\(\int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1133]

Optimal result

Integrand size = 28, antiderivative size = 109 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {2 i a \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}-\frac {2 a}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {2 i a}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}} \] Output:

-2*I*a*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))/(c-I*d)^(5/2)/f-2/3*a 
/(I*c+d)/f/(c+d*tan(f*x+e))^(3/2)+2*I*a/(c-I*d)^2/f/(c+d*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 = 0.36 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.56 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {2 i a \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {c+d \tan (e+f x)}{c-i d}\right )}{3 (c-i d) f (c+d \tan (e+f x))^{3/2}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.70 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.33, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3042, 4012, 3042, 4012, 3042, 4020, 25, 27, 73, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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 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 4012

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

rule 4020

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2596 vs. \(2 (91 ) = 182\).

Time = 0.41 (sec) , antiderivative size = 2597, normalized size of antiderivative = 23.83

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2597\)
default	\(\text {Expression too large to display}\)	\(2597\)
parts	\(\text {Expression too large to display}\)	\(4468\)

Input:

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

Output:

1/2/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^ 
(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*d^3-1/ 
2/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d 
*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*d^3-1/f* 
a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e)) 
^(1/2)+(2*(c^2+d^2)^(1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^3+1 
/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1 
/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e))^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*d^ 
3-4/f*a/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c*d+2/3*I/f*a/(c^2+d^2)/(c+d*ta 
n(f*x+e))^(3/2)*c+2*I/f*a/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c^2-2*I/f*a/( 
c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*d^2-3*I/f*a/(c^2+d^2)^(5/2)/(2*(c^2+d^2) 
^(1/2)-2*c)^(1/2)*arctan(((2*(c^2+d^2)^(1/2)+2*c)^(1/2)-2*(c+d*tan(f*x+e)) 
^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c*d^2-3/2*I/f*a/(c^2+d^2)^(5/2)/(2* 
(c^2+d^2)^(1/2)+2*c)^(1/2)*ln((c+d*tan(f*x+e))^(1/2)*(2*(c^2+d^2)^(1/2)+2* 
c)^(1/2)-d*tan(f*x+e)-c-(c^2+d^2)^(1/2))*c*d^2+3*I/f*a/(c^2+d^2)^(5/2)/(2* 
(c^2+d^2)^(1/2)-2*c)^(1/2)*arctan((2*(c+d*tan(f*x+e))^(1/2)+(2*(c^2+d^2)^( 
1/2)+2*c)^(1/2))/(2*(c^2+d^2)^(1/2)-2*c)^(1/2))*c*d^2+3/2*I/f*a/(c^2+d^2)^ 
(5/2)/(2*(c^2+d^2)^(1/2)+2*c)^(1/2)*ln(d*tan(f*x+e)+c+(c+d*tan(f*x+e))^(1/ 
2)*(2*(c^2+d^2)^(1/2)+2*c)^(1/2)+(c^2+d^2)^(1/2))*c*d^2-2/3/f*a/(c^2+d^2)/ 
(c+d*tan(f*x+e))^(3/2)*d-1/2*I/f*a/(c^2+d^2)^2/(2*(c^2+d^2)^(1/2)+2*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. 823 vs. \(2 (85) = 170\).

Time = 0.19 (sec) , antiderivative size = 823, normalized size of antiderivative = 7.55 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/12*(3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4* 
I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^4 
+ 2*c^2*d^2 + d^4)*f)*sqrt(-4*I*a^2/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10* 
c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log((2*a*c + ((I*c^3 + 3*c^2*d - 3*I*c*d^ 
2 - d^3)*f*e^(2*I*f*x + 2*I*e) + (I*c^3 + 3*c^2*d - 3*I*c*d^2 - d^3)*f)*sq 
rt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sq 
rt(-4*I*a^2/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^ 
5)*f^2)) + 2*(a*c - I*a*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a) - 
3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*I*e) + 
 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) + (c^4 + 2*c^ 
2*d^2 + d^4)*f)*sqrt(-4*I*a^2/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^ 
3 + 5*I*c*d^4 + d^5)*f^2))*log((2*a*c + ((-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d 
^3)*f*e^(2*I*f*x + 2*I*e) + (-I*c^3 - 3*c^2*d + 3*I*c*d^2 + d^3)*f)*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(- 
4*I*a^2/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f 
^2)) + 2*(a*c - I*a*d)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a) - 16*( 
-2*I*a*c + a*d + 2*(-I*a*c - a*d)*e^(4*I*f*x + 4*I*e) + (-4*I*a*c - a*d)*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)))/((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^ 
(4*I*f*x + 4*I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x ...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

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(((-(2*c*d^4)/((c^2-d^2)^2>0)', s 
ee `assume
 

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 222 vs. \(2 (85) = 170\).

