\(\int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [1016]

Optimal result

Integrand size = 35, antiderivative size = 204 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {15 i a^{7/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a (a+i a \tan (e+f x))^{5/2}}{f \sqrt {c-i c \tan (e+f x)}}-\frac {15 i a^3 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 c f}-\frac {5 i a^2 (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}{2 c f} \] Output:

15*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^(1/2)/f-2*I*a*(a+I*a*tan(f*x+e))^(5/2)/f/(c-I*c*tan(f*x+e)) 
^(1/2)-15/2*I*a^3*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(1/2)/c/f-5/ 
2*I*a^2*(a+I*a*tan(f*x+e))^(3/2)*(c-I*c*tan(f*x+e))^(1/2)/c/f
 

Mathematica [A] (verified)

Time = 3.36 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.01 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {a^{7/2} \left (\frac {16 i \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c}}+\frac {14 i \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {1-i \tan (e+f x)}}{\sqrt {c-i c \tan (e+f x)}}+\frac {\sqrt {a} \left (-24 i+17 \tan (e+f x)-8 i \tan ^2(e+f x)+\tan ^3(e+f x)\right )}{\sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}\right )}{2 f} \] Input:

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

Output:

(a^(7/2)*(((16*I)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqr 
t[c - I*c*Tan[e + f*x]])])/Sqrt[c] + ((14*I)*ArcSin[Sqrt[a + I*a*Tan[e + f 
*x]]/(Sqrt[2]*Sqrt[a])]*Sqrt[1 - I*Tan[e + f*x]])/Sqrt[c - I*c*Tan[e + f*x 
]] + (Sqrt[a]*(-24*I + 17*Tan[e + f*x] - (8*I)*Tan[e + f*x]^2 + Tan[e + f* 
x]^3))/(Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])))/(2*f)
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.00, 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, 60, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

Time = 0.36 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.61

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (30 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+6 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}+15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3} \sqrt {a c}-24 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -31 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{2 f c \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(328\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{3} \left (30 i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+6 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}+15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3} \sqrt {a c}-24 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-15 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -31 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{2 f c \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \left (i+\tan \left (f x +e \right )\right )^{2}}\)	\(328\)

Input:

int((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

-1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^3/c*(30*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)+6*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2+1 
5*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-(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3*(a*c)^(1/2)-2 
4*I*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-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-31*tan(f*x+e)*(a*c*(1+ 
tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+e)^2))^(1/2)/(a*c)^(1/2) 
/(I+tan(f*x+e))^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. 396 vs. \(2 (152) = 304\).

Time = 0.13 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.94 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {15 \, \sqrt {\frac {a^{7}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{7}}{c f^{2}}} {\left (i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c f\right )}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) - 15 \, \sqrt {\frac {a^{7}}{c f^{2}}} {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{7}}{c f^{2}}} {\left (-i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c f\right )}\right )}}{a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{3}}\right ) + 4 \, {\left (8 i \, a^{3} e^{\left (5 i \, f x + 5 i \, e\right )} + 25 i \, a^{3} e^{\left (3 i \, f x + 3 i \, e\right )} + 15 i \, a^{3} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (c f e^{\left (2 i \, f x + 2 i \, e\right )} + c f\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm=" 
fricas")
 

Output:

-1/4*(15*sqrt(a^7/(c*f^2))*(c*f*e^(2*I*f*x + 2*I*e) + c*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)) - sqrt(a^7/(c*f^2))*(I*c*f*e^(2*I*f*x 
+ 2*I*e) - I*c*f))/(a^3*e^(2*I*f*x + 2*I*e) + a^3)) - 15*sqrt(a^7/(c*f^2)) 
*(c*f*e^(2*I*f*x + 2*I*e) + c*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)) - sqrt(a^7/(c*f^2))*(-I*c*f*e^(2*I*f*x + 2*I*e) + I*c*f))/(a^3*e 
^(2*I*f*x + 2*I*e) + a^3)) + 4*(8*I*a^3*e^(5*I*f*x + 5*I*e) + 25*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*f*e^(2*I*f*x + 2*I*e) + c*f)
 

Sympy [F(-1)]

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

integrate((a+I*a*tan(f*x+e))**(7/2)/(c-I*c*tan(f*x+e))**(1/2),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. 787 vs. \(2 (152) = 304\).

Time = 0.22 (sec) , antiderivative size = 787, normalized size of antiderivative = 3.86 \[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm=" 
maxima")
 

Output:

-2*(36*a^3*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 36*I*a^3 
*sin(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 30*(a^3*cos(4*f*x 
+ 4*e) + 2*a^3*cos(2*f*x + 2*e) + I*a^3*sin(4*f*x + 4*e) + 2*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*c 
os(4*f*x + 4*e) + 2*a^3*cos(2*f*x + 2*e) + I*a^3*sin(4*f*x + 4*e) + 2*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) + 
 4*(8*a^3*cos(4*f*x + 4*e) + 16*a^3*cos(2*f*x + 2*e) + 8*I*a^3*sin(4*f*x + 
 4*e) + 16*I*a^3*sin(2*f*x + 2*e) + 15*a^3)*cos(1/2*arctan2(sin(2*f*x + 2* 
e), cos(2*f*x + 2*e))) - 15*(I*a^3*cos(4*f*x + 4*e) + 2*I*a^3*cos(2*f*x + 
2*e) - a^3*sin(4*f*x + 4*e) - 2*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*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(4*f*x + 4*e) - 2*I*a^3*cos(2*f*x + 2*e) 
 + a^3*sin(4*f*x + 4*e) + 2*a^3*sin(2*f*x + 2*e) - I*a^3)*log(cos(1/2*arct 
an2(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) - 4*(-8*I*a^3*cos(4*f*x + 4*e) - 16*I*a^3*cos(2*f*x + 2*e) + 
 8*a^3*sin(4*f*x + 4*e) + 16*a^3*sin(2*f*x + 2*e) - 15*I*a^3)*sin(1/2*a...
 

Giac [F]

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

integrate((a+I*a*tan(f*x+e))^(7/2)/(c-I*c*tan(f*x+e))^(1/2),x, algorithm=" 
giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^{7/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{3} \left (-4 \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2} i +\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )-12 \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, i +\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f +\left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{2}+1}d x \right ) f -6 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) \tan \left (f x +e \right )^{2} f -6 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{2}+1}d x \right ) f \right )}{c f \left (\tan \left (f x +e \right )^{2}+1\right )} \] Input:

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

Output:

(sqrt(c)*sqrt(a)*a**3*( - 4*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)* 
i + 1)*tan(e + f*x)**2*i + sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i 
 + 1)*tan(e + f*x) - 12*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 
1)*i + int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f 
*x)**4)/(tan(e + f*x)**2 + 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)/(tan(e + f*x)**2 + 1 
),x)*f - 6*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e 
 + f*x)**2)/(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f - 6*int((sqrt(tan(e 
 + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)* 
*2 + 1),x)*f))/(c*f*(tan(e + f*x)**2 + 1))