3.5.41 \(\int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx\) [441]

3.5.41.1 Optimal result

Integrand size = 30, antiderivative size = 83 \[ \int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {5}{3},\frac {5}{6},\frac {1}{2} (1-i \tan (c+d x))\right ) (1+i \tan (c+d x))^{2/3}}{2^{2/3} d \sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

output

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

3.5.41.2 Mathematica [A] (verified)

Time = 1.35 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {12 i-\frac {30 i e^{2 i (c+d x)} \operatorname {Hypergeometric2F1}\left (\frac {1}{6},\frac {1}{3},\frac {4}{3},-e^{2 i (c+d x)}\right )}{\left (1+e^{2 i (c+d x)}\right )^{5/6}}}{16 d \sqrt [3]{e \sec (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

input

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

output

(12*I - ((30*I)*E^((2*I)*(c + d*x))*Hypergeometric2F1[1/6, 1/3, 4/3, -E^(( 
2*I)*(c + d*x))])/(1 + E^((2*I)*(c + d*x)))^(5/6))/(16*d*(e*Sec[c + d*x])^ 
(1/3)*Sqrt[a + I*a*Tan[c + d*x]])
 

3.5.41.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.233, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3986

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 80

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 79

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

input

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

output

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

3.5.41.3.1 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]
 

3.5.41.4 Maple [F]

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

input

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

output

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

3.5.41.5 Fricas [F]

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

input

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

output

-1/8*(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)*(4*I*e^(6*I*d*x + 6*I*c) + 9*I*e^(4*I*d*x + 4*I*c) + 6*I*e^(2* 
I*d*x + 2*I*c) + I)*e^(2/3*I*d*x + 2/3*I*c) - 8*(a*d*e*e^(4*I*d*x + 4*I*c) 
 - a*d*e*e^(2*I*d*x + 2*I*c))*integral(-15/16*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)*(3*I*e^(4*I*d*x + 4*I*c) 
 + 4*I*e^(2*I*d*x + 2*I*c) + I)*e^(2/3*I*d*x + 2/3*I*c)/(a*d*e*e^(6*I*d*x 
+ 6*I*c) - 2*a*d*e*e^(4*I*d*x + 4*I*c) + a*d*e*e^(2*I*d*x + 2*I*c)), x))/( 
a*d*e*e^(4*I*d*x + 4*I*c) - a*d*e*e^(2*I*d*x + 2*I*c))
 

3.5.41.6 Sympy [F]

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

input

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

output

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

3.5.41.7 Maxima [F]

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

input

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

output

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

3.5.41.8 Giac [F]

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

input

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

output

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

3.5.41.9 Mupad [F(-1)]

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

input

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

output

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