\(\int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx\) [173]

Optimal result

Integrand size = 28, antiderivative size = 241 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2} \arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}-\frac {\left (\frac {9}{16}+\frac {5 i}{16}\right ) d^{5/2} \arctan \left (1+\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} a^2 f}+\frac {\left (\frac {9}{16}-\frac {5 i}{16}\right ) d^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}+\sqrt {d} \tan (e+f x)}\right )}{\sqrt {2} a^2 f}+\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{8 a^2 f (1+i \tan (e+f x))}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2} \] Output:

(9/32+5/32*I)*d^(5/2)*arctan(1-2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/2))*2^(1/ 
2)/a^2/f-(9/32+5/32*I)*d^(5/2)*arctan(1+2^(1/2)*(d*tan(f*x+e))^(1/2)/d^(1/ 
2))*2^(1/2)/a^2/f+(9/32-5/32*I)*d^(5/2)*arctanh(2^(1/2)*(d*tan(f*x+e))^(1/ 
2)/(d^(1/2)+d^(1/2)*tan(f*x+e)))*2^(1/2)/a^2/f+5/8*I*d^2*(d*tan(f*x+e))^(1 
/2)/a^2/f/(1+I*tan(f*x+e))-1/4*d*(d*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e)) 
^2
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.78 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {d^2 \sec ^2(e+f x) \left (4 (-1)^{3/4} \sqrt {d} \arctan \left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+14 \sqrt [4]{-1} \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right ) (-i \cos (2 (e+f x))+\sin (2 (e+f x)))+(-5 i-5 i \cos (2 (e+f x))+7 \sin (2 (e+f x))) \sqrt {d \tan (e+f x)}\right )}{16 a^2 f (-i+\tan (e+f x))^2} \] Input:

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

Output:

(d^2*Sec[e + f*x]^2*(4*(-1)^(3/4)*Sqrt[d]*ArcTan[((-1)^(3/4)*Sqrt[d*Tan[e 
+ f*x]])/Sqrt[d]]*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + 14*(-1)^(1/4)* 
Sqrt[d]*ArcTanh[((-1)^(3/4)*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]*((-I)*Cos[2*(e 
+ f*x)] + Sin[2*(e + f*x)]) + (-5*I - (5*I)*Cos[2*(e + f*x)] + 7*Sin[2*(e 
+ f*x)])*Sqrt[d*Tan[e + f*x]]))/(16*a^2*f*(-I + Tan[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.84 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.16, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4041, 27, 3042, 4078, 3042, 4017, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4041

\(\displaystyle -\frac {\int -\frac {\sqrt {d \tan (e+f x)} \left (3 a d^2-7 i a d^2 \tan (e+f x)\right )}{2 (i \tan (e+f x) a+a)}dx}{4 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sqrt {d \tan (e+f x)} \left (3 a d^2-7 i a d^2 \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\sqrt {d \tan (e+f x)} \left (3 a d^2-7 i a d^2 \tan (e+f x)\right )}{i \tan (e+f x) a+a}dx}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {\int \frac {5 i a^2 d^3+9 a^2 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {\int \frac {5 i a^2 d^3+9 a^2 \tan (e+f x) d^3}{\sqrt {d \tan (e+f x)}}dx}{2 a^2}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 4017

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {\int \frac {a^2 d^3 (9 \tan (e+f x) d+5 i d)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{a^2 f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \int \frac {9 \tan (e+f x) d+5 i d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1482

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \left (\left (\frac {9}{2}+\frac {5 i}{2}\right ) \int \frac {\tan (e+f x) d+d}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}-\left (\frac {9}{2}-\frac {5 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1476

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \left (\left (\frac {9}{2}+\frac {5 i}{2}\right ) \left (\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}+\frac {1}{2} \int \frac {1}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}\right )-\left (\frac {9}{2}-\frac {5 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \left (\left (\frac {9}{2}+\frac {5 i}{2}\right ) \left (\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}-\frac {\int \frac {1}{-d \tan (e+f x)-1}d\left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {9}{2}-\frac {5 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \left (\left (\frac {9}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {9}{2}-\frac {5 i}{2}\right ) \int \frac {d-d \tan (e+f x)}{\tan ^2(e+f x) d^2+d^2}d\sqrt {d \tan (e+f x)}\right )}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1479

