\(\int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx\) [442]

Optimal result

Integrand size = 30, antiderivative size = 86 \[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 i a \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {13}{6},\frac {1}{3},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{13/6}}{8 \sqrt [6]{2} d (e \sec (c+d x))^{4/3} (a+i a \tan (c+d x))^{3/2}} \] Output:

-3/16*I*a*hypergeom([-2/3, 13/6],[1/3],1/2-1/2*I*tan(d*x+c))*(1+I*tan(d*x+ 
c))^(13/6)*2^(5/6)/d/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30 \[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 i \sec ^2(c+d x) \left (3+3 \cos (2 (c+d x))-55 \sqrt [6]{1+e^{2 i (c+d x)}} \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{6},\frac {5}{6},-e^{2 i (c+d x)}\right )+11 i \sin (2 (c+d x))\right )}{112 d (e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \] Input:

Integrate[1/((e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c + d*x]]),x]
 

Output:

(((-3*I)/112)*Sec[c + d*x]^2*(3 + 3*Cos[2*(c + d*x)] - 55*(1 + E^((2*I)*(c 
 + d*x)))^(1/6)*Hypergeometric2F1[-1/6, 1/6, 5/6, -E^((2*I)*(c + d*x))] + 
(11*I)*Sin[2*(c + d*x)]))/(d*(e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c + d 
*x]])
 

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, 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.

\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{4/3}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (c+d x)} (e \sec (c+d x))^{4/3}}dx\)

\(\Big \downarrow \) 3986

\(\displaystyle \frac {(a-i a \tan (c+d x))^{2/3} (a+i a \tan (c+d x))^{2/3} \int \frac {1}{(a-i a \tan (c+d x))^{2/3} (i \tan (c+d x) a+a)^{7/6}}dx}{(e \sec (c+d x))^{4/3}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a-i a \tan (c+d x))^{2/3} (a+i a \tan (c+d x))^{2/3} \int \frac {1}{(a-i a \tan (c+d x))^{2/3} (i \tan (c+d x) a+a)^{7/6}}dx}{(e \sec (c+d x))^{4/3}}\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{2/3} (a+i a \tan (c+d x))^{2/3} \int \frac {1}{(a-i a \tan (c+d x))^{5/3} (i \tan (c+d x) a+a)^{13/6}}d\tan (c+d x)}{d (e \sec (c+d x))^{4/3}}\)

\(\Big \downarrow \) 80

\(\displaystyle \frac {\sqrt [6]{1+i \tan (c+d x)} (a-i a \tan (c+d x))^{2/3} \sqrt {a+i a \tan (c+d x)} \int \frac {4 \sqrt [6]{2}}{(i \tan (c+d x)+1)^{13/6} (a-i a \tan (c+d x))^{5/3}}d\tan (c+d x)}{4 \sqrt [6]{2} d (e \sec (c+d x))^{4/3}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\sqrt [6]{1+i \tan (c+d x)} (a-i a \tan (c+d x))^{2/3} \sqrt {a+i a \tan (c+d x)} \int \frac {1}{(i \tan (c+d x)+1)^{13/6} (a-i a \tan (c+d x))^{5/3}}d\tan (c+d x)}{d (e \sec (c+d x))^{4/3}}\)

\(\Big \downarrow \) 79

\(\displaystyle -\frac {3 i \sqrt [6]{1+i \tan (c+d x)} \sqrt {a+i a \tan (c+d x)} \operatorname {Hypergeometric2F1}\left (-\frac {2}{3},\frac {13}{6},\frac {1}{3},\frac {1}{2} (1-i \tan (c+d x))\right )}{8 \sqrt [6]{2} a d (e \sec (c+d x))^{4/3}}\)

Input:

Int[1/((e*Sec[c + d*x])^(4/3)*Sqrt[a + I*a*Tan[c + d*x]]),x]
 

Output:

(((-3*I)/8)*Hypergeometric2F1[-2/3, 13/6, 1/3, (1 - I*Tan[c + d*x])/2]*(1 
+ I*Tan[c + d*x])^(1/6)*Sqrt[a + I*a*Tan[c + d*x]])/(2^(1/6)*a*d*(e*Sec[c 
+ d*x])^(4/3))
 

