\(\int \frac {(d \sec (e+f x))^{5/3}}{(a+i a \tan (e+f x))^2} \, dx\) [275]

Optimal result

Integrand size = 28, antiderivative size = 84 \[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+i a \tan (e+f x))^2} \, dx=\frac {3 i a \operatorname {Hypergeometric2F1}\left (\frac {5}{6},\frac {13}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3} (1+i \tan (e+f x))^{13/6}}{10 \sqrt [6]{2} f (a+i a \tan (e+f x))^3} \] Output:

3/20*I*a*hypergeom([5/6, 13/6],[11/6],1/2-1/2*I*tan(f*x+e))*(d*sec(f*x+e)) 
^(5/3)*(1+I*tan(f*x+e))^(13/6)*2^(5/6)/f/(a+I*a*tan(f*x+e))^3
 

Mathematica [A] (verified)

Time = 1.49 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.52 \[ \int \frac {(d \sec (e+f x))^{5/3}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {3 e^{-i (4 e+5 f x)} \left (1+e^{2 i (e+f x)}\right ) \left (1+e^{2 i (e+f x)}+2 e^{2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {2}{3},\frac {5}{6},-e^{2 i (e+f x)}\right )\right ) (d \sec (e+f x))^{5/3} (-i \cos (f x)+\sin (f x))}{28 a^2 f} \] Input:

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

Output:

(-3*(1 + E^((2*I)*(e + f*x)))*(1 + E^((2*I)*(e + f*x)) + 2*E^((2*I)*(e + f 
*x))*(1 + E^((2*I)*(e + f*x)))^(2/3)*Hypergeometric2F1[-1/6, 2/3, 5/6, -E^ 
((2*I)*(e + f*x))])*(d*Sec[e + f*x])^(5/3)*((-I)*Cos[f*x] + Sin[f*x]))/(28 
*a^2*E^(I*(4*e + 5*f*x))*f)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.02, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int \frac {(a-i a \tan (e+f x))^{5/6}}{(i \tan (e+f x) a+a)^{7/6}}dx}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int \frac {(a-i a \tan (e+f x))^{5/6}}{(i \tan (e+f x) a+a)^{7/6}}dx}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 (d \sec (e+f x))^{5/3} \int \frac {1}{\sqrt [6]{a-i a \tan (e+f x)} (i \tan (e+f x) a+a)^{13/6}}d\tan (e+f x)}{f (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {\sqrt [6]{1+i \tan (e+f x)} (d \sec (e+f x))^{5/3} \int \frac {4 \sqrt [6]{2}}{(i \tan (e+f x)+1)^{13/6} \sqrt [6]{a-i a \tan (e+f x)}}d\tan (e+f x)}{4 \sqrt [6]{2} f (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [6]{1+i \tan (e+f x)} (d \sec (e+f x))^{5/3} \int \frac {1}{(i \tan (e+f x)+1)^{13/6} \sqrt [6]{a-i a \tan (e+f x)}}d\tan (e+f x)}{f (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {3 i \sqrt [6]{1+i \tan (e+f x)} (d \sec (e+f x))^{5/3} \operatorname {Hypergeometric2F1}\left (\frac {5}{6},\frac {13}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{10 \sqrt [6]{2} a f (a+i a \tan (e+f x))}\)

Input:

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

Output:

(((3*I)/10)*Hypergeometric2F1[5/6, 13/6, 11/6, (1 - I*Tan[e + f*x])/2]*(d* 
Sec[e + f*x])^(5/3)*(1 + I*Tan[e + f*x])^(1/6))/(2^(1/6)*a*f*(a + I*a*Tan[ 
e + f*x]))
 

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]

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

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]

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

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

Output:

1/14*(14*a^2*f*e^(3*I*f*x + 3*I*e)*integral(-1/7*I*2^(2/3)*d*(d/(e^(2*I*f* 
x + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x + 2/3*I*e)/(a^2*f), x) - 3*2^(2/3)*(-2 
*I*d*e^(4*I*f*x + 4*I*e) - 3*I*d*e^(2*I*f*x + 2*I*e) - I*d)*(d/(e^(2*I*f*x 
 + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x + 2/3*I*e))*e^(-3*I*f*x - 3*I*e)/(a^2*f 
)
 

Sympy [F]

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

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

Output:

-Integral((d*sec(e + f*x))**(5/3)/(tan(e + f*x)**2 - 2*I*tan(e + f*x) - 1) 
, x)/a**2
 

Maxima [F(-2)]

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

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

Output:

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un 
defined.
 

Giac [F]

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

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

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

Output:

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