\(\int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{7/3}} \, dx\) [443]

Optimal result

Integrand size = 30, antiderivative size = 437 \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{7/3}} \, dx=\frac {i (d \sec (e+f x))^{2/3}}{4 f (a+i a \tan (e+f x))^{7/3}}-\frac {5 x (d \sec (e+f x))^{2/3}}{72\ 2^{2/3} a^{5/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {5 i \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt {3} \sqrt [3]{a}}\right ) (d \sec (e+f x))^{2/3}}{12\ 2^{2/3} \sqrt {3} a^{5/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {5 i \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{72\ 2^{2/3} a^{5/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {5 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right ) (d \sec (e+f x))^{2/3}}{24\ 2^{2/3} a^{5/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {5 i (d \sec (e+f x))^{2/3}}{24 f \sqrt [3]{a+i a \tan (e+f x)} \left (a^2+i a^2 \tan (e+f x)\right )} \] Output:

1/4*I*(d*sec(f*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(7/3)-5/144*x*(d*sec(f*x+e 
))^(2/3)*2^(1/3)/a^(5/3)/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3) 
+5/72*I*arctan(1/3*(a^(1/3)+2^(2/3)*(a-I*a*tan(f*x+e))^(1/3))*3^(1/2)/a^(1 
/3))*(d*sec(f*x+e))^(2/3)*2^(1/3)*3^(1/2)/a^(5/3)/f/(a-I*a*tan(f*x+e))^(1/ 
3)/(a+I*a*tan(f*x+e))^(1/3)-5/144*I*ln(cos(f*x+e))*(d*sec(f*x+e))^(2/3)*2^ 
(1/3)/a^(5/3)/f/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)-5/48*I*l 
n(2^(1/3)*a^(1/3)-(a-I*a*tan(f*x+e))^(1/3))*(d*sec(f*x+e))^(2/3)*2^(1/3)/a 
^(5/3)/f/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)+5/24*I*(d*sec(f 
*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(1/3)/(a^2+I*a^2*tan(f*x+e))
 

Mathematica [A] (verified)

Time = 2.34 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.55 \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{7/3}} \, dx=\frac {e^{-2 i (e+f x)} \left (9 i+33 i e^{2 i (e+f x)}+24 i e^{4 i (e+f x)}-10 e^{4 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} f x-10 i \sqrt {3} e^{4 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \arctan \left (\frac {1+2 \sqrt [3]{1+e^{2 i (e+f x)}}}{\sqrt {3}}\right )-15 i e^{4 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \log \left (1-\sqrt [3]{1+e^{2 i (e+f x)}}\right )\right ) \sec ^2(e+f x) (d \sec (e+f x))^{2/3}}{144 f (a+i a \tan (e+f x))^{7/3}} \] Input:

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

Output:

((9*I + (33*I)*E^((2*I)*(e + f*x)) + (24*I)*E^((4*I)*(e + f*x)) - 10*E^((4 
*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*f*x - (10*I)*Sqrt[3]*E^((4* 
I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*ArcTan[(1 + 2*(1 + E^((2*I)* 
(e + f*x)))^(1/3))/Sqrt[3]] - (15*I)*E^((4*I)*(e + f*x))*(1 + E^((2*I)*(e 
+ f*x)))^(1/3)*Log[1 - (1 + E^((2*I)*(e + f*x)))^(1/3)])*Sec[e + f*x]^2*(d 
*Sec[e + f*x])^(2/3))/(144*E^((2*I)*(e + f*x))*f*(a + I*a*Tan[e + f*x])^(7 
/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.71 (sec) , antiderivative size = 248, normalized size of antiderivative = 0.57, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3986, 3042, 4005, 3042, 3968, 52, 52, 69, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

\(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \frac {\sqrt [3]{a-i a \tan (e+f x)}}{(i \tan (e+f x) a+a)^2}dx}{\sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \frac {\sqrt [3]{a-i a \tan (e+f x)}}{(i \tan (e+f x) a+a)^2}dx}{\sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 4005

\(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \cos ^4(e+f x) (a-i a \tan (e+f x))^{7/3}dx}{a^4 \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \frac {(a-i a \tan (e+f x))^{7/3}}{\sec (e+f x)^4}dx}{a^4 \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 3968

\(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \int \frac {1}{(a-i a \tan (e+f x))^{2/3} (i \tan (e+f x) a+a)^3}d(-i a \tan (e+f x))}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {5 \int \frac {1}{(a-i a \tan (e+f x))^{2/3} (i \tan (e+f x) a+a)^2}d(-i a \tan (e+f x))}{12 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{4 a (a+i a \tan (e+f x))^2}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 52

