\(\int \frac {\tan ^4(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx\) [292]

Optimal result

Integrand size = 26, antiderivative size = 282 \[ \int \frac {\tan ^4(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {x}{4 \sqrt [3]{2} \sqrt [3]{a}}+\frac {i \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {i \log (\cos (c+d x))}{4 \sqrt [3]{2} \sqrt [3]{a} d}+\frac {3 i \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right )}{4 \sqrt [3]{2} \sqrt [3]{a} d}-\frac {15 i \tan ^2(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}+\frac {45 i (a+i a \tan (c+d x))^{2/3}}{8 a d}-\frac {39 i (a+i a \tan (c+d x))^{5/3}}{20 a^2 d} \] 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-15/8*I*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/3)+3/8*tan(d*x+c)^ 
3/d/(a+I*a*tan(d*x+c))^(1/3)+45/8*I*(a+I*a*tan(d*x+c))^(2/3)/a/d-39/20*I*( 
a+I*a*tan(d*x+c))^(5/3)/a^2/d
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.89 \[ \int \frac {\tan ^4(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {-69 \sqrt [3]{a} \tan (c+d x)+3 i \sqrt [3]{a} \tan ^2(c+d x)+15 \sqrt [3]{a} \tan ^3(c+d x)+10 i 2^{2/3} \sqrt {3} \arctan \left (\frac {\sqrt [3]{a}+2^{2/3} \sqrt [3]{a+i a \tan (c+d x)}}{\sqrt {3} \sqrt [3]{a}}\right ) \sqrt [3]{a+i a \tan (c+d x)}+i \left (147 \sqrt [3]{a}-5\ 2^{2/3} \log (i+\tan (c+d x)) \sqrt [3]{a+i a \tan (c+d x)}+15\ 2^{2/3} \log \left (\sqrt [3]{2} \sqrt [3]{a}-\sqrt [3]{a+i a \tan (c+d x)}\right ) \sqrt [3]{a+i a \tan (c+d x)}\right )}{40 \sqrt [3]{a} d \sqrt [3]{a+i a \tan (c+d x)}} \] Input:

Integrate[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(1/3),x]
 

Output:

(-69*a^(1/3)*Tan[c + d*x] + (3*I)*a^(1/3)*Tan[c + d*x]^2 + 15*a^(1/3)*Tan[ 
c + d*x]^3 + (10*I)*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 + I*a*Tan[c + d*x])^(1/3) + I*(147 
*a^(1/3) - 5*2^(2/3)*Log[I + Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(1/3) + 
15*2^(2/3)*Log[2^(1/3)*a^(1/3) - (a + I*a*Tan[c + d*x])^(1/3)]*(a + I*a*Ta 
n[c + d*x])^(1/3)))/(40*a^(1/3)*d*(a + I*a*Tan[c + d*x])^(1/3))
 

Rubi [A] (warning: unable to verify)

Time = 0.97 (sec) , antiderivative size = 246, normalized size of antiderivative = 0.87, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.615, Rules used = {3042, 4043, 27, 3042, 4078, 27, 3042, 4075, 3042, 4010, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4043

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \int \frac {\tan ^2(c+d x) (9 a-i a \tan (c+d x))}{3 \sqrt [3]{i \tan (c+d x) a+a}}dx}{8 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\int \frac {\tan ^2(c+d x) (9 a-i a \tan (c+d x))}{\sqrt [3]{i \tan (c+d x) a+a}}dx}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\int \frac {\tan (c+d x)^2 (9 a-i a \tan (c+d x))}{\sqrt [3]{i \tan (c+d x) a+a}}dx}{8 a}\)

\(\Big \downarrow \) 4078

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {3 \int \frac {4}{3} \tan (c+d x) (i \tan (c+d x) a+a)^{2/3} \left (13 \tan (c+d x) a^2+15 i a^2\right )dx}{2 a^2}}{8 a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \int \tan (c+d x) (i \tan (c+d x) a+a)^{2/3} \left (13 \tan (c+d x) a^2+15 i a^2\right )dx}{a^2}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \int \tan (c+d x) (i \tan (c+d x) a+a)^{2/3} \left (13 \tan (c+d x) a^2+15 i a^2\right )dx}{a^2}}{8 a}\)

\(\Big \downarrow \) 4075

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (\int (i \tan (c+d x) a+a)^{2/3} \left (15 i a^2 \tan (c+d x)-13 a^2\right )dx-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (\int (i \tan (c+d x) a+a)^{2/3} \left (15 i a^2 \tan (c+d x)-13 a^2\right )dx-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

\(\Big \downarrow \) 4010

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (2 a^2 \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {45 i a^2 (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (2 a^2 \int (i \tan (c+d x) a+a)^{2/3}dx+\frac {45 i a^2 (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

