\(\int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx\) [1161]

Optimal result

Integrand size = 32, antiderivative size = 209 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\frac {2 \sqrt [4]{-1} a^{5/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{d^{3/2} f}-\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{3/2} f}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{(c-i d) d f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1510\) vs. \(2(209)=418\).

Time = 6.76 (sec) , antiderivative size = 1510, normalized size of antiderivative = 7.22 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(-2*d*(a + I*a*Tan[e + f*x])^(5/2))/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x] 
]) + (2*((((-3*I)/4)*(I*a*c - a*d)^3*Sqrt[(I*a)/(-((a^2*c)/(I*a*c - a*d)) 
- (I*a^2*d)/(I*a*c - a*d))]*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - 
 a*d))^3*Sqrt[(I*a*(c + d*Tan[e + f*x]))/(I*a*c - a*d)]*Sqrt[1 + (I*a*d*(a 
 + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d) 
/(I*a*c - a*d)))]*(((2*I)*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a 
^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))) + (4*a^2*d^2*(a + I*a*Tan 
[e + f*x])^2)/(3*(I*a*c - a*d)^2*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I* 
a*c - a*d))^2) - (2*(-1)^(1/4)*Sqrt[a]*Sqrt[d]*ArcSinh[((-1)^(1/4)*Sqrt[a] 
*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d]*Sqrt[-((a^2*c)/(I* 
a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt 
[I*a*c - a*d]*Sqrt[-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)]*Sqr 
t[1 + (I*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a* 
d)) - (I*a^2*d)/(I*a*c - a*d)))])))/(a^2*d^2*f*Sqrt[a + I*a*Tan[e + f*x]]* 
Sqrt[c + d*Tan[e + f*x]]) + (a*((I/2)*a^2*(c + (5*I)*d) + 2*a^2*d)*(2*a*(( 
-2*Sqrt[2]*ArcTan[(Sqrt[-(a*c) + I*a*d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[ 
2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a*c) + I*a*d] - (2*(-1)^(3/4)*Sqrt 
[c + I*d]*Sqrt[(c/(c + I*d) + (I*d)/(c + I*d))^(-1)]*Sqrt[c/(c + I*d) + (I 
*d)/(c + I*d)]*ArcSin[((-1)^(1/4)*Sqrt[d]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqr 
t[a]*Sqrt[c + I*d]*Sqrt[c/(c + I*d) + (I*d)/(c + I*d)])]*Sqrt[(c + d*Ta...
 

Rubi [A] (verified)

Time = 1.24 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {3042, 4036, 27, 3042, 4084, 3042, 4027, 221, 4082, 66, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4036

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

\(\displaystyle \frac {4 i a^2 d \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx+a (c-i d) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{d (d+i c)}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {4 i a^2 d \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx+a (c-i d) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{d (d+i c)}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {\frac {8 a^4 d \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f}+a (c-i d) \int \frac {(a-i a \tan (e+f x)) \sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{d (d+i c)}+\frac {2 a^2 (c+i d) \sqrt {a+i a \tan (e+f x)}}{d f (c-i d) \sqrt {c+d \tan (e+f x)}}\)

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 66

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 
2   Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre 
eQ[{a, b, c, d}, x] &&  !GtQ[c - a*(d/b), 0]
 

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 4036

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x] 
)^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] + Si 
mp[a/(d*(b*c + a*d)*(n + 1))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[ 
e + f*x])^(n + 1)*Simp[b*(b*c*(m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) 
 + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4082

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

rule 4084

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1814 vs. \(2 (165 ) = 330\).

Time = 0.63 (sec) , antiderivative size = 1815, normalized size of antiderivative = 8.68

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1815\)
default	\(\text {Expression too large to display}\)	\(1815\)

Input:

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

Output:

1/2/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(2*I*2^(1/2)*ln(1/2*(2*I*a*d*tan( 
f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a 
*d)/(I*a*d)^(1/2))*a*c*d*(-a*(I*d-c))^(1/2)+2*I*2^(1/2)*c^3*(I*a*d)^(1/2)* 
(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)+2*I*2^(1/2) 
*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^ 
(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e 
)+5*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I* 
tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^3*d*(-a*(I*d-c))^ 
(1/2)+2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+ 
d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a* 
c^3*d*tan(f*x+e)-7*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I 
*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a 
*d)^(1/2))*a*c*d^3*tan(f*x+e)-3*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c 
+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^ 
(1/2))*a*d^4*(-a*(I*d-c))^(1/2)*tan(f*x+e)-6*I*(I*a*d)^(1/2)*2^(1/2)*(-a*( 
I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c*d^2-2*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)))*a*c*d*(I*a*d)^ 
(1/2)+2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I* 
tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^4*(-a*(I*d-c))...
 

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. 940 vs. \(2 (157) = 314\).

Time = 0.12 (sec) , antiderivative size = 940, normalized size of antiderivative = 4.50 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(4*sqrt(2)*((a^2*c + I*a^2*d)*e^(3*I*f*x + 3*I*e) + (a^2*c + I*a^2*d)* 
e^(I*f*x + 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)) + ((c^2*d - 2*I*c*d^2 - 
 d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(4*I*a^5/(d^3*f^2))*log 
((I*d^2*f*sqrt(4*I*a^5/(d^3*f^2))*e^(I*f*x + I*e) + sqrt(2)*(a^2*e^(2*I*f* 
x + 2*I*e) + a^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^(-I*f*x - I*e)/a^2) 
 - ((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqr 
t(4*I*a^5/(d^3*f^2))*log((-I*d^2*f*sqrt(4*I*a^5/(d^3*f^2))*e^(I*f*x + I*e) 
 + sqrt(2)*(a^2*e^(2*I*f*x + 2*I*e) + a^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^(-I*f*x - I*e)/a^2) + ((c^2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I* 
e) + (c^2*d + d^3)*f)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)* 
f^2))*log(1/4*((I*c^2 + 2*c*d - I*d^2)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 
3*I*c*d^2 + d^3)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2*I 
*e) + a^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^(-I*f*x - I*e)/a^2) - ((c^ 
2*d - 2*I*c*d^2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (c^2*d + d^3)*f)*sqrt(32*I* 
a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*log(1/4*((-I*c^2 - 2*c*d + 
 I*d^2)*sqrt(32*I*a^5/((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f^2))*f*e^(...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume((d^2-2*c*d-c^2)>0)', see `assume 
?` for mor
 

Giac [F(-2)]

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

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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