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]

\[\int \frac {1}{\left (e \sec \left (d x +c \right )\right )^{\frac {4}{3}} \sqrt {a +i a \tan \left (d x +c \right )}}d x\]

Input:

int(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x)
 

Output:

int(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x)
 

Fricas [F]

\[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fr 
icas")
 

Output:

-1/112*(3*2^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c 
) + 1))^(2/3)*(7*I*e^(8*I*d*x + 8*I*c) - 14*I*e^(7*I*d*x + 7*I*c) - 38*I*e 
^(6*I*d*x + 6*I*c) - 20*I*e^(5*I*d*x + 5*I*c) - 101*I*e^(4*I*d*x + 4*I*c) 
+ 2*I*e^(3*I*d*x + 3*I*c) - 60*I*e^(2*I*d*x + 2*I*c) + 8*I*e^(I*d*x + I*c) 
 - 4*I)*e^(2/3*I*d*x + 2/3*I*c) - 112*(a*d*e^2*e^(5*I*d*x + 5*I*c) - 2*a*d 
*e^2*e^(4*I*d*x + 4*I*c) + a*d*e^2*e^(3*I*d*x + 3*I*c))*integral(-55/112*2 
^(1/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c) + 1))^(2/ 
3)*(-I*e^(4*I*d*x + 4*I*c) - 7*I*e^(3*I*d*x + 3*I*c) - 5*I*e^(2*I*d*x + 2* 
I*c) - 7*I*e^(I*d*x + I*c) - 4*I)*e^(2/3*I*d*x + 2/3*I*c)/(a*d*e^2*e^(4*I* 
d*x + 4*I*c) - 3*a*d*e^2*e^(3*I*d*x + 3*I*c) + 3*a*d*e^2*e^(2*I*d*x + 2*I* 
c) - a*d*e^2*e^(I*d*x + I*c)), x))/(a*d*e^2*e^(5*I*d*x + 5*I*c) - 2*a*d*e^ 
2*e^(4*I*d*x + 4*I*c) + a*d*e^2*e^(3*I*d*x + 3*I*c))
 

Sympy [F]

\[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\left (e \sec {\left (c + d x \right )}\right )^{\frac {4}{3}} \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \] Input:

integrate(1/(e*sec(d*x+c))**(4/3)/(a+I*a*tan(d*x+c))**(1/2),x)
 

Output:

Integral(1/((e*sec(c + d*x))**(4/3)*sqrt(I*a*(tan(c + d*x) - I))), x)
 

Maxima [F]

\[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\left (e \sec \left (d x + c\right )\right )^{\frac {4}{3}} \sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \] Input:

integrate(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="ma 
xima")
 

Output:

integrate(1/((e*sec(d*x + c))^(4/3)*sqrt(I*a*tan(d*x + c) + a)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="gi 
ac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{4/3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \] Input:

int(1/((e/cos(c + d*x))^(4/3)*(a + a*tan(c + d*x)*1i)^(1/2)),x)
 

Output:

int(1/((e/cos(c + d*x))^(4/3)*(a + a*tan(c + d*x)*1i)^(1/2)), x)
 

Reduce [F]

\[ \int \frac {1}{(e \sec (c+d x))^{4/3} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}}{\sec \left (d x +c \right )^{\frac {4}{3}} \tan \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{\frac {4}{3}}}d x -\left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{\frac {4}{3}} \tan \left (d x +c \right )^{2}+\sec \left (d x +c \right )^{\frac {4}{3}}}d x \right ) i \right )}{e^{\frac {4}{3}} a} \] Input:

int(1/(e*sec(d*x+c))^(4/3)/(a+I*a*tan(d*x+c))^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(a)*(int(sqrt(tan(c + d*x)*i + 1)/(sec(c + d*x)**(1/3)*sec(c + d*x)*t 
an(c + d*x)**2 + sec(c + d*x)**(1/3)*sec(c + d*x)),x) - int((sqrt(tan(c + 
d*x)*i + 1)*tan(c + d*x))/(sec(c + d*x)**(1/3)*sec(c + d*x)*tan(c + d*x)** 
2 + sec(c + d*x)**(1/3)*sec(c + d*x)),x)*i))/(e**(1/3)*a*e)