3.9.38 \(\int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx\) [838]

3.9.38.1 Optimal result
3.9.38.2 Mathematica [A] (verified)
3.9.38.3 Rubi [A] (verified)
3.9.38.4 Maple [B] (verified)
3.9.38.5 Fricas [B] (verification not implemented)
3.9.38.6 Sympy [F]
3.9.38.7 Maxima [A] (verification not implemented)
3.9.38.8 Giac [F]
3.9.38.9 Mupad [F(-1)]

3.9.38.1 Optimal result

Integrand size = 45, antiderivative size = 157 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {2 B c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{a^{3/2} f}+\frac {2 B c \sqrt {c-i c \tan (e+f x)}}{a f \sqrt {a+i a \tan (e+f x)}}+\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{3 f (a+i a \tan (e+f x))^{3/2}} \]

output
2*B*c^(3/2)*arctan(c^(1/2)*(a+I*a*tan(f*x+e))^(1/2)/a^(1/2)/(c-I*c*tan(f*x 
+e))^(1/2))/a^(3/2)/f+2*B*c*(c-I*c*tan(f*x+e))^(1/2)/a/f/(a+I*a*tan(f*x+e) 
)^(1/2)+1/3*(I*A-B)*(c-I*c*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^(3/2)
 
3.9.38.2 Mathematica [A] (verified)

Time = 5.23 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {c \sqrt {c-i c \tan (e+f x)} \left (\frac {6 B \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {1-i \tan (e+f x)}}+\frac {\sqrt {a} (A-5 i B+(-i A+7 B) \tan (e+f x))}{(-i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}\right )}{3 a^{3/2} f} \]

input
Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2))/(a + I*a*Tan 
[e + f*x])^(3/2),x]
 
output
(c*Sqrt[c - I*c*Tan[e + f*x]]*((6*B*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqr 
t[2]*Sqrt[a])])/Sqrt[1 - I*Tan[e + f*x]] + (Sqrt[a]*(A - (5*I)*B + ((-I)*A 
 + 7*B)*Tan[e + f*x]))/((-I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])))/ 
(3*a^(3/2)*f)
 
3.9.38.3 Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {3042, 4071, 87, 57, 45, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4071

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

\(\Big \downarrow \) 87

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

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

input
Int[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2))/(a + I*a*Tan[e + f 
*x])^(3/2),x]
 
output
(a*c*(((I*A - B)*(c - I*c*Tan[e + f*x])^(3/2))/(3*a*c*(a + I*a*Tan[e + f*x 
])^(3/2)) - (I*B*(((2*I)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x] 
])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/a^(3/2) + ((2*I)*Sqrt[c - I*c*Ta 
n[e + f*x]])/(a*Sqrt[a + I*a*Tan[e + f*x]])))/a))/f
 

3.9.38.3.1 Defintions of rubi rules used

rule 45
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] && EqQ[b*c + a*d, 0] &&  !GtQ[c, 0]
 

rule 57
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, x]
 

rule 87
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p 
+ 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p 
+ 1)))/(f*(p + 1)*(c*f - d*e))   Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] 
/; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege 
rQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ[p, n]))))
 

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 4071
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si 
mp[a*(c/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x 
, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c 
+ a*d, 0] && EqQ[a^2 + b^2, 0]
 
3.9.38.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 407 vs. \(2 (128 ) = 256\).

Time = 0.34 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.60

method result size
derivativedivides \(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (-3 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}+9 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+7 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-9 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+A \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}-5 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+3 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +12 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{3 f \,a^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{3} \sqrt {a c}}\) \(408\)
default \(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (-3 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}+9 i B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+7 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}-9 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+A \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}-5 i B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}+3 B \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c +12 B \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+A \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{3 f \,a^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{3} \sqrt {a c}}\) \(408\)
parts \(\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan \left (f x +e \right )^{2}\right )}{3 f \,a^{2} \left (i-\tan \left (f x +e \right )\right )^{3}}+\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (9 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{2} a c -3 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right )^{3} a c -3 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) a c -12 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )+9 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}{\sqrt {a c}}\right ) \tan \left (f x +e \right ) a c +7 \tan \left (f x +e \right )^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-5 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\right )}{3 f \,a^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i-\tan \left (f x +e \right )\right )^{3} \sqrt {a c}}\) \(415\)

input
int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/2),x,m 
ethod=_RETURNVERBOSE)
 
output
1/3/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^2*c*(-3*I*B 
*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1/2)) 
*a*c*tan(f*x+e)^3+9*I*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^ 
2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)+7*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e 
)^2))^(1/2)*tan(f*x+e)^2-9*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f* 
x+e)^2))^(1/2))/(a*c)^(1/2))*a*c*tan(f*x+e)^2+A*(a*c*(1+tan(f*x+e)^2))^(1/ 
2)*(a*c)^(1/2)*tan(f*x+e)^2-5*I*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2) 
+3*B*ln((a*c*tan(f*x+e)+(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c)^(1 
/2))*a*c+12*B*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)+A*(a*c)^ 
(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(I-tan(f* 
x+e))^3/(a*c)^(1/2)
 
3.9.38.5 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. 400 vs. \(2 (119) = 238\).

