\(\int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\) [276]

Optimal result

Integrand size = 28, antiderivative size = 84 \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {3 i a \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {17}{6},\frac {7}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \sqrt [3]{d \sec (e+f x)} (1+i \tan (e+f x))^{17/6}}{2\ 2^{5/6} f (a+i a \tan (e+f x))^3} \] Output:

3/4*I*a*hypergeom([1/6, 17/6],[7/6],1/2-1/2*I*tan(f*x+e))*(d*sec(f*x+e))^( 
1/3)*(1+I*tan(f*x+e))^(17/6)*2^(1/6)/f/(a+I*a*tan(f*x+e))^3
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.27 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {3 \sec ^2(e+f x) \sqrt [3]{d \sec (e+f x)} \left (-2 i-2 i \cos (2 (e+f x))+4 i e^{2 i (e+f x)} \sqrt [3]{1+e^{2 i (e+f x)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {7}{6},-e^{2 i (e+f x)}\right )+\sin (2 (e+f x))\right )}{22 a^2 f (-i+\tan (e+f x))^2} \] Input:

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

Output:

(3*Sec[e + f*x]^2*(d*Sec[e + f*x])^(1/3)*(-2*I - (2*I)*Cos[2*(e + f*x)] + 
(4*I)*E^((2*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^(1/3)*Hypergeometric2F 
1[1/6, 1/3, 7/6, -E^((2*I)*(e + f*x))] + Sin[2*(e + f*x)]))/(22*a^2*f*(-I 
+ Tan[e + f*x])^2)
 

Rubi [A] (verified)

Time = 0.46 (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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 \sqrt [3]{d \sec (e+f x)} \int \frac {1}{(a-i a \tan (e+f x))^{5/6} (i \tan (e+f x) a+a)^{17/6}}d\tan (e+f x)}{f \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {(1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)} \int \frac {4\ 2^{5/6}}{(i \tan (e+f x)+1)^{17/6} (a-i a \tan (e+f x))^{5/6}}d\tan (e+f x)}{4\ 2^{5/6} f \sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x))}\)

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 79

\(\displaystyle \frac {3 i (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {17}{6},\frac {7}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{2\ 2^{5/6} a f (a+i a \tan (e+f x))}\)

Input:

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

Output:

(((3*I)/2)*Hypergeometric2F1[1/6, 17/6, 7/6, (1 - I*Tan[e + f*x])/2]*(d*Se 
c[e + f*x])^(1/3)*(1 + I*Tan[e + f*x])^(5/6))/(2^(5/6)*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

1/44*(44*a^2*f*e^(4*I*f*x + 4*I*e)*integral(-2/11*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*f), x) - 3*2^(1/3)*(d/ 
(e^(2*I*f*x + 2*I*e) + 1))^(1/3)*(-3*I*e^(4*I*f*x + 4*I*e) - 4*I*e^(2*I*f* 
x + 2*I*e) - I)*e^(1/3*I*f*x + 1/3*I*e))*e^(-4*I*f*x - 4*I*e)/(a^2*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((d*sec(f*x+e))^(1/3)/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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