\(\int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx\) [154]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 28, antiderivative size = 93 \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=\frac {4 \sqrt [4]{-1} a^2 \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{d^{5/2} f}-\frac {2 a^2}{3 d f (d \tan (e+f x))^{3/2}}-\frac {4 i a^2}{d^2 f \sqrt {d \tan (e+f x)}} \] Output:

4*(-1)^(1/4)*a^2*arctan((-1)^(3/4)*(d*tan(f*x+e))^(1/2)/d^(1/2))/d^(5/2)/f 
-2/3*a^2/d/f/(d*tan(f*x+e))^(3/2)-4*I*a^2/d^2/f/(d*tan(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 1.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.89 \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=\frac {(-6+6 i) \sqrt {2} a^2 \sqrt {d} \text {arctanh}\left (\frac {(1+i) \sqrt {d \tan (e+f x)}}{\sqrt {2} \sqrt {d}}\right )-\frac {2 a^2 d (6 i+\cot (e+f x))}{\sqrt {d \tan (e+f x)}}}{3 d^3 f} \] Input:

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

Output:

((-6 + 6*I)*Sqrt[2]*a^2*Sqrt[d]*ArcTanh[((1 + I)*Sqrt[d*Tan[e + f*x]])/(Sq 
rt[2]*Sqrt[d])] - (2*a^2*d*(6*I + Cot[e + f*x]))/Sqrt[d*Tan[e + f*x]])/(3* 
d^3*f)
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4025, 27, 3042, 4012, 25, 3042, 4016, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4025

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4016

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(-2*a^2)/(3*d*f*(d*Tan[e + f*x])^(3/2)) + (2*((2*(-1)^(1/4)*a^2*ArcTan[((- 
1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]])/(Sqrt[d]*f) - ((2*I)*a^2)/(f*Sqrt 
[d*Tan[e + f*x]])))/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 218
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[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 4016
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2*(c^2/f)   Subst[Int[1/(b*c - d*x^2), x], x, Sqrt[b 
*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]
 

rule 4025
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*c - a*d)^2*((a + b*Tan[e + f*x])^(m + 
 1)/(b*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^2 + 2*b*c*d - a*d^2 - (b*c^2 - 2*a*c*d - b*d^2)*Ta 
n[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] 
 && LtQ[m, -1] && NeQ[a^2 + b^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. 311 vs. \(2 (76 ) = 152\).

Time = 1.61 (sec) , antiderivative size = 312, normalized size of antiderivative = 3.35

method result size
derivativedivides \(\frac {2 a^{2} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d}-\frac {1}{3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2 i}{d \sqrt {d \tan \left (f x +e \right )}}\right )}{f d}\) \(312\)
default \(\frac {2 a^{2} \left (\frac {-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d}-\frac {i \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 \left (d^{2}\right )^{\frac {1}{4}}}}{d}-\frac {1}{3 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {2 i}{d \sqrt {d \tan \left (f x +e \right )}}\right )}{f d}\) \(312\)
parts \(\frac {2 a^{2} d \left (-\frac {1}{3 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 d^{4}}\right )}{f}+\frac {2 i a^{2} \left (-\frac {2}{d^{2} \sqrt {d \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 d^{2} \left (d^{2}\right )^{\frac {1}{4}}}\right )}{f}-\frac {a^{2} \left (d^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {d \tan \left (f x +e \right )+\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}{d \tan \left (f x +e \right )-\left (d^{2}\right )^{\frac {1}{4}} \sqrt {d \tan \left (f x +e \right )}\, \sqrt {2}+\sqrt {d^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {d \tan \left (f x +e \right )}}{\left (d^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{4 f \,d^{3}}\) \(460\)

Input:

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

Output:

2/f*a^2/d*(1/d*(-1/4/d*(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)+(d^2)^(1/4)*( 
d*tan(f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)-(d^2)^(1/4)*(d*tan( 
f*x+e))^(1/2)*2^(1/2)+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f* 
x+e))^(1/2)+1)-2*arctan(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1))-1/4* 
I/(d^2)^(1/4)*2^(1/2)*(ln((d*tan(f*x+e)-(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2 
^(1/2)+(d^2)^(1/2))/(d*tan(f*x+e)+(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)*2^(1/2) 
+(d^2)^(1/2)))+2*arctan(2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)-2*arct 
an(-2^(1/2)/(d^2)^(1/4)*(d*tan(f*x+e))^(1/2)+1)))-1/3/(d*tan(f*x+e))^(3/2) 
-2*I/d/(d*tan(f*x+e))^(1/2))
 

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. 403 vs. \(2 (75) = 150\).

