\(\int \frac {1}{\sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2} \, dx\) [277]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int \frac {1}{\sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2} \, dx=-\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {19}{6},\frac {5}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \sqrt [6]{1+i \tan (e+f x)}}{4 \sqrt [6]{2} a^2 f \sqrt [3]{d \sec (e+f x)}} \] Output:

-3/8*I*hypergeom([-1/6, 19/6],[5/6],1/2-1/2*I*tan(f*x+e))*(1+I*tan(f*x+e)) 
^(1/6)*2^(5/6)/a^2/f/(d*sec(f*x+e))^(1/3)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 2.00 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.99 \[ \int \frac {1}{\sqrt [3]{d \sec (e+f x)} (a+i a \tan (e+f x))^2} \, dx=\frac {(d \sec (e+f x))^{2/3} \left (16 e^{3 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {11}{6},-e^{2 i (e+f x)}\right )-10 \left (7 \cos (e+f x)+5 \cos (3 (e+f x))+18 i \cos ^2(e+f x) \sin (e+f x)\right )\right ) (-3 i \cos (2 (e+f x))-3 \sin (2 (e+f x)))}{260 a^2 d f} \] Input:

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

Output:

((d*Sec[e + f*x])^(2/3)*(16*E^((3*I)*(e + f*x))*(1 + E^((2*I)*(e + f*x)))^ 
(2/3)*Hypergeometric2F1[2/3, 5/6, 11/6, -E^((2*I)*(e + f*x))] - 10*(7*Cos[ 
e + f*x] + 5*Cos[3*(e + f*x)] + (18*I)*Cos[e + f*x]^2*Sin[e + f*x]))*((-3* 
I)*Cos[2*(e + f*x)] - 3*Sin[2*(e + f*x)]))/(260*a^2*d*f)
 

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)} \int \frac {1}{(a-i a \tan (e+f x))^{7/6} (i \tan (e+f x) a+a)^{19/6}}d\tan (e+f x)}{f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {\sqrt [6]{1+i \tan (e+f x)} \sqrt [6]{a-i a \tan (e+f x)} \int \frac {8 \sqrt [6]{2}}{(i \tan (e+f x)+1)^{19/6} (a-i a \tan (e+f x))^{7/6}}d\tan (e+f x)}{8 \sqrt [6]{2} a f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [6]{1+i \tan (e+f x)} \sqrt [6]{a-i a \tan (e+f x)} \int \frac {1}{(i \tan (e+f x)+1)^{19/6} (a-i a \tan (e+f x))^{7/6}}d\tan (e+f x)}{a f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {3 i \sqrt [6]{1+i \tan (e+f x)} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {19}{6},\frac {5}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{4 \sqrt [6]{2} a^2 f \sqrt [3]{d \sec (e+f x)}}\)

Input:

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

Output:

(((-3*I)/4)*Hypergeometric2F1[-1/6, 19/6, 5/6, (1 - I*Tan[e + f*x])/2]*(1 
+ I*Tan[e + f*x])^(1/6))/(2^(1/6)*a^2*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

-1/104*(3*2^(2/3)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*(13*I*e^(7*I*f*x + 7 
*I*e) + 19*I*e^(6*I*f*x + 6*I*e) + 9*I*e^(5*I*f*x + 5*I*e) + 23*I*e^(4*I*f 
*x + 4*I*e) - 5*I*e^(3*I*f*x + 3*I*e) + 5*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) - 104*(a^2*d*f*e^(6*I*f*x + 6*I*e) 
- a^2*d*f*e^(5*I*f*x + 5*I*e))*integral(-8/13*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^2*d*f*e^(3*I*f*x + 3*I*e) - 2*a^2*d*f*e^(2*I*f*x + 2*I*e) 
 + a^2*d*f*e^(I*f*x + I*e)), x))/(a^2*d*f*e^(6*I*f*x + 6*I*e) - a^2*d*f*e^ 
(5*I*f*x + 5*I*e))
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(1/(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]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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