\(\int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [440]

Optimal result

Integrand size = 30, antiderivative size = 84 \[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 i a \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {4}{3},\frac {7}{6},\frac {1}{2} (1-i \tan (c+d x))\right ) \sqrt [3]{e \sec (c+d x)} (1+i \tan (c+d x))^{4/3}}{\sqrt [3]{2} d (a+i a \tan (c+d x))^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 \left (8 i-\frac {2 i e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {2}{3},\frac {5}{6},\frac {5}{3},-e^{2 i (c+d x)}\right )}{\sqrt [6]{1+e^{2 i (c+d x)}}}\right ) \sqrt [3]{e \sec (c+d x)}}{16 d \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(3*(8*I - ((2*I)*E^((2*I)*(c + d*x))*Hypergeometric2F1[2/3, 5/6, 5/3, -E^( 
(2*I)*(c + d*x))])/(1 + E^((2*I)*(c + d*x)))^(1/6))*(e*Sec[c + d*x])^(1/3) 
)/(16*d*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 79

\(\displaystyle \frac {3 i \sqrt [3]{1+i \tan (c+d x)} \sqrt [3]{e \sec (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {4}{3},\frac {7}{6},\frac {1}{2} (1-i \tan (c+d x))\right )}{\sqrt [3]{2} d \sqrt {a+i a \tan (c+d x)}}\)

Input:

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

Output:

((3*I)*Hypergeometric2F1[1/6, 4/3, 7/6, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + 
 d*x])^(1/3)*(1 + I*Tan[c + d*x])^(1/3))/(2^(1/3)*d*Sqrt[a + I*a*Tan[c + d 
*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]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

1/4*(4*a*d*e^(I*d*x + I*c)*integral(-1/4*I*2^(5/6)*sqrt(a/(e^(2*I*d*x + 2* 
I*c) + 1))*(e/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(1/3*I*d*x + 1/3*I*c)/(a* 
d), x) - 3*2^(5/6)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e/(e^(2*I*d*x + 2*I* 
c) + 1))^(1/3)*(-I*e^(2*I*d*x + 2*I*c) - I)*e^(1/3*I*d*x + 1/3*I*c))*e^(-I 
*d*x - I*c)/(a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {\sqrt [3]{e \sec (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 e^{\frac {1}{3}} \sqrt {a}\, i \left (-3 \sec \left (d x +c \right )^{\frac {1}{3}} \sqrt {\tan \left (d x +c \right ) i +1}-5 \left (\int \frac {\sec \left (d x +c \right )^{\frac {1}{3}} \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) \tan \left (d x +c \right )^{2} d -5 \left (\int \frac {\sec \left (d x +c \right )^{\frac {1}{3}} \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )}{\tan \left (d x +c \right )^{2}+1}d x \right ) d \right )}{3 a d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(2*e**(1/3)*sqrt(a)*i*( - 3*sec(c + d*x)**(1/3)*sqrt(tan(c + d*x)*i + 1) - 
 5*int((sec(c + d*x)**(1/3)*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x))/(tan(c 
+ d*x)**2 + 1),x)*tan(c + d*x)**2*d - 5*int((sec(c + d*x)**(1/3)*sqrt(tan( 
c + d*x)*i + 1)*tan(c + d*x))/(tan(c + d*x)**2 + 1),x)*d))/(3*a*d*(tan(c + 
 d*x)**2 + 1))