\(\int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx\) [1025]

Optimal result

Integrand size = 35, antiderivative size = 204 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {10 i a^{7/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2} f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{3 f (c-i c \tan (e+f x))^{3/2}}+\frac {10 i a^2 (a+i a \tan (e+f x))^{3/2}}{3 c f \sqrt {c-i c \tan (e+f x)}}+\frac {5 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c^2 f} \] Output:

-10*I*a^(7/2)*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))/c^(3/2)/f-2/3*I*a*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+ 
e))^(3/2)+10/3*I*a^2*(a+I*a*tan(f*x+e))^(3/2)/c/f/(c-I*c*tan(f*x+e))^(1/2) 
+5*I*a^3*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c^2/f
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 4.00 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.26 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {-3 i \sqrt {2} a^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5}{2},\frac {7}{2},\frac {1}{2} (1+i \tan (e+f x))\right ) \sec ^4(e+f x) (-i+\tan (e+f x))+40 a^{7/2} \sec ^2(e+f x) \left (\sqrt {a} (-3+\cos (2 (e+f x))+i \sin (2 (e+f x))) \sqrt {1-i \tan (e+f x)}+3 \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) (\cos (2 (e+f x))-i \sin (2 (e+f x))) \sqrt {a+i a \tan (e+f x)}\right )}{30 c f \sqrt {1-i \tan (e+f x)} (i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:

Integrate[(a + I*a*Tan[e + f*x])^(7/2)/(c - I*c*Tan[e + f*x])^(3/2),x]
 

Output:

((-3*I)*Sqrt[2]*a^4*Hypergeometric2F1[3/2, 5/2, 7/2, (1 + I*Tan[e + f*x])/ 
2]*Sec[e + f*x]^4*(-I + Tan[e + f*x]) + 40*a^(7/2)*Sec[e + f*x]^2*(Sqrt[a] 
*(-3 + Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*Sqrt[1 - I*Tan[e + f*x]] + 3 
*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])]*(Cos[2*(e + f*x)] - 
I*Sin[2*(e + f*x)])*Sqrt[a + I*a*Tan[e + f*x]]))/(30*c*f*Sqrt[1 - I*Tan[e 
+ f*x]]*(I + Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + 
 f*x]])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4006, 57, 57, 60, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}}dx\)

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

\(\displaystyle \frac {a c \left (-\frac {5 a \left (-\frac {3 a \left (a \int \frac {1}{\sqrt {i \tan (e+f x) a+a} \sqrt {c-i c \tan (e+f x)}}d\tan (e+f x)+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\)

\(\Big \downarrow \) 45

\(\displaystyle \frac {a c \left (-\frac {5 a \left (-\frac {3 a \left (2 a \int \frac {1}{i a+\frac {i c (i \tan (e+f x) a+a)}{c-i c \tan (e+f x)}}d\frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c-i c \tan (e+f x)}}+\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {a c \left (-\frac {5 a \left (-\frac {3 a \left (\frac {i \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{c}-\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c}}\right )}{c}-\frac {2 i (a+i a \tan (e+f x))^{3/2}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{3 c}-\frac {2 i (a+i a \tan (e+f x))^{5/2}}{3 c (c-i c \tan (e+f x))^{3/2}}\right )}{f}\)

Input:

Int[(a + I*a*Tan[e + f*x])^(7/2)/(c - I*c*Tan[e + f*x])^(3/2),x]
 

Output:

(a*c*((((-2*I)/3)*(a + I*a*Tan[e + f*x])^(5/2))/(c*(c - I*c*Tan[e + f*x])^ 
(3/2)) - (5*a*(((-2*I)*(a + I*a*Tan[e + f*x])^(3/2))/(c*Sqrt[c - I*c*Tan[e 
 + f*x]]) - (3*a*(((-2*I)*Sqrt[a]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x 
]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])])/Sqrt[c] + (I*Sqrt[a + I*a*Tan[e 
 + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/c))/c))/(3*c)))/f
 

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, 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 4006

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 378 vs. \(2 (164 ) = 328\).

Time = 0.35 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.86

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (45 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+3 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{3}+15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}-15 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -57 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-45 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-37 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+23 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{3 f \,c^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{3} \sqrt {a c}}\)	\(379\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (45 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}+3 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{3}+15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{3}-15 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -57 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-45 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )-37 \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{2}+23 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{3 f \,c^{2} \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{3} \sqrt {a c}}\)	\(379\)

Input:

int((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/3/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c^2*(45*I 
*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2)) 
*a*c*tan(f*x+e)^2+3*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^ 
3+15*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1 
/2))*a*c*tan(f*x+e)^3-15*I*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2) 
*(a*c)^(1/2))/(a*c)^(1/2))*a*c-57*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/ 
2)*tan(f*x+e)-45*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/ 
2))/(a*c)^(1/2))*a*c*tan(f*x+e)-37*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2 
)*tan(f*x+e)^2+23*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f* 
x+e)^2))^(1/2)/(I+tan(f*x+e))^3/(a*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. 368 vs. \(2 (152) = 304\).

