\(\int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx\) [448]

Optimal result

Integrand size = 30, antiderivative size = 122 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx=\frac {54 i a^3 (d \sec (e+f x))^{2/3}}{7 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {9 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{7 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{7 f} \]

[Out]

54/7*I*a^3*(d*sec(f*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(1/3)+9/7*I*a^2*(d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(
2/3)/f+3/7*I*a*(d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(5/3)/f

Rubi [A] (verified)

Time = 0.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3575, 3574} \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx=\frac {54 i a^3 (d \sec (e+f x))^{2/3}}{7 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {9 i a^2 (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{7 f}+\frac {3 i a (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{7 f} \]

[In]

Int[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(8/3),x]

[Out]

(((54*I)/7)*a^3*(d*Sec[e + f*x])^(2/3))/(f*(a + I*a*Tan[e + f*x])^(1/3)) + (((9*I)/7)*a^2*(d*Sec[e + f*x])^(2/
3)*(a + I*a*Tan[e + f*x])^(2/3))/f + (((3*I)/7)*a*(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3))/f

Rule 3574

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

Rule 3575

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

Rubi steps \begin{align*} \text {integral}& = \frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{7 f}+\frac {1}{7} (12 a) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3} \, dx \\ & = \frac {9 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{7 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{7 f}+\frac {1}{7} \left (18 a^2\right ) \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3} \, dx \\ & = \frac {54 i a^3 (d \sec (e+f x))^{2/3}}{7 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {9 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{7 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{7 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.50 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.82 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx=\frac {3 a^2 (d \sec (e+f x))^{5/3} (i \cos (e-f x)+\sin (e-f x)) (21+23 \cos (2 (e+f x))+5 i \sin (2 (e+f x))) (a+i a \tan (e+f x))^{2/3}}{14 d f (\cos (f x)+i \sin (f x))^2} \]

[In]

Integrate[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(8/3),x]

[Out]

(3*a^2*(d*Sec[e + f*x])^(5/3)*(I*Cos[e - f*x] + Sin[e - f*x])*(21 + 23*Cos[2*(e + f*x)] + (5*I)*Sin[2*(e + f*x
)])*(a + I*a*Tan[e + f*x])^(2/3))/(14*d*f*(Cos[f*x] + I*Sin[f*x])^2)

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{\frac {8}{3}}d x\]

[In]

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(8/3),x)

[Out]

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(8/3),x)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.88 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx=-\frac {6 \cdot 2^{\frac {1}{3}} {\left (-14 i \, a^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 21 i \, a^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 9 i \, a^{2}\right )} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}}{7 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \]

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(8/3),x, algorithm="fricas")

[Out]

-6/7*2^(1/3)*(-14*I*a^2*e^(4*I*f*x + 4*I*e) - 21*I*a^2*e^(2*I*f*x + 2*I*e) - 9*I*a^2)*(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)/(f*e^(4*I*f*x + 4*I*e) + f*e^(2*I*f*x + 2*
I*e))

Sympy [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx=\text {Timed out} \]

[In]

integrate((d*sec(f*x+e))**(2/3)*(a+I*a*tan(f*x+e))**(8/3),x)

[Out]

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. 402 vs. \(2 (92) = 184\).

Time = 0.89 (sec) , antiderivative size = 402, normalized size of antiderivative = 3.30 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx=\frac {6 \, {\left (7 \, {\left (-i \cdot 2^{\frac {1}{3}} a^{2} \cos \left (\frac {4}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) - 2^{\frac {1}{3}} a^{2} \sin \left (\frac {4}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )} \sqrt {\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1} a^{\frac {2}{3}} d^{\frac {2}{3}} + 2 \, {\left (i \cdot 2^{\frac {1}{3}} a^{2} \cos \left (\frac {7}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + 2^{\frac {1}{3}} a^{2} \sin \left (\frac {7}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + 7 \, {\left (i \cdot 2^{\frac {1}{3}} a^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + i \cdot 2^{\frac {1}{3}} a^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 i \cdot 2^{\frac {1}{3}} a^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \cdot 2^{\frac {1}{3}} a^{2}\right )} \cos \left (\frac {1}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right ) + 7 \, {\left (2^{\frac {1}{3}} a^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + 2^{\frac {1}{3}} a^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \cdot 2^{\frac {1}{3}} a^{2} \cos \left (2 \, f x + 2 \, e\right ) + 2^{\frac {1}{3}} a^{2}\right )} \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right ) + 1\right )\right )\right )} a^{\frac {2}{3}} d^{\frac {2}{3}}\right )}}{7 \, {\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} + 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )}^{\frac {7}{6}} f} \]

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(8/3),x, algorithm="maxima")

[Out]

6/7*(7*(-I*2^(1/3)*a^2*cos(4/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 2^(1/3)*a^2*sin(4/3*arctan2(
sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*sqrt(cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) +
1)*a^(2/3)*d^(2/3) + 2*(I*2^(1/3)*a^2*cos(7/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 2^(1/3)*a^2*s
in(7/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 7*(I*2^(1/3)*a^2*cos(2*f*x + 2*e)^2 + I*2^(1/3)*a^2*
sin(2*f*x + 2*e)^2 + 2*I*2^(1/3)*a^2*cos(2*f*x + 2*e) + I*2^(1/3)*a^2)*cos(1/3*arctan2(sin(2*f*x + 2*e), cos(2
*f*x + 2*e) + 1)) + 7*(2^(1/3)*a^2*cos(2*f*x + 2*e)^2 + 2^(1/3)*a^2*sin(2*f*x + 2*e)^2 + 2*2^(1/3)*a^2*cos(2*f
*x + 2*e) + 2^(1/3)*a^2)*sin(1/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)))*a^(2/3)*d^(2/3))/((cos(2*f*
x + 2*e)^2 + sin(2*f*x + 2*e)^2 + 2*cos(2*f*x + 2*e) + 1)^(7/6)*f)

Giac [F]

\[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {8}{3}} \,d x } \]

[In]

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(8/3),x, algorithm="giac")

[Out]

integrate((d*sec(f*x + e))^(2/3)*(I*a*tan(f*x + e) + a)^(8/3), x)

Mupad [B] (verification not implemented)

Time = 6.36 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3} \, dx=\frac {3\,a^2\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{2/3}\,{\left (\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}\right )}^{2/3}\,\left (\cos \left (2\,e+2\,f\,x\right )\,44{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,9{}\mathrm {i}+16\,\sin \left (2\,e+2\,f\,x\right )+9\,\sin \left (4\,e+4\,f\,x\right )+35{}\mathrm {i}\right )}{14\,f\,\left (\cos \left (2\,e+2\,f\,x\right )+1\right )} \]

[In]

int((d/cos(e + f*x))^(2/3)*(a + a*tan(e + f*x)*1i)^(8/3),x)

[Out]

(3*a^2*(d/cos(e + f*x))^(2/3)*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(2/3)*
(cos(2*e + 2*f*x)*44i + cos(4*e + 4*f*x)*9i + 16*sin(2*e + 2*f*x) + 9*sin(4*e + 4*f*x) + 35i))/(14*f*(cos(2*e
+ 2*f*x) + 1))