\(\int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx\) [985]

Optimal result

Integrand size = 33, antiderivative size = 242 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {105 i \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{256 \sqrt {2} a^3 c^{3/2} f}-\frac {35 i}{128 a^3 f (c-i c \tan (e+f x))^{3/2}}+\frac {i}{6 a^3 f (1+i \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}}+\frac {3 i}{16 a^3 f (1+i \tan (e+f x))^2 (c-i c \tan (e+f x))^{3/2}}+\frac {21 i}{64 a^3 f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {105 i}{256 a^3 c f \sqrt {c-i c \tan (e+f x)}} \] Output:

105/512*I*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))*2^(1/2)/a^ 
3/c^(3/2)/f-35/128*I/a^3/f/(c-I*c*tan(f*x+e))^(3/2)+1/6*I/a^3/f/(1+I*tan(f 
*x+e))^3/(c-I*c*tan(f*x+e))^(3/2)+3/16*I/a^3/f/(1+I*tan(f*x+e))^2/(c-I*c*t 
an(f*x+e))^(3/2)+21/64*I/a^3/f/(1+I*tan(f*x+e))/(c-I*c*tan(f*x+e))^(3/2)-1 
05/256*I/a^3/c/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 = 0.90 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.22 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},4,-\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )}{24 a^3 f (c-i c \tan (e+f x))^{3/2}} \] Input:

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

Output:

((-1/24*I)*Hypergeometric2F1[-3/2, 4, -1/2, (-1/2*I)*(I + Tan[e + f*x])])/ 
(a^3*f*(c - I*c*Tan[e + f*x])^(3/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 241, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4005, 3042, 3968, 52, 52, 52, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4005

\(\displaystyle \frac {\int \cos ^6(e+f x) (c-i c \tan (e+f x))^{3/2}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {(c-i c \tan (e+f x))^{3/2}}{\sec (e+f x)^6}dx}{a^3 c^3}\)

\(\Big \downarrow \) 3968

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

\(\Big \downarrow \) 52

\(\displaystyle \frac {i c^4 \left (\frac {3 \int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)^3}d(-i c \tan (e+f x))}{4 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {i c^4 \left (\frac {3 \left (\frac {7 \int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)^2}d(-i c \tan (e+f x))}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{4 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {i c^4 \left (\frac {3 \left (\frac {7 \left (\frac {5 \int \frac {1}{(c-i c \tan (e+f x))^{5/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{4 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {i c^4 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {\int \frac {1}{(c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{4 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 61

\(\displaystyle \frac {i c^4 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{\sqrt {c-i c \tan (e+f x)} (i \tan (e+f x) c+c)}d(-i c \tan (e+f x))}{2 c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{4 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {i c^4 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {\frac {\int \frac {1}{c^2 \tan ^2(e+f x)+2 c}d\sqrt {c-i c \tan (e+f x)}}{c}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{4 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {i c^4 \left (\frac {3 \left (\frac {7 \left (\frac {5 \left (\frac {-\frac {i \arctan \left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2}}\right )}{\sqrt {2} c^{3/2}}-\frac {1}{c \sqrt {c-i c \tan (e+f x)}}}{2 c}-\frac {1}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{4 c}+\frac {1}{2 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))}\right )}{8 c}+\frac {1}{4 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^2}\right )}{4 c}+\frac {1}{6 c (c-i c \tan (e+f x))^{3/2} (c+i c \tan (e+f x))^3}\right )}{a^3 f}\)

Input:

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

Output:

(I*c^4*(1/(6*c*(c - I*c*Tan[e + f*x])^(3/2)*(c + I*c*Tan[e + f*x])^3) + (3 
*(1/(4*c*(c - I*c*Tan[e + f*x])^(3/2)*(c + I*c*Tan[e + f*x])^2) + (7*(1/(2 
*c*(c - I*c*Tan[e + f*x])^(3/2)*(c + I*c*Tan[e + f*x])) + (5*(-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)))/(4*c)))/( 
8*c)))/(4*c)))/(a^3*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 = 0.34 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {2 i c^{4} \left (\frac {\frac {\frac {41 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {35 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+\frac {55 \sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {105 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{16 c^{5}}-\frac {1}{8 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{48 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{3}}\)	\(158\)
default	\(\frac {2 i c^{4} \left (\frac {\frac {\frac {41 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{32}-\frac {35 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{6}+\frac {55 \sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}}{8}}{\left (c +i c \tan \left (f x +e \right )\right )^{3}}+\frac {105 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{64 \sqrt {c}}}{16 c^{5}}-\frac {1}{8 c^{5} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {1}{48 c^{4} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f \,a^{3}}\)	\(158\)

Input:

int(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERBOS 
E)
 

Output:

2*I/f/a^3*c^4*(1/16/c^5*(8*(41/256*(c-I*c*tan(f*x+e))^(5/2)-35/48*c*(c-I*c 
*tan(f*x+e))^(3/2)+55/64*(c-I*c*tan(f*x+e))^(1/2)*c^2)/(c+I*c*tan(f*x+e))^ 
3+105/64*2^(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1 
/2)))-1/8/c^5/(c-I*c*tan(f*x+e))^(1/2)-1/48/c^4/(c-I*c*tan(f*x+e))^(3/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 331, normalized size of antiderivative = 1.37 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (-315 i \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {105 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) + 315 i \, \sqrt {\frac {1}{2}} a^{3} c^{2} f \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} e^{\left (6 i \, f x + 6 i \, e\right )} \log \left (-\frac {105 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{3} c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{3} c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {1}{a^{6} c^{3} f^{2}}} - i\right )} e^{\left (-i \, f x - i \, e\right )}}{128 \, a^{3} c f}\right ) + \sqrt {2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} {\left (-16 i \, e^{\left (10 i \, f x + 10 i \, e\right )} - 224 i \, e^{\left (8 i \, f x + 8 i \, e\right )} - 43 i \, e^{\left (6 i \, f x + 6 i \, e\right )} + 215 i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 58 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + 8 i\right )}\right )} e^{\left (-6 i \, f x - 6 i \, e\right )}}{1536 \, a^{3} c^{2} f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

