\(\int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx\) [267]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\frac {12 i 2^{5/6} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3}}{5 f (1+i \tan (e+f x))^{5/6}} \] Output:

12/5*I*2^(5/6)*a^2*hypergeom([-11/6, 5/6],[11/6],1/2-1/2*I*tan(f*x+e))*(d* 
sec(f*x+e))^(5/3)/f/(1+I*tan(f*x+e))^(5/6)
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.56 \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\frac {3 i a^2 (d \sec (e+f x))^{5/3} \left (i \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+i \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+2 \sqrt {-\tan ^2(e+f x)}\right )}{5 f \sqrt {-\tan ^2(e+f x)}} \] Input:

Integrate[(d*Sec[e + f*x])^(5/3)*(a + I*a*Tan[e + f*x])^2,x]
 

Output:

(((3*I)/5)*a^2*(d*Sec[e + f*x])^(5/3)*(I*Hypergeometric2F1[-1/2, 5/6, 11/6 
, Sec[e + f*x]^2]*Tan[e + f*x] + I*Hypergeometric2F1[1/2, 5/6, 11/6, Sec[e 
 + f*x]^2]*Tan[e + f*x] + 2*Sqrt[-Tan[e + f*x]^2]))/(f*Sqrt[-Tan[e + f*x]^ 
2])
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3986, 3042, 4006, 80, 27, 79}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int (a-i a \tan (e+f x))^{5/6} (i \tan (e+f x) a+a)^{17/6}dx}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int (a-i a \tan (e+f x))^{5/6} (i \tan (e+f x) a+a)^{17/6}dx}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 (d \sec (e+f x))^{5/3} \int \frac {(i \tan (e+f x) a+a)^{11/6}}{\sqrt [6]{a-i a \tan (e+f x)}}d\tan (e+f x)}{f (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {2\ 2^{5/6} a^3 (d \sec (e+f x))^{5/3} \int \frac {(i \tan (e+f x)+1)^{11/6}}{2\ 2^{5/6} \sqrt [6]{a-i a \tan (e+f x)}}d\tan (e+f x)}{f (1+i \tan (e+f x))^{5/6} (a-i a \tan (e+f x))^{5/6}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a^3 (d \sec (e+f x))^{5/3} \int \frac {(i \tan (e+f x)+1)^{11/6}}{\sqrt [6]{a-i a \tan (e+f x)}}d\tan (e+f x)}{f (1+i \tan (e+f x))^{5/6} (a-i a \tan (e+f x))^{5/6}}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {12 i 2^{5/6} a^2 (d \sec (e+f x))^{5/3} \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}}\)

Input:

Int[(d*Sec[e + f*x])^(5/3)*(a + I*a*Tan[e + f*x])^2,x]
 

Output:

(((12*I)/5)*2^(5/6)*a^2*Hypergeometric2F1[-11/6, 5/6, 11/6, (1 - I*Tan[e + 
 f*x])/2]*(d*Sec[e + f*x])^(5/3))/(f*(1 + I*Tan[e + f*x])^(5/6))
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 80

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c 
 + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) 
^FracPart[n])   Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) 
), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] &&  !IntegerQ[m] &&  !Integ 
erQ[n] && (RationalQ[m] ||  !SimplerQ[n + 1, m + 1])
 

rule 3042

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

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 4006

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

Maple [F]

\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{2}d x\]

Input:

int((d*sec(f*x+e))^(5/3)*(a+I*a*tan(f*x+e))^2,x)
 

Output:

int((d*sec(f*x+e))^(5/3)*(a+I*a*tan(f*x+e))^2,x)
 

Fricas [F]

\[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \] Input:

integrate((d*sec(f*x+e))^(5/3)*(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

-1/80*(3*2^(2/3)*(55*I*a^2*d*e^(5*I*f*x + 5*I*e) + 26*I*a^2*d*e^(3*I*f*x + 
 3*I*e) + 11*I*a^2*d*e^(I*f*x + I*e))*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)* 
e^(2/3*I*f*x + 2/3*I*e) - 80*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I 
*e) + f)*integral(11/16*I*2^(2/3)*a^2*d*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3 
)*e^(2/3*I*f*x + 2/3*I*e)/f, x))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 
 2*I*e) + f)
 

Sympy [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\text {Timed out} \] Input:

integrate((d*sec(f*x+e))**(5/3)*(a+I*a*tan(f*x+e))**2,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \] Input:

integrate((d*sec(f*x+e))^(5/3)*(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

integrate((d*sec(f*x + e))^(5/3)*(I*a*tan(f*x + e) + a)^2, x)
 

Giac [F]

\[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \] Input:

integrate((d*sec(f*x+e))^(5/3)*(a+I*a*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*sec(f*x + e))^(5/3)*(I*a*tan(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \] Input:

int((d/cos(e + f*x))^(5/3)*(a + a*tan(e + f*x)*1i)^2,x)
 

Output:

int((d/cos(e + f*x))^(5/3)*(a + a*tan(e + f*x)*1i)^2, x)
 

Reduce [F]

\[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\frac {d^{\frac {5}{3}} a^{2} \left (6 \sec \left (f x +e \right )^{\frac {5}{3}} i -5 \left (\int \sec \left (f x +e \right )^{\frac {5}{3}} \tan \left (f x +e \right )^{2}d x \right ) f +5 \left (\int \sec \left (f x +e \right )^{\frac {5}{3}}d x \right ) f \right )}{5 f} \] Input:

int((d*sec(f*x+e))^(5/3)*(a+I*a*tan(f*x+e))^2,x)
                                                                                    
                                                                                    
 

Output:

(d**(2/3)*a**2*d*(6*sec(e + f*x)**(2/3)*sec(e + f*x)*i - 5*int(sec(e + f*x 
)**(2/3)*sec(e + f*x)*tan(e + f*x)**2,x)*f + 5*int(sec(e + f*x)**(2/3)*sec 
(e + f*x),x)*f))/(5*f)