\(\Big \downarrow \) 3962

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (-\frac {2 i a^3 \int \frac {1}{(a-i a \tan (c+d x)) \sqrt [3]{i \tan (c+d x) a+a}}d(i a \tan (c+d x))}{d}+\frac {45 i a^2 (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

\(\Big \downarrow \) 67

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (-\frac {2 i a^3 \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}+\frac {3 \int \frac {1}{\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)}d\sqrt [3]{i \tan (c+d x) a+a}}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}+\frac {45 i a^2 (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (-\frac {2 i a^3 \left (-\frac {3}{2} \int \frac {1}{-a^2 \tan ^2(c+d x)+i \sqrt [3]{2} a^{4/3} \tan (c+d x)+2^{2/3} a^{2/3}}d\sqrt [3]{i \tan (c+d x) a+a}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}+\frac {45 i a^2 (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (-\frac {2 i a^3 \left (\frac {3 \int \frac {1}{a^2 \tan ^2(c+d x)-3}d\left (i 2^{2/3} a^{2/3} \tan (c+d x)+1\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}+\frac {45 i a^2 (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 \tan ^3(c+d x)}{8 d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {\frac {15 i a \tan ^2(c+d x)}{d \sqrt [3]{a+i a \tan (c+d x)}}-\frac {2 \left (-\frac {2 i a^3 \left (-\frac {i \sqrt {3} \text {arctanh}\left (\frac {a \tan (c+d x)}{\sqrt {3}}\right )}{\sqrt [3]{2} \sqrt [3]{a}}-\frac {3 \log \left (\sqrt [3]{2} \sqrt [3]{a}-i a \tan (c+d x)\right )}{2 \sqrt [3]{2} \sqrt [3]{a}}+\frac {\log (a-i a \tan (c+d x))}{2 \sqrt [3]{2} \sqrt [3]{a}}\right )}{d}+\frac {45 i a^2 (a+i a \tan (c+d x))^{2/3}}{2 d}-\frac {39 i a (a+i a \tan (c+d x))^{5/3}}{5 d}\right )}{a^2}}{8 a}\)

Input:

Int[Tan[c + d*x]^4/(a + I*a*Tan[c + d*x])^(1/3),x]
 

Output:

(3*Tan[c + d*x]^3)/(8*d*(a + I*a*Tan[c + d*x])^(1/3)) - (((15*I)*a*Tan[c + 
 d*x]^2)/(d*(a + I*a*Tan[c + d*x])^(1/3)) - (2*(((-2*I)*a^3*(((-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 + (((45*I)/2)*a^2*(a + I*a*Tan[c + d*x])^(2/3))/d 
- (((39*I)/5)*a*(a + I*a*Tan[c + d*x])^(5/3))/d))/a^2)/(8*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 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]
 

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]
 

rule 4078

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

Maple [A] (verified)

Time = 1.03 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {3 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\frac {a^{3} \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right )}{2}+\frac {a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d \,a^{3}}\)	\(211\)
default	\(\frac {3 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {8}{3}}}{8}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{3}}}{5}+a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+\frac {a^{3} \left (\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}-2^{\frac {1}{3}} a^{\frac {1}{3}}\right )}{6 a^{\frac {1}{3}}}-\frac {2^{\frac {2}{3}} \ln \left (\left (a +i a \tan \left (d x +c \right )\right )^{\frac {2}{3}}+2^{\frac {1}{3}} a^{\frac {1}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}+2^{\frac {2}{3}} a^{\frac {2}{3}}\right )}{12 a^{\frac {1}{3}}}+\frac {\sqrt {3}\, 2^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}{a^{\frac {1}{3}}}+1\right )}{3}\right )}{6 a^{\frac {1}{3}}}\right )}{2}+\frac {a^{3}}{2 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {1}{3}}}\right )}{d \,a^{3}}\)	\(211\)

Input:

int(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/3),x,method=_RETURNVERBOSE)
 

Output:

3*I/d/a^3*(1/8*(a+I*a*tan(d*x+c))^(8/3)-2/5*a*(a+I*a*tan(d*x+c))^(5/3)+a^2 
*(a+I*a*tan(d*x+c))^(2/3)+1/2*a^3*(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^3/(a+I*a*tan(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. 479 vs. \(2 (201) = 402\).