\(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {5 \left (\frac {\int \frac {1}{(a-i a \tan (e+f x))^{2/3} (i \tan (e+f x) a+a)}d(-i a \tan (e+f x))}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{12 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{4 a (a+i a \tan (e+f x))^2}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 69

\(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {5 \left (\frac {\frac {3 \int \frac {1}{i a \tan (e+f x)+\sqrt [3]{2} \sqrt [3]{a}}d\sqrt [3]{a-i a \tan (e+f x)}}{2\ 2^{2/3} a^{2/3}}+\frac {3 \int \frac {1}{-a^2 \tan ^2(e+f x)-i \sqrt [3]{2} a^{4/3} \tan (e+f x)+2^{2/3} a^{2/3}}d\sqrt [3]{a-i a \tan (e+f x)}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{12 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{4 a (a+i a \tan (e+f x))^2}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {5 \left (\frac {\frac {3 \int \frac {1}{-a^2 \tan ^2(e+f x)-i \sqrt [3]{2} a^{4/3} \tan (e+f x)+2^{2/3} a^{2/3}}d\sqrt [3]{a-i a \tan (e+f x)}}{2 \sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{12 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{4 a (a+i a \tan (e+f x))^2}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {5 \left (\frac {-\frac {3 \int \frac {1}{a^2 \tan ^2(e+f x)-3}d\left (1-i 2^{2/3} a^{2/3} \tan (e+f x)\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{12 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{4 a (a+i a \tan (e+f x))^2}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {5 \left (\frac {-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (e+f x)}{\sqrt {3}}\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{12 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{4 a (a+i a \tan (e+f x))^2}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\)

Input:

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

Output:

(I*a*(d*Sec[e + f*x])^(2/3)*((a - I*a*Tan[e + f*x])^(1/3)/(4*a*(a + I*a*Ta 
n[e + f*x])^2) + (5*((((-I)*Sqrt[3]*ArcTanh[(a*Tan[e + f*x])/Sqrt[3]])/(2^ 
(2/3)*a^(2/3)) - (3*Log[2^(1/3)*a^(1/3) + I*a*Tan[e + f*x]])/(2*2^(2/3)*a^ 
(2/3)) + Log[a + I*a*Tan[e + f*x]]/(2*2^(2/3)*a^(2/3)))/(3*a) + (a - I*a*T 
an[e + f*x])^(1/3)/(2*a*(a + I*a*Tan[e + f*x]))))/(12*a)))/(f*(a - I*a*Tan 
[e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 52

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*(( 
m + n + 2)/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], 
x] /; FreeQ[{a, b, c, d, n}, x] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
 

rule 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], 
 x])] /; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3968

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

rule 3986

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

rule 4005

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

Maple [F]

\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}}}{\left (a +i a \tan \left (f x +e \right )\right )^{\frac {7}{3}}}d x\]

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 528, normalized size of antiderivative = 1.21 \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{7/3}} \, dx =\text {Too large to display} \] Input:

integrate((d*sec(f*x+e))^(2/3)/(a+I*a*tan(f*x+e))^(7/3),x, algorithm="fric 
as")
 

Output:

1/48*(48*a^3*f*(125/186624*I*d^2/(a^7*f^3))^(1/3)*e^(6*I*f*x + 6*I*e)*log( 
-2/5*(72*I*a^3*f*(125/186624*I*d^2/(a^7*f^3))^(1/3)*e^(2*I*f*x + 2*I*e) - 
5*2^(1/3)*(a/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1) 
)^(2/3)*(e^(2*I*f*x + 2*I*e) + 1)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e 
)) + 2^(1/3)*(a/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 
 1))^(2/3)*(8*I*e^(6*I*f*x + 6*I*e) + 19*I*e^(4*I*f*x + 4*I*e) + 14*I*e^(2 
*I*f*x + 2*I*e) + 3*I)*e^(2*I*f*x + 2*I*e) - 24*(-I*sqrt(3)*a^3*f + a^3*f) 
*(125/186624*I*d^2/(a^7*f^3))^(1/3)*e^(6*I*f*x + 6*I*e)*log(2/5*(5*2^(1/3) 
*(a/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*( 
e^(2*I*f*x + 2*I*e) + 1)*e^(2*I*f*x + 2*I*e) + 36*(sqrt(3)*a^3*f + I*a^3*f 
)*(125/186624*I*d^2/(a^7*f^3))^(1/3)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2* 
I*e)) - 24*(I*sqrt(3)*a^3*f + a^3*f)*(125/186624*I*d^2/(a^7*f^3))^(1/3)*e^ 
(6*I*f*x + 6*I*e)*log(2/5*(5*2^(1/3)*(a/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*( 
d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(e^(2*I*f*x + 2*I*e) + 1)*e^(2*I*f*x + 
2*I*e) - 36*(sqrt(3)*a^3*f - I*a^3*f)*(125/186624*I*d^2/(a^7*f^3))^(1/3)*e 
^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)))*e^(-6*I*f*x - 6*I*e)/(a^3*f)
 

