\(\int \frac {x}{a+b \tan (c+d \sqrt [3]{x})} \, dx\) [58]

Optimal result

Integrand size = 18, antiderivative size = 352 \[ \int \frac {x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\frac {x^2}{2 (a+i b)}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d}-\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^2}+\frac {15 b x \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\left (a^2+b^2\right ) d^3}+\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^4}-\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 \left (a^2+b^2\right ) d^5}-\frac {45 i b \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 \left (a^2+b^2\right ) d^6} \] Output:

x^2/(2*a+2*I*b)+3*b*x^(5/3)*ln(1+(a^2+b^2)*exp(2*I*(c+d*x^(1/3)))/(a+I*b)^ 
2)/(a^2+b^2)/d-15/2*I*b*x^(4/3)*polylog(2,-(a^2+b^2)*exp(2*I*(c+d*x^(1/3)) 
)/(a+I*b)^2)/(a^2+b^2)/d^2+15*b*x*polylog(3,-(a^2+b^2)*exp(2*I*(c+d*x^(1/3 
)))/(a+I*b)^2)/(a^2+b^2)/d^3+45/2*I*b*x^(2/3)*polylog(4,-(a^2+b^2)*exp(2*I 
*(c+d*x^(1/3)))/(a+I*b)^2)/(a^2+b^2)/d^4-45/2*b*x^(1/3)*polylog(5,-(a^2+b^ 
2)*exp(2*I*(c+d*x^(1/3)))/(a+I*b)^2)/(a^2+b^2)/d^5-45/4*I*b*polylog(6,-(a^ 
2+b^2)*exp(2*I*(c+d*x^(1/3)))/(a+I*b)^2)/(a^2+b^2)/d^6
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 310, normalized size of antiderivative = 0.88 \[ \int \frac {x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\frac {2 a d^6 x^2+2 i b d^6 x^2+12 b d^5 x^{5/3} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+30 i b d^4 x^{4/3} \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+60 b d^3 x \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-90 i b d^2 x^{2/3} \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-90 b d \sqrt [3]{x} \operatorname {PolyLog}\left (5,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+45 i b \operatorname {PolyLog}\left (6,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{4 \left (a^2+b^2\right ) d^6} \] Input:

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

Output:

(2*a*d^6*x^2 + (2*I)*b*d^6*x^2 + 12*b*d^5*x^(5/3)*Log[1 + (a + I*b)/((a - 
I*b)*E^((2*I)*(c + d*x^(1/3))))] + (30*I)*b*d^4*x^(4/3)*PolyLog[2, (-a - I 
*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + 60*b*d^3*x*PolyLog[3, (-a - I 
*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - (90*I)*b*d^2*x^(2/3)*PolyLog[ 
4, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] - 90*b*d*x^(1/3)*Poly 
Log[5, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + (45*I)*b*PolyLo 
g[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))])/(4*(a^2 + b^2)*d^6 
)
 

Rubi [A] (verified)

Time = 1.41 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4234, 3042, 4215, 2620, 3011, 7163, 7163, 7163, 2720, 7143}

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.

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

\(\Big \downarrow \) 4234

\(\displaystyle 3 \int \frac {x^{5/3}}{a+b \tan \left (c+d \sqrt [3]{x}\right )}d\sqrt [3]{x}\)

\(\Big \downarrow \) 3042

\(\displaystyle 3 \int \frac {x^{5/3}}{a+b \tan \left (c+d \sqrt [3]{x}\right )}d\sqrt [3]{x}\)

\(\Big \downarrow \) 4215

\(\displaystyle 3 \left (2 i b \int \frac {e^{2 i \left (c+d \sqrt [3]{x}\right )} x^{5/3}}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}d\sqrt [3]{x}+\frac {x^2}{6 (a+i b)}\right )\)

\(\Big \downarrow \) 2620

\(\displaystyle 3 \left (2 i b \left (\frac {5 i \int x^{4/3} \log \left (\frac {e^{2 i \left (c+d \sqrt [3]{x}\right )} \left (a^2+b^2\right )}{(a+i b)^2}+1\right )d\sqrt [3]{x}}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^2}{6 (a+i b)}\right )\)

\(\Big \downarrow \) 3011

\(\displaystyle 3 \left (2 i b \left (\frac {5 i \left (\frac {i x^{4/3} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \int x \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )d\sqrt [3]{x}}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^2}{6 (a+i b)}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 3 \left (2 i b \left (\frac {5 i \left (\frac {i x^{4/3} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \int x^{2/3} \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )d\sqrt [3]{x}}{2 d}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^2}{6 (a+i b)}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 3 \left (2 i b \left (\frac {5 i \left (\frac {i x^{4/3} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \left (\frac {i \int \sqrt [3]{x} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )d\sqrt [3]{x}}{d}-\frac {i x^{2/3} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^2}{6 (a+i b)}\right )\)

\(\Big \downarrow \) 7163

\(\displaystyle 3 \left (2 i b \left (\frac {5 i \left (\frac {i x^{4/3} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )d\sqrt [3]{x}}{2 d}-\frac {i \sqrt [3]{x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x^{2/3} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^2}{6 (a+i b)}\right )\)

\(\Big \downarrow \) 2720

\(\displaystyle 3 \left (2 i b \left (\frac {5 i \left (\frac {i x^{4/3} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {\int \frac {\operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{\sqrt [3]{x}}de^{2 i \left (c+d \sqrt [3]{x}\right )}}{4 d^2}-\frac {i \sqrt [3]{x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x^{2/3} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^2}{6 (a+i b)}\right )\)