Time = 0.12 (sec) , antiderivative size = 368, normalized size of antiderivative = 1.80 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {15 \, \sqrt {\frac {a^{7}}{c^{3} f^{2}}} c^{2} f \log \left (\frac {4 \, {\left (2 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c^{2} f\right )} \sqrt {\frac {a^{7}}{c^{3} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, \sqrt {\frac {a^{7}}{c^{3} f^{2}}} c^{2} f \log \left (\frac {4 \, {\left (2 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (-i \, c^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{2} f\right )} \sqrt {\frac {a^{7}}{c^{3} f^{2}}}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 4 \, {\left (2 i \, a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} - 10 i \, a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} - 15 i \, a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{6 \, c^{2} f} \] Input:

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" 
fricas")
 

Output:

1/6*(15*sqrt(a^7/(c^3*f^2))*c^2*f*log(4*(2*(a^3*e^(3*I*f*x + 3*I*e) + a^3* 
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)) - (I*c^2*f*e^(2*I*f*x + 2*I*e) - I*c^2*f)*sqrt(a^7/(c^3*f^2)))/ 
(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - 15*sqrt(a^7/(c^3*f^2))*c^2*f*log(4*(2*( 
a^3*e^(3*I*f*x + 3*I*e) + a^3*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)) - (-I*c^2*f*e^(2*I*f*x + 2*I*e) + 
 I*c^2*f)*sqrt(a^7/(c^3*f^2)))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - 4*(2*I*a 
^3*e^(5*I*f*x + 5*I*e) - 10*I*a^3*e^(3*I*f*x + 3*I*e) - 15*I*a^3*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) 
))/(c^2*f)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))**(7/2)/(c-I*c*tan(f*x+e))**(3/2),x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 660 vs. \(2 (152) = 304\).

Time = 0.21 (sec) , antiderivative size = 660, normalized size of antiderivative = 3.24 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" 
maxima")
 

Output:

-3*(30*(a^3*cos(2*f*x + 2*e) + I*a^3*sin(2*f*x + 2*e) + a^3)*arctan2(cos(1 
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), sin(1/2*arctan2(sin(2*f*x 
 + 2*e), cos(2*f*x + 2*e))) + 1) + 30*(a^3*cos(2*f*x + 2*e) + I*a^3*sin(2* 
f*x + 2*e) + a^3)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2* 
e))), -sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 8*(a^3* 
cos(2*f*x + 2*e) + I*a^3*sin(2*f*x + 2*e) + a^3)*cos(3/2*arctan2(sin(2*f*x 
 + 2*e), cos(2*f*x + 2*e))) - 12*(4*a^3*cos(2*f*x + 2*e) + 4*I*a^3*sin(2*f 
*x + 2*e) + 5*a^3)*cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 
15*(I*a^3*cos(2*f*x + 2*e) - a^3*sin(2*f*x + 2*e) + I*a^3)*log(cos(1/2*arc 
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 
2*e), cos(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x 
 + 2*e))) + 1) + 15*(-I*a^3*cos(2*f*x + 2*e) + a^3*sin(2*f*x + 2*e) - I*a^ 
3)*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*ar 
ctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x 
 + 2*e), cos(2*f*x + 2*e))) + 1) + 8*(I*a^3*cos(2*f*x + 2*e) - a^3*sin(2*f 
*x + 2*e) + I*a^3)*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 
12*(-4*I*a^3*cos(2*f*x + 2*e) + 4*a^3*sin(2*f*x + 2*e) - 5*I*a^3)*sin(1/2* 
arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/((-18*I*c^2* 
cos(2*f*x + 2*e) + 18*c^2*sin(2*f*x + 2*e) - 18*I*c^2)*f)
 

Giac [F]

\[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" 
giac")
 

Output:

integrate((I*a*tan(f*x + e) + a)^(7/2)/(-I*c*tan(f*x + e) + c)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}{{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:

int((a + a*tan(e + f*x)*1i)^(7/2)/(c - c*tan(e + f*x)*1i)^(3/2),x)
                                                                                    
                                                                                    
 

Output:

int((a + a*tan(e + f*x)*1i)^(7/2)/(c - c*tan(e + f*x)*1i)^(3/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

int((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(3/2),x)
 

Output:

(sqrt(c)*sqrt(a)*a**3*( - 10*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x) 
*i + 1)*tan(e + f*x) + 6*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 
 1)*i - 4*int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(t 
an(e + f*x)**3 + tan(e + f*x)**2*i + tan(e + f*x) + i),x)*tan(e + f*x)**2* 
f*i - 4*int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan 
(e + f*x)**3 + tan(e + f*x)**2*i + tan(e + f*x) + i),x)*f*i - int((sqrt(ta 
n(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(tan(e + f* 
x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*f - int 
((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**4)/(t 
an(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*f + 4*int((sqr 
t(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**3)/(tan(e 
+ f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*f*i 
 + 4*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x 
)**3)/(tan(e + f*x)**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*f*i - 
7*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x) 
**3*i - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*f - 7*int 
((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**3*i 
 - tan(e + f*x)**2 + tan(e + f*x)*i - 1),x)*f))/(c**2*f*(tan(e + f*x)**2 + 
 1))