\(\int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [296]

Optimal result

Integrand size = 17, antiderivative size = 184 \[ \int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}} \] Output:

-1/8*x*2^(2/3)/a^(1/3)+1/4*I*3^(1/2)*arctan(1/3*(a^(1/3)+2^(2/3)*(a+I*a*ta 
n(d*x+c))^(1/3))*3^(1/2)/a^(1/3))*2^(2/3)/a^(1/3)/d+1/8*I*ln(cos(d*x+c))*2 
^(2/3)/a^(1/3)/d+3/8*I*ln(2^(1/3)*a^(1/3)-(a+I*a*tan(d*x+c))^(1/3))*2^(2/3 
)/a^(1/3)/d+3/2*I/d/(a+I*a*tan(d*x+c))^(1/3)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.06 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.27 \[ \int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {3 i \operatorname {Hypergeometric2F1}\left (-\frac {1}{3},1,\frac {2}{3},\frac {1}{2} (1+i \tan (c+d x))\right )}{2 d \sqrt [3]{a+i a \tan (c+d x)}} \] Input:

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

Output:

(((3*I)/2)*Hypergeometric2F1[-1/3, 1, 2/3, (1 + I*Tan[c + d*x])/2])/(d*(a 
+ I*a*Tan[c + d*x])^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.74, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.471, Rules used = {3042, 3960, 3042, 3962, 67, 16, 1082, 217}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3960

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3962

\(\displaystyle \frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 d}\)

\(\Big \downarrow \) 67

\(\displaystyle \frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 i}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {i \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}\)

Input:

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

Output:

((-1/2*I)*(((-I)*Sqrt[3]*ArcTanh[(a*Tan[c + d*x])/Sqrt[3]])/(2^(1/3)*a^(1/ 
3)) - (3*Log[2^(1/3)*a^(1/3) - I*a*Tan[c + d*x]])/(2*2^(1/3)*a^(1/3)) + Lo 
g[a - I*a*Tan[c + d*x]]/(2*2^(1/3)*a^(1/3))))/d + ((3*I)/2)/(d*(a + I*a*Ta 
n[c + d*x])^(1/3))
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 67

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3960

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[a*((a + 
b*Tan[c + d*x])^n/(2*b*d*n)), x] + Simp[1/(2*a)   Int[(a + b*Tan[c + d*x])^ 
(n + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && LtQ[n, 0]
 

rule 3962

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[-b/d   S 
ubst[Int[(a + x)^(n - 1)/(a - x), x], x, b*Tan[c + d*x]], x] /; FreeQ[{a, b 
, c, d, n}, x] && EqQ[a^2 + b^2, 0]
 

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.86

method	result	size
derivativedivides	\(\frac {3 i a \left (\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}}{2 a}+\frac {1}{2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d}\)	\(158\)
default	\(\frac {3 i a \left (\frac {\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}}{2 a}+\frac {1}{2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d}\)	\(158\)

Input:

int(1/(a+I*a*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)
 

Output:

3*I/d*a*(1/2/a*(1/6*2^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^ 
(1/3))-1/12*2^(2/3)/a^(1/3)*ln((a+I*a*tan(d*x+c))^(2/3)+2^(1/3)*a^(1/3)*(a 
+I*a*tan(d*x+c))^(1/3)+2^(2/3)*a^(2/3))+1/6*3^(1/2)*2^(2/3)/a^(1/3)*arctan 
(1/3*3^(1/2)*(2^(2/3)/a^(1/3)*(a+I*a*tan(d*x+c))^(1/3)+1)))+1/2/a/(a+I*a*t 
an(d*x+c))^(1/3))
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (125) = 250\).

