\(\int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx\) [193]

Optimal result

Integrand size = 38, antiderivative size = 194 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(13 A+3 i B) \sqrt {\tan (c+d x)}}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(67 A-3 i B) \sqrt {\tan (c+d x)}}{60 a^2 d \sqrt {a+i a \tan (c+d x)}} \] 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*(A+I*B)*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(5/ 
2)+1/30*(13*A+3*I*B)*tan(d*x+c)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(3/2)+1/60*(6 
7*A-3*I*B)*tan(d*x+c)^(1/2)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 2.82 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.25 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {\tan (c+d x)} \left (24 a^5 (A+i B) \sqrt {i a \tan (c+d x)}-(a+i a \tan (c+d x)) \left (-4 a^4 (13 A+3 i B) \sqrt {i a \tan (c+d x)}+(a+i a \tan (c+d x)) \left (2 a^3 (-67 A+3 i B) \sqrt {i a \tan (c+d x)}-15 \sqrt {2} a^3 (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)}\right )\right )\right )}{120 a^5 d \sqrt {i a \tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \] Input:

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

Output:

(Sqrt[Tan[c + d*x]]*(24*a^5*(A + I*B)*Sqrt[I*a*Tan[c + d*x]] - (a + I*a*Ta 
n[c + d*x])*(-4*a^4*(13*A + (3*I)*B)*Sqrt[I*a*Tan[c + d*x]] + (a + I*a*Tan 
[c + d*x])*(2*a^3*(-67*A + (3*I)*B)*Sqrt[I*a*Tan[c + d*x]] - 15*Sqrt[2]*a^ 
3*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + 
d*x]]]*Sqrt[a + I*a*Tan[c + d*x]]))))/(120*a^5*d*Sqrt[I*a*Tan[c + d*x]]*(a 
 + I*a*Tan[c + d*x])^(5/2))
 

Rubi [A] (verified)

Time = 1.07 (sec) , antiderivative size = 205, 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, 4079, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\int \frac {a (9 A-i B)-4 a (i A-B) \tan (c+d x)}{2 \sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {a (9 A-i B)-4 a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {a (9 A-i B)-4 a (i A-B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {a^2 (41 A-9 i B)-2 a^2 (13 i A-3 B) \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {a^2 (41 A-9 i B)-2 a^2 (13 i A-3 B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {a^2 (41 A-9 i B)-2 a^2 (13 i A-3 B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\frac {\int \frac {15 a^3 (A-i B) \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}+\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\frac {15}{2} a (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {15}{2} a (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {\frac {\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {15 i a^3 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\frac {\left (\frac {15}{2}-\frac {15 i}{2}\right ) a^{3/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {a^2 (67 A-3 i B) \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {a (13 A+3 i B) \sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \sqrt {\tan (c+d x)}}{5 d (a+i a \tan (c+d x))^{5/2}}\)

Input:

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

Output:

((A + I*B)*Sqrt[Tan[c + d*x]])/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((a*(1 
3*A + (3*I)*B)*Sqrt[Tan[c + d*x]])/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (( 
(15/2 - (15*I)/2)*a^(3/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + 
d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + (a^2*(67*A - (3*I)*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 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. \(2 (156 ) = 312\).

Time = 0.25 (sec) , antiderivative size = 1096, normalized size of antiderivative = 5.65

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((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(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)*(15*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-60*I*B*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I 
*a)^(1/2)+15*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)^4-12*I*B* 
(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^2+60*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*tan(d*x+c)+60*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+268*I*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2 
)*(-I*a)^(1/2)*tan(d*x+c)^3-90*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)^2-90*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)^2+12*B 
*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^3+15*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)^4-1060*I*A*(a*tan(d*x+c)*( 
1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)-60*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)+908*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(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. 460 vs. \(2 (146) = 292\).

Time = 0.12 (sec) , antiderivative size = 460, normalized size of antiderivative = 2.37 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {2 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {-2 i \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 i \, A + 4 \, B}\right ) + \sqrt {2} {\left ({\left (83 \, A + 3 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, {\left (17 \, A + 2 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, {\left (11 \, A + 6 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, A + 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \] Input:

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(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*I*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)*s 
qrt(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*I*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)*((83*A + 3*I*B)*e^(6*I*d*x + 6*I*c) + 6*(17*A + 2*I*B)*e^(4*I*d*x + 4* 
I*c) + 2*(11*A + 6*I*B)*e^(2*I*d*x + 2*I*c) + 3*A + 3*I*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(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(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. \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((A+B*tan(d*x+c))/tan(d*x+c)^(1/2)/(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 TypeError: Bad Ar 
gument Ty
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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