Time = 0.28 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.55 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=-\frac {{\left (3 \, a^{2} f \sqrt {-\frac {B^{2} c^{3}}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B c e^{\left (3 i \, f x + 3 i \, e\right )} + B c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} + {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{2} f\right )} \sqrt {-\frac {B^{2} c^{3}}{a^{3} f^{2}}}\right )}}{B c e^{\left (2 i \, f x + 2 i \, e\right )} + B c}\right ) - 3 \, a^{2} f \sqrt {-\frac {B^{2} c^{3}}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (\frac {4 \, {\left (2 \, {\left (B c e^{\left (3 i \, f x + 3 i \, e\right )} + B c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - a^{2} f\right )} \sqrt {-\frac {B^{2} c^{3}}{a^{3} f^{2}}}\right )}}{B c e^{\left (2 i \, f x + 2 i \, e\right )} + B c}\right ) - 2 \, {\left (6 \, B c e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-i \, A - 5 \, B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} - {\left (-i \, A + B\right )} c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{6 \, a^{2} f} \]

input
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/ 
2),x, algorithm="fricas")
 
output
-1/6*(3*a^2*f*sqrt(-B^2*c^3/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(4*(2*(B*c*e 
^(3*I*f*x + 3*I*e) + B*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1) 
)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)) + (a^2*f*e^(2*I*f*x + 2*I*e) - a^2*f)* 
sqrt(-B^2*c^3/(a^3*f^2)))/(B*c*e^(2*I*f*x + 2*I*e) + B*c)) - 3*a^2*f*sqrt( 
-B^2*c^3/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(4*(2*(B*c*e^(3*I*f*x + 3*I*e) 
+ B*c*e^(I*f*x + I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f* 
x + 2*I*e) + 1)) - (a^2*f*e^(2*I*f*x + 2*I*e) - a^2*f)*sqrt(-B^2*c^3/(a^3* 
f^2)))/(B*c*e^(2*I*f*x + 2*I*e) + B*c)) - 2*(6*B*c*e^(4*I*f*x + 4*I*e) - ( 
-I*A - 5*B)*c*e^(2*I*f*x + 2*I*e) - (-I*A + B)*c)*sqrt(a/(e^(2*I*f*x + 2*I 
*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-3*I*f*x - 3*I*e)/(a^2*f)
 
3.9.38.6 Sympy [F]

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

input
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))**(3/2)/(a+I*a*tan(f*x+e))**( 
3/2),x)
 
output
Integral((-I*c*(tan(e + f*x) + I))**(3/2)*(A + B*tan(e + f*x))/(I*a*(tan(e 
 + f*x) - I))**(3/2), x)
 
3.9.38.7 Maxima [A] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {{\left (6 \, B c \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 6 \, B c \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (-i \, A + B\right )} c \cos \left (3 \, f x + 3 \, e\right ) + 12 \, B c \cos \left (f x + e\right ) + 3 i \, B c \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - 3 i \, B c \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (A + i \, B\right )} c \sin \left (3 \, f x + 3 \, e\right ) - 12 i \, B c \sin \left (f x + e\right )\right )} \sqrt {c}}{6 \, a^{\frac {3}{2}} f} \]

input
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/ 
2),x, algorithm="maxima")
 
output
1/6*(6*B*c*arctan2(cos(f*x + e), sin(f*x + e) + 1) + 6*B*c*arctan2(cos(f*x 
 + e), -sin(f*x + e) + 1) - 2*(-I*A + B)*c*cos(3*f*x + 3*e) + 12*B*c*cos(f 
*x + e) + 3*I*B*c*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1 
) - 3*I*B*c*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) + 2* 
(A + I*B)*c*sin(3*f*x + 3*e) - 12*I*B*c*sin(f*x + e))*sqrt(c)/(a^(3/2)*f)
 
3.9.38.8 Giac [F]

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

input
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(3/ 
2),x, algorithm="giac")
 
output
integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(3/2)/(I*a*tan(f*x 
+ e) + a)^(3/2), x)
 
3.9.38.9 Mupad [F(-1)]

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

input
int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^(3/2))/(a + a*tan(e + f* 
x)*1i)^(3/2),x)
 
output
int(((A + B*tan(e + f*x))*(c - c*tan(e + f*x)*1i)^(3/2))/(a + a*tan(e + f* 
x)*1i)^(3/2), x)