\(\int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx\) [282]

Optimal result

Integrand size = 24, antiderivative size = 192 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\frac {i a^{4/3} x}{2^{2/3}}-\frac {\sqrt [3]{2} \sqrt {3} a^{4/3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{d}+\frac {a^{4/3} \log (\cos (c+d x))}{2^{2/3} d}+\frac {3 a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{2^{2/3} d}+\frac {3 a \sqrt [3]{a+i a \tan (c+d x)}}{d}+\frac {3 (a+i a \tan (c+d x))^{4/3}}{4 d} \] Output:

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

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.17 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\frac {-4 \sqrt [3]{2} \sqrt {3} a^{4/3} \arctan \left (\frac {1+\frac {2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 \sqrt [3]{2} a^{4/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )-2 \sqrt [3]{2} a^{4/3} \log \left (2^{2/3} a^{2/3}+\sqrt [3]{2} \sqrt [3]{a} \sqrt [3]{a+i a \tan (c+d x)}+(a+i a \tan (c+d x))^{2/3}\right )+15 a \sqrt [3]{a+i a \tan (c+d x)}+3 i a \tan (c+d x) \sqrt [3]{a+i a \tan (c+d x)}}{4 d} \] Input:

Integrate[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(4/3),x]
 

Output:

(-4*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]] + 4*2^(1/3)*a^(4/3)*Log[2^(1/3)*a^(1/3) - (a + I*a* 
Tan[c + d*x])^(1/3)] - 2*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)] + 15*a*( 
a + I*a*Tan[c + d*x])^(1/3) + (3*I)*a*Tan[c + d*x]*(a + I*a*Tan[c + d*x])^ 
(1/3))/(4*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.47 (sec) , antiderivative size = 165, normalized size of antiderivative = 0.86, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.417, Rules used = {3042, 4010, 3042, 3959, 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.

\(\displaystyle \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3}dx\)

\(\Big \downarrow \) 4010

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3959

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3962

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

\(\Big \downarrow \) 69

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[Tan[c + d*x]*(a + I*a*Tan[c + d*x])^(4/3),x]
 

Output:

(3*(a + I*a*Tan[c + d*x])^(4/3))/(4*d) - I*(((-2*I)*a^2*((I*Sqrt[3]*ArcTan 
h[(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)*a*(a + I*a*Tan[c + d*x])^(1/3))/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 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 3959

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((a + 
b*Tan[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[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] && GtQ[n 
, 1]
 

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 4010

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

Maple [A] (verified)

Time = 0.86 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.90

method	result	size
derivativedivides	\(\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4 d}+\frac {3 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}+\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{d}-\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{2 d}-\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{d}\)	\(173\)
default	\(\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {4}{3}}}{4 d}+\frac {3 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{d}+\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{d}-\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{2 d}-\frac {a^{\frac {4}{3}} 2^{\frac {1}{3}} \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{d}\)	\(173\)

Input:

int(tan(d*x+c)*(a+I*a*tan(d*x+c))^(4/3),x,method=_RETURNVERBOSE)
 

Output:

3/4*(a+I*a*tan(d*x+c))^(4/3)/d+3*a*(a+I*a*tan(d*x+c))^(1/3)/d+1/d*a^(4/3)* 
2^(1/3)*ln((a+I*a*tan(d*x+c))^(1/3)-2^(1/3)*a^(1/3))-1/2/d*a^(4/3)*2^(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/d*a^(4/3)*2^(1/3)*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. 350 vs. \(2 (144) = 288\).