\(\Big \downarrow \) 7143

\(\displaystyle 3 \left (2 i b \left (\frac {5 i \left (\frac {i x^{4/3} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}-\frac {2 i \left (\frac {3 i \left (\frac {i \left (\frac {\operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 d^2}-\frac {i \sqrt [3]{x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}-\frac {i x^{2/3} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{2 d}-\frac {i x \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d}\right )}{d}\right )}{2 d \left (a^2+b^2\right )}-\frac {i x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d \left (a^2+b^2\right )}\right )+\frac {x^2}{6 (a+i b)}\right )\)

Input:

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

Output:

3*(x^2/(6*(a + I*b)) + (2*I)*b*(((-1/2*I)*x^(5/3)*Log[1 + ((a^2 + b^2)*E^( 
(2*I)*(c + d*x^(1/3))))/(a + I*b)^2])/((a^2 + b^2)*d) + (((5*I)/2)*(((I/2) 
*x^(4/3)*PolyLog[2, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2) 
])/d - ((2*I)*(((-1/2*I)*x*PolyLog[3, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/ 
3))))/(a + I*b)^2)])/d + (((3*I)/2)*(((-1/2*I)*x^(2/3)*PolyLog[4, -(((a^2 
+ b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)])/d + (I*(((-1/2*I)*x^(1/3) 
*PolyLog[5, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)])/d + P 
olyLog[6, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)]/(4*d^2)) 
)/d))/d))/d))/((a^2 + b^2)*d)))
 

Defintions of rubi rules used

rule 2620

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/ 
((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp 
[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x] - Si 
mp[d*(m/(b*f*g*n*Log[F]))   Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x 
)))^n/a)], x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
 

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4215

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[(c + d*x)^(m + 1)/(d*(m + 1)*(a + I*b)), x] + Simp[2*I*b   In 
t[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Simp[2 
*I*(e + f*x), x])), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2 
, 0] && IGtQ[m, 0]
 

rule 4234

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol 
] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*Tan[c + d*x])^ 
p, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p}, x] && IGtQ[Simplify[(m + 
 1)/n], 0] && IntegerQ[p]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

rule 7163

Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_. 
)*(x_))))^(p_.)], x_Symbol] :> Simp[(e + f*x)^m*(PolyLog[n + 1, d*(F^(c*(a 
+ b*x)))^p]/(b*c*p*Log[F])), x] - Simp[f*(m/(b*c*p*Log[F]))   Int[(e + f*x) 
^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c 
, d, e, f, n, p}, x] && GtQ[m, 0]
 

Maple [F]

\[\int \frac {x}{a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )}d x\]

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(x/(b*tan(d*x^(1/3) + c) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 813 vs. \(2 (289) = 578\).

Time = 0.36 (sec) , antiderivative size = 813, normalized size of antiderivative = 2.31 \[ \int \frac {x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/10*(15*(2*(d*x^(1/3) + c)*a/(a^2 + b^2) + 2*b*log(b*tan(d*x^(1/3) + c) 
+ a)/(a^2 + b^2) - b*log(tan(d*x^(1/3) + c)^2 + 1)/(a^2 + b^2))*c^5 - (5*( 
d*x^(1/3) + c)^6*(a - I*b) - 30*(d*x^(1/3) + c)^5*(a - I*b)*c + 75*(d*x^(1 
/3) + c)^4*(a - I*b)*c^2 - 100*(d*x^(1/3) + c)^3*(a - I*b)*c^3 + 75*(d*x^( 
1/3) + c)^2*(a - I*b)*c^4 - 2*(48*I*(d*x^(1/3) + c)^5*b - 150*I*(d*x^(1/3) 
 + c)^4*b*c + 200*I*(d*x^(1/3) + c)^3*b*c^2 - 150*I*(d*x^(1/3) + c)^2*b*c^ 
3 + 75*I*(d*x^(1/3) + c)*b*c^4)*arctan2((2*a*b*cos(2*d*x^(1/3) + 2*c) - (a 
^2 - b^2)*sin(2*d*x^(1/3) + 2*c))/(a^2 + b^2), (2*a*b*sin(2*d*x^(1/3) + 2* 
c) + a^2 + b^2 + (a^2 - b^2)*cos(2*d*x^(1/3) + 2*c))/(a^2 + b^2)) - 15*(16 
*I*(d*x^(1/3) + c)^4*b - 40*I*(d*x^(1/3) + c)^3*b*c + 40*I*(d*x^(1/3) + c) 
^2*b*c^2 - 20*I*(d*x^(1/3) + c)*b*c^3 + 5*I*b*c^4)*dilog((I*a + b)*e^(2*I* 
d*x^(1/3) + 2*I*c)/(-I*a + b)) + (48*(d*x^(1/3) + c)^5*b - 150*(d*x^(1/3) 
+ c)^4*b*c + 200*(d*x^(1/3) + c)^3*b*c^2 - 150*(d*x^(1/3) + c)^2*b*c^3 + 7 
5*(d*x^(1/3) + c)*b*c^4)*log(((a^2 + b^2)*cos(2*d*x^(1/3) + 2*c)^2 + 4*a*b 
*sin(2*d*x^(1/3) + 2*c) + (a^2 + b^2)*sin(2*d*x^(1/3) + 2*c)^2 + a^2 + b^2 
 + 2*(a^2 - b^2)*cos(2*d*x^(1/3) + 2*c))/(a^2 + b^2)) - 360*I*b*polylog(6, 
 (I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I*a + b)) - 90*(8*(d*x^(1/3) + c)*b 
 - 5*b*c)*polylog(5, (I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I*a + b)) - 60* 
(-12*I*(d*x^(1/3) + c)^2*b + 15*I*(d*x^(1/3) + c)*b*c - 5*I*b*c^2)*polylog 
(4, (I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I*a + b)) + 30*(16*(d*x^(1/3)...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\int \frac {x}{a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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