\(\int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx\) [1166]

Optimal result

Integrand size = 32, antiderivative size = 349 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} (c-i d)^{3/2} f}-\frac {1}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}+\frac {5 i c-17 d}{30 a (c+i d)^2 f (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {15 c^2+70 i c d-151 d^2}{60 a^2 (i c-d)^3 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {d \left (15 c^3+65 i c^2 d-117 c d^2+317 i d^3\right ) \sqrt {a+i a \tan (e+f x)}}{60 a^3 (c-i d) (c+i d)^4 f \sqrt {c+d \tan (e+f x)}} \] Output:

-1/8*I*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2)/(a+I*a 
*tan(f*x+e))^(1/2))*2^(1/2)/a^(5/2)/(c-I*d)^(3/2)/f-1/5/(I*c-d)/f/(a+I*a*t 
an(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(1/2)+1/30*(5*I*c-17*d)/a/(c+I*d)^2/f/(a 
+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2)+1/60*(15*c^2+70*I*c*d-151*d^ 
2)/a^2/(I*c-d)^3/f/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(1/2)+1/60*d* 
(15*c^3+65*I*c^2*d-117*c*d^2+317*I*d^3)*(a+I*a*tan(f*x+e))^(1/2)/a^3/(c-I* 
d)/(c+I*d)^4/f/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 3.70 (sec) , antiderivative size = 345, normalized size of antiderivative = 0.99 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {\frac {24 a^6 (-i c+d)}{(a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}+\frac {4 a^5 (-5 i c+17 d)}{(a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}+\frac {2 a^4 \left (-15 i c^2+70 c d+151 i d^2\right )}{(c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {i a^3 \left (\frac {15 \sqrt {2} a (c+i d)^4 \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {-a (c-i d)}}-\frac {2 d \left (-15 i c^3+65 c^2 d+117 i c d^2+317 d^3\right ) \sqrt {a+i a \tan (e+f x)}}{\sqrt {c+d \tan (e+f x)}}\right )}{(c+i d) \left (c^2+d^2\right )}}{120 a^6 (c+i d)^2 f} \] Input:

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

Output:

-1/120*((24*a^6*((-I)*c + d))/((a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan 
[e + f*x]]) + (4*a^5*((-5*I)*c + 17*d))/((a + I*a*Tan[e + f*x])^(3/2)*Sqrt 
[c + d*Tan[e + f*x]]) + (2*a^4*((-15*I)*c^2 + 70*c*d + (151*I)*d^2))/((c + 
 I*d)*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) + (I*a^3*((15*S 
qrt[2]*a*(c + I*d)^4*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I*a*Tan[e + f*x 
]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a*(c - I*d))] - (2*d*((- 
15*I)*c^3 + 65*c^2*d + (117*I)*c*d^2 + 317*d^3)*Sqrt[a + I*a*Tan[e + f*x]] 
)/Sqrt[c + d*Tan[e + f*x]]))/((c + I*d)*(c^2 + d^2)))/(a^6*(c + I*d)^2*f)
 

Rubi [A] (verified)

Time = 1.97 (sec) , antiderivative size = 404, normalized size of antiderivative = 1.16, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.469, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4081, 27, 3042, 4027, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

\(\displaystyle -\frac {\int -\frac {a (5 i c-11 d)+6 i a d \tan (e+f x)}{2 (i \tan (e+f x) a+a)^{3/2} (c+d \tan (e+f x))^{3/2}}dx}{5 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (5 i c-11 d)+6 i a d \tan (e+f x)}{(i \tan (e+f x) a+a)^{3/2} (c+d \tan (e+f x))^{3/2}}dx}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (5 i c-11 d)+6 i a d \tan (e+f x)}{(i \tan (e+f x) a+a)^{3/2} (c+d \tan (e+f x))^{3/2}}dx}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {-\frac {\int \frac {\left (15 c^2+50 i d c-83 d^2\right ) a^2+4 (5 c+17 i d) d \tan (e+f x) a^2}{2 \sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{3/2}}dx}{3 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\int \frac {\left (15 c^2+50 i d c-83 d^2\right ) a^2+4 (5 c+17 i d) d \tan (e+f x) a^2}{\sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{3/2}}dx}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\int \frac {\left (15 c^2+50 i d c-83 d^2\right ) a^2+4 (5 c+17 i d) d \tan (e+f x) a^2}{\sqrt {i \tan (e+f x) a+a} (c+d \tan (e+f x))^{3/2}}dx}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {-\frac {\frac {a^2 \left (15 i c^2-70 c d-151 i d^2\right )}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}-\frac {\int -\frac {\sqrt {i \tan (e+f x) a+a} \left (\left (15 i c^3-75 d c^2-185 i d^2 c+317 d^3\right ) a^3+2 d \left (15 i c^2-70 d c-151 i d^2\right ) \tan (e+f x) a^3\right )}{2 (c+d \tan (e+f x))^{3/2}}dx}{a^2 (-d+i c)}}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (15 i c^3-75 d c^2-185 i d^2 c+317 d^3\right ) a^3+2 d \left (15 i c^2-70 d c-151 i d^2\right ) \tan (e+f x) a^3\right )}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}+\frac {a^2 \left (15 i c^2-70 c d-151 i d^2\right )}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\int \frac {\sqrt {i \tan (e+f x) a+a} \left (\left (15 i c^3-75 d c^2-185 i d^2 c+317 d^3\right ) a^3+2 d \left (15 i c^2-70 d c-151 i d^2\right ) \tan (e+f x) a^3\right )}{(c+d \tan (e+f x))^{3/2}}dx}{2 a^2 (-d+i c)}+\frac {a^2 \left (15 i c^2-70 c d-151 i d^2\right )}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4081

