\(\int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1132]

Optimal result

Integrand size = 30, antiderivative size = 127 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {4 i a^2 \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2} f}+\frac {2 a^2 (i c-d)}{3 d (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {4 i a^2}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 1.73 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.83 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {2 a^2 \left (-\frac {6 i \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{(c-i d)^{5/2}}+\frac {c^2+6 i c d+d^2+6 i d^2 \tan (e+f x)}{(c-i d)^2 d (c+d \tan (e+f x))^{3/2}}\right )}{3 f} \] Input:

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

Output:

(2*a^2*(((-6*I)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/(c - I*d) 
^(5/2) + (c^2 + (6*I)*c*d + d^2 + (6*I)*d^2*Tan[e + f*x])/((c - I*d)^2*d*( 
c + d*Tan[e + f*x])^(3/2))))/(3*f)
 

Rubi [A] (warning: unable to verify)

Time = 0.78 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.31, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {3042, 4025, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4025

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4012

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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. 2684 vs. \(2 (108 ) = 216\).

Time = 0.43 (sec) , antiderivative size = 2685, normalized size of antiderivative = 21.14

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2685\)
default	\(\text {Expression too large to display}\)	\(2685\)
parts	\(\text {Expression too large to display}\)	\(6797\)

Input:

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

Output:

-3*I/f*a^2*d^2/(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+6*I/f*a^2*d^2/(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+6*I/f*a^2*d^2/(c^2+d^2)^(5/2)/(2*(c^2+d^2)^(1/2)-2*c)^(1/2)*ar 
ctan((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+3*I/f*a^2*d^2/(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+4/3*I/f*a^2/(c^2+d^2)/(c+d*tan(f*x+e))^(3/2)*c-8/f* 
a^2*d/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c+2/3/f*a^2/d/(c^2+d^2)/(c+d*tan( 
f*x+e))^(3/2)*c^2-2/f*a^2*d^3/(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))+1/f*a^2*d^3/(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))-1/f*a^2*d^3/(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))-2/f*a^2*d^3/(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))-4*I/f*a^2*d^2/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)+4 
*I/f*a^2/(c^2+d^2)^2/(c+d*tan(f*x+e))^(1/2)*c^2-2/3/f*a^2*d/(c^2+d^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. 909 vs. \(2 (101) = 202\).

Time = 0.17 (sec) , antiderivative size = 909, normalized size of antiderivative = 7.16 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/12*(3*((c^4*d - 4*I*c^3*d^2 - 6*c^2*d^3 + 4*I*c*d^4 + d^5)*f*e^(4*I*f*x 
+ 4*I*e) + 2*(c^4*d - 2*I*c^3*d^2 - 2*I*c*d^4 - d^5)*f*e^(2*I*f*x + 2*I*e) 
 + (c^4*d + 2*c^2*d^3 + d^5)*f)*sqrt(-16*I*a^4/((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(1/2*(4*a^2*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(-16*I*a^4/((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))*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/ 
(e^(2*I*f*x + 2*I*e) + 1)) + 4*(a^2*c - I*a^2*d)*e^(2*I*f*x + 2*I*e))*e^(- 
2*I*f*x - 2*I*e)/a^2) - 3*((c^4*d - 4*I*c^3*d^2 - 6*c^2*d^3 + 4*I*c*d^4 + 
d^5)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4*d - 2*I*c^3*d^2 - 2*I*c*d^4 - d^5)*f*e 
^(2*I*f*x + 2*I*e) + (c^4*d + 2*c^2*d^3 + d^5)*f)*sqrt(-16*I*a^4/((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(1/2*(4*a 
^2*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(-16*I*a^4/((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))*sqrt(((c - I*d)*e^(2*I*f*x 
+ 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1)) + 4*(a^2*c - I*a^2*d)*e^(2* 
I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/a^2) + 8*(a^2*c^2 + 6*I*a^2*c*d - 5*a 
^2*d^2 + (a^2*c^2 + 6*I*a^2*c*d + 7*a^2*d^2)*e^(4*I*f*x + 4*I*e) + 2*(a^2* 
c^2 + 6*I*a^2*c*d + a^2*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)))/((c^4*d - 4*I*c^3*d^...
 

