3.1.63 \(\int \frac {x}{(a+b \tan (c+d \sqrt [3]{x}))^2} \, dx\) [63]

Optimal. Leaf size=1155 \[ -\frac {6 i b^2 x^{5/3}}{\left (a^2+b^2\right )^2 d}+\frac {6 b^2 x^{5/3}}{(a+i b) (i a+b)^2 d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {15 b^2 x^{4/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}-\frac {30 i b^2 x \text {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {15 b x^{4/3} \text {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {15 b^2 x^{4/3} \text {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {45 b^2 x^{2/3} \text {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}+\frac {30 b x \text {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {30 i b^2 x \text {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {45 i b^2 \sqrt [3]{x} \text {PolyLog}\left (4,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^5}-\frac {45 b x^{2/3} \text {PolyLog}\left (4,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^4}+\frac {45 b^2 x^{2/3} \text {PolyLog}\left (4,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}-\frac {45 b^2 \text {PolyLog}\left (5,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{2 \left (a^2+b^2\right )^2 d^6}-\frac {45 b \sqrt [3]{x} \text {PolyLog}\left (5,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^5}+\frac {45 i b^2 \sqrt [3]{x} \text {PolyLog}\left (5,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^5}+\frac {45 b \text {PolyLog}\left (6,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{2 (i a-b) (a-i b)^2 d^6}-\frac {45 b^2 \text {PolyLog}\left (6,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{2 \left (a^2+b^2\right )^2 d^6} \]

[Out]

-30*I*b^2*x*polylog(3,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^3+6*b^2*x^(5/3)/(a+I*b)/(I*a+b)^2
/d/(I*a-b+(I*a+b)*exp(2*I*(c+d*x^(1/3))))+1/2*x^2/(a-I*b)^2+2*b*x^2/(I*a-b)/(a-I*b)^2-2*b^2*x^2/(a^2+b^2)^2+15
*b^2*x^(4/3)*ln(1+(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^2+6*b*x^(5/3)*ln(1+(a-I*b)*exp(2*I*(c+
d*x^(1/3)))/(a+I*b))/(a-I*b)^2/(a+I*b)/d-6*I*b^2*x^(5/3)/(a^2+b^2)^2/d-30*I*b^2*x*polylog(2,-(a-I*b)*exp(2*I*(
c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^3+15*b*x^(4/3)*polylog(2,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(I*a-b)
/(a-I*b)^2/d^2-15*b^2*x^(4/3)*polylog(2,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^2+45*b^2*x^(2/3
)*polylog(3,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^4+30*b*x*polylog(3,-(a-I*b)*exp(2*I*(c+d*x^
(1/3)))/(a+I*b))/(a-I*b)^2/(a+I*b)/d^3-6*I*b^2*x^(5/3)*ln(1+(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^
2/d+45*I*b^2*x^(1/3)*polylog(5,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^5-45*b*x^(2/3)*polylog(4
,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(I*a-b)/(a-I*b)^2/d^4+45*b^2*x^(2/3)*polylog(4,-(a-I*b)*exp(2*I*(c+d
*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^4-45/2*b^2*polylog(5,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^
6-45*b*x^(1/3)*polylog(5,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a-I*b)^2/(a+I*b)/d^5+45*I*b^2*x^(1/3)*polyl
og(4,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^5+45/2*b*polylog(6,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))
/(a+I*b))/(I*a-b)/(a-I*b)^2/d^6-45/2*b^2*polylog(6,-(a-I*b)*exp(2*I*(c+d*x^(1/3)))/(a+I*b))/(a^2+b^2)^2/d^6

