\(\int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx\) [989]

Optimal result

Integrand size = 33, antiderivative size = 188 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {7 i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{16 \sqrt {2} a c^{5/2} f}-\frac {7 i}{20 a f (c-i c \tan (e+f x))^{5/2}}+\frac {i}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}-\frac {7 i}{24 a c f (c-i c \tan (e+f x))^{3/2}}-\frac {7 i}{16 a c^2 f \sqrt {c-i c \tan (e+f x)}} \] Output:

7/32*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a/c^( 
5/2)/f-7/20*I/a/f/(c-I*c*tan(f*x+e))^(5/2)+1/2*I/a/f/(1+I*tan(f*x+e))/(c-I 
*c*tan(f*x+e))^(5/2)-7/24*I/a/c/f/(c-I*c*tan(f*x+e))^(3/2)-7/16*I/a/c^2/f/ 
(c-I*c*tan(f*x+e))^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.74 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.28 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},2,-\frac {3}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )}{10 a f (c-i c \tan (e+f x))^{5/2}} \] Input:

Integrate[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)),x]
 

Output:

((-1/10*I)*Hypergeometric2F1[-5/2, 2, -3/2, (-1/2*I)*(I + Tan[e + f*x])])/ 
(a*f*(c - I*c*Tan[e + f*x])^(5/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.44 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.95, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4005, 3042, 3968, 52, 61, 61, 61, 73, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 4005

\(\displaystyle \frac {\int \frac {\cos ^2(e+f x)}{(c-i c \tan (e+f x))^{3/2}}dx}{a c}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {1}{\sec (e+f x)^2 (c-i c \tan (e+f x))^{3/2}}dx}{a c}\)

\(\Big \downarrow \) 3968

\(\displaystyle \frac {i c^2 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{a f}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {i c^2 \left (\frac {7 \int \frac {1}{(c-i c \tan (e+f x))^{7/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{a f}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {i c^2 \left (\frac {7 \left (\frac {\int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{a f}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {i c^2 \left (\frac {7 \left (\frac {\frac {\int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{a f}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {i c^2 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{a f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {i c^2 \left (\frac {7 \left (\frac {\frac {\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{a f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {i c^2 \left (\frac {7 \left (\frac {\frac {-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} c^{3/2}}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}}{2 c}-\frac {1}{5 c (c-i c \tan (e+f x))^{5/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{5/2} (c+i c \tan (e+f x))}\right )}{a f}\)

Input:

Int[1/((a + I*a*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(5/2)),x]
 

Output:

(I*c^2*(1/(2*c*(c - I*c*Tan[e + f*x])^(5/2)*(c + I*c*Tan[e + f*x])) + (7*( 
-1/5*1/(c*(c - I*c*Tan[e + f*x])^(5/2)) + (-1/3*1/(c*(c - I*c*Tan[e + f*x] 
)^(3/2)) + (((-I)*ArcTan[(Sqrt[c]*Tan[e + f*x])/Sqrt[2]])/(Sqrt[2]*c^(3/2) 
) - 1/(c*Sqrt[c - I*c*Tan[e + f*x]]))/(2*c))/(2*c)))/(4*c)))/(a*f)
 

Defintions of rubi rules used

rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 61

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0 
] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d 
, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3968

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ 
), x_Symbol] :> Simp[1/(a^(m - 2)*b*f)   Subst[Int[(a - x)^(m/2 - 1)*(a + x 
)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && 
 EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
 

rule 4005

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m   Int[Sec[e + f*x]^(2*m)*(c + 
 d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ 
b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] &&  !(IGtQ[n, 0] && (LtQ[ 
m, 0] || GtQ[m, n]))
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {2 i c^{2} \left (-\frac {3}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{12 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{20 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}+\frac {7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{16 c^{4}}\right )}{f a}\)	\(140\)
default	\(\frac {2 i c^{2} \left (-\frac {3}{16 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{12 c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {1}{20 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}+\frac {\frac {\sqrt {c -i c \tan \left (f x +e \right )}}{2 c +2 i c \tan \left (f x +e \right )}+\frac {7 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{16 c^{4}}\right )}{f a}\)	\(140\)

Input:

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

Output:

2*I/f/a*c^2*(-3/16/c^4/(c-I*c*tan(f*x+e))^(1/2)-1/12/c^3/(c-I*c*tan(f*x+e) 
)^(3/2)-1/20/c^2/(c-I*c*tan(f*x+e))^(5/2)+1/16/c^4*(1/4*(c-I*c*tan(f*x+e)) 
^(1/2)/(1/2*c+1/2*I*c*tan(f*x+e))+7/4*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*t 
an(f*x+e))^(1/2)*2^(1/2)/c^(1/2))))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 316 vs. \(2 (139) = 278\).

