\(\int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx\) [449]

Optimal result

Integrand size = 30, antiderivative size = 163 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\frac {486 i a^4 (d \sec (e+f x))^{2/3}}{35 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {81 i a^3 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{2/3}}{35 f}+\frac {27 i a^2 (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{5/3}}{35 f}+\frac {3 i a (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{8/3}}{10 f} \] Output:

486/35*I*a^4*(d*sec(f*x+e))^(2/3)/f/(a+I*a*tan(f*x+e))^(1/3)+81/35*I*a^3*( 
d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(2/3)/f+27/35*I*a^2*(d*sec(f*x+e))^ 
(2/3)*(a+I*a*tan(f*x+e))^(5/3)/f+3/10*I*a*(d*sec(f*x+e))^(2/3)*(a+I*a*tan( 
f*x+e))^(8/3)/f
 

Mathematica [A] (verified)

Time = 1.96 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.71 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\frac {3 a^3 (d \sec (e+f x))^{5/3} (i \cos (e-2 f x)+\sin (e-2 f x)) (364+442 \cos (2 (e+f x))+59 i \sec (e+f x) \sin (3 (e+f x))+45 i \tan (e+f x)) (a+i a \tan (e+f x))^{2/3}}{140 d f (\cos (f x)+i \sin (f x))^3} \] Input:

Integrate[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(11/3),x]
 

Output:

(3*a^3*(d*Sec[e + f*x])^(5/3)*(I*Cos[e - 2*f*x] + Sin[e - 2*f*x])*(364 + 4 
42*Cos[2*(e + f*x)] + (59*I)*Sec[e + f*x]*Sin[3*(e + f*x)] + (45*I)*Tan[e 
+ f*x])*(a + I*a*Tan[e + f*x])^(2/3))/(140*d*f*(Cos[f*x] + I*Sin[f*x])^3)
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3042, 3975, 3042, 3975, 3042, 3975, 3042, 3974}

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.

\(\displaystyle \int (a+i a \tan (e+f x))^{11/3} (d \sec (e+f x))^{2/3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (e+f x))^{11/3} (d \sec (e+f x))^{2/3}dx\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {9}{5} a \int (d \sec (e+f x))^{2/3} (i \tan (e+f x) a+a)^{8/3}dx+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \int (d \sec (e+f x))^{2/3} (i \tan (e+f x) a+a)^{8/3}dx+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {9}{5} a \left (\frac {12}{7} a \int (d \sec (e+f x))^{2/3} (i \tan (e+f x) a+a)^{5/3}dx+\frac {3 i a (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{7 f}\right )+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \left (\frac {12}{7} a \int (d \sec (e+f x))^{2/3} (i \tan (e+f x) a+a)^{5/3}dx+\frac {3 i a (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{7 f}\right )+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f}\)

\(\Big \downarrow \) 3975

\(\displaystyle \frac {9}{5} a \left (\frac {12}{7} a \left (\frac {3}{2} a \int (d \sec (e+f x))^{2/3} (i \tan (e+f x) a+a)^{2/3}dx+\frac {3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f}\right )+\frac {3 i a (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{7 f}\right )+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {9}{5} a \left (\frac {12}{7} a \left (\frac {3}{2} a \int (d \sec (e+f x))^{2/3} (i \tan (e+f x) a+a)^{2/3}dx+\frac {3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f}\right )+\frac {3 i a (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{7 f}\right )+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f}\)

\(\Big \downarrow \) 3974

\(\displaystyle \frac {9}{5} a \left (\frac {12}{7} a \left (\frac {9 i a^2 (d \sec (e+f x))^{2/3}}{2 f \sqrt [3]{a+i a \tan (e+f x)}}+\frac {3 i a (a+i a \tan (e+f x))^{2/3} (d \sec (e+f x))^{2/3}}{4 f}\right )+\frac {3 i a (a+i a \tan (e+f x))^{5/3} (d \sec (e+f x))^{2/3}}{7 f}\right )+\frac {3 i a (a+i a \tan (e+f x))^{8/3} (d \sec (e+f x))^{2/3}}{10 f}\)

Input:

Int[(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(11/3),x]
 

Output:

