3.3.65 \(\int \frac {a+i a \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx\) [265]

3.3.65.1 Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int \frac {a+i a \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 i 2^{5/6} a \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{6},\frac {5}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \sqrt [6]{1+i \tan (e+f x)}}{f \sqrt [3]{d \sec (e+f x)}} \]

output

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

3.3.65.2 Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.91 \[ \int \frac {a+i a \tan (e+f x)}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {3 a \left (i+\cot (e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(e+f x)\right ) \sqrt {-\tan ^2(e+f x)}\right )}{f \sqrt [3]{d \sec (e+f x)}} \]

input

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

output

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

3.3.65.3 Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 79

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

input

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

output

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

3.3.65.3.1 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]
 

3.3.65.4 Maple [F]

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

input

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

output

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

3.3.65.5 Fricas [F]

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

input

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

output

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

3.3.65.6 Sympy [F]

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

input

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

output

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

3.3.65.7 Maxima [F]

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

input

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

output

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

3.3.65.8 Giac [F]

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

input

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

output

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

3.3.65.9 Mupad [F(-1)]

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

input

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

output

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