________________________________________________________________________________________

Rubi [A]
time = 1.46, antiderivative size = 1155, normalized size of antiderivative = 1.00, number of steps used = 28, number of rules used = 10, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3832, 3815, 2216, 2215, 2221, 2611, 6744, 2320, 6724, 2222} \begin {gather*} -\frac {2 x^2 b^2}{\left (a^2+b^2\right )^2}-\frac {6 i x^{5/3} b^2}{\left (a^2+b^2\right )^2 d}+\frac {6 x^{5/3} b^2}{(a+i b) (i a+b)^2 d \left (i a+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}-b\right )}-\frac {6 i x^{5/3} \log \left (\frac {e^{2 i \left (c+d \sqrt [3]{x}\right )} (a-i b)}{a+i b}+1\right ) b^2}{\left (a^2+b^2\right )^2 d}+\frac {15 x^{4/3} \log \left (\frac {e^{2 i \left (c+d \sqrt [3]{x}\right )} (a-i b)}{a+i b}+1\right ) b^2}{\left (a^2+b^2\right )^2 d^2}-\frac {15 x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^2}-\frac {30 i x \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^3}-\frac {30 i x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^3}+\frac {45 x^{2/3} \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^4}+\frac {45 x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^4}+\frac {45 i \sqrt [3]{x} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^5}+\frac {45 i \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{\left (a^2+b^2\right )^2 d^5}-\frac {45 \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{2 \left (a^2+b^2\right )^2 d^6}-\frac {45 \text {Li}_6\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b^2}{2 \left (a^2+b^2\right )^2 d^6}+\frac {2 x^2 b}{(i a-b) (a-i b)^2}+\frac {6 x^{5/3} \log \left (\frac {e^{2 i \left (c+d \sqrt [3]{x}\right )} (a-i b)}{a+i b}+1\right ) b}{(a-i b)^2 (a+i b) d}+\frac {15 x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b}{(i a-b) (a-i b)^2 d^2}+\frac {30 x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b}{(a-i b)^2 (a+i b) d^3}-\frac {45 x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b}{(i a-b) (a-i b)^2 d^4}-\frac {45 \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b}{(a-i b)^2 (a+i b) d^5}+\frac {45 \text {Li}_6\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right ) b}{2 (i a-b) (a-i b)^2 d^6}+\frac {x^2}{2 (a-i b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a + b*Tan[c + d*x^(1/3)])^2,x]

[Out]

((-6*I)*b^2*x^(5/3))/((a^2 + b^2)^2*d) + (6*b^2*x^(5/3))/((a + I*b)*(I*a + b)^2*d*(I*a - b + (I*a + b)*E^((2*I
)*(c + d*x^(1/3))))) + x^2/(2*(a - I*b)^2) + (2*b*x^2)/((I*a - b)*(a - I*b)^2) - (2*b^2*x^2)/(a^2 + b^2)^2 + (
15*b^2*x^(4/3)*Log[1 + ((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)])/((a^2 + b^2)^2*d^2) + (6*b*x^(5/3)*Lo
g[1 + ((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)])/((a - I*b)^2*(a + I*b)*d) - ((6*I)*b^2*x^(5/3)*Log[1 +
((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)])/((a^2 + b^2)^2*d) - ((30*I)*b^2*x*PolyLog[2, -(((a - I*b)*E
^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/((a^2 + b^2)^2*d^3) + (15*b*x^(4/3)*PolyLog[2, -(((a - I*b)*E^((2*I)*(c
+ d*x^(1/3))))/(a + I*b))])/((I*a - b)*(a - I*b)^2*d^2) - (15*b^2*x^(4/3)*PolyLog[2, -(((a - I*b)*E^((2*I)*(c
+ d*x^(1/3))))/(a + I*b))])/((a^2 + b^2)^2*d^2) + (45*b^2*x^(2/3)*PolyLog[3, -(((a - I*b)*E^((2*I)*(c + d*x^(
1/3))))/(a + I*b))])/((a^2 + b^2)^2*d^4) + (30*b*x*PolyLog[3, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b
))])/((a - I*b)^2*(a + I*b)*d^3) - ((30*I)*b^2*x*PolyLog[3, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))
])/((a^2 + b^2)^2*d^3) + ((45*I)*b^2*x^(1/3)*PolyLog[4, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/(
(a^2 + b^2)^2*d^5) - (45*b*x^(2/3)*PolyLog[4, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/((I*a - b)*
(a - I*b)^2*d^4) + (45*b^2*x^(2/3)*PolyLog[4, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/((a^2 + b^2
)^2*d^4) - (45*b^2*PolyLog[5, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/(2*(a^2 + b^2)^2*d^6) - (45
*b*x^(1/3)*PolyLog[5, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/((a - I*b)^2*(a + I*b)*d^5) + ((45*
I)*b^2*x^(1/3)*PolyLog[5, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/((a^2 + b^2)^2*d^5) + (45*b*Pol
yLog[6, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/(2*(I*a - b)*(a - I*b)^2*d^6) - (45*b^2*PolyLog[6
, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))])/(2*(a^2 + b^2)^2*d^6)