Time = 0.08 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.82 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\frac {3 \cdot 2^{\frac {1}{3}} {\left (3 \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, a\right )} \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + 2^{\frac {1}{3}} {\left ({\left (-i \, \sqrt {3} d - d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, \sqrt {3} d - d\right )} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \cdot 2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + 2^{\frac {1}{3}} {\left (i \, \sqrt {3} d + d\right )} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) + 2^{\frac {1}{3}} {\left ({\left (i \, \sqrt {3} d - d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, \sqrt {3} d - d\right )} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2 \cdot 2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} + 2^{\frac {1}{3}} {\left (-i \, \sqrt {3} d + d\right )} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}}}{2 \, a}\right ) + 2 \cdot 2^{\frac {1}{3}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}} \log \left (\frac {2^{\frac {1}{3}} a \left (\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{\frac {1}{3}} e^{\left (\frac {2}{3} i \, d x + \frac {2}{3} i \, c\right )} - 2^{\frac {1}{3}} \left (\frac {a^{4}}{d^{3}}\right )^{\frac {1}{3}} d}{a}\right )}{2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {4}{3}} \tan {\left (c + d x \right )}\, dx \] Input:

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))**(4/3),x)
 

Output:

Integral((I*a*(tan(c + d*x) - I))**(4/3)*tan(c + d*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.90 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=-\frac {4 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {10}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2 \cdot 2^{\frac {1}{3}} a^{\frac {10}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 4 \cdot 2^{\frac {1}{3}} a^{\frac {10}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} a^{2} - 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{3}}{4 \, a^{2} d} \] Input:

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="maxima")
 

Output:

-1/4*(4*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*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)) - 4*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 - 12*(I*a*tan(d*x + 
 c) + a)^(1/3)*a^3)/(a^2*d)
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.85 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=-\frac {4 \, \sqrt {3} 2^{\frac {1}{3}} a^{\frac {4}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 2 \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 4 \cdot 2^{\frac {1}{3}} a^{\frac {4}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {4}{3}} - 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a}{4 \, d} \] Input:

integrate(tan(d*x+c)*(a+I*a*tan(d*x+c))^(4/3),x, algorithm="giac")
 

Output:

-1/4*(4*sqrt(3)*2^(1/3)*a^(4/3)*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^(1/3)*a^(4/3)*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)) - 4*2^(1/3)*a^(4/3)*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) - 12*(I*a*tan(d*x + c) + a 
)^(1/3)*a)/d
 

Mupad [B] (verification not implemented)

Time = 1.52 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.03 \[ \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{4/3}}{4\,d}+\frac {3\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {2^{1/3}\,a^{4/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-2^{1/3}\,a^{1/3}\right )}{d}+\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {18\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}-\frac {9\,2^{1/3}\,a^{7/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{d}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d}-\frac {2^{1/3}\,a^{4/3}\,\ln \left (\frac {18\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{d}+\frac {9\,2^{1/3}\,a^{7/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{d}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{d} \] Input:

int(tan(c + d*x)*(a + a*tan(c + d*x)*1i)^(4/3),x)
 

Output:

(3*(a + a*tan(c + d*x)*1i)^(4/3))/(4*d) + (3*a*(a + a*tan(c + d*x)*1i)^(1/ 
3))/d + (2^(1/3)*a^(4/3)*log((a*(tan(c + d*x)*1i + 1))^(1/3) - 2^(1/3)*a^( 
1/3)))/d + (2^(1/3)*a^(4/3)*log((18*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d - 
 (9*2^(1/3)*a^(7/3)*(3^(1/2)*1i - 1))/d)*((3^(1/2)*1i)/2 - 1/2))/d - (2^(1 
/3)*a^(4/3)*log((18*a^2*(a + a*tan(c + d*x)*1i)^(1/3))/d + (9*2^(1/3)*a^(7 
/3)*(3^(1/2)*1i + 1))/d)*((3^(1/2)*1i)/2 + 1/2))/d
 

Reduce [F]

\[ \int \tan (c+d x) (a+i a \tan (c+d x))^{4/3} \, dx=a^{\frac {4}{3}} \left (\left (\int \left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}} \tan \left (d x +c \right )^{2}d x \right ) i +\int \left (\tan \left (d x +c \right ) i +1\right )^{\frac {1}{3}} \tan \left (d x +c \right )d x \right ) \] Input:

int(tan(d*x+c)*(a+I*a*tan(d*x+c))^(4/3),x)
 

Output:

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