\(\int \frac {1}{\sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))} \, dx\) [273]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int \frac {1}{\sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))} \, dx=-\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {13}{6},\frac {5}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \sqrt [6]{1+i \tan (e+f x)}}{2 \sqrt [6]{2} a f \sqrt [3]{d \sec (e+f x)}} \] Output:

-3/4*I*hypergeom([-1/6, 13/6],[5/6],1/2-1/2*I*tan(f*x+e))*(1+I*tan(f*x+e)) 
^(1/6)*2^(5/6)/a/f/(d*sec(f*x+e))^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.58 \[ \int \frac {1}{\sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))} \, dx=\frac {3 \left (-8 e^{2 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {11}{6},-e^{2 i (e+f x)}\right )+5 (5+5 \cos (2 (e+f x))+4 i \sin (2 (e+f x)))\right ) (i+\tan (e+f x))}{70 a f \sqrt [3]{d \sec (e+f x)}} \] Input:

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

Output:

(3*(-8*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(2/3)*Hypergeometric2 
F1[2/3, 5/6, 11/6, -E^((2*I)*(e + f*x))] + 5*(5 + 5*Cos[2*(e + f*x)] + (4* 
I)*Sin[2*(e + f*x)]))*(I + Tan[e + f*x]))/(70*a*f*(d*Sec[e + f*x])^(1/3))
 

Rubi [A] (verified)

Time = 0.44 (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.

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) \sqrt [3]{d \sec (e+f x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+i a \tan (e+f x)) \sqrt [3]{d \sec (e+f x)}}dx\)

\(\Big \downarrow \) 3986

\(\displaystyle \frac {\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)} \int \frac {1}{\sqrt [6]{a-i a \tan (e+f x)} (i \tan (e+f x) a+a)^{7/6}}dx}{\sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)} \int \frac {1}{\sqrt [6]{a-i a \tan (e+f x)} (i \tan (e+f x) a+a)^{7/6}}dx}{\sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)} \int \frac {1}{(a-i a \tan (e+f x))^{7/6} (i \tan (e+f x) a+a)^{13/6}}d\tan (e+f x)}{f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {\sqrt [6]{1+i \tan (e+f x)} \sqrt [6]{a-i a \tan (e+f x)} \int \frac {4 \sqrt [6]{2}}{(i \tan (e+f x)+1)^{13/6} (a-i a \tan (e+f x))^{7/6}}d\tan (e+f x)}{4 \sqrt [6]{2} f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [6]{1+i \tan (e+f x)} \sqrt [6]{a-i a \tan (e+f x)} \int \frac {1}{(i \tan (e+f x)+1)^{13/6} (a-i a \tan (e+f x))^{7/6}}d\tan (e+f x)}{f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

-1/28*(3*2^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(7*I*e^(5*I*f*x + 5*I 
*e) + 9*I*e^(4*I*f*x + 4*I*e) + 6*I*e^(3*I*f*x + 3*I*e) + 10*I*e^(2*I*f*x 
+ 2*I*e) - I*e^(I*f*x + I*e) + I)*e^(2/3*I*f*x + 2/3*I*e) - 28*(a*d*f*e^(4 
*I*f*x + 4*I*e) - a*d*f*e^(3*I*f*x + 3*I*e))*integral(-8/7*2^(2/3)*(d/(e^( 
2*I*f*x + 2*I*e) + 1))^(2/3)*(I*e^(2*I*f*x + 2*I*e) + I*e^(I*f*x + I*e) + 
I)*e^(2/3*I*f*x + 2/3*I*e)/(a*d*f*e^(3*I*f*x + 3*I*e) - 2*a*d*f*e^(2*I*f*x 
 + 2*I*e) + a*d*f*e^(I*f*x + I*e)), x))/(a*d*f*e^(4*I*f*x + 4*I*e) - a*d*f 
*e^(3*I*f*x + 3*I*e))
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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