\(\displaystyle \frac {-\frac {\frac {\frac {2 \int \frac {15 i a^4 (c+i d)^4 \sqrt {i \tan (e+f x) a+a}}{2 \sqrt {c+d \tan (e+f x)}}dx}{a \left (c^2+d^2\right )}+\frac {2 a^3 d \left (15 i c^3-65 c^2 d-117 i c d^2-317 d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}+\frac {a^2 \left (15 i c^2-70 c d-151 i d^2\right )}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {\frac {\frac {15 i a^3 (c+i d)^4 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^3 d \left (15 i c^3-65 c^2 d-117 i c d^2-317 d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}+\frac {a^2 \left (15 i c^2-70 c d-151 i d^2\right )}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {-\frac {\frac {\frac {15 i a^3 (c+i d)^4 \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c^2+d^2}+\frac {2 a^3 d \left (15 i c^3-65 c^2 d-117 i c d^2-317 d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}+\frac {a^2 \left (15 i c^2-70 c d-151 i d^2\right )}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {-\frac {\frac {\frac {30 a^5 (c+i d)^4 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f \left (c^2+d^2\right )}+\frac {2 a^3 d \left (15 i c^3-65 c^2 d-117 i c d^2-317 d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}+\frac {a^2 \left (15 i c^2-70 c d-151 i d^2\right )}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {-\frac {\frac {a^2 \left (15 i c^2-70 c d-151 i d^2\right )}{f (c+i d) \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {\frac {15 \sqrt {2} a^{7/2} (c+i d)^4 \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f \sqrt {c-i d} \left (c^2+d^2\right )}+\frac {2 a^3 d \left (15 i c^3-65 c^2 d-117 i c d^2-317 d^3\right ) \sqrt {a+i a \tan (e+f x)}}{f \left (c^2+d^2\right ) \sqrt {c+d \tan (e+f x)}}}{2 a^2 (-d+i c)}}{6 a^2 (-d+i c)}-\frac {a (5 c+17 i d)}{3 f (c+i d) (a+i a \tan (e+f x))^{3/2} \sqrt {c+d \tan (e+f x)}}}{10 a^2 (-d+i c)}-\frac {1}{5 f (-d+i c) (a+i a \tan (e+f x))^{5/2} \sqrt {c+d \tan (e+f x)}}\)

Input:

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

Output:

-1/5*1/((I*c - d)*f*(a + I*a*Tan[e + f*x])^(5/2)*Sqrt[c + d*Tan[e + f*x]]) 
 + (-1/3*(a*(5*c + (17*I)*d))/((c + I*d)*f*(a + I*a*Tan[e + f*x])^(3/2)*Sq 
rt[c + d*Tan[e + f*x]]) - ((a^2*((15*I)*c^2 - 70*c*d - (151*I)*d^2))/((c + 
 I*d)*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]]) + ((15*Sqrt[2 
]*a^(7/2)*(c + I*d)^4*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])/( 
Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/(Sqrt[c - I*d]*(c^2 + d^2)*f) 
+ (2*a^3*d*((15*I)*c^3 - 65*c^2*d - (117*I)*c*d^2 - 317*d^3)*Sqrt[a + I*a* 
Tan[e + f*x]])/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]))/(2*a^2*(I*c - d)) 
)/(6*a^2*(I*c - d)))/(10*a^2*(I*c - d))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 221

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

rule 3042

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

rule 4027

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f)   Subst[Int[1/(a*c - b*d - 2* 
a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N 
eQ[c^2 + d^2, 0]
 

rule 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   In 
t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m 
 + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, 
e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] 
 && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( 
b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m 
 + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m 
- b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free 
Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] 
 && LtQ[m, 0] &&  !GtQ[n, 0]
 

rule 4081

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 
1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2))   Int[(a + b*Tan[e + 
f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* 
m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr 
eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 
0] && LtQ[n, -1]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 7869 vs. \(2 (290 ) = 580\).