Time = 0.09 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.71 \[ \int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {{\left (4 \, a d \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (8 \, a d^{2} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (-i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )} - 2 \, {\left (-i \, \sqrt {3} a d + a d\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-4 \, {\left (i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 2 \, {\left (i \, \sqrt {3} a d + a d\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-4 \, {\left (-i \, \sqrt {3} a d^{2} + a d^{2}\right )} \left (-\frac {i}{16 \, a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right )\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \] Input:

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

Output:

1/4*(4*a*d*(-1/16*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(8*a*d^2*(-1/16* 
I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d* 
x + 2/3*I*c)) - 3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(-I*e^(2*I*d 
*x + 2*I*c) - I)*e^(4/3*I*d*x + 4/3*I*c) - 2*(-I*sqrt(3)*a*d + a*d)*(-1/16 
*I/(a*d^3))^(1/3)*e^(2*I*d*x + 2*I*c)*log(-4*(I*sqrt(3)*a*d^2 + a*d^2)*(-1 
/16*I/(a*d^3))^(2/3) + 2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3* 
I*d*x + 2/3*I*c)) - 2*(I*sqrt(3)*a*d + a*d)*(-1/16*I/(a*d^3))^(1/3)*e^(2*I 
*d*x + 2*I*c)*log(-4*(-I*sqrt(3)*a*d^2 + a*d^2)*(-1/16*I/(a*d^3))^(2/3) + 
2^(1/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(1/3)*e^(2/3*I*d*x + 2/3*I*c)))*e^(- 
2*I*d*x - 2*I*c)/(a*d)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {12 \, a}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{8 \, a d} \] Input:

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

Output:

1/8*I*(2*sqrt(3)*2^(2/3)*a^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/ 
3) + 2*(I*a*tan(d*x + c) + a)^(1/3))/a^(1/3)) - 2^(2/3)*a^(2/3)*log(2^(2/3 
)*a^(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + (I*a*tan(d*x + 
c) + a)^(2/3)) + 2*2^(2/3)*a^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c 
) + a)^(1/3)) + 12*a/(I*a*tan(d*x + c) + a)^(1/3))/(a*d)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.80 \[ \int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {-\frac {2 i \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} + \frac {i \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {1}{3}}} - \frac {2 i \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {1}{3}}} - \frac {12 i}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}}{8 \, d} \] Input:

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

Output:

-1/8*(-2*I*sqrt(3)*2^(2/3)*arctan(1/6*sqrt(3)*2^(2/3)*(2^(1/3)*a^(1/3) + 2 
*(I*a*tan(d*x + c) + a)^(1/3))/a^(1/3))/a^(1/3) + I*2^(2/3)*log(2^(2/3)*a^ 
(2/3) + 2^(1/3)*(I*a*tan(d*x + c) + a)^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + 
 a)^(2/3))/a^(1/3) - 2*I*2^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) 
+ a)^(1/3))/a^(1/3) - 12*I/(I*a*tan(d*x + c) + a)^(1/3))/d
 

Mupad [B] (verification not implemented)

Time = 1.66 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.07 \[ \int \frac {1}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {3{}\mathrm {i}}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\right )}{{\left (-a\right )}^{1/3}\,d}-\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{4\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{1/3}\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{4\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{1/3}\,d} \] Input:

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

Output:

3i/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) + ((1i/16)^(1/3)*log((a*(tan(c + d* 
x)*1i + 1))^(1/3) - (-1)^(1/3)*2^(1/3)*(-a)^(1/3)))/((-a)^(1/3)*d) - ((1i/ 
16)^(1/3)*log((9*(-1)^(1/3)*2^(1/3)*(-a)^(1/3)*(3^(1/2)*1i - 1))/(8*d^2) - 
 (9*(a + a*tan(c + d*x)*1i)^(1/3))/(4*d^2))*((3^(1/2)*1i)/2 + 1/2))/((-a)^ 
(1/3)*d) + ((1i/16)^(1/3)*log(- (9*(a + a*tan(c + d*x)*1i)^(1/3))/(4*d^2) 
- (9*(-1)^(1/3)*2^(1/3)*(-a)^(1/3)*(3^(1/2)*1i + 1))/(8*d^2))*((3^(1/2)*1i 
)/2 - 1/2))/((-a)^(1/3)*d)
 

Reduce [F]

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

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

Output:

int(1/(tan(c + d*x)*i + 1)**(1/3),x)/a**(1/3)