Time = 0.09 (sec) , antiderivative size = 403, normalized size of antiderivative = 4.33 \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {-\frac {16 i \, a^{4}}{d^{5} f^{2}}} \log \left (\frac {{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} + {\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {16 i \, a^{4}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 3 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {-\frac {16 i \, a^{4}}{d^{5} f^{2}}} \log \left (\frac {{\left (-4 i \, a^{2} d e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {16 i \, a^{4}}{d^{5} f^{2}}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{2 \, a^{2}}\right ) - 8 \, {\left (7 \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 5 \, a^{2}\right )} \sqrt {\frac {-i \, d e^{\left (2 i \, f x + 2 i \, e\right )} + i \, d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (d^{3} f e^{\left (4 i \, f x + 4 i \, e\right )} - 2 \, d^{3} f e^{\left (2 i \, f x + 2 i \, e\right )} + d^{3} f\right )}} \] Input:

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

Output:

-1/12*(3*(d^3*f*e^(4*I*f*x + 4*I*e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f) 
*sqrt(-16*I*a^4/(d^5*f^2))*log(1/2*(-4*I*a^2*d*e^(2*I*f*x + 2*I*e) + (d^3* 
f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2 
*I*f*x + 2*I*e) + 1))*sqrt(-16*I*a^4/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) 
 - 3*(d^3*f*e^(4*I*f*x + 4*I*e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)*sqr 
t(-16*I*a^4/(d^5*f^2))*log(1/2*(-4*I*a^2*d*e^(2*I*f*x + 2*I*e) - (d^3*f*e^ 
(2*I*f*x + 2*I*e) + d^3*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f 
*x + 2*I*e) + 1))*sqrt(-16*I*a^4/(d^5*f^2)))*e^(-2*I*f*x - 2*I*e)/a^2) - 8 
*(7*a^2*e^(4*I*f*x + 4*I*e) + 2*a^2*e^(2*I*f*x + 2*I*e) - 5*a^2)*sqrt((-I* 
d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1)))/(d^3*f*e^(4*I*f*x 
 + 4*I*e) - 2*d^3*f*e^(2*I*f*x + 2*I*e) + d^3*f)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (75) = 150\).

Time = 0.12 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.13 \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=\frac {\frac {3 \, a^{2} {\left (-\frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} - \frac {\left (2 i + 2\right ) \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (f x + e\right )}\right )}}{2 \, \sqrt {d}}\right )}{\sqrt {d}} + \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) + \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}} - \frac {\left (i - 1\right ) \, \sqrt {2} \log \left (d \tan \left (f x + e\right ) - \sqrt {2} \sqrt {d \tan \left (f x + e\right )} \sqrt {d} + d\right )}{\sqrt {d}}\right )}}{d} - \frac {4 \, {\left (6 i \, a^{2} d \tan \left (f x + e\right ) + a^{2} d\right )}}{\left (d \tan \left (f x + e\right )\right )^{\frac {3}{2}} d}}{6 \, d f} \] Input:

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

Output:

1/6*(3*a^2*(-(2*I + 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqr 
t(d*tan(f*x + e)))/sqrt(d))/sqrt(d) - (2*I + 2)*sqrt(2)*arctan(-1/2*sqrt(2 
)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(f*x + e)))/sqrt(d))/sqrt(d) + (I - 1)*sq 
rt(2)*log(d*tan(f*x + e) + sqrt(2)*sqrt(d*tan(f*x + e))*sqrt(d) + d)/sqrt( 
d) - (I - 1)*sqrt(2)*log(d*tan(f*x + e) - sqrt(2)*sqrt(d*tan(f*x + e))*sqr 
t(d) + d)/sqrt(d))/d - 4*(6*I*a^2*d*tan(f*x + e) + a^2*d)/((d*tan(f*x + e) 
)^(3/2)*d))/(d*f)
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.03 \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {2 \, a^{2} {\left (\frac {6 i \, \sqrt {2} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{d^{\frac {5}{2}} {\left (\frac {i \, d}{{\left | d \right |}} + 1\right )}} + \frac {6 i \, d \tan \left (f x + e\right ) + d}{\sqrt {d \tan \left (f x + e\right )} d^{3} \tan \left (f x + e\right )}\right )}}{3 \, f} \] Input:

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

Output:

-2/3*a^2*(6*I*sqrt(2)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt(2)*d^(3 
/2) + sqrt(2)*sqrt(d)*abs(d)))/(d^(5/2)*(I*d/abs(d) + 1)) + (6*I*d*tan(f*x 
 + e) + d)/(sqrt(d*tan(f*x + e))*d^3*tan(f*x + e)))/f
 

Mupad [B] (verification not implemented)

Time = 1.42 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \frac {(a+i a \tan (e+f x))^2}{(d \tan (e+f x))^{5/2}} \, dx=-\frac {\frac {2\,a^2}{3\,d\,f}+\frac {a^2\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}}{d\,f}}{{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}-\frac {\sqrt {4{}\mathrm {i}}\,a^2\,\mathrm {atan}\left (\frac {\sqrt {4{}\mathrm {i}}\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}}{2\,\sqrt {-d}}\right )\,2{}\mathrm {i}}{{\left (-d\right )}^{5/2}\,f} \] Input:

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

Output:

- ((2*a^2)/(3*d*f) + (a^2*tan(e + f*x)*4i)/(d*f))/(d*tan(e + f*x))^(3/2) - 
 (4i^(1/2)*a^2*atan((4i^(1/2)*(d*tan(e + f*x))^(1/2))/(2*(-d)^(1/2)))*2i)/ 
((-d)^(5/2)*f)
 

Reduce [F]

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

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

Output:

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