\(\int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [192]

Optimal result

Integrand size = 38, antiderivative size = 196 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {(i A-B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(3 i A+7 B) \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {(3 i A-13 B) \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}} \] Output:

(-1/8-1/8*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x 
+c))^(1/2))/a^(5/2)/d+1/5*(I*A-B)*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(5 
/2)+1/30*(3*I*A+7*B)*tan(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)-1/60*(3 
*I*A-13*B)*tan(d*x+c)^(1/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 1.71 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sec ^2(c+d x) \sqrt {\tan (c+d x)} \left (-2 i (9 A-i B+2 (3 A-7 i B) \cos (2 (c+d x))+20 B \sin (2 (c+d x))) \sqrt {i a \tan (c+d x)}+15 \sqrt {2} (i A+B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x))) \sqrt {a+i a \tan (c+d x)}\right )}{120 a^2 d \sqrt {i a \tan (c+d x)} (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[(Sqrt[Tan[c + d*x]]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x]) 
^(5/2),x]
 

Output:

(Sec[c + d*x]^2*Sqrt[Tan[c + d*x]]*((-2*I)*(9*A - I*B + 2*(3*A - (7*I)*B)* 
Cos[2*(c + d*x)] + 20*B*Sin[2*(c + d*x)])*Sqrt[I*a*Tan[c + d*x]] + 15*Sqrt 
[2]*(I*A + B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c 
+ d*x]]]*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*Sqrt[a + I*a*Tan[c + d*x] 
]))/(120*a^2*d*Sqrt[I*a*Tan[c + d*x]]*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*T 
an[c + d*x]])
 

Rubi [A] (verified)

Time = 1.05 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.06, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 4078, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4027, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4078

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {a (i A-B)-2 a (2 A-3 i B) \tan (c+d x)}{2 \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {a (i A-B)-2 a (2 A-3 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {a (i A-B)-2 a (2 A-3 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {a^2 (9 i A+B)-2 a^2 (3 A-7 i B) \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}-\frac {a (7 B+3 i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {a^2 (9 i A+B)-2 a^2 (3 A-7 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}-\frac {a (7 B+3 i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\int \frac {a^2 (9 i A+B)-2 a^2 (3 A-7 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}-\frac {a (7 B+3 i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {\int \frac {15 a^3 (i A+B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}+\frac {a^2 (-13 B+3 i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (7 B+3 i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {15}{2} a (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {a^2 (-13 B+3 i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (7 B+3 i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {15}{2} a (B+i A) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {a^2 (-13 B+3 i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (7 B+3 i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {a^2 (-13 B+3 i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^3 (B+i A) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{6 a^2}-\frac {a (7 B+3 i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {(-B+i A) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\frac {\frac {\left (\frac {15}{2}-\frac {15 i}{2}\right ) a^{3/2} (B+i A) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (-13 B+3 i A) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}-\frac {a (7 B+3 i A) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}\)

Input:

Int[(Sqrt[Tan[c + d*x]]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2) 
,x]
 

Output:

((I*A - B)*Sqrt[Tan[c + d*x]])/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) - (-1/3* 
(a*((3*I)*A + 7*B)*Sqrt[Tan[c + d*x]])/(d*(a + I*a*Tan[c + d*x])^(3/2)) + 
(((15/2 - (15*I)/2)*a^(3/2)*(I*A + B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c 
+ d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + (a^2*((3*I)*A - 13*B)*Sqrt[Tan[c 
 + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(6*a^2))/(10*a^2)
 

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 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 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 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]
 

rule 4079

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1095 vs. \(2 (158 ) = 316\).

Time = 0.23 (sec) , antiderivative size = 1096, normalized size of antiderivative = 5.59

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1096\)
default	\(\text {Expression too large to display}\)	\(1096\)
parts	\(\text {Expression too large to display}\)	\(1143\)

Input:

int(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x,method=_R 
ETURNVERBOSE)
 

Output:

1/240/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(15*I*B*2^(1/2)*ln 
(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*t 
an(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^4-12*A*(-I*a)^(1/2)*(a*tan(d*x+c)* 
(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+60*I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a) 
^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+ 
c)+I))*a*tan(d*x+c)^3-15*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x 
+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+ 
c)^4-212*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2 
-60*I*A*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c) 
))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)-52*I*B*(a*tan(d* 
x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+60*B*2^(1/2)*ln(-(- 
2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d 
*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^3+15*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a 
)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x 
+c)+I))*a-90*I*B*2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t 
an(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^2+90*A* 
2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2) 
+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^2+220*I*B*(a*tan(d*x+c)* 
(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+12*I*A*(a*tan(d*x+c)*(1+I* 
tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-60*B*2^(1/2)*ln(-(-2*2^(1/...
 

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

Time = 0.15 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.35 \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {2 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {2 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - \sqrt {2} {\left ({\left (3 i \, A + 17 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 6 \, {\left (-2 i \, A - 3 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (-6 i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \] Input:

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, al 
gorithm="fricas")
 

Output:

-1/120*(15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(5*I* 
d*x + 5*I*c)*log((2*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B - I*B^2)/(a^5*d^2) 
)*e^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt 
(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d* 
x + 2*I*c) + 1)))/(4*I*A + 4*B)) - 15*sqrt(1/2)*a^3*d*sqrt((I*A^2 + 2*A*B 
- I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-(2*sqrt(1/2)*a^3*d*sqrt((I*A^ 
2 + 2*A*B - I*B^2)/(a^5*d^2))*e^(I*d*x + I*c) - sqrt(2)*((I*A + B)*e^(2*I* 
d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I* 
d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(4*I*A + 4*B)) - sqrt(2)*((3 
*I*A + 17*B)*e^(6*I*d*x + 6*I*c) - 6*(-2*I*A - 3*B)*e^(4*I*d*x + 4*I*c) - 
2*(-6*I*A + B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*sqrt(a/(e^(2*I*d*x + 2*I 
*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))*e^ 
(-5*I*d*x - 5*I*c)/(a^3*d)
 

Sympy [F]

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

integrate(tan(d*x+c)**(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)
 

Output:

Integral((A + B*tan(c + d*x))*sqrt(tan(c + d*x))/(I*a*(tan(c + d*x) - I))* 
*(5/2), x)
 

Maxima [F(-2)]

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

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, al 
gorithm="maxima")
 

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F(-2)]

Exception generated. \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, al 
gorithm="giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeError: Bad Argument TypeError: Bad Argument TypeWarning, need 
 to choos
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {\tan (c+d x)} (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

int(tan(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)
 

Output:

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