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \left (\left (\frac {9}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {9}{2}-\frac {5 i}{2}\right ) \left (-\frac {\int -\frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \left (\left (\frac {9}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {9}{2}-\frac {5 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}\right )}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \left (\left (\frac {9}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {9}{2}-\frac {5 i}{2}\right ) \left (\frac {\int \frac {\sqrt {2} \sqrt {d}-2 \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d-\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {2} \sqrt {d}}+\frac {\int \frac {\sqrt {d}+\sqrt {2} \sqrt {d \tan (e+f x)}}{\tan (e+f x) d+d+\sqrt {2} \sqrt {d \tan (e+f x)} \sqrt {d}}d\sqrt {d \tan (e+f x)}}{2 \sqrt {d}}\right )\right )}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\frac {5 i d^2 \sqrt {d \tan (e+f x)}}{f (1+i \tan (e+f x))}-\frac {d^3 \left (\left (\frac {9}{2}+\frac {5 i}{2}\right ) \left (\frac {\arctan \left (\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}+1\right )}{\sqrt {2} \sqrt {d}}-\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt {d \tan (e+f x)}}{\sqrt {d}}\right )}{\sqrt {2} \sqrt {d}}\right )-\left (\frac {9}{2}-\frac {5 i}{2}\right ) \left (\frac {\log \left (d \tan (e+f x)+\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}-\frac {\log \left (d \tan (e+f x)-\sqrt {2} \sqrt {d} \sqrt {d \tan (e+f x)}+d\right )}{2 \sqrt {2} \sqrt {d}}\right )\right )}{f}}{8 a^2}-\frac {d (d \tan (e+f x))^{3/2}}{4 f (a+i a \tan (e+f x))^2}\)

Input:

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

Output:

-1/4*(d*(d*Tan[e + f*x])^(3/2))/(f*(a + I*a*Tan[e + f*x])^2) + (-((d^3*((9 
/2 + (5*I)/2)*(-(ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[ 
2]*Sqrt[d])) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[e + f*x]])/Sqrt[d]]/(Sqrt[2] 
*Sqrt[d])) - (9/2 - (5*I)/2)*(-1/2*Log[d + d*Tan[e + f*x] - Sqrt[2]*Sqrt[d 
]*Sqrt[d*Tan[e + f*x]]]/(Sqrt[2]*Sqrt[d]) + Log[d + d*Tan[e + f*x] + Sqrt[ 
2]*Sqrt[d]*Sqrt[d*Tan[e + f*x]]]/(2*Sqrt[2]*Sqrt[d]))))/f) + ((5*I)*d^2*Sq 
rt[d*Tan[e + f*x]])/(f*(1 + I*Tan[e + f*x])))/(8*a^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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

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

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[c^2 + d^2, 0]
 

rule 4041

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

rule 4078

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

Maple [A] (verified)

Time = 1.47 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.45

method	result	size
derivativedivides	\(\frac {2 d^{3} \left (-\frac {\frac {7 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {5 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{8 \left (i d \tan \left (f x +e \right )+d \right )^{2}}-\frac {7 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{16 \sqrt {-i d}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\)	\(108\)
default	\(\frac {2 d^{3} \left (-\frac {\frac {7 \left (d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2}-\frac {5 i d \sqrt {d \tan \left (f x +e \right )}}{2}}{8 \left (i d \tan \left (f x +e \right )+d \right )^{2}}-\frac {7 \arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {-i d}}\right )}{16 \sqrt {-i d}}-\frac {\arctan \left (\frac {\sqrt {d \tan \left (f x +e \right )}}{\sqrt {i d}}\right )}{8 \sqrt {i d}}\right )}{f \,a^{2}}\)	\(108\)

Input:

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

Output:

2/f/a^2*d^3*(-1/8*(7/2*(d*tan(f*x+e))^(3/2)-5/2*I*d*(d*tan(f*x+e))^(1/2))/ 
(I*d*tan(f*x+e)+d)^2-7/16/(-I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(-I*d)^ 
(1/2))-1/8/(I*d)^(1/2)*arctan((d*tan(f*x+e))^(1/2)/(I*d)^(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. 577 vs. \(2 (177) = 354\).

Time = 0.09 (sec) , antiderivative size = 577, normalized size of antiderivative = 2.39 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

