\(\int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \, dx\) [1145]

Optimal result

Integrand size = 32, antiderivative size = 196 \[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \, dx=-\frac {\sqrt [4]{-1} \sqrt {a} (3 c-i d) \sqrt {d} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{f}-\frac {i \sqrt {2} \sqrt {a} (c-i d)^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f}+\frac {d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}{f} \] Output:

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

Mathematica [A] (verified)

Time = 2.81 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.35 \[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \, dx=\frac {\sqrt {2} \sqrt {-a (c-i d)} (i c+d) \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )+d \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}+\frac {(-1)^{3/4} (i a)^{3/2} \sqrt {i a (c+i d)} \sqrt {d} (3 i c+d) \text {arcsinh}\left (\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {i a} \sqrt {i a (c+i d)}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{a^{3/2} \sqrt {c+d \tan (e+f x)}}}{f} \] Input:

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

Output:

(Sqrt[2]*Sqrt[-(a*(c - I*d))]*(I*c + d)*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[ 
a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])] + d*Sqrt[a + 
I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]] + ((-1)^(3/4)*(I*a)^(3/2)*Sqrt[ 
I*a*(c + I*d)]*Sqrt[d]*((3*I)*c + d)*ArcSinh[((-1)^(1/4)*Sqrt[a]*Sqrt[d]*S 
qrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a]*Sqrt[I*a*(c + I*d)])]*Sqrt[(c + d*Ta 
n[e + f*x])/(c + I*d)])/(a^(3/2)*Sqrt[c + d*Tan[e + f*x]]))/f
 

Rubi [A] (verified)

Time = 1.18 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.03, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.344, Rules used = {3042, 4043, 27, 3042, 4084, 3042, 4027, 221, 4082, 66, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4043

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4084

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((2*(-1)^(3/4)*a^(3/2)*Sqrt[d]*((3*I)*c + d)*ArcTanh[((-1)^(3/4)*Sqrt[d]*S 
qrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c + d*Tan[e + f*x]])])/f - ((2*I) 
*Sqrt[2]*a^(3/2)*(c - I*d)^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*Tan[e 
 + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/f)/(2*a) + (d*Sqrt[ 
a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])/f
 

Defintions of rubi rules used

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 66

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] &&  !GtQ[c - a*(d/b), 0]
 

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 4027

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

rule 4043

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

rule 4082

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

rule 4084

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1701 vs. \(2 (153 ) = 306\).

Time = 0.43 (sec) , antiderivative size = 1702, normalized size of antiderivative = 8.68

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1702\)
default	\(\text {Expression too large to display}\)	\(1702\)

Input:

int((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERBOS 
E)
 

Output:

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

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. 768 vs. \(2 (146) = 292\).

Time = 0.11 (sec) , antiderivative size = 768, normalized size of antiderivative = 3.92 \[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \, dx =\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="fr 
icas")
 

Output:

1/2*(2*sqrt(2)*d*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(a/(e^(2*I*f*x + 2*I*e) + 1))*e^(I*f*x + I*e) - sqrt( 
2)*f*sqrt(-(a*c^3 - 3*I*a*c^2*d - 3*a*c*d^2 + I*a*d^3)/f^2)*log((sqrt(2)*f 
*sqrt(-(a*c^3 - 3*I*a*c^2*d - 3*a*c*d^2 + I*a*d^3)/f^2)*e^(I*f*x + I*e) + 
sqrt(2)*((I*c + d)*e^(2*I*f*x + 2*I*e) + I*c + d)*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(a/(e^(2*I*f*x + 2*I 
*e) + 1)))*e^(-I*f*x - I*e)/(I*c + d)) + sqrt(2)*f*sqrt(-(a*c^3 - 3*I*a*c^ 
2*d - 3*a*c*d^2 + I*a*d^3)/f^2)*log(-(sqrt(2)*f*sqrt(-(a*c^3 - 3*I*a*c^2*d 
 - 3*a*c*d^2 + I*a*d^3)/f^2)*e^(I*f*x + I*e) - sqrt(2)*((I*c + d)*e^(2*I*f 
*x + 2*I*e) + I*c + d)*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(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e) 
/(I*c + d)) + f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*log((sqrt(2) 
*((3*I*c + d)*e^(2*I*f*x + 2*I*e) + 3*I*c + d)*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(a/(e^(2*I*f*x + 2*I*e) 
 + 1)) + 2*f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/f^2)*e^(I*f*x + I*e) 
)*e^(-I*f*x - I*e)/(3*I*c + d)) - f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^ 
3)/f^2)*log((sqrt(2)*((3*I*c + d)*e^(2*I*f*x + 2*I*e) + 3*I*c + d)*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(a/ 
(e^(2*I*f*x + 2*I*e) + 1)) - 2*f*sqrt((9*I*a*c^2*d + 6*a*c*d^2 - I*a*d^3)/ 
f^2)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)/(3*I*c + d)))/f
 

Sympy [F]

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

integrate((a+I*a*tan(f*x+e))**(1/2)*(c+d*tan(f*x+e))**(3/2),x)
 

Output:

Integral(sqrt(I*a*(tan(e + f*x) - I))*(c + d*tan(e + f*x))**(3/2), x)
 

Maxima [F]

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

integrate((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="ma 
xima")
 

Output:

integrate(sqrt(I*a*tan(f*x + e) + a)*(d*tan(f*x + e) + c)^(3/2), x)
 

Giac [F(-1)]

Timed out. \[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2),x, algorithm="gi 
ac")
 

Output:

Timed out
 

Mupad [F(-1)]

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

int((a + a*tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(3/2),x)
 

Output:

int((a + a*tan(e + f*x)*1i)^(1/2)*(c + d*tan(e + f*x))^(3/2), x)
 

Reduce [F]

\[ \int \sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2} \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}\, \tan \left (f x +e \right )d x \right ) d +\left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {d \tan \left (f x +e \right )+c}d x \right ) c \right ) \] Input:

int((a+I*a*tan(f*x+e))^(1/2)*(c+d*tan(f*x+e))^(3/2),x)
 

Output:

sqrt(a)*(int(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x 
),x)*d + int(sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c),x)*c)