Integrand size = 30, antiderivative size = 378 \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx=\frac {i (d \sec (e+f x))^{2/3}}{2 f (a+i a \tan (e+f x))^{4/3}}-\frac {x (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}+\frac {i \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a-i a \tan (e+f x)}}{\sqrt {3} \sqrt [3]{a}}\right ) (d \sec (e+f x))^{2/3}}{2^{2/3} \sqrt {3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log (\cos (e+f x)) (d \sec (e+f x))^{2/3}}{6\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}-\frac {i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a-i a \tan (e+f x)}\right ) (d \sec (e+f x))^{2/3}}{2\ 2^{2/3} a^{2/3} f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}} \]
1/2*I*(d*sec(f*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(4/3)-1/12*x*(d*sec(f*x+e) )^(2/3)*2^(1/3)/a^(2/3)/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)- 1/12*I*ln(cos(f*x+e))*(d*sec(f*x+e))^(2/3)*2^(1/3)/a^(2/3)/f/(a-I*a*tan(f* x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)-1/4*I*ln(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^(2/3)/f/(a-I*a*tan(f*x+e))^(1 /3)/(a+I*a*tan(f*x+e))^(1/3)+1/6*I*arctan(1/3*(a^(1/3)+2^(2/3)*(a-I*a*tan( f*x+e))^(1/3))/a^(1/3)*3^(1/2))*(d*sec(f*x+e))^(2/3)*2^(1/3)/a^(2/3)/f*3^( 1/2)/(a-I*a*tan(f*x+e))^(1/3)/(a+I*a*tan(f*x+e))^(1/3)
Time = 1.94 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.58 \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx=\frac {e^{-i (e+f x)} \left (3 i+3 i e^{2 i (e+f x)}-2 e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} f x-2 i \sqrt {3} e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \arctan \left (\frac {1+2 \sqrt [3]{1+e^{2 i (e+f x)}}}{\sqrt {3}}\right )-3 i e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \log \left (1-\sqrt [3]{1+e^{2 i (e+f x)}}\right )\right ) (d \sec (e+f x))^{5/3}}{12 d f (a+i a \tan (e+f x))^{4/3}} \]
((3*I + (3*I)*E^((2*I)*(e + f*x)) - 2*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*f*x - (2*I)*Sqrt[3]*E^((2*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]] - (3 *I)*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*Log[1 - (1 + E^((2 *I)*(e + f*x)))^(1/3)])*(d*Sec[e + f*x])^(5/3))/(12*d*E^(I*(e + f*x))*f*(a + I*a*Tan[e + f*x])^(4/3))
Time = 0.65 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.53, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.367, Rules used = {3042, 3986, 3042, 4005, 3042, 3968, 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.
| \(\displaystyle \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}}dx\) |
| \(\Big \downarrow \) 3986 |
| \(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \frac {\sqrt [3]{a-i a \tan (e+f x)}}{i \tan (e+f x) a+a}dx}{\sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \frac {\sqrt [3]{a-i a \tan (e+f x)}}{i \tan (e+f x) a+a}dx}{\sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 4005 |
| \(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \cos ^2(e+f x) (a-i a \tan (e+f x))^{4/3}dx}{a^2 \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(d \sec (e+f x))^{2/3} \int \frac {(a-i a \tan (e+f x))^{4/3}}{\sec (e+f x)^2}dx}{a^2 \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \int \frac {1}{(a-i a \tan (e+f x))^{2/3} (i \tan (e+f x) a+a)^2}d(-i a \tan (e+f x))}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 52 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {\int \frac {1}{(a-i a \tan (e+f x))^{2/3} (i \tan (e+f x) a+a)}d(-i a \tan (e+f x))}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 69 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {\frac {3 \int \frac {1}{i a \tan (e+f x)+\sqrt [3]{2} \sqrt [3]{a}}d\sqrt [3]{a-i a \tan (e+f x)}}{2\ 2^{2/3} a^{2/3}}+\frac {3 \int \frac {1}{-a^2 \tan ^2(e+f x)-i \sqrt [3]{2} a^{4/3} \tan (e+f x)+2^{2/3} a^{2/3}}d\sqrt [3]{a-i a \tan (e+f x)}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {\frac {3 \int \frac {1}{-a^2 \tan ^2(e+f x)-i \sqrt [3]{2} a^{4/3} \tan (e+f x)+2^{2/3} a^{2/3}}d\sqrt [3]{a-i a \tan (e+f x)}}{2 \sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {-\frac {3 \int \frac {1}{a^2 \tan ^2(e+f x)-3}d\left (1-i 2^{2/3} a^{2/3} \tan (e+f x)\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {i a (d \sec (e+f x))^{2/3} \left (\frac {-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (e+f x)}{\sqrt {3}}\right )}{2^{2/3} a^{2/3}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}+i a \tan (e+f x)\right )}{2\ 2^{2/3} a^{2/3}}+\frac {\log (a+i a \tan (e+f x))}{2\ 2^{2/3} a^{2/3}}}{3 a}+\frac {\sqrt [3]{a-i a \tan (e+f x)}}{2 a (a+i a \tan (e+f x))}\right )}{f \sqrt [3]{a-i a \tan (e+f x)} \sqrt [3]{a+i a \tan (e+f x)}}\) |
(I*a*(d*Sec[e + f*x])^(2/3)*((((-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*Tan[e + f*x])^(1/3)/(2*a*(a + I*a*Tan[e + f*x]))))/(f*(a - I*a*Tan[ e + f*x])^(1/3)*(a + I*a*Tan[e + f*x])^(1/3))
