\(\int \frac {\tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [295]

Optimal result

Integrand size = 24, antiderivative size = 178 \[ \int \frac {\tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3}{2 d \sqrt [3]{a+i a \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {\frac {2\ 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{a}}-\frac {2^{2/3} \log (i+\tan (c+d x))}{\sqrt [3]{a}}+\frac {3\ 2^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{\sqrt [3]{a}}-\frac {12}{\sqrt [3]{a+i a \tan (c+d x)}}}{8 d} \] Input:

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

Output:

((2*2^(2/3)*Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3) 
)/(Sqrt[3]*a^(1/3))])/a^(1/3) - (2^(2/3)*Log[I + Tan[c + d*x]])/a^(1/3) + 
(3*2^(2/3)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/a^(1/3) - 
12/(a + I*a*Tan[c + d*x])^(1/3))/(8*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.38 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.74, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4009, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4009

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3962

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

rule 4009

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-(b*c - a*d))*((a + b*Tan[e + f*x])^m/(2*a 
*f*m)), x] + Simp[(b*c + a*d)/(2*a*b)   Int[(a + b*Tan[e + f*x])^(m + 1), x 
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2 
, 0] && LtQ[m, 0]
 

Maple [A] (verified)

Time = 0.92 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.82

method	result	size
derivativedivides	\(\frac {-\frac {3}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 a^{\frac {1}{3}}}}{d}\)	\(146\)
default	\(\frac {-\frac {3}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 a^{\frac {1}{3}}}}{d}\)	\(146\)

Input:

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

Output:

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

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. 327 vs. \(2 (125) = 250\).

Time = 0.09 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.84 \[ \int \frac {\tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {{\left (2 \, \left (\frac {1}{2}\right )^{\frac {1}{3}} a d \left (\frac {1}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-2 \, \left (\frac {1}{2}\right )^{\frac {2}{3}} a d^{2} \left (\frac {1}{a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (i \, \sqrt {3} a d + a d\right )} \left (\frac {1}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (i \, \sqrt {3} a d^{2} - a d^{2}\right )} \left (\frac {1}{a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - \left (\frac {1}{2}\right )^{\frac {1}{3}} {\left (-i \, \sqrt {3} a d + a d\right )} \left (\frac {1}{a d^{3}}\right )^{\frac {1}{3}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\left (\frac {1}{2}\right )^{\frac {2}{3}} {\left (-i \, \sqrt {3} a d^{2} - a d^{2}\right )} \left (\frac {1}{a d^{3}}\right )^{\frac {2}{3}} + 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )}\right ) - 3 \cdot 2^{\frac {2}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {2}{3}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (\frac {4}{3} i \, d x + \frac {4}{3} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.87 \[ \int \frac {\tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {2 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {5}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 2 \cdot 2^{\frac {2}{3}} a^{\frac {5}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - \frac {12 \, a^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}}{8 \, a^{2} d} \] Input:

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

Output:

1/8*(2*sqrt(3)*2^(2/3)*a^(5/3)*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^(5/3)*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^(5/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) 
+ a)^(1/3)) - 12*a^2/(I*a*tan(d*x + c) + a)^(1/3))/(a^2*d)
 

Giac [A] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.85 \[ \int \frac {\tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (-2 i \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + i \cdot 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 i \cdot 2^{\frac {2}{3}} a^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + \frac {12 i \, a}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{8 \, a d} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.97 \[ \int \frac {\tan (c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {3}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}+\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-9\,4^{2/3}\,a^{1/3}\,d\right )}{4\,a^{1/3}\,d}+\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-144\,4^{2/3}\,a^{1/3}\,d\,{\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}{a^{1/3}\,d}-\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}-144\,4^{2/3}\,a^{1/3}\,d\,{\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}{a^{1/3}\,d} \] Input:

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

Output:

(4^(1/3)*log(18*d*(a + a*tan(c + d*x)*1i)^(1/3) - 9*4^(2/3)*a^(1/3)*d))/(4 
*a^(1/3)*d) - 3/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) + (4^(1/3)*log(18*d*(a 
 + a*tan(c + d*x)*1i)^(1/3) - 144*4^(2/3)*a^(1/3)*d*((3^(1/2)*1i)/8 - 1/8) 
^2)*((3^(1/2)*1i)/8 - 1/8))/(a^(1/3)*d) - (4^(1/3)*log(18*d*(a + a*tan(c + 
 d*x)*1i)^(1/3) - 144*4^(2/3)*a^(1/3)*d*((3^(1/2)*1i)/8 + 1/8)^2)*((3^(1/2 
)*1i)/8 + 1/8))/(a^(1/3)*d)
 

Reduce [F]

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

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

Output:

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