\(\int \frac {a+i a \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx\) [266]

Optimal result

Integrand size = 26, antiderivative size = 69 \[ \int \frac {a+i a \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx=-\frac {3 i \sqrt [6]{2} a \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {5}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{5 f (d \sec (e+f x))^{5/3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int \frac {a+i a \tan (e+f x)}{(d \sec (e+f x))^{5/3}} \, dx=-\frac {3 a \left (i+\cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {1}{2},\frac {1}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{5 f (d \sec (e+f x))^{5/3}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

1/10*(10*d^2*f*integral(-2/5*I*2^(1/3)*a*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/ 
3)*e^(-2/3*I*f*x - 2/3*I*e)/(d^2*f), x) - 3*2^(1/3)*(I*a*e^(2*I*f*x + 2*I* 
e) + I*a)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(1/3*I*f*x + 1/3*I*e))/(d^ 
2*f)
 

Sympy [F]

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

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

Output:

I*a*(Integral(-I/(d*sec(e + f*x))**(5/3), x) + Integral(tan(e + f*x)/(d*se 
c(e + f*x))**(5/3), x))
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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