\(\int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx\) [815]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F(-1)]
Maxima [A] (verification not implemented)
Giac [F(-2)]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 45, antiderivative size = 261 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx=-\frac {(i A+B) (a+i a \tan (e+f x))^{5/2}}{13 f (c-i c \tan (e+f x))^{13/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{143 c f (c-i c \tan (e+f x))^{11/2}}-\frac {(4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{429 c^2 f (c-i c \tan (e+f x))^{9/2}}-\frac {2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{3003 c^3 f (c-i c \tan (e+f x))^{7/2}}-\frac {2 (4 i A-9 B) (a+i a \tan (e+f x))^{5/2}}{15015 c^4 f (c-i c \tan (e+f x))^{5/2}} \] Output:

-1/13*(I*A+B)*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e))^(13/2)-1/143*( 
4*I*A-9*B)*(a+I*a*tan(f*x+e))^(5/2)/c/f/(c-I*c*tan(f*x+e))^(11/2)-1/429*(4 
*I*A-9*B)*(a+I*a*tan(f*x+e))^(5/2)/c^2/f/(c-I*c*tan(f*x+e))^(9/2)-2/3003*( 
4*I*A-9*B)*(a+I*a*tan(f*x+e))^(5/2)/c^3/f/(c-I*c*tan(f*x+e))^(7/2)-2/15015 
*(4*I*A-9*B)*(a+I*a*tan(f*x+e))^(5/2)/c^4/f/(c-I*c*tan(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 16.34 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.70 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx=-\frac {i a^2 \cos (e+f x) (5005 A+780 (9 A+i B) \cos (2 (e+f x))+231 (9 A+4 i B) \cos (4 (e+f x))-1560 i A \sin (2 (e+f x))+3510 B \sin (2 (e+f x))-924 i A \sin (4 (e+f x))+2079 B \sin (4 (e+f x))) (\cos (9 e+11 f x)+i \sin (9 e+11 f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{120120 c^7 f (\cos (f x)+i \sin (f x))^2} \] Input:

Integrate[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan 
[e + f*x])^(13/2),x]
 

Output:

((-1/120120*I)*a^2*Cos[e + f*x]*(5005*A + 780*(9*A + I*B)*Cos[2*(e + f*x)] 
 + 231*(9*A + (4*I)*B)*Cos[4*(e + f*x)] - (1560*I)*A*Sin[2*(e + f*x)] + 35 
10*B*Sin[2*(e + f*x)] - (924*I)*A*Sin[4*(e + f*x)] + 2079*B*Sin[4*(e + f*x 
)])*(Cos[9*e + 11*f*x] + I*Sin[9*e + 11*f*x])*Sqrt[a + I*a*Tan[e + f*x]]*S 
qrt[c - I*c*Tan[e + f*x]])/(c^7*f*(Cos[f*x] + I*Sin[f*x])^2)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {3042, 4071, 87, 55, 55, 55, 48}

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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}}dx\)

\(\Big \downarrow \) 4071

\(\displaystyle \frac {a c \int \frac {(i \tan (e+f x) a+a)^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{15/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 87

\(\displaystyle \frac {a c \left (\frac {(4 A+9 i B) \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c-i c \tan (e+f x))^{13/2}}d\tan (e+f x)}{13 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{13 a c (c-i c \tan (e+f x))^{13/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {(4 A+9 i B) \left (\frac {3 \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c-i c \tan (e+f x))^{11/2}}d\tan (e+f x)}{11 c}-\frac {i (a+i a \tan (e+f x))^{5/2}}{11 a c (c-i c \tan (e+f x))^{11/2}}\right )}{13 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{13 a c (c-i c \tan (e+f x))^{13/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {(4 A+9 i B) \left (\frac {3 \left (\frac {2 \int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c-i c \tan (e+f x))^{9/2}}d\tan (e+f x)}{9 c}-\frac {i (a+i a \tan (e+f x))^{5/2}}{9 a c (c-i c \tan (e+f x))^{9/2}}\right )}{11 c}-\frac {i (a+i a \tan (e+f x))^{5/2}}{11 a c (c-i c \tan (e+f x))^{11/2}}\right )}{13 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{13 a c (c-i c \tan (e+f x))^{13/2}}\right )}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {(4 A+9 i B) \left (\frac {3 \left (\frac {2 \left (\frac {\int \frac {(i \tan (e+f x) a+a)^{3/2}}{(c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{7 c}-\frac {i (a+i a \tan (e+f x))^{5/2}}{7 a c (c-i c \tan (e+f x))^{7/2}}\right )}{9 c}-\frac {i (a+i a \tan (e+f x))^{5/2}}{9 a c (c-i c \tan (e+f x))^{9/2}}\right )}{11 c}-\frac {i (a+i a \tan (e+f x))^{5/2}}{11 a c (c-i c \tan (e+f x))^{11/2}}\right )}{13 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{13 a c (c-i c \tan (e+f x))^{13/2}}\right )}{f}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {a c \left (\frac {(4 A+9 i B) \left (\frac {3 \left (\frac {2 \left (-\frac {i (a+i a \tan (e+f x))^{5/2}}{35 a c^2 (c-i c \tan (e+f x))^{5/2}}-\frac {i (a+i a \tan (e+f x))^{5/2}}{7 a c (c-i c \tan (e+f x))^{7/2}}\right )}{9 c}-\frac {i (a+i a \tan (e+f x))^{5/2}}{9 a c (c-i c \tan (e+f x))^{9/2}}\right )}{11 c}-\frac {i (a+i a \tan (e+f x))^{5/2}}{11 a c (c-i c \tan (e+f x))^{11/2}}\right )}{13 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{5/2}}{13 a c (c-i c \tan (e+f x))^{13/2}}\right )}{f}\)

