\(\int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [293]

Optimal result

Integrand size = 26, antiderivative size = 237 \[ \int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {i x}{4 \sqrt [3]{2} \sqrt [3]{a}}-\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {\log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d} \] 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*ta 
n(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+21/10/d/(a+I*a*tan(d*x+c))^(1/3)+3/5*tan(d*x+c)^2/d/(a+I*a*tan(d*x 
+c))^(1/3)+3/10*(a+I*a*tan(d*x+c))^(2/3)/a/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.95 \[ \int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {\log (i+\tan (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {21}{10 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 (a+i a \tan (c+d x))^{2/3}}{10 a d} \] Input:

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

Output:

-1/2*(Sqrt[3]*ArcTan[(a^(1/3) + 2^(2/3)*(a + I*a*Tan[c + d*x])^(1/3))/(Sqr 
t[3]*a^(1/3))])/(2^(1/3)*a^(1/3)*d) + Log[I + Tan[c + d*x]]/(4*2^(1/3)*a^( 
1/3)*d) - (3*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)])/(4*2^(1/ 
3)*a^(1/3)*d) + 21/(10*d*(a + I*a*Tan[c + d*x])^(1/3)) + (3*Tan[c + d*x]^2 
)/(5*d*(a + I*a*Tan[c + d*x])^(1/3)) + (3*(a + I*a*Tan[c + d*x])^(2/3))/(1 
0*a*d)
 

Rubi [A] (warning: unable to verify)

Time = 0.71 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.84, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4043, 27, 3042, 4075, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4043

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4075

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4009

\(\displaystyle \frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {-\frac {5}{2} i \int (i \tan (c+d x) a+a)^{2/3}dx-\frac {21 a}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 (a+i a \tan (c+d x))^{2/3}}{2 d}}{5 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {-\frac {5}{2} i \int (i \tan (c+d x) a+a)^{2/3}dx-\frac {21 a}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 (a+i a \tan (c+d x))^{2/3}}{2 d}}{5 a}\)

\(\Big \downarrow \) 3962

\(\displaystyle \frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {-\frac {5 a \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{2 d}-\frac {21 a}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 (a+i a \tan (c+d x))^{2/3}}{2 d}}{5 a}\)

\(\Big \downarrow \) 67

\(\displaystyle \frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {-\frac {5 a \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}-\frac {21 a}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 (a+i a \tan (c+d x))^{2/3}}{2 d}}{5 a}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {-\frac {5 a \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}-\frac {21 a}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 (a+i a \tan (c+d x))^{2/3}}{2 d}}{5 a}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {-\frac {5 a \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}-\frac {21 a}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 (a+i a \tan (c+d x))^{2/3}}{2 d}}{5 a}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 \tan ^2(c+d x)}{5 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {-\frac {5 a \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{2 d}-\frac {21 a}{2 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 (a+i a \tan (c+d x))^{2/3}}{2 d}}{5 a}\)

Input:

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

Output:

(3*Tan[c + d*x]^2)/(5*d*(a + I*a*Tan[c + d*x])^(1/3)) - ((-5*a*(((-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))))/(2*d) - (21*a)/(2*d*(a + I*a*Tan[c + d*x])^(1/3)) 
 - (3*(a + I*a*Tan[c + d*x])^(2/3))/(2*d))/(5*a)
 

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 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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]
 

rule 4043

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

rule 4075

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

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.84

method	result	size
derivativedivides	\(-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5 d \,a^{2}}+\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 a d}+\frac {3}{2 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 d \,a^{\frac {1}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 d \,a^{\frac {1}{3}}}-\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 d \,a^{\frac {1}{3}}}\)	\(198\)
default	\(-\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5 d \,a^{2}}+\frac {3 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}}{2 a d}+\frac {3}{2 d \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{4 d \,a^{\frac {1}{3}}}+\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{8 d \,a^{\frac {1}{3}}}-\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{4 d \,a^{\frac {1}{3}}}\)	\(198\)

Input:

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

Output:

-3/5/d/a^2*(a+I*a*tan(d*x+c))^(5/3)+3/2*(a+I*a*tan(d*x+c))^(2/3)/a/d+3/2/d 
/(a+I*a*tan(d*x+c))^(1/3)-1/4/d/a^(1/3)*2^(2/3)*ln((a+I*a*tan(d*x+c))^(1/3 
)-2^(1/3)*a^(1/3))+1/8/d/a^(1/3)*2^(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))-1/4/d/a^(1/3)*3^(1/2) 
*2^(2/3)*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. 418 vs. \(2 (172) = 344\).