Rule 2215

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

Rule 2216

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

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 2222

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

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 3815

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandIntegrand[(
c + d*x)^m, (1/(a - I*b) - 2*I*(b/(a^2 + b^2 + (a - I*b)^2*E^(2*I*(e + f*x)))))^(-n), x], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[n, 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*} \int \frac {x}{\left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2} \, dx &=3 \text {Subst}\left (\int \frac {x^5}{(a+b \tan (c+d x))^2} \, dx,x,\sqrt [3]{x}\right )\\ &=3 \text {Subst}\left (\int \left (\frac {x^5}{(a-i b)^2}-\frac {4 b^2 x^5}{(i a+b)^2 \left (i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}\right )^2}+\frac {4 b x^5}{(a-i b)^2 \left (i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}\right )}\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {x^2}{2 (a-i b)^2}+\frac {(12 b) \text {Subst}\left (\int \frac {x^5}{i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(a-i b)^2}-\frac {\left (12 b^2\right ) \text {Subst}\left (\int \frac {x^5}{\left (i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}\right )^2} \, dx,x,\sqrt [3]{x}\right )}{(i a+b)^2}\\ &=\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}+\frac {\left (12 b^2\right ) \text {Subst}\left (\int \frac {x^5}{i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(i a-b) (a-i b)^2}-\frac {(12 b) \text {Subst}\left (\int \frac {e^{2 i c+2 i d x} x^5}{i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{a^2+b^2}-\frac {\left (12 b^2\right ) \text {Subst}\left (\int \frac {e^{2 i c+2 i d x} x^5}{\left (i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}\right )^2} \, dx,x,\sqrt [3]{x}\right )}{a^2+b^2}\\ &=-\frac {6 b^2 x^{5/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {\left (12 b^2\right ) \text {Subst}\left (\int \frac {e^{2 i c+2 i d x} x^5}{i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(a+i b)^2 (i a+b)}-\frac {(30 b) \text {Subst}\left (\int x^4 \log \left (1+\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{(a-i b)^2 (a+i b) d}+\frac {\left (30 b^2\right ) \text {Subst}\left (\int \frac {x^4}{i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(a-i b)^2 (a+i b) d}\\ &=-\frac {6 i b^2 x^{5/3}}{\left (a^2+b^2\right )^2 d}-\frac {6 b^2 x^{5/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}+\frac {15 b x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {(60 b) \text {Subst}\left (\int x^3 \text {Li}_2\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {\left (30 b^2\right ) \text {Subst}\left (\int \frac {e^{2 i c+2 i d x} x^4}{i a \left (1+\frac {i b}{a}\right )+i a \left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}} \, dx,x,\sqrt [3]{x}\right )}{(a-i b) (a+i b)^2 d}+\frac {\left (30 i b^2\right ) \text {Subst}\left (\int x^4 \log \left (1+\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d}\\ &=-\frac {6 i b^2 x^{5/3}}{\left (a^2+b^2\right )^2 d}-\frac {6 b^2 x^{5/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {15 b^2 x^{4/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}+\frac {15 b x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {15 b^2 x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {30 b x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {(90 b) \text {Subst}\left (\int x^2 \text {Li}_3\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {\left (60 b^2\right ) \text {Subst}\left (\int x^3 \log \left (1+\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {\left (60 b^2\right ) \text {Subst}\left (\int x^3 \text {Li}_2\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^2}\\ &=-\frac {6 i b^2 x^{5/3}}{\left (a^2+b^2\right )^2 d}-\frac {6 b^2 x^{5/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {15 b^2 x^{4/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}-\frac {30 i b^2 x \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {15 b x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {15 b^2 x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {30 b x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {30 i b^2 x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}-\frac {45 b x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^4}+\frac {(90 b) \text {Subst}\left (\int x \text {Li}_4\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{(i a-b) (a-i b)^2 d^4}+\frac {\left (90 i b^2\right ) \text {Subst}\left (\int x^2 \text {Li}_2\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {\left (90 i b^2\right ) \text {Subst}\left (\int x^2 \text {Li}_3\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^3}\\ &=-\frac {6 i b^2 x^{5/3}}{\left (a^2+b^2\right )^2 d}-\frac {6 b^2 x^{5/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {15 b^2 x^{4/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}-\frac {30 i b^2 x \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {15 b x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {15 b^2 x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}+\frac {30 b x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {30 i b^2 x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}-\frac {45 b x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^4}+\frac {45 b^2 x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^5}+\frac {(45 b) \text {Subst}\left (\int \text {Li}_5\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{(a-i b)^2 (a+i b) d^5}-\frac {\left (90 b^2\right ) \text {Subst}\left (\int x \text {Li}_3\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^4}-\frac {\left (90 b^2\right ) \text {Subst}\left (\int x \text {Li}_4\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^4}\\ &=-\frac {6 i b^2 x^{5/3}}{\left (a^2+b^2\right )^2 d}-\frac {6 b^2 x^{5/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {15 b^2 x^{4/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}-\frac {30 i b^2 x \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {15 b x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {15 b^2 x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}+\frac {30 b x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {30 i b^2 x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^5}-\frac {45 b x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^4}+\frac {45 b^2 x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^5}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^5}+\frac {(45 b) \text {Subst}\left (\int \frac {\text {Li}_5\left (-\frac {(a-i b) x}{a+i b}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 (i a-b) (a-i b)^2 d^6}-\frac {\left (45 i b^2\right ) \text {Subst}\left (\int \text {Li}_4\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^5}-\frac {\left (45 i b^2\right ) \text {Subst}\left (\int \text {Li}_5\left (-\frac {\left (1-\frac {i b}{a}\right ) e^{2 i c+2 i d x}}{1+\frac {i b}{a}}\right ) \, dx,x,\sqrt [3]{x}\right )}{\left (a^2+b^2\right )^2 d^5}\\ &=-\frac {6 i b^2 x^{5/3}}{\left (a^2+b^2\right )^2 d}-\frac {6 b^2 x^{5/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {15 b^2 x^{4/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}-\frac {30 i b^2 x \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {15 b x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {15 b^2 x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}+\frac {30 b x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {30 i b^2 x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^5}-\frac {45 b x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^4}+\frac {45 b^2 x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^5}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^5}+\frac {45 b \text {Li}_6\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{2 (i a-b) (a-i b)^2 d^6}-\frac {\left (45 b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_4\left (-\frac {(a-i b) x}{a+i b}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 \left (a^2+b^2\right )^2 d^6}-\frac {\left (45 b^2\right ) \text {Subst}\left (\int \frac {\text {Li}_5\left (-\frac {(a-i b) x}{a+i b}\right )}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 \left (a^2+b^2\right )^2 d^6}\\ &=-\frac {6 i b^2 x^{5/3}}{\left (a^2+b^2\right )^2 d}-\frac {6 b^2 x^{5/3}}{(a-i b)^2 (a+i b) d \left (i a-b+(i a+b) e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}+\frac {x^2}{2 (a-i b)^2}+\frac {2 b x^2}{(i a-b) (a-i b)^2}-\frac {2 b^2 x^2}{\left (a^2+b^2\right )^2}+\frac {15 b^2 x^{4/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {6 b x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d}-\frac {6 i b^2 x^{5/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d}-\frac {30 i b^2 x \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {15 b x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^2}-\frac {15 b^2 x^{4/3} \text {Li}_2\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^2}+\frac {45 b^2 x^{2/3} \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}+\frac {30 b x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^3}-\frac {30 i b^2 x \text {Li}_3\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^3}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^5}-\frac {45 b x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(i a-b) (a-i b)^2 d^4}+\frac {45 b^2 x^{2/3} \text {Li}_4\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^4}-\frac {45 b^2 \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{2 \left (a^2+b^2\right )^2 d^6}-\frac {45 b \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{(a-i b)^2 (a+i b) d^5}+\frac {45 i b^2 \sqrt [3]{x} \text {Li}_5\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{\left (a^2+b^2\right )^2 d^5}+\frac {45 b \text {Li}_6\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{2 (i a-b) (a-i b)^2 d^6}-\frac {45 b^2 \text {Li}_6\left (-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )}{2 \left (a^2+b^2\right )^2 d^6}\\ \end {align*}

