\(\int \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx\) [273]

Optimal result

Integrand size = 26, antiderivative size = 185 \[ \int \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {\sqrt [3]{a} x}{2\ 2^{2/3}}+\frac {i \sqrt {3} \sqrt [3]{a} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2^{2/3} d}-\frac {i \sqrt [3]{a} \log (\cos (c+d x))}{2\ 2^{2/3} d}-\frac {3 i \sqrt [3]{a} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2\ 2^{2/3} d}-\frac {3 i (a+i a \tan (c+d x))^{4/3}}{4 a d} \] Output:

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

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.08 \[ \int \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {i \left (2 \sqrt [3]{2} \sqrt {3} a^{4/3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )-2 \sqrt [3]{2} a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )+\sqrt [3]{2} a^{4/3} \log \left (2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} \sqrt [3]{a+i a \tan (c+d x)}+(a+i a \tan (c+d x))^{2/3}\right )-3 (a+i a \tan (c+d x))^{4/3}\right )}{4 a d} \] Input:

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

Output:

((I/4)*(2*2^(1/3)*Sqrt[3]*a^(4/3)*ArcTan[(1 + (2^(2/3)*(a + I*a*Tan[c + d* 
x])^(1/3))/a^(1/3))/Sqrt[3]] - 2*2^(1/3)*a^(4/3)*Log[2^(1/3)*a^(1/3) - (a 
+ I*a*Tan[c + d*x])^(1/3)] + 2^(1/3)*a^(4/3)*Log[2^(2/3)*a^(2/3) + 2^(1/3) 
*a^(1/3)*(a + I*a*Tan[c + d*x])^(1/3) + (a + I*a*Tan[c + d*x])^(2/3)] - 3* 
(a + I*a*Tan[c + d*x])^(4/3)))/(a*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.41 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.75, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 4026, 25, 3042, 3962, 69, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4026

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3962

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

(I*a*((I*Sqrt[3]*ArcTanh[(a*Tan[c + d*x])/Sqrt[3]])/(2^(2/3)*a^(2/3)) - (3 
*Log[2^(1/3)*a^(1/3) - I*a*Tan[c + d*x]])/(2*2^(2/3)*a^(2/3)) + Log[a - I* 
a*Tan[c + d*x]]/(2*2^(2/3)*a^(2/3))))/d - (((3*I)/4)*(a + I*a*Tan[c + d*x] 
)^(4/3))/(a*d)
 

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 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 69

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q^2), 
 x] + (-Simp[3/(2*b*q)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1 
/3)], x] - Simp[3/(2*b*q^2)   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 4026

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

Maple [A] (verified)

Time = 0.97 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.85

method	result	size
derivativedivides	\(\frac {3 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4}-a^{2} \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right )\right )}{d a}\)	\(157\)
default	\(\frac {3 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4}-a^{2} \left (\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {2}{3}}}-\frac {2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {2}{3}}}\right )\right )}{d a}\)	\(157\)

Input:

int(tan(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)
 

Output:

3*I/d/a*(-1/4*(a+I*a*tan(d*x+c))^(4/3)-a^2*(1/6*2^(1/3)/a^(2/3)*ln((a+I*a* 
tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/12*2^(1/3)/a^(2/3)*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*2^ 
(1/3)/a^(2/3)*3^(1/2)*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. 294 vs. \(2 (128) = 256\).

Time = 0.09 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.59 \[ \int \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {{\left ({\left (i \, \sqrt {3} d - d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {3} d - d\right )} \left (\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + {\left (\sqrt {3} d + i \, d\right )} \left (\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}}\right ) + {\left ({\left (-i \, \sqrt {3} d - d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {3} d - d\right )} \left (\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - {\left (\sqrt {3} d - i \, d\right )} \left (\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}}\right ) + 2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}} \log \left (2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - 2 i \, d \left (\frac {i \, a}{4 \, d^{3}}\right )^{\frac {1}{3}}\right ) - 3 i \cdot 2^{\frac {1}{3}} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {8}{3} i \, d x + \frac {8}{3} i \, c\right )}}{2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((I*a*(tan(c + d*x) - I))**(1/3)*tan(c + d*x)**2, x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.83 \[ \int \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {i \, {\left (2 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {10}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2^{\frac {1}{3}} a^{\frac {10}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 \cdot 2^{\frac {1}{3}} a^{\frac {10}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a^{2}\right )}}{4 \, a^{3} d} \] Input:

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

Output:

1/4*I*(2*sqrt(3)*2^(1/3)*a^(10/3)*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^(1/3)*a^(10/3)*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^(1/3)*a^(10/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x 
+ c) + a)^(1/3)) - 3*(I*a*tan(d*x + c) + a)^(4/3)*a^2)/(a^3*d)
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.82 \[ \int \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=\frac {2 i \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {7}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + i \cdot 2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 2 i \cdot 2^{\frac {1}{3}} a^{\frac {7}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 3 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a}{4 \, a^{2} d} \] Input:

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

Output:

1/4*(2*I*sqrt(3)*2^(1/3)*a^(7/3)*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^(1/3)*a^(7/3)*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^(1/3)*a^(7/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x 
 + c) + a)^(1/3)) - 3*I*(I*a*tan(d*x + c) + a)^(4/3)*a)/(a^2*d)
 

Mupad [B] (verification not implemented)

Time = 1.24 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.05 \[ \int \tan ^2(c+d x) \sqrt [3]{a+i a \tan (c+d x)} \, dx=-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}\,3{}\mathrm {i}}{4\,a\,d}+\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,a^{1/3}\,\ln \left (18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,d^2+a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}\right )}{d}+\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,a^{1/3}\,\ln \left (a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}+18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,d^2\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}-\frac {{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,a^{1/3}\,\ln \left (a\,d^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}\,9{}\mathrm {i}-18\,{\left (\frac {1}{4}{}\mathrm {i}\right )}^{1/3}\,a^{4/3}\,d^2\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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