(((3*I)/10)*a*(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(8/3))/f + (9* 
a*((((3*I)/7)*a*(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(5/3))/f + ( 
12*a*((((9*I)/2)*a^2*(d*Sec[e + f*x])^(2/3))/(f*(a + I*a*Tan[e + f*x])^(1/ 
3)) + (((3*I)/4)*a*(d*Sec[e + f*x])^(2/3)*(a + I*a*Tan[e + f*x])^(2/3))/f) 
)/7))/5
 

Defintions of rubi rules used

rule 3042

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

rule 3974

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 3975

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] + Simp[a*((m + 2*n - 2)/(m + n - 1))   Int[(d*Se 
c[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] &&  !Inte 
gerQ[n]
 

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {2}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{\frac {11}{3}}d x\]

Input:

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x)
 

Output:

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.82 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=-\frac {6 \cdot 2^{\frac {1}{3}} {\left (-140 i \, a^{3} e^{\left (6 i \, f x + 6 i \, e\right )} - 315 i \, a^{3} e^{\left (4 i \, f x + 4 i \, e\right )} - 270 i \, a^{3} e^{\left (2 i \, f x + 2 i \, e\right )} - 81 i \, a^{3}\right )} \left (\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} \left (\frac {d}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}\right )^{\frac {2}{3}} e^{\left (2 i \, f x + 2 i \, e\right )}}{35 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 2 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + f e^{\left (2 i \, f x + 2 i \, e\right )}\right )}} \] Input:

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x, algorithm="fri 
cas")
 

Output:

-6/35*2^(1/3)*(-140*I*a^3*e^(6*I*f*x + 6*I*e) - 315*I*a^3*e^(4*I*f*x + 4*I 
*e) - 270*I*a^3*e^(2*I*f*x + 2*I*e) - 81*I*a^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)/(f*e^(6* 
I*f*x + 6*I*e) + 2*f*e^(4*I*f*x + 4*I*e) + f*e^(2*I*f*x + 2*I*e))
 

Sympy [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))**(2/3)*(a+I*a*tan(f*x+e))**(11/3),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. 983 vs. \(2 (123) = 246\).

Time = 0.24 (sec) , antiderivative size = 983, normalized size of antiderivative = 6.03 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\text {Too large to display} \] Input:

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x, algorithm="max 
ima")
 

Output:

6/35*(7*(-2*I*2^(1/3)*a^3*cos(10/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2 
*e) + 1)) - 2*2^(1/3)*a^3*sin(10/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2 
*e) + 1)) + 15*(-I*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 - I*2^(1/3)*a^3*sin(2*f* 
x + 2*e)^2 - 2*I*2^(1/3)*a^3*cos(2*f*x + 2*e) - I*2^(1/3)*a^3)*cos(4/3*arc 
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) - 15*(2^(1/3)*a^3*cos(2*f*x 
+ 2*e)^2 + 2^(1/3)*a^3*sin(2*f*x + 2*e)^2 + 2*2^(1/3)*a^3*cos(2*f*x + 2*e) 
 + 2^(1/3)*a^3)*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) + 20*(3*(I*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + I*2^(1/3)*a^3*sin 
(2*f*x + 2*e)^2 + 2*I*2^(1/3)*a^3*cos(2*f*x + 2*e) + I*2^(1/3)*a^3)*cos(7/ 
3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 7*(I*2^(1/3)*a^3*cos( 
2*f*x + 2*e)^4 + I*2^(1/3)*a^3*sin(2*f*x + 2*e)^4 + 4*I*2^(1/3)*a^3*cos(2* 
f*x + 2*e)^3 + 6*I*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + 4*I*2^(1/3)*a^3*cos(2* 
f*x + 2*e) + I*2^(1/3)*a^3 + 2*(I*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + 2*I*2^( 
1/3)*a^3*cos(2*f*x + 2*e) + I*2^(1/3)*a^3)*sin(2*f*x + 2*e)^2)*cos(1/3*arc 
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 3*(2^(1/3)*a^3*cos(2*f*x + 
 2*e)^2 + 2^(1/3)*a^3*sin(2*f*x + 2*e)^2 + 2*2^(1/3)*a^3*cos(2*f*x + 2*e) 
+ 2^(1/3)*a^3)*sin(7/3*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e) + 1)) + 
7*(2^(1/3)*a^3*cos(2*f*x + 2*e)^4 + 2^(1/3)*a^3*sin(2*f*x + 2*e)^4 + 4*2^( 
1/3)*a^3*cos(2*f*x + 2*e)^3 + 6*2^(1/3)*a^3*cos(2*f*x + 2*e)^2 + 4*2^(1...
 

