\(\int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [1148]

Optimal result

Integrand size = 32, antiderivative size = 225 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {i (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{4 \sqrt {2} a^{5/2} f}+\frac {(i c+d) \sqrt {c+d \tan (e+f x)}}{4 a^2 f \sqrt {a+i a \tan (e+f x)}}+\frac {i (c+d \tan (e+f x))^{3/2}}{6 a f (a+i a \tan (e+f x))^{3/2}}-\frac {(c+d \tan (e+f x))^{5/2}}{5 (i c-d) f (a+i a \tan (e+f x))^{5/2}} \] Output:

-1/8*I*(c-I*d)^(3/2)*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)/f+1/4*(I*c+d)*(c+d*tan(f 
*x+e))^(1/2)/a^2/f/(a+I*a*tan(f*x+e))^(1/2)+1/6*I*(c+d*tan(f*x+e))^(3/2)/a 
/f/(a+I*a*tan(f*x+e))^(3/2)-1/5*(c+d*tan(f*x+e))^(5/2)/(I*c-d)/f/(a+I*a*ta 
n(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 3.38 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.25 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {24 i (c+i d) (c+d \tan (e+f x))^{5/2}-\frac {5 i (i c-d) (-i+\tan (e+f x)) \left (4 a d (-i+\tan (e+f x)) (c+d \tan (e+f x))^{3/2}-4 a (c+d \tan (e+f x))^{5/2}-3 i \sqrt {-a (c-i d)} (c+i d) (-i+\tan (e+f x)) \left (\sqrt {2} (c-i d) \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right ) \sqrt {a+i a \tan (e+f x)}-2 \sqrt {-a (c-i d)} \sqrt {c+d \tan (e+f x)}\right )\right )}{a}}{120 (c+i d)^2 f (a+i a \tan (e+f x))^{5/2}} \] Input:

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

Output:

((24*I)*(c + I*d)*(c + d*Tan[e + f*x])^(5/2) - ((5*I)*(I*c - d)*(-I + Tan[ 
e + f*x])*(4*a*d*(-I + Tan[e + f*x])*(c + d*Tan[e + f*x])^(3/2) - 4*a*(c + 
 d*Tan[e + f*x])^(5/2) - (3*I)*Sqrt[-(a*(c - I*d))]*(c + I*d)*(-I + Tan[e 
+ f*x])*(Sqrt[2]*(c - I*d)*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 + I*a*Tan[e + f*x]] 
 - 2*Sqrt[-(a*(c - I*d))]*Sqrt[c + d*Tan[e + f*x]])))/a)/(120*(c + I*d)^2* 
f*(a + I*a*Tan[e + f*x])^(5/2))
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3042, 4030, 3042, 4029, 3042, 4029, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4029

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4029

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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 4029

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/(2*b*f*m)), x] - Simp[(a*c - b*d)/(2*b^2)   Int[(a + b*Tan[e + f 
*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, 
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Eq 
Q[m + n, 0] && LeQ[m, -2^(-1)]
 

rule 4030

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)   Int[(a + b*Tan[e 
 + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f} 
, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[ 
m + n + 1, 0] && LtQ[m, -1]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1656 vs. \(2 (179 ) = 358\).

Time = 0.53 (sec) , antiderivative size = 1657, normalized size of antiderivative = 7.36

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1657\)
default	\(\text {Expression too large to display}\)	\(1657\)

Input:

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

Output:

-1/240/f/a^3*(-15*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1 
/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan 
(f*x+e)))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2-15*ln((3*a*c+I*a*tan(f*x+e)*c-I*a 
*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I* 
tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^2+60*I*ln 
((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/ 
2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*2^(1/2)*(- 
a*(I*d-c))^(1/2)*d^2*tan(f*x+e)^3+308*I*c^2*(a*(c+d*tan(f*x+e))*(1+I*tan(f 
*x+e)))^(1/2)*tan(f*x+e)-52*I*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)* 
d^2*tan(f*x+e)^3-60*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2* 
2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I 
+tan(f*x+e)))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2*tan(f*x+e)-60*I*ln((3*a*c+I*a 
*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d* 
tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*2^(1/2)*(-a*(I*d-c))^ 
(1/2)*d^2*tan(f*x+e)-212*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^2*t 
an(f*x+e)^2-40*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d*tan(f*x+e)^ 
3+56*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d*tan(f*x+e)-15*ln((3*a 
*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a 
*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*2^(1/2)*(-a*(I* 
d-c))^(1/2)*c^2*tan(f*x+e)^4-15*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*...
 

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. 571 vs. \(2 (171) = 342\).

Time = 0.12 (sec) , antiderivative size = 571, normalized size of antiderivative = 2.54 \[ \int \frac {(c+d \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{3} c + a^{3} d\right )} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} + \sqrt {2} {\left ({\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c + d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{i \, c + d}\right ) + 15 \, \sqrt {\frac {1}{2}} {\left (i \, a^{3} c - a^{3} d\right )} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (5 i \, f x + 5 i \, e\right )} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{3} f \sqrt {-\frac {c^{3} - 3 i \, c^{2} d - 3 \, c d^{2} + i \, d^{3}}{a^{5} f^{2}}} e^{\left (i \, f x + i \, e\right )} - \sqrt {2} {\left ({\left (i \, c + d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c + d\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{i \, c + d}\right ) - \sqrt {2} {\left (3 \, c^{2} + 6 i \, c d - 3 \, d^{2} + {\left (23 \, c^{2} - 6 i \, c d + 17 \, d^{2}\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, {\left (17 \, c^{2} + 2 i \, c d + 9 \, d^{2}\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (7 \, c^{2} + 8 i \, c d - d^{2}\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {{\left (c - i \, d\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + c + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-5 i \, f x - 5 i \, e\right )}}{120 \, {\left (i \, a^{3} c - a^{3} d\right )} f} \] Input:

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

Output:

1/120*(15*sqrt(1/2)*(-I*a^3*c + a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^2 
+ I*d^3)/(a^5*f^2))*e^(5*I*f*x + 5*I*e)*log((2*sqrt(1/2)*a^3*f*sqrt(-(c^3 
- 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(I*f*x + I*e) + sqrt(2)*((I*c 
+ d)*e^(2*I*f*x + 2*I*e) + I*c + d)*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)))/(I* 
c + d)) + 15*sqrt(1/2)*(I*a^3*c - a^3*d)*f*sqrt(-(c^3 - 3*I*c^2*d - 3*c*d^ 
2 + I*d^3)/(a^5*f^2))*e^(5*I*f*x + 5*I*e)*log(-(2*sqrt(1/2)*a^3*f*sqrt(-(c 
^3 - 3*I*c^2*d - 3*c*d^2 + I*d^3)/(a^5*f^2))*e^(I*f*x + I*e) - sqrt(2)*((I 
*c + d)*e^(2*I*f*x + 2*I*e) + I*c + d)*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)))/ 
(I*c + d)) - sqrt(2)*(3*c^2 + 6*I*c*d - 3*d^2 + (23*c^2 - 6*I*c*d + 17*d^2 
)*e^(6*I*f*x + 6*I*e) + 2*(17*c^2 + 2*I*c*d + 9*d^2)*e^(4*I*f*x + 4*I*e) + 
 2*(7*c^2 + 8*I*c*d - d^2)*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)))*e^(-5*I*f*x - 5*I*e)/((I*a^3*c - a^3*d)*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [F(-2)]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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