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

3.5.49.1 Optimal result
3.5.49.2 Mathematica [A] (verified)
3.5.49.3 Rubi [A] (verified)
3.5.49.4 Maple [F]
3.5.49.5 Fricas [A] (verification not implemented)
3.5.49.6 Sympy [F(-1)]
3.5.49.7 Maxima [B] (verification not implemented)
3.5.49.8 Giac [F]
3.5.49.9 Mupad [B] (verification not implemented)

3.5.49.1 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
 
3.5.49.2 Mathematica [A] (verified)

Time = 2.04 (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)
 
3.5.49.3 Rubi [A] (verified)

Time = 0.78 (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
 

3.5.49.3.1 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]
 
3.5.49.4 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)
 
3.5.49.5 Fricas [A] (verification not implemented)

Time = 0.24 (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))
 
3.5.49.6 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
 
3.5.49.7 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 = 0.40 (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...
 
3.5.49.8 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)
 
3.5.49.9 Mupad [B] (verification not implemented)

Time = 8.95 (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))