1/16*(4*a^2*f*sqrt(1/16*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-2*(I*d^3 
*e^(2*I*f*x + 2*I*e) + 4*(I*a^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*f)*sqrt((-I* 
d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(1/16*I*d^5/(a 
^4*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) - 4*a^2*f*sqrt(1/16*I*d^5/(a^4*f^2))*e 
^(4*I*f*x + 4*I*e)*log(-2*(I*d^3*e^(2*I*f*x + 2*I*e) + 4*(-I*a^2*f*e^(2*I* 
f*x + 2*I*e) - I*a^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x 
+ 2*I*e) + 1))*sqrt(1/16*I*d^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I*e)/d^2) + 4*a 
^2*f*sqrt(-49/64*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(1/8*(7*d^3 + 8*( 
a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/( 
e^(2*I*f*x + 2*I*e) + 1))*sqrt(-49/64*I*d^5/(a^4*f^2)))*e^(-2*I*f*x - 2*I* 
e)/(a^2*f)) - 4*a^2*f*sqrt(-49/64*I*d^5/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log 
(1/8*(7*d^3 - 8*(a^2*f*e^(2*I*f*x + 2*I*e) + a^2*f)*sqrt((-I*d*e^(2*I*f*x 
+ 2*I*e) + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-49/64*I*d^5/(a^4*f^2)))*e 
^(-2*I*f*x - 2*I*e)/(a^2*f)) + (6*I*d^2*e^(4*I*f*x + 4*I*e) + 5*I*d^2*e^(2 
*I*f*x + 2*I*e) - I*d^2)*sqrt((-I*d*e^(2*I*f*x + 2*I*e) + I*d)/(e^(2*I*f*x 
 + 2*I*e) + 1)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.72 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {d^{2} {\left (\frac {2 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{\frac {i \, d}{{\left | d \right |}} + 1} + \frac {7 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {2 \, \sqrt {d \tan \left (f x + e\right )} {\left | d \right |}}{-i \, \sqrt {2} d^{\frac {3}{2}} + \sqrt {2} \sqrt {d} {\left | d \right |}}\right )}{-\frac {i \, d}{{\left | d \right |}} + 1} - \frac {7 \, \sqrt {d \tan \left (f x + e\right )} d^{2} \tan \left (f x + e\right ) - 5 i \, \sqrt {d \tan \left (f x + e\right )} d^{2}}{{\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}}\right )}}{8 \, a^{2} f} \] Input:

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

Output:

-1/8*d^2*(2*sqrt(2)*sqrt(d)*arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(I*sqrt(2 
)*d^(3/2) + sqrt(2)*sqrt(d)*abs(d)))/(I*d/abs(d) + 1) + 7*sqrt(2)*sqrt(d)* 
arctan(2*sqrt(d*tan(f*x + e))*abs(d)/(-I*sqrt(2)*d^(3/2) + sqrt(2)*sqrt(d) 
*abs(d)))/(-I*d/abs(d) + 1) - (7*sqrt(d*tan(f*x + e))*d^2*tan(f*x + e) - 5 
*I*sqrt(d*tan(f*x + e))*d^2)/(d*tan(f*x + e) - I*d)^2)/(a^2*f)
 

Mupad [B] (verification not implemented)

Time = 2.89 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.73 \[ \int \frac {(d \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^2} \, dx=\frac {-\frac {7\,d^3\,{\left (d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}{8\,a^2\,f}+\frac {d^4\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,5{}\mathrm {i}}{8\,a^2\,f}}{-d^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+d^2\,\mathrm {tan}\left (e+f\,x\right )\,2{}\mathrm {i}+d^2}+2\,\mathrm {atanh}\left (\frac {8\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {\frac {d^5\,1{}\mathrm {i}}{64\,a^4\,f^2}}}{d^3}\right )\,\sqrt {\frac {d^5\,1{}\mathrm {i}}{64\,a^4\,f^2}}+2\,\mathrm {atanh}\left (\frac {16\,a^2\,f\,\sqrt {d\,\mathrm {tan}\left (e+f\,x\right )}\,\sqrt {-\frac {d^5\,49{}\mathrm {i}}{256\,a^4\,f^2}}}{7\,d^3}\right )\,\sqrt {-\frac {d^5\,49{}\mathrm {i}}{256\,a^4\,f^2}} \] Input:

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

Output:

((d^4*(d*tan(e + f*x))^(1/2)*5i)/(8*a^2*f) - (7*d^3*(d*tan(e + f*x))^(3/2) 
)/(8*a^2*f))/(d^2*tan(e + f*x)*2i + d^2 - d^2*tan(e + f*x)^2) + 2*atanh((8 
*a^2*f*(d*tan(e + f*x))^(1/2)*((d^5*1i)/(64*a^4*f^2))^(1/2))/d^3)*((d^5*1i 
)/(64*a^4*f^2))^(1/2) + 2*atanh((16*a^2*f*(d*tan(e + f*x))^(1/2)*(-(d^5*49 
i)/(256*a^4*f^2))^(1/2))/(7*d^3))*(-(d^5*49i)/(256*a^4*f^2))^(1/2)
 

Reduce [F]

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

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

Output:

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