3.5.44.3.1 Defintions of rubi rules used
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
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]
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]
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])
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]
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]
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]
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]))
\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}}}{\left (a +i a \tan \left (f x +e \right )\right )^{\frac {4}{3}}}d x\]
Time = 0.26 (sec) , antiderivative size = 515, normalized size of antiderivative = 1.36 \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx=\frac {{\left (4 \, a^{2} f \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (-2 \, {\left (6 i \, a^{2} f \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )} - 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (i \, e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, e^{\left (2 i \, f x + 2 i \, e\right )} + i\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 \, {\left (-i \, \sqrt {3} a^{2} f + a^{2} f\right )} \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (2 \, {\left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + 3 \, {\left (\sqrt {3} a^{2} f + i \, a^{2} f\right )} \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right ) - 2 \, {\left (i \, \sqrt {3} a^{2} f + a^{2} f\right )} \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (2 \, {\left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 \, {\left (\sqrt {3} a^{2} f - i \, a^{2} f\right )} \left (\frac {i \, d^{2}}{108 \, a^{4} f^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}\right )\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{4 \, a^{2} f} \]
1/4*(4*a^2*f*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(4*I*f*x + 4*I*e)*log(-2*(6*I *a^2*f*(1/108*I*d^2/(a^4*f^3))^(1/3)*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)*(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)*(I*e^( 4*I*f*x + 4*I*e) + 2*I*e^(2*I*f*x + 2*I*e) + I)*e^(2*I*f*x + 2*I*e) - 2*(- I*sqrt(3)*a^2*f + a^2*f)*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(4*I*f*x + 4*I*e) *log(2*(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) + 3*(sqrt(3)*a ^2*f + I*a^2*f)*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(2*I*f*x + 2*I*e))*e^(-2*I *f*x - 2*I*e)) - 2*(I*sqrt(3)*a^2*f + a^2*f)*(1/108*I*d^2/(a^4*f^3))^(1/3) *e^(4*I*f*x + 4*I*e)*log(2*(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) - 3*(sqrt(3)*a^2*f - I*a^2*f)*(1/108*I*d^2/(a^4*f^3))^(1/3)*e^(2*I*f *x + 2*I*e))*e^(-2*I*f*x - 2*I*e)))*e^(-4*I*f*x - 4*I*e)/(a^2*f)
\[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx=\int \frac {\left (d \sec {\left (e + f x \right )}\right )^{\frac {2}{3}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {4}{3}}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1906 vs. \(2 (279) = 558\).
Time = 0.44 (sec) , antiderivative size = 1906, normalized size of antiderivative = 5.04 \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx=\text {Too large to display} \]
-1/24*(6*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1 )^(1/3)*((-I*2^(1/3)*cos(2*f*x + 2*e) - 2^(1/3)*sin(2*f*x + 2*e))*cos(2/3* arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + (2^(1/3)*cos(2*f*x + 2* e) - I*2^(1/3)*sin(2*f*x + 2*e))*sin(2/3*arctan2(sin(2*f*x + 2*e), cos(2*f *x + 2*e) + 1)))*d^(2/3) - (-2*I*sqrt(3)*2^(1/3)*arctan2(2/3*sqrt(3)*(cos( 2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*cos(1/ 3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 1/3*sqrt(3), 1/3*sqrt (3)*(2*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^ (1/6)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + sqrt(3))) - 2*I*sqrt(3)*2^(1/3)*arctan2(2/3*sqrt(3)*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*cos(1/3*arctan2(sin(2*f*x + 2*e) , cos(2*f*x + 2*e) + 1)) + 1/3*sqrt(3), -1/3*sqrt(3)*(2*(cos(2*f*x + 2*e)^ 2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*sin(1/3*arctan2(sin (2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - sqrt(3))) + sqrt(3)*2^(1/3)*log(4/ 3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/3) *(cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2 + sin(1/3*arc tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))^2) + 4/3*(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(1/6)*(sqrt(3)*sin(1/3*arc tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1))) + 4/3) - sqrt(3)*2^(1/3)*log(4/3*(cos(2...
\[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {4}{3}}} \,d x } \]
Timed out. \[ \int \frac {(d \sec (e+f x))^{2/3}}{(a+i a \tan (e+f x))^{4/3}} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{2/3}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{4/3}} \,d x \]