Time = 0.13 (sec) , antiderivative size = 418, normalized size of antiderivative = 1.76 \[ \int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

1/20*(3*2^(2/3)*(a/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(7*e^(4*I*d*x + 4*I*c) 
 + 20*e^(2*I*d*x + 2*I*c) + 5)*e^(4/3*I*d*x + 4/3*I*c) + 10*(1/2)^(1/3)*(a 
*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d*x + 2*I*c))*(-1/(a*d^3))^(1/3)*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)) - 5*(1/2)^(1/3)*((I*sqrt(3)*a*d + a*d) 
*e^(4*I*d*x + 4*I*c) + (I*sqrt(3)*a*d + a*d)*e^(2*I*d*x + 2*I*c))*(-1/(a*d 
^3))^(1/3)*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)) - 5* 
(1/2)^(1/3)*((-I*sqrt(3)*a*d + a*d)*e^(4*I*d*x + 4*I*c) + (-I*sqrt(3)*a*d 
+ a*d)*e^(2*I*d*x + 2*I*c))*(-1/(a*d^3))^(1/3)*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)))/(a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d* 
x + 2*I*c))
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.80 \[ \int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {10 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {11}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 5 \cdot 2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 10 \cdot 2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 24 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} a^{2} - 60 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{3} - \frac {60 \, a^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}}{40 \, a^{4} d} \] Input:

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

Output:

-1/40*(10*sqrt(3)*2^(2/3)*a^(11/3)*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)) - 5*2^(2/3)*a^(11/3)*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)) + 10*2^(2/3)*a^(11/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan( 
d*x + c) + a)^(1/3)) + 24*(I*a*tan(d*x + c) + a)^(5/3)*a^2 - 60*(I*a*tan(d 
*x + c) + a)^(2/3)*a^3 - 60*a^4/(I*a*tan(d*x + c) + a)^(1/3))/(a^4*d)
 

Giac [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.80 \[ \int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {\frac {10 \, \sqrt {3} 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right )}{a^{\frac {1}{3}}} - \frac {5 \cdot 2^{\frac {2}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right )}{a^{\frac {1}{3}}} + \frac {10 \cdot 2^{\frac {2}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}{a^{\frac {1}{3}}} - \frac {60}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}} + \frac {12 \, {\left (2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} a^{8} - 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{9}\right )}}{a^{10}}}{40 \, d} \] Input:

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

Output:

-1/40*(10*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) - 5*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) + 10*2^(2/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan(d*x + c) + 
a)^(1/3))/a^(1/3) - 60/(I*a*tan(d*x + c) + a)^(1/3) + 12*(2*(I*a*tan(d*x + 
 c) + a)^(5/3)*a^8 - 5*(I*a*tan(d*x + c) + a)^(2/3)*a^9)/a^10)/d
                                                                                    
                                                                                    
 

Mupad [B] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.96 \[ \int \frac {\tan ^3(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {3}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}+\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}}{2\,a\,d}-\frac {3\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}}{5\,a^2\,d}+\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}+9\,4^{2/3}\,{\left (-a\right )}^{1/3}\,d\right )}{4\,{\left (-a\right )}^{1/3}\,d}+\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}+144\,4^{2/3}\,{\left (-a\right )}^{1/3}\,d\,{\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (-\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}{{\left (-a\right )}^{1/3}\,d}-\frac {4^{1/3}\,\ln \left (18\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}+144\,4^{2/3}\,{\left (-a\right )}^{1/3}\,d\,{\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}^2\right )\,\left (\frac {1}{8}+\frac {\sqrt {3}\,1{}\mathrm {i}}{8}\right )}{{\left (-a\right )}^{1/3}\,d} \] Input:

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

Output:

3/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) + (3*(a + a*tan(c + d*x)*1i)^(2/3))/ 
(2*a*d) - (3*(a + a*tan(c + d*x)*1i)^(5/3))/(5*a^2*d) + (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) 
+ (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]

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

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

Output:

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