Input:

Int[((a + I*a*Tan[e + f*x])^(5/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f 
*x])^(13/2),x]
 

Output:

(a*c*(-1/13*((I*A + B)*(a + I*a*Tan[e + f*x])^(5/2))/(a*c*(c - I*c*Tan[e + 
 f*x])^(13/2)) + ((4*A + (9*I)*B)*(((-1/11*I)*(a + I*a*Tan[e + f*x])^(5/2) 
)/(a*c*(c - I*c*Tan[e + f*x])^(11/2)) + (3*(((-1/9*I)*(a + I*a*Tan[e + f*x 
])^(5/2))/(a*c*(c - I*c*Tan[e + f*x])^(9/2)) + (2*(((-1/7*I)*(a + I*a*Tan[ 
e + f*x])^(5/2))/(a*c*(c - I*c*Tan[e + f*x])^(7/2)) - ((I/35)*(a + I*a*Tan 
[e + f*x])^(5/2))/(a*c^2*(c - I*c*Tan[e + f*x])^(5/2))))/(9*c)))/(11*c)))/ 
(13*c)))/f
 

Defintions of rubi rules used

rule 48
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp 
[(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ 
a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
 

rule 55
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

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 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]
 
Maple [A] (verified)

Time = 0.63 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.65

method result size
risch \(-\frac {a^{2} \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (1155 i A \,{\mathrm e}^{12 i \left (f x +e \right )}+1155 B \,{\mathrm e}^{12 i \left (f x +e \right )}+5460 i A \,{\mathrm e}^{10 i \left (f x +e \right )}+2730 B \,{\mathrm e}^{10 i \left (f x +e \right )}+10010 i A \,{\mathrm e}^{8 i \left (f x +e \right )}+8580 i A \,{\mathrm e}^{6 i \left (f x +e \right )}-4290 B \,{\mathrm e}^{6 i \left (f x +e \right )}+3003 i A \,{\mathrm e}^{4 i \left (f x +e \right )}-3003 B \,{\mathrm e}^{4 i \left (f x +e \right )}\right )}{240240 c^{6} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) \(169\)
derivativedivides \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (18 i B \tan \left (f x +e \right )^{5}+64 i A \tan \left (f x +e \right )^{4}+8 A \tan \left (f x +e \right )^{5}-531 i B \tan \left (f x +e \right )^{3}-144 B \tan \left (f x +e \right )^{4}-544 i A \tan \left (f x +e \right )^{2}-236 A \tan \left (f x +e \right )^{3}-1704 i B \tan \left (f x +e \right )+1224 B \tan \left (f x +e \right )^{2}-1763 i A +911 A \tan \left (f x +e \right )+213 B \right )}{15015 f \,c^{7} \left (i+\tan \left (f x +e \right )\right )^{8}}\) \(183\)
default \(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (18 i B \tan \left (f x +e \right )^{5}+64 i A \tan \left (f x +e \right )^{4}+8 A \tan \left (f x +e \right )^{5}-531 i B \tan \left (f x +e \right )^{3}-144 B \tan \left (f x +e \right )^{4}-544 i A \tan \left (f x +e \right )^{2}-236 A \tan \left (f x +e \right )^{3}-1704 i B \tan \left (f x +e \right )+1224 B \tan \left (f x +e \right )^{2}-1763 i A +911 A \tan \left (f x +e \right )+213 B \right )}{15015 f \,c^{7} \left (i+\tan \left (f x +e \right )\right )^{8}}\) \(183\)
parts \(-\frac {i A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (8 i \tan \left (f x +e \right )^{5}-236 i \tan \left (f x +e \right )^{3}-64 \tan \left (f x +e \right )^{4}+911 i \tan \left (f x +e \right )+544 \tan \left (f x +e \right )^{2}+1763\right )}{15015 f \,c^{7} \left (i+\tan \left (f x +e \right )\right )^{8}}+\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (48 i \tan \left (f x +e \right )^{4}+6 \tan \left (f x +e \right )^{5}-408 i \tan \left (f x +e \right )^{2}-177 \tan \left (f x +e \right )^{3}-71 i-568 \tan \left (f x +e \right )\right )}{5005 f \,c^{7} \left (i+\tan \left (f x +e \right )\right )^{8}}\) \(238\)