Time = 0.67 (sec) , antiderivative size = 222, normalized size of antiderivative = 2.04 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {2 i \, a {\left (\frac {3 \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{{\left (c^{2} - 2 i \, c d - d^{2}\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {3 \, d \tan \left (f x + e\right ) + 4 \, c - i \, d}{{\left (c^{2} - 2 i \, c d - d^{2}\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\right )}}{3 \, f} \] Input:

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

Output:

2/3*I*a*(3*sqrt(2)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)* 
sqrt(d*tan(f*x + e) + c))/(sqrt(2)*c*sqrt(-c + sqrt(c^2 + d^2)) - I*sqrt(2 
)*sqrt(-c + sqrt(c^2 + d^2))*d - sqrt(2)*sqrt(c^2 + d^2)*sqrt(-c + sqrt(c^ 
2 + d^2))))/((c^2 - 2*I*c*d - d^2)*sqrt(-c + sqrt(c^2 + d^2))*(-I*d/(c - s 
qrt(c^2 + d^2)) + 1)) + (3*d*tan(f*x + e) + 4*c - I*d)/((c^2 - 2*I*c*d - d 
^2)*(d*tan(f*x + e) + c)^(3/2)))/f
 

Mupad [B] (verification not implemented)

Time = 21.71 (sec) , antiderivative size = 7068, normalized size of antiderivative = 64.84 \[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

((a*c*2i)/(3*f*(c^2 + d^2)) + (a*(c^2 - d^2)*(c + d*tan(e + f*x))*2i)/(f*( 
c^2 + d^2)^2))/(c + d*tan(e + f*x))^(3/2) + (log(((((320*a^4*c^2*d^8*f^4 - 
 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c 
^8*d^2*f^4)^(1/2) - 4*a^2*c^5*f^2 - 20*a^2*c*d^4*f^2 + 40*a^2*c^3*d^2*f^2) 
/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 
5*c^8*d^2*f^4))^(1/2)*((c + d*tan(e + f*x))^(1/2)*(320*a^2*c^4*d^14*f^3 - 
16*a^2*d^18*f^3 + 1024*a^2*c^6*d^12*f^3 + 1440*a^2*c^8*d^10*f^3 + 1024*a^2 
*c^10*d^8*f^3 + 320*a^2*c^12*d^6*f^3 - 16*a^2*c^16*d^2*f^3) - ((((320*a^4* 
c^2*d^8*f^4 - 16*a^4*d^10*f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^ 
4 - 400*a^4*c^8*d^2*f^4)^(1/2) - 4*a^2*c^5*f^2 - 20*a^2*c*d^4*f^2 + 40*a^2 
*c^3*d^2*f^2)/(c^10*f^4 + d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c 
^6*d^4*f^4 + 5*c^8*d^2*f^4))^(1/2)*(((((320*a^4*c^2*d^8*f^4 - 16*a^4*d^10* 
f^4 - 1760*a^4*c^4*d^6*f^4 + 1600*a^4*c^6*d^4*f^4 - 400*a^4*c^8*d^2*f^4)^( 
1/2) - 4*a^2*c^5*f^2 - 20*a^2*c*d^4*f^2 + 40*a^2*c^3*d^2*f^2)/(c^10*f^4 + 
d^10*f^4 + 5*c^2*d^8*f^4 + 10*c^4*d^6*f^4 + 10*c^6*d^4*f^4 + 5*c^8*d^2*f^4 
))^(1/2)*(c + d*tan(e + f*x))^(1/2)*(64*c*d^22*f^5 + 640*c^3*d^20*f^5 + 28 
80*c^5*d^18*f^5 + 7680*c^7*d^16*f^5 + 13440*c^9*d^14*f^5 + 16128*c^11*d^12 
*f^5 + 13440*c^13*d^10*f^5 + 7680*c^15*d^8*f^5 + 2880*c^17*d^6*f^5 + 640*c 
^19*d^4*f^5 + 64*c^21*d^2*f^5))/4 - 32*a*d^21*f^4 - 160*a*c^2*d^19*f^4 - 1 
28*a*c^4*d^17*f^4 + 896*a*c^6*d^15*f^4 + 3136*a*c^8*d^13*f^4 + 4928*a*c...
 

Reduce [F]

\[ \int \frac {a+i a \tan (e+f x)}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {a \left (-2 \sqrt {d \tan \left (f x +e \right )+c}-3 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \tan \left (f x +e \right )^{2} d^{3} f -6 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \tan \left (f x +e \right ) c \,d^{2} f -3 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) c^{2} d f +3 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \tan \left (f x +e \right )^{2} d^{3} f i +6 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) \tan \left (f x +e \right ) c \,d^{2} f i +3 \left (\int \frac {\sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{3} d^{3}+3 \tan \left (f x +e \right )^{2} c \,d^{2}+3 \tan \left (f x +e \right ) c^{2} d +c^{3}}d x \right ) c^{2} d f i \right )}{3 d f \left (\tan \left (f x +e \right )^{2} d^{2}+2 \tan \left (f x +e \right ) c d +c^{2}\right )} \] Input:

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

Output:

(a*( - 2*sqrt(tan(e + f*x)*d + c) - 3*int((sqrt(tan(e + f*x)*d + c)*tan(e 
+ f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 3*tan(e + f* 
x)*c**2*d + c**3),x)*tan(e + f*x)**2*d**3*f - 6*int((sqrt(tan(e + f*x)*d + 
 c)*tan(e + f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 3* 
tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)*c*d**2*f - 3*int((sqrt(tan(e + 
 f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c* 
d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*c**2*d*f + 3*int((sqrt(tan(e + f*x 
)*d + c)*tan(e + f*x))/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2*c*d**2 + 
3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)**2*d**3*f*i + 6*int((sqrt(ta 
n(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)**3*d**3 + 3*tan(e + f*x)**2* 
c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*tan(e + f*x)*c*d**2*f*i + 3*int( 
(sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)**3*d**3 + 3*tan(e + 
f*x)**2*c*d**2 + 3*tan(e + f*x)*c**2*d + c**3),x)*c**2*d*f*i))/(3*d*f*(tan 
(e + f*x)**2*d**2 + 2*tan(e + f*x)*c*d + c**2))