\(\int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx\) [269]

Optimal result

Integrand size = 28, antiderivative size = 83 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {6 i 2^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{6},\frac {5}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \left (a^2+i a^2 \tan (e+f x)\right )}{f \sqrt [3]{d \sec (e+f x)} (1+i \tan (e+f x))^{5/6}} \] Output:

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\frac {3 a^2 \left (\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{6},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+\operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)-2 i \sqrt {-\tan ^2(e+f x)}\right )}{f \sqrt [3]{d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)} \int \frac {(i \tan (e+f x) a+a)^{5/6}}{(a-i a \tan (e+f x))^{7/6}}d\tan (e+f x)}{f \sqrt [3]{d \sec (e+f x)}}\)

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 79

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

Input:

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

Output:

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

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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

-1/2*(3*2^(2/3)*(4*I*a^2*e^(2*I*f*x + 2*I*e) + I*a^2*e^(I*f*x + I*e) + 5*I 
*a^2)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x + 2/3*I*e) - 2*(d*f 
*e^(I*f*x + I*e) - d*f)*integral(-5*2^(2/3)*(I*a^2*e^(2*I*f*x + 2*I*e) + I 
*a^2*e^(I*f*x + I*e) + I*a^2)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2/3*I 
*f*x + 2/3*I*e)/(d*f*e^(3*I*f*x + 3*I*e) - 2*d*f*e^(2*I*f*x + 2*I*e) + d*f 
*e^(I*f*x + I*e)), x))/(d*f*e^(I*f*x + I*e) - d*f)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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