Sympy [F(-1)]

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

integrate((d*sec(f*x+e))**(2/3)/(a+I*a*tan(f*x+e))**(7/3),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. 3902 vs. \(2 (324) = 648\).

Time = 0.43 (sec) , antiderivative size = 3902, normalized size of antiderivative = 8.93 \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{7/3}} \, dx=\text {Too large to display} \] Input:

integrate((d*sec(f*x+e))^(2/3)/(a+I*a*tan(f*x+e))^(7/3),x, algorithm="maxi 
ma")
 

Output:

1/288*(48*(cos(1/2*arctan2(sin(4*f*x + 4*e), cos(4*f*x + 4*e)))^2 + sin(1/ 
2*arctan2(sin(4*f*x + 4*e), cos(4*f*x + 4*e)))^2 + 2*cos(1/2*arctan2(sin(4 
*f*x + 4*e), cos(4*f*x + 4*e))) + 1)^(5/6)*((I*2^(1/3)*cos(4*f*x + 4*e) + 
2^(1/3)*sin(4*f*x + 4*e))*cos(5/3*arctan2(sin(1/2*arctan2(sin(4*f*x + 4*e) 
, cos(4*f*x + 4*e))), cos(1/2*arctan2(sin(4*f*x + 4*e), cos(4*f*x + 4*e))) 
 + 1)) - (2^(1/3)*cos(4*f*x + 4*e) - I*2^(1/3)*sin(4*f*x + 4*e))*sin(5/3*a 
rctan2(sin(1/2*arctan2(sin(4*f*x + 4*e), cos(4*f*x + 4*e))), cos(1/2*arcta 
n2(sin(4*f*x + 4*e), cos(4*f*x + 4*e))) + 1)))*d^(2/3) + 30*(cos(1/2*arcta 
n2(sin(4*f*x + 4*e), cos(4*f*x + 4*e)))^2 + sin(1/2*arctan2(sin(4*f*x + 4* 
e), cos(4*f*x + 4*e)))^2 + 2*cos(1/2*arctan2(sin(4*f*x + 4*e), cos(4*f*x + 
 4*e))) + 1)^(1/3)*((-I*2^(1/3)*cos(4*f*x + 4*e) - 2^(1/3)*sin(4*f*x + 4*e 
))*cos(2/3*arctan2(sin(1/2*arctan2(sin(4*f*x + 4*e), cos(4*f*x + 4*e))), c 
os(1/2*arctan2(sin(4*f*x + 4*e), cos(4*f*x + 4*e))) + 1)) + (2^(1/3)*cos(4 
*f*x + 4*e) - I*2^(1/3)*sin(4*f*x + 4*e))*sin(2/3*arctan2(sin(1/2*arctan2( 
sin(4*f*x + 4*e), cos(4*f*x + 4*e))), cos(1/2*arctan2(sin(4*f*x + 4*e), co 
s(4*f*x + 4*e))) + 1)))*d^(2/3) + 5*(-2*I*sqrt(3)*2^(1/3)*arctan2(2/3*sqrt 
(3)*(cos(1/2*arctan2(sin(4*f*x + 4*e), cos(4*f*x + 4*e)))^2 + sin(1/2*arct 
an2(sin(4*f*x + 4*e), cos(4*f*x + 4*e)))^2 + 2*cos(1/2*arctan2(sin(4*f*x + 
 4*e), cos(4*f*x + 4*e))) + 1)^(1/6)*cos(1/3*arctan2(sin(1/2*arctan2(sin(4 
*f*x + 4*e), cos(4*f*x + 4*e))), cos(1/2*arctan2(sin(4*f*x + 4*e), cos(...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - d**(2/3)*int(sec(e + f*x)**(2/3)/((tan(e + f*x)*i + 1)**(1/3)*tan(e + 
f*x)**2 - 2*(tan(e + f*x)*i + 1)**(1/3)*tan(e + f*x)*i - (tan(e + f*x)*i + 
 1)**(1/3)),x))/(a**(1/3)*a**2)