Giac [F]

\[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {2}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {11}{3}} \,d x } \] Input:

integrate((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x, algorithm="gia 
c")
 

Output:

integrate((d*sec(f*x + e))^(2/3)*(I*a*tan(f*x + e) + a)^(11/3), x)
 

Mupad [B] (verification not implemented)

Time = 5.31 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.86 \[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=\frac {{\left (-\frac {d}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-1\right )\,\left (\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,243{}\mathrm {i}}{35\,f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (e+f\,x\right )}^2+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}+1\right )\,162{}\mathrm {i}}{7\,f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (2\,e+2\,f\,x\right )}^2+\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}+1\right )\,27{}\mathrm {i}}{f}+\frac {a^3\,{\left (a-\frac {a\,\sin \left (e+f\,x\right )\,1{}\mathrm {i}}{2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-1}\right )}^{2/3}\,\left (-2\,{\sin \left (3\,e+3\,f\,x\right )}^2+\sin \left (6\,e+6\,f\,x\right )\,1{}\mathrm {i}+1\right )\,12{}\mathrm {i}}{f}\right )}{4\,\left ({\sin \left (e+f\,x\right )}^2-1\right )} \] Input:

int((d/cos(e + f*x))^(2/3)*(a + a*tan(e + f*x)*1i)^(11/3),x)
 

Output:

((-d/(2*sin(e/2 + (f*x)/2)^2 - 1))^(2/3)*(sin(4*e + 4*f*x)*1i + 2*sin(2*e 
+ 2*f*x)^2 - 1)*((a^3*(a - (a*sin(e + f*x)*1i)/(2*sin(e/2 + (f*x)/2)^2 - 1 
))^(2/3)*243i)/(35*f) + (a^3*(a - (a*sin(e + f*x)*1i)/(2*sin(e/2 + (f*x)/2 
)^2 - 1))^(2/3)*(sin(2*e + 2*f*x)*1i - 2*sin(e + f*x)^2 + 1)*162i)/(7*f) + 
 (a^3*(a - (a*sin(e + f*x)*1i)/(2*sin(e/2 + (f*x)/2)^2 - 1))^(2/3)*(sin(4* 
e + 4*f*x)*1i - 2*sin(2*e + 2*f*x)^2 + 1)*27i)/f + (a^3*(a - (a*sin(e + f* 
x)*1i)/(2*sin(e/2 + (f*x)/2)^2 - 1))^(2/3)*(sin(6*e + 6*f*x)*1i - 2*sin(3* 
e + 3*f*x)^2 + 1)*12i)/f))/(4*(sin(e + f*x)^2 - 1))
 

Reduce [F]

\[ \int (d \sec (e+f x))^{2/3} (a+i a \tan (e+f x))^{11/3} \, dx=d^{\frac {2}{3}} a^{\frac {11}{3}} \left (-\left (\int \sec \left (f x +e \right )^{\frac {2}{3}} \left (\tan \left (f x +e \right ) i +1\right )^{\frac {2}{3}} \tan \left (f x +e \right )^{3}d x \right ) i -3 \left (\int \sec \left (f x +e \right )^{\frac {2}{3}} \left (\tan \left (f x +e \right ) i +1\right )^{\frac {2}{3}} \tan \left (f x +e \right )^{2}d x \right )+3 \left (\int \sec \left (f x +e \right )^{\frac {2}{3}} \left (\tan \left (f x +e \right ) i +1\right )^{\frac {2}{3}} \tan \left (f x +e \right )d x \right ) i +\int \sec \left (f x +e \right )^{\frac {2}{3}} \left (\tan \left (f x +e \right ) i +1\right )^{\frac {2}{3}}d x \right ) \] Input:

int((d*sec(f*x+e))^(2/3)*(a+I*a*tan(f*x+e))^(11/3),x)
 

Output:

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