Sympy [F]

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

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

Output:

-a**2*(Integral(tan(e + f*x)**2/(c**2*sqrt(c + d*tan(e + f*x)) + 2*c*d*sqr 
t(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(-2*I*tan(e + f*x)/(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(-1/(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))
 

Maxima [F(-1)]

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

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

Output:

Timed out
 

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. 240 vs. \(2 (101) = 202\).

Time = 0.70 (sec) , antiderivative size = 240, normalized size of antiderivative = 1.89 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {2 \, a^{2} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{\sqrt {2} c \sqrt {-c + \sqrt {c^{2} + d^{2}}} - i \, \sqrt {2} \sqrt {-c + \sqrt {c^{2} + d^{2}}} d - \sqrt {2} \sqrt {c^{2} + d^{2}} \sqrt {-c + \sqrt {c^{2} + d^{2}}}}\right )}{{\left (-i \, c^{2} - 2 \, c d + i \, d^{2}\right )} \sqrt {-c + \sqrt {c^{2} + d^{2}}} {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} - \frac {-i \, c^{2} + 6 \, {\left (d \tan \left (f x + e\right ) + c\right )} d - i \, d^{2}}{-3 \, {\left (-i \, c^{2} d - 2 \, c d^{2} + i \, d^{3}\right )} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}\right )}}{f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 5.43 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.74 \[ \int \frac {(a+i a \tan (e+f x))^2}{(c+d \tan (e+f x))^{5/2}} \, dx=\frac {\frac {a^2\,\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )\,4{}\mathrm {i}}{f\,{\left (c-d\,1{}\mathrm {i}\right )}^2}+\frac {2\,a^2\,\left (c+d\,1{}\mathrm {i}\right )}{3\,d\,f\,\left (c-d\,1{}\mathrm {i}\right )}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{3/2}}+\frac {a^2\,\mathrm {atan}\left (\frac {\sqrt {c+d\,\mathrm {tan}\left (e+f\,x\right )}\,\left (2\,c^8\,f^2+8\,c^6\,d^2\,f^2+12\,c^4\,d^4\,f^2+8\,c^2\,d^6\,f^2+2\,d^8\,f^2\right )}{2\,f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{5/2}\,\left (f\,c^6+2{}\mathrm {i}\,f\,c^5\,d+f\,c^4\,d^2+4{}\mathrm {i}\,f\,c^3\,d^3-f\,c^2\,d^4+2{}\mathrm {i}\,f\,c\,d^5-f\,d^6\right )}\right )\,4{}\mathrm {i}}{f\,{\left (-c+d\,1{}\mathrm {i}\right )}^{5/2}} \] Input:

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

Output:

((a^2*(c + d*tan(e + f*x))*4i)/(f*(c - d*1i)^2) + (2*a^2*(c + d*1i))/(3*d* 
f*(c - d*1i)))/(c + d*tan(e + f*x))^(3/2) + (a^2*atan(((c + d*tan(e + f*x) 
)^(1/2)*(2*c^8*f^2 + 2*d^8*f^2 + 8*c^2*d^6*f^2 + 12*c^4*d^4*f^2 + 8*c^6*d^ 
2*f^2))/(2*f*(d*1i - c)^(5/2)*(c^6*f - d^6*f - c^2*d^4*f + c^3*d^3*f*4i + 
c^4*d^2*f + c*d^5*f*2i + c^5*d*f*2i)))*4i)/(f*(d*1i - c)^(5/2))
 

Reduce [F]

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

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

Output:

(2*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 
(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)*c*d**2*f*i + 3*i 
nt((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))