Time = 0.12 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.68 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-105 i \, \sqrt {\frac {1}{2}} a c^{3} f \sqrt {\frac {1}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {7 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a c^{2} f}\right ) + 105 i \, \sqrt {\frac {1}{2}} a c^{3} f \sqrt {\frac {1}{a^{2} c^{5} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {7 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a c^{2} f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{2} c^{5} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{8 \, a c^{2} f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-6 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 38 i \, e^{\left (6 i \, f x + 6 i \, e\right )} - 148 i \, e^{\left (4 i \, f x + 4 i \, e\right )} - 101 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 15 i\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{480 \, a c^{3} f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fric 
as")
 

Output:

1/480*(-105*I*sqrt(1/2)*a*c^3*f*sqrt(1/(a^2*c^5*f^2))*e^(2*I*f*x + 2*I*e)* 
log(-7/8*(sqrt(2)*sqrt(1/2)*(I*a*c^2*f*e^(2*I*f*x + 2*I*e) + I*a*c^2*f)*sq 
rt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^2*c^5*f^2)) - I)*e^(-I*f*x - I*e 
)/(a*c^2*f)) + 105*I*sqrt(1/2)*a*c^3*f*sqrt(1/(a^2*c^5*f^2))*e^(2*I*f*x + 
2*I*e)*log(-7/8*(sqrt(2)*sqrt(1/2)*(-I*a*c^2*f*e^(2*I*f*x + 2*I*e) - I*a*c 
^2*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^2*c^5*f^2)) - I)*e^(-I*f 
*x - I*e)/(a*c^2*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*(-6*I*e^( 
8*I*f*x + 8*I*e) - 38*I*e^(6*I*f*x + 6*I*e) - 148*I*e^(4*I*f*x + 4*I*e) - 
101*I*e^(2*I*f*x + 2*I*e) + 15*I))*e^(-2*I*f*x - 2*I*e)/(a*c^3*f)
 

Sympy [F]

\[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=- \frac {i \int \frac {1}{- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )} - c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} - i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a} \] Input:

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

Output:

-I*Integral(1/(-c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - I*c**2* 
sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - c**2*sqrt(-I*c*tan(e + f*x) 
+ c)*tan(e + f*x) - I*c**2*sqrt(-I*c*tan(e + f*x) + c)), x)/a
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.86 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (105 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} - 140 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c - 56 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{2} - 48 \, c^{3}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a c - 2 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a c^{2}} + \frac {105 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a c^{\frac {3}{2}}}\right )}}{960 \, c f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxi 
ma")
 

Output:

-1/960*I*(4*(105*(-I*c*tan(f*x + e) + c)^3 - 140*(-I*c*tan(f*x + e) + c)^2 
*c - 56*(-I*c*tan(f*x + e) + c)*c^2 - 48*c^3)/((-I*c*tan(f*x + e) + c)^(7/ 
2)*a*c - 2*(-I*c*tan(f*x + e) + c)^(5/2)*a*c^2) + 105*sqrt(2)*log(-(sqrt(2 
)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I*c*tan( 
f*x + e) + c)))/(a*c^(3/2)))/(c*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [B] (verification not implemented)

Time = 0.67 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.88 \[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {\frac {c\,1{}\mathrm {i}}{5\,a\,f}+\frac {\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,7{}\mathrm {i}}{30\,a\,f}+\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,7{}\mathrm {i}}{12\,a\,c\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,7{}\mathrm {i}}{16\,a\,c^2\,f}}{2\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,7{}\mathrm {i}}{32\,a\,{\left (-c\right )}^{5/2}\,f} \] Input:

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

Output:

- ((c*1i)/(5*a*f) + ((c - c*tan(e + f*x)*1i)*7i)/(30*a*f) + ((c - c*tan(e 
+ f*x)*1i)^2*7i)/(12*a*c*f) - ((c - c*tan(e + f*x)*1i)^3*7i)/(16*a*c^2*f)) 
/(2*c*(c - c*tan(e + f*x)*1i)^(5/2) - (c - c*tan(e + f*x)*1i)^(7/2)) - (2^ 
(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*7i)/(32 
*a*(-c)^(5/2)*f)
 

Reduce [F]

\[ \int \frac {1}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

int(1/(a+I*a*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x)
 

Output:

(sqrt(c)*( - sqrt( - tan(e + f*x)*i + 1)*i - 165888*int(sqrt( - tan(e + f* 
x)*i + 1)/(36864*tan(e + f*x)**5*i - 36864*tan(e + f*x)**4 + 73728*tan(e + 
 f*x)**3*i - 73728*tan(e + f*x)**2 + 36864*tan(e + f*x)*i - 36864),x)*tan( 
e + f*x)**4*f - 331776*int(sqrt( - tan(e + f*x)*i + 1)/(36864*tan(e + f*x) 
**5*i - 36864*tan(e + f*x)**4 + 73728*tan(e + f*x)**3*i - 73728*tan(e + f* 
x)**2 + 36864*tan(e + f*x)*i - 36864),x)*tan(e + f*x)**2*f - 165888*int(sq 
rt( - tan(e + f*x)*i + 1)/(36864*tan(e + f*x)**5*i - 36864*tan(e + f*x)**4 
 + 73728*tan(e + f*x)**3*i - 73728*tan(e + f*x)**2 + 36864*tan(e + f*x)*i 
- 36864),x)*f + 129024*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/( 
36864*tan(e + f*x)**5*i - 36864*tan(e + f*x)**4 + 73728*tan(e + f*x)**3*i 
- 73728*tan(e + f*x)**2 + 36864*tan(e + f*x)*i - 36864),x)*tan(e + f*x)**4 
*f + 258048*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(36864*tan(e 
 + f*x)**5*i - 36864*tan(e + f*x)**4 + 73728*tan(e + f*x)**3*i - 73728*tan 
(e + f*x)**2 + 36864*tan(e + f*x)*i - 36864),x)*tan(e + f*x)**2*f + 129024 
*int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(36864*tan(e + f*x)**5* 
i - 36864*tan(e + f*x)**4 + 73728*tan(e + f*x)**3*i - 73728*tan(e + f*x)** 
2 + 36864*tan(e + f*x)*i - 36864),x)*f))/(4*a*c**3*f*(tan(e + f*x)**4 + 2* 
tan(e + f*x)**2 + 1))