Time = 0.09 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.70 \[ \int \frac {\tan ^4(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

integrate(tan(d*x+c)^4/(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)*(-19*I*e^(6*I*d*x + 6 
*I*c) - 39*I*e^(4*I*d*x + 4*I*c) - 35*I*e^(2*I*d*x + 2*I*c) - 5*I)*e^(4/3* 
I*d*x + 4/3*I*c) - 20*(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) 
 + a*d*e^(2*I*d*x + 2*I*c))*(-1/16*I/(a*d^3))^(1/3)*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)) + 10*((-I*sqrt(3)*a*d + a*d)*e^(6*I*d*x + 6*I*c) + 2*(-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))*(-1/16*I/(a*d^3))^(1/3)*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)) + 10*((I*sqrt(3)*a*d + a*d)*e^(6*I*d*x + 6*I*c) + 2*(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))* 
(-1/16*I/(a*d^3))^(1/3)*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)))/(a*d*e^(6*I*d*x + 6*I*c) + 2*a*d*e^(4*I*d*x + 4*I*c) + a*d*e^(2*I*d* 
x + 2*I*c))
 

Sympy [F]

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

integrate(tan(d*x+c)**4/(a+I*a*tan(d*x+c))**(1/3),x)
 

Output:

Integral(tan(c + d*x)**4/(I*a*(tan(c + d*x) - I))**(1/3), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.74 \[ \int \frac {\tan ^4(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {i \, {\left (10 \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {14}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) - 5 \cdot 2^{\frac {2}{3}} a^{\frac {14}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) + 10 \cdot 2^{\frac {2}{3}} a^{\frac {14}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) + 15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {8}{3}} a^{2} - 48 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} a^{3} + 120 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{4} + \frac {60 \, a^{5}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}\right )}}{40 \, a^{5} d} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="maxima")
 

Output:

1/40*I*(10*sqrt(3)*2^(2/3)*a^(14/3)*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^(14/3)*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^(14/3)*log(-2^(1/3)*a^(1/3) + (I*a*tan 
(d*x + c) + a)^(1/3)) + 15*(I*a*tan(d*x + c) + a)^(8/3)*a^2 - 48*(I*a*tan( 
d*x + c) + a)^(5/3)*a^3 + 120*(I*a*tan(d*x + c) + a)^(2/3)*a^4 + 60*a^5/(I 
*a*tan(d*x + c) + a)^(1/3))/(a^5*d)
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.73 \[ \int \frac {\tan ^4(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=-\frac {-10 i \, \sqrt {3} 2^{\frac {2}{3}} a^{\frac {11}{3}} \arctan \left (\frac {\sqrt {3} 2^{\frac {2}{3}} {\left (2^{\frac {1}{3}} a^{\frac {1}{3}} + 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right )}}{6 \, a^{\frac {1}{3}}}\right ) + 5 i \cdot 2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (2^{\frac {2}{3}} a^{\frac {2}{3}} + 2^{\frac {1}{3}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}}\right ) - 10 i \cdot 2^{\frac {2}{3}} a^{\frac {11}{3}} \log \left (-2^{\frac {1}{3}} a^{\frac {1}{3}} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}\right ) - 15 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {8}{3}} a + 48 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{3}} a^{2} - 120 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {2}{3}} a^{3} - \frac {60 i \, a^{4}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {1}{3}}}}{40 \, a^{4} d} \] Input:

integrate(tan(d*x+c)^4/(a+I*a*tan(d*x+c))^(1/3),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

-1/40*(-10*I*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*I*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*I*2^(2/3)*a^(11/3)*log(-2^(1/3)*a^(1/3) + (I 
*a*tan(d*x + c) + a)^(1/3)) - 15*I*(I*a*tan(d*x + c) + a)^(8/3)*a + 48*I*( 
I*a*tan(d*x + c) + a)^(5/3)*a^2 - 120*I*(I*a*tan(d*x + c) + a)^(2/3)*a^3 - 
 60*I*a^4/(I*a*tan(d*x + c) + a)^(1/3))/(a^4*d)
 

Mupad [B] (verification not implemented)

Time = 1.96 (sec) , antiderivative size = 266, normalized size of antiderivative = 0.94 \[ \int \frac {\tan ^4(c+d x)}{\sqrt [3]{a+i a \tan (c+d x)}} \, dx=\frac {3{}\mathrm {i}}{2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{2/3}\,3{}\mathrm {i}}{a\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/3}\,6{}\mathrm {i}}{5\,a^2\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{8/3}\,3{}\mathrm {i}}{8\,a^3\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left ({\left (a\,\left (1+\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\right )}^{1/3}-{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\right )}{{\left (-a\right )}^{1/3}\,d}-\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{4\,d^2}+\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{1/3}\,d}+\frac {{\left (\frac {1}{16}{}\mathrm {i}\right )}^{1/3}\,\ln \left (-\frac {9\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{1/3}}{4\,d^2}-\frac {9\,{\left (-1\right )}^{1/3}\,2^{1/3}\,{\left (-a\right )}^{1/3}\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{8\,d^2}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{{\left (-a\right )}^{1/3}\,d} \] Input:

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

Output:

3i/(2*d*(a + a*tan(c + d*x)*1i)^(1/3)) + ((a + a*tan(c + d*x)*1i)^(2/3)*3i 
)/(a*d) - ((a + a*tan(c + d*x)*1i)^(5/3)*6i)/(5*a^2*d) + ((a + a*tan(c + d 
*x)*1i)^(8/3)*3i)/(8*a^3*d) + ((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)*l 
og((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]

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

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

Output:

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