Time = 0.68 (sec) , antiderivative size = 7870, normalized size of antiderivative = 22.55

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(7870\)
default	\(\text {Expression too large to display}\)	\(7870\)

Input:

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

Output:

result too large to display
 

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. 1074 vs. \(2 (273) = 546\).

Time = 0.23 (sec) , antiderivative size = 1074, normalized size of antiderivative = 3.08 \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/120*(sqrt(2)*(-3*I*c^4 + 6*c^3*d + 6*c*d^3 + 3*I*d^4 + (-23*I*c^4 + 68* 
c^3*d + 18*I*c^2*d^2 + 332*c*d^3 - 463*I*d^4)*e^(8*I*f*x + 8*I*e) + (-57*I 
*c^4 + 200*c^3*d + 178*I*c^2*d^2 + 464*c*d^3 - 269*I*d^4)*e^(6*I*f*x + 6*I 
*e) - 4*(12*I*c^4 - 43*c^3*d - 43*I*c^2*d^2 - 43*c*d^3 - 55*I*d^4)*e^(4*I* 
f*x + 4*I*e) + (-17*I*c^4 + 46*c^3*d + 12*I*c^2*d^2 + 46*c*d^3 + 29*I*d^4) 
*e^(2*I*f*x + 2*I*e))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2 
*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)) + 30*((a^3*c^6 + 2 
*I*a^3*c^5*d + a^3*c^4*d^2 + 4*I*a^3*c^3*d^3 - a^3*c^2*d^4 + 2*I*a^3*c*d^5 
 - a^3*d^6)*f*e^(7*I*f*x + 7*I*e) + (a^3*c^6 + 4*I*a^3*c^5*d - 5*a^3*c^4*d 
^2 - 5*a^3*c^2*d^4 - 4*I*a^3*c*d^5 + a^3*d^6)*f*e^(5*I*f*x + 5*I*e))*sqrt( 
-1/8*I/((I*a^5*c^3 + 3*a^5*c^2*d - 3*I*a^5*c*d^2 - a^5*d^3)*f^2))*log(-4*( 
I*a^3*c^2 + 2*a^3*c*d - I*a^3*d^2)*f*sqrt(-1/8*I/((I*a^5*c^3 + 3*a^5*c^2*d 
 - 3*I*a^5*c*d^2 - a^5*d^3)*f^2))*e^(I*f*x + I*e) + sqrt(2)*sqrt(((c - I*d 
)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I 
*f*x + 2*I*e) + 1))*(e^(2*I*f*x + 2*I*e) + 1)) - 30*((a^3*c^6 + 2*I*a^3*c^ 
5*d + a^3*c^4*d^2 + 4*I*a^3*c^3*d^3 - a^3*c^2*d^4 + 2*I*a^3*c*d^5 - a^3*d^ 
6)*f*e^(7*I*f*x + 7*I*e) + (a^3*c^6 + 4*I*a^3*c^5*d - 5*a^3*c^4*d^2 - 5*a^ 
3*c^2*d^4 - 4*I*a^3*c*d^5 + a^3*d^6)*f*e^(5*I*f*x + 5*I*e))*sqrt(-1/8*I/(( 
I*a^5*c^3 + 3*a^5*c^2*d - 3*I*a^5*c*d^2 - a^5*d^3)*f^2))*log(-4*(-I*a^3*c^ 
2 - 2*a^3*c*d + I*a^3*d^2)*f*sqrt(-1/8*I/((I*a^5*c^3 + 3*a^5*c^2*d - 3*...
 

Sympy [F]

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

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

Output:

Integral(1/((I*a*(tan(e + f*x) - I))**(5/2)*(c + d*tan(e + f*x))**(3/2)), 
x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(a+i a \tan (e+f x))^{5/2} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(3/2),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 TypeError: Bad Argument TypeError: Bad Argument TypeRecursive ass 
umption s
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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