Input:

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x, 
method=_RETURNVERBOSE)
 

Output:

-1/240240*a^2/c^6*(a*exp(2*I*(f*x+e))/(exp(2*I*(f*x+e))+1))^(1/2)/(c/(exp( 
2*I*(f*x+e))+1))^(1/2)/f*(1155*I*A*exp(12*I*(f*x+e))+1155*B*exp(12*I*(f*x+ 
e))+5460*I*A*exp(10*I*(f*x+e))+2730*B*exp(10*I*(f*x+e))+10010*I*A*exp(8*I* 
(f*x+e))+8580*I*A*exp(6*I*(f*x+e))-4290*B*exp(6*I*(f*x+e))+3003*I*A*exp(4* 
I*(f*x+e))-3003*B*exp(4*I*(f*x+e)))
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.64 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx=-\frac {{\left (1155 \, {\left (i \, A + B\right )} a^{2} e^{\left (15 i \, f x + 15 i \, e\right )} + 105 \, {\left (63 i \, A + 37 \, B\right )} a^{2} e^{\left (13 i \, f x + 13 i \, e\right )} + 910 \, {\left (17 i \, A + 3 \, B\right )} a^{2} e^{\left (11 i \, f x + 11 i \, e\right )} + 1430 \, {\left (13 i \, A - 3 \, B\right )} a^{2} e^{\left (9 i \, f x + 9 i \, e\right )} + 429 \, {\left (27 i \, A - 17 \, B\right )} a^{2} e^{\left (7 i \, f x + 7 i \, e\right )} + 3003 \, {\left (i \, A - B\right )} a^{2} e^{\left (5 i \, f x + 5 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{240240 \, c^{7} f} \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13 
/2),x, algorithm="fricas")
 

Output:

-1/240240*(1155*(I*A + B)*a^2*e^(15*I*f*x + 15*I*e) + 105*(63*I*A + 37*B)* 
a^2*e^(13*I*f*x + 13*I*e) + 910*(17*I*A + 3*B)*a^2*e^(11*I*f*x + 11*I*e) + 
 1430*(13*I*A - 3*B)*a^2*e^(9*I*f*x + 9*I*e) + 429*(27*I*A - 17*B)*a^2*e^( 
7*I*f*x + 7*I*e) + 3003*(I*A - B)*a^2*e^(5*I*f*x + 5*I*e))*sqrt(a/(e^(2*I* 
f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(c^7*f)
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))**(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**( 
13/2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.27 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx=\frac {{\left (1155 \, {\left (-i \, A - B\right )} a^{2} \cos \left (\frac {13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 2730 \, {\left (-2 i \, A - B\right )} a^{2} \cos \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 10010 i \, A a^{2} \cos \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 4290 \, {\left (-2 i \, A + B\right )} a^{2} \cos \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3003 \, {\left (-i \, A + B\right )} a^{2} \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1155 \, {\left (A - i \, B\right )} a^{2} \sin \left (\frac {13}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 2730 \, {\left (2 \, A - i \, B\right )} a^{2} \sin \left (\frac {11}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 10010 \, A a^{2} \sin \left (\frac {9}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 4290 \, {\left (2 \, A + i \, B\right )} a^{2} \sin \left (\frac {7}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3003 \, {\left (A + i \, B\right )} a^{2} \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{240240 \, c^{\frac {13}{2}} f} \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13 
/2),x, algorithm="maxima")
 

Output:

1/240240*(1155*(-I*A - B)*a^2*cos(13/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x 
 + 2*e))) + 2730*(-2*I*A - B)*a^2*cos(11/2*arctan2(sin(2*f*x + 2*e), cos(2 
*f*x + 2*e))) - 10010*I*A*a^2*cos(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x 
+ 2*e))) + 4290*(-2*I*A + B)*a^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f 
*x + 2*e))) + 3003*(-I*A + B)*a^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2* 
f*x + 2*e))) + 1155*(A - I*B)*a^2*sin(13/2*arctan2(sin(2*f*x + 2*e), cos(2 
*f*x + 2*e))) + 2730*(2*A - I*B)*a^2*sin(11/2*arctan2(sin(2*f*x + 2*e), co 
s(2*f*x + 2*e))) + 10010*A*a^2*sin(9/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x 
 + 2*e))) + 4290*(2*A + I*B)*a^2*sin(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f 
*x + 2*e))) + 3003*(A + I*B)*a^2*sin(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f 
*x + 2*e))))*sqrt(a)/(c^(13/2)*f)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13 
/2),x, algorithm="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 TypeDone
 

Mupad [B] (verification not implemented)

Time = 9.97 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.73 \[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx=-\frac {\sqrt {a+\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}\,\left (\frac {a^2\,{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,\left (2\,A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{56\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,10{}\mathrm {i}+f\,x\,10{}\mathrm {i}}\,\left (2\,A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{88\,c^6\,f}+\frac {A\,a^2\,{\mathrm {e}}^{e\,8{}\mathrm {i}+f\,x\,8{}\mathrm {i}}\,1{}\mathrm {i}}{24\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (A+B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{80\,c^6\,f}+\frac {a^2\,{\mathrm {e}}^{e\,12{}\mathrm {i}+f\,x\,12{}\mathrm {i}}\,\left (A-B\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{208\,c^6\,f}\right )}{\sqrt {c-\frac {c\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{\cos \left (e+f\,x\right )}}} \] Input:

int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^(5/2))/(c - c*tan(e + f* 
x)*1i)^(13/2),x)
 

Output:

-((a + (a*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)*((a^2*exp(e*6i + f*x*6i)*(2 
*A + B*1i)*1i)/(56*c^6*f) + (a^2*exp(e*10i + f*x*10i)*(2*A - B*1i)*1i)/(88 
*c^6*f) + (A*a^2*exp(e*8i + f*x*8i)*1i)/(24*c^6*f) + (a^2*exp(e*4i + f*x*4 
i)*(A + B*1i)*1i)/(80*c^6*f) + (a^2*exp(e*12i + f*x*12i)*(A - B*1i)*1i)/(2 
08*c^6*f)))/(c - (c*sin(e + f*x)*1i)/cos(e + f*x))^(1/2)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^{5/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{13/2}} \, dx =\text {Too large to display} \] Input:

int((a+I*a*tan(f*x+e))^(5/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(13/2),x)
 

Output:

(sqrt(c)*sqrt(a)*a**2*( - int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + 
 f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**8 + 6*tan(e + f*x)**7*i - 14* 
tan(e + f*x)**6 - 14*tan(e + f*x)**5*i - 14*tan(e + f*x)**3*i + 14*tan(e + 
 f*x)**2 + 6*tan(e + f*x)*i - 1),x)*a*i - 3*int(( - sqrt(tan(e + f*x)*i + 
1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/(tan(e + f*x)**8 + 6*tan(e 
 + f*x)**7*i - 14*tan(e + f*x)**6 - 14*tan(e + f*x)**5*i - 14*tan(e + f*x) 
**3*i + 14*tan(e + f*x)**2 + 6*tan(e + f*x)*i - 1),x)*b + 3*int(( - sqrt(t 
an(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x) 
**8 + 6*tan(e + f*x)**7*i - 14*tan(e + f*x)**6 - 14*tan(e + f*x)**5*i - 14 
*tan(e + f*x)**3*i + 14*tan(e + f*x)**2 + 6*tan(e + f*x)*i - 1),x)*a*i + i 
nt(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/ 
(tan(e + f*x)**8 + 6*tan(e + f*x)**7*i - 14*tan(e + f*x)**6 - 14*tan(e + f 
*x)**5*i - 14*tan(e + f*x)**3*i + 14*tan(e + f*x)**2 + 6*tan(e + f*x)*i - 
1),x)*b + int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(t 
an(e + f*x)**8 + 6*tan(e + f*x)**7*i - 14*tan(e + f*x)**6 - 14*tan(e + f*x 
)**5*i - 14*tan(e + f*x)**3*i + 14*tan(e + f*x)**2 + 6*tan(e + f*x)*i - 1) 
,x)*a + int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + 
f*x)**4)/(tan(e + f*x)**8 + 6*tan(e + f*x)**7*i - 14*tan(e + f*x)**6 - 14* 
tan(e + f*x)**5*i - 14*tan(e + f*x)**3*i + 14*tan(e + f*x)**2 + 6*tan(e + 
f*x)*i - 1),x)*b*i + 3*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*...