x\right )+45\,\sin \left (6\,c+6\,d\,x\right )+2340{}\mathrm {i}\right )}{4680\,a^2\,d\,e^2\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}} \]

[In]

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

[Out]

((e/cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*507i + cos(4*c + 4*d*x)*260i + cos(6*c + 6*d*x)*45i + 897*sin(2*c +
2*d*x) + 260*sin(4*c + 4*d*x) + 45*sin(6*c + 6*d*x) + 2340i))/(4680*a^2*d*e^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))