\(\int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2} \, dx\) [278]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2} \, dx=-\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {23}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{5/6}}{20\ 2^{5/6} a^2 f (d \sec (e+f x))^{5/3}} \] Output:

-3/40*I*hypergeom([-5/6, 23/6],[1/6],1/2-1/2*I*tan(f*x+e))*(1+I*tan(f*x+e) 
)^(5/6)*2^(1/6)/a^2/f/(d*sec(f*x+e))^(5/3)
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(143\) vs. \(2(71)=142\).

Time = 1.74 (sec) , antiderivative size = 143, normalized size of antiderivative = 2.01 \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2} \, dx=\frac {3 i \sec ^4(e+f x) \left (-46-40 \cos (2 (e+f x))+6 \cos (4 (e+f x))+128 e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-e^{2 i (e+f x)}\right )-10 i \sin (2 (e+f x))+11 i \sin (4 (e+f x))\right )}{680 a^2 f (d \sec (e+f x))^{5/3} (-i+\tan (e+f x))^2} \] Input:

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

Output:

(((3*I)/680)*Sec[e + f*x]^4*(-46 - 40*Cos[2*(e + f*x)] + 6*Cos[4*(e + f*x) 
] + 128*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*Hypergeometric 
2F1[1/6, 1/3, 7/6, -E^((2*I)*(e + f*x))] - (10*I)*Sin[2*(e + f*x)] + (11*I 
)*Sin[4*(e + f*x)]))/(a^2*f*(d*Sec[e + f*x])^(5/3)*(-I + Tan[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.47 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

\(\displaystyle \frac {(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6} \int \frac {1}{(a-i a \tan (e+f x))^{5/6} (i \tan (e+f x) a+a)^{17/6}}dx}{(d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6} \int \frac {1}{(a-i a \tan (e+f x))^{5/6} (i \tan (e+f x) a+a)^{17/6}}dx}{(d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6} \int \frac {1}{(a-i a \tan (e+f x))^{11/6} (i \tan (e+f x) a+a)^{23/6}}d\tan (e+f x)}{f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {(1+i \tan (e+f x))^{5/6} (a-i a \tan (e+f x))^{5/6} \int \frac {8\ 2^{5/6}}{(i \tan (e+f x)+1)^{23/6} (a-i a \tan (e+f x))^{11/6}}d\tan (e+f x)}{8\ 2^{5/6} a f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(1+i \tan (e+f x))^{5/6} (a-i a \tan (e+f x))^{5/6} \int \frac {1}{(i \tan (e+f x)+1)^{23/6} (a-i a \tan (e+f x))^{11/6}}d\tan (e+f x)}{a f (d \sec (e+f x))^{5/3}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {3 i (1+i \tan (e+f x))^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},\frac {23}{6},\frac {1}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{20\ 2^{5/6} a^2 f (d \sec (e+f x))^{5/3}}\)

Input:

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

Output:

(((-3*I)/20)*Hypergeometric2F1[-5/6, 23/6, 1/6, (1 - I*Tan[e + f*x])/2]*(1 
 + I*Tan[e + f*x])^(5/6))/(2^(5/6)*a^2*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]

\[\int \frac {1}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{2}}d x\]

Input:

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

Output:

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

Fricas [F]

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

integrate(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas 
")
 

Output:

1/1360*(1360*a^2*d^2*f*e^(6*I*f*x + 6*I*e)*integral(-16/85*I*2^(1/3)*(d/(e 
^(2*I*f*x + 2*I*e) + 1))^(1/3)*e^(-2/3*I*f*x - 2/3*I*e)/(a^2*d^2*f), x) - 
3*2^(1/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*(17*I*e^(8*I*f*x + 8*I*e) - 
50*I*e^(6*I*f*x + 6*I*e) - 92*I*e^(4*I*f*x + 4*I*e) - 30*I*e^(2*I*f*x + 2* 
I*e) - 5*I)*e^(1/3*I*f*x + 1/3*I*e))*e^(-6*I*f*x - 6*I*e)/(a^2*d^2*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \] Input:

integrate(1/(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]

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

integrate(1/(d*sec(f*x+e))^(5/3)/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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