Output:

1/1536*(-315*I*sqrt(1/2)*a^3*c^2*f*sqrt(1/(a^6*c^3*f^2))*e^(6*I*f*x + 6*I* 
e)*log(-105/128*(sqrt(2)*sqrt(1/2)*(I*a^3*c*f*e^(2*I*f*x + 2*I*e) + I*a^3* 
c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c^3*f^2)) - I)*e^(-I*f* 
x - I*e)/(a^3*c*f)) + 315*I*sqrt(1/2)*a^3*c^2*f*sqrt(1/(a^6*c^3*f^2))*e^(6 
*I*f*x + 6*I*e)*log(-105/128*(sqrt(2)*sqrt(1/2)*(-I*a^3*c*f*e^(2*I*f*x + 2 
*I*e) - I*a^3*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/(a^6*c^3*f^2)) 
 - I)*e^(-I*f*x - I*e)/(a^3*c*f)) + sqrt(2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 
1))*(-16*I*e^(10*I*f*x + 10*I*e) - 224*I*e^(8*I*f*x + 8*I*e) - 43*I*e^(6*I 
*f*x + 6*I*e) + 215*I*e^(4*I*f*x + 4*I*e) + 58*I*e^(2*I*f*x + 2*I*e) + 8*I 
))*e^(-6*I*f*x - 6*I*e)/(a^3*c^2*f)
 

Sympy [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {i \int \frac {1}{- i c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )} - 2 c \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )} + i c \sqrt {- i c \tan {\left (e + f x \right )} + c}}\, dx}{a^{3}} \] Input:

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

Output:

I*Integral(1/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4 - 2*c*sqrt( 
-I*c*tan(e + f*x) + c)*tan(e + f*x)**3 - 2*c*sqrt(-I*c*tan(e + f*x) + c)*t 
an(e + f*x) + I*c*sqrt(-I*c*tan(e + f*x) + c)), x)/a**3
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.92 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \, {\left (\frac {4 \, {\left (315 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{4} - 1680 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{3} c + 2772 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} c^{2} - 1152 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} c^{3} - 256 \, c^{4}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} - 6 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c + 12 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{2} - 8 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a^{3} c^{3}} + \frac {315 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a^{3} \sqrt {c}}\right )}}{3072 \, c f} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

Output:

-1/3072*I*(4*(315*(-I*c*tan(f*x + e) + c)^4 - 1680*(-I*c*tan(f*x + e) + c) 
^3*c + 2772*(-I*c*tan(f*x + e) + c)^2*c^2 - 1152*(-I*c*tan(f*x + e) + c)*c 
^3 - 256*c^4)/((-I*c*tan(f*x + e) + c)^(9/2)*a^3 - 6*(-I*c*tan(f*x + e) + 
c)^(7/2)*a^3*c + 12*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c^2 - 8*(-I*c*tan(f* 
x + e) + c)^(3/2)*a^3*c^3) + 315*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^3* 
sqrt(c)))/(c*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^3/(c-I*c*tan(f*x+e))^(3/2),x, algorithm="gi 
ac")
 

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 = 2.20 (sec) , antiderivative size = 229, normalized size of antiderivative = 0.95 \[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^3\,35{}\mathrm {i}}{16\,a^3\,f}+\frac {c^3\,1{}\mathrm {i}}{3\,a^3\,f}-\frac {{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^4\,105{}\mathrm {i}}{256\,a^3\,c\,f}-\frac {c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,231{}\mathrm {i}}{64\,a^3\,f}+\frac {c^2\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,3{}\mathrm {i}}{2\,a^3\,f}}{6\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{9/2}+8\,c^3\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-12\,c^2\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,105{}\mathrm {i}}{512\,a^3\,{\left (-c\right )}^{3/2}\,f} \] Input:

int(1/((a + a*tan(e + f*x)*1i)^3*(c - c*tan(e + f*x)*1i)^(3/2)),x)
 

Output:

(2^(1/2)*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*105i 
)/(512*a^3*(-c)^(3/2)*f) - (((c - c*tan(e + f*x)*1i)^3*35i)/(16*a^3*f) + ( 
c^3*1i)/(3*a^3*f) - ((c - c*tan(e + f*x)*1i)^4*105i)/(256*a^3*c*f) - (c*(c 
 - c*tan(e + f*x)*1i)^2*231i)/(64*a^3*f) + (c^2*(c - c*tan(e + f*x)*1i)*3i 
)/(2*a^3*f))/(6*c*(c - c*tan(e + f*x)*1i)^(7/2) - (c - c*tan(e + f*x)*1i)^ 
(9/2) + 8*c^3*(c - c*tan(e + f*x)*1i)^(3/2) - 12*c^2*(c - c*tan(e + f*x)*1 
i)^(5/2))
 

Reduce [F]

\[ \int \frac {1}{(a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {\int \frac {1}{\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{4}-2 \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3} i -2 \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {-\tan \left (f x +e \right ) i +1}}d x}{\sqrt {c}\, a^{3} c} \] Input:

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

Output:

( - int(1/(sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4 - 2*sqrt( - tan(e + 
 f*x)*i + 1)*tan(e + f*x)**3*i - 2*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x 
)*i - sqrt( - tan(e + f*x)*i + 1)),x))/(sqrt(c)*a**3*c)