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

Optimal result

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

[Out]

1/2*x^2/(a+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)*poly
log(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

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 352, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {3832, 3813, 2221, 2611, 6744, 2320, 6724} \[ \int \frac {x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=-\frac {45 i b \operatorname {PolyLog}\left (6,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{4 d^6 \left (a^2+b^2\right )}-\frac {45 b \sqrt [3]{x} \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^5 \left (a^2+b^2\right )}+\frac {45 i b x^{2/3} \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^4 \left (a^2+b^2\right )}+\frac {15 b x \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d^3 \left (a^2+b^2\right )}-\frac {15 i b x^{4/3} \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{2 d^2 \left (a^2+b^2\right )}+\frac {3 b x^{5/3} \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{(a+i b)^2}\right )}{d \left (a^2+b^2\right )}+\frac {x^2}{2 (a+i b)} \]

[In]

Int[x/(a + b*Tan[c + d*x^(1/3)]),x]

[Out]

x^2/(2*(a + I*b)) + (3*b*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)
- (((15*I)/2)*b*x^(4/3)*PolyLog[2, -(((a^2 + b^2)*E^((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)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)])/((a^2 + b^2)*d^3) + (((45*I)/2)
*b*x^(2/3)*PolyLog[4, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)])/((a^2 + b^2)*d^4) - (45*b*x^(1/
3)*PolyLog[5, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)])/(2*(a^2 + b^2)*d^5) - (((45*I)/4)*b*Pol
yLog[6, -(((a^2 + b^2)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)^2)])/((a^2 + b^2)*d^6)

Rule 2221

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]
- Dist[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 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2611

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] + Dist[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 3813

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c + d*x)^(m + 1)/(d*
(m + 1)*(a + I*b)), x] + Dist[2*I*b, Int[(c + d*x)^m*(E^Simp[2*I*(e + f*x), x]/((a + I*b)^2 + (a^2 + b^2)*E^Si
mp[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 3832

Int[(x_)^(m_.)*((a_.) + (b_.)*Tan[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(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 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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 6744

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = 3 \text {Subst}\left (\int \frac {x^5}{a+b \tan (c+d x)} \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {x^2}{2 (a+i b)}+(6 i b) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^5}{(a+i b)^2+\left (a^2+b^2\right ) e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right ) \\ & = \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 b) \text {Subst}\left (\int x^4 \log \left (1+\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d} \\ & = \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 {(30 i b) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (2,-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^2} \\ & = \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 b) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (3,-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^3} \\ & = \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 i b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (4,-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right ) d^4} \\ & = \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 b) \text {Subst}\left (\int \operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) e^{2 i (c+d x)}}{(a+i b)^2}\right ) \, dx,x,\sqrt [3]{x}\right )}{2 \left (a^2+b^2\right ) d^5} \\ & = \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) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (5,-\frac {\left (a^2+b^2\right ) x}{(a+i b)^2}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{4 \left (a^2+b^2\right ) d^6} \\ & = \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} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.12 (sec) , antiderivative size = 310, normalized size of antiderivative = 0.88 \[ \int \frac {x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\frac {2 a d^6 x^2+2 i b d^6 x^2+12 b d^5 x^{5/3} \log \left (1+\frac {(a+i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+30 i b d^4 x^{4/3} \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+60 b d^3 x \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-90 i b d^2 x^{2/3} \operatorname {PolyLog}\left (4,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )-90 b d \sqrt [3]{x} \operatorname {PolyLog}\left (5,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )+45 i b \operatorname {PolyLog}\left (6,\frac {(-a-i b) e^{-2 i \left (c+d \sqrt [3]{x}\right )}}{a-i b}\right )}{4 \left (a^2+b^2\right ) d^6} \]

[In]

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

[Out]

(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)*PolyLog[5, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))] + (45*I)*b*Po
lyLog[6, (-a - I*b)/((a - I*b)*E^((2*I)*(c + d*x^(1/3))))])/(4*(a^2 + b^2)*d^6)

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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 = 0.51 (sec) , antiderivative size = 813, normalized size of antiderivative = 2.31 \[ \int \frac {x}{a+b \tan \left (c+d \sqrt [3]{x}\right )} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-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 + 75*(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) + c)^3*b - 30*(d*x^(1/3) + c)^2*b*c + 20*(d*x^(1/3) + c)*b*c^2
- 5*b*c^3)*polylog(3, (I*a + b)*e^(2*I*d*x^(1/3) + 2*I*c)/(-I*a + b)))/(a^2 + b^2))/d^6

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(x/(a + b*tan(c + d*x^(1/3))),x)

[Out]

int(x/(a + b*tan(c + d*x^(1/3))), x)