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Mathematica [A]
time = 4.87, size = 644, normalized size = 0.56 \begin {gather*} \frac {\frac {b e^{2 i c} \left (-12 b x^{5/3}-4 a d x^2+\frac {3 e^{-2 i c} \left (-i b \left (-1+e^{2 i c}\right )+a \left (1+e^{2 i c}\right )\right ) \left (10 b d^4 x^{4/3} \log \left (1+\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )+4 a 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 )-10 i d^3 \left (2 b+a d \sqrt [3]{x}\right ) x \text {PolyLog}\left (2,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )+10 \left (3 b d^2 x^{2/3}+2 a d^3 x\right ) \text {PolyLog}\left (3,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )+30 i b d \sqrt [3]{x} \text {PolyLog}\left (4,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )+30 i a d^2 x^{2/3} \text {PolyLog}\left (4,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )-15 b \text {PolyLog}\left (5,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )-30 a d \sqrt [3]{x} \text {PolyLog}\left (5,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )-15 i a \text {PolyLog}\left (6,-\frac {(a-i b) e^{2 i \left (c+d \sqrt [3]{x}\right )}}{a+i b}\right )\right )}{(i a+b) d^5}\right )}{d \left (b-b e^{2 i c}-i a \left (1+e^{2 i c}\right )\right )}+\frac {(a+i b) x^2 (a \cos (c)-b \sin (c))}{a \cos (c)+b \sin (c)}+\frac {6 (a+i b) b^2 x^{5/3} \sin \left (d \sqrt [3]{x}\right )}{d (a \cos (c)+b \sin (c)) \left (a \cos \left (c+d \sqrt [3]{x}\right )+b \sin \left (c+d \sqrt [3]{x}\right )\right )}}{2 (a-i b) (a+i b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a + b*Tan[c + d*x^(1/3)])^2,x]

[Out]

((b*E^((2*I)*c)*(-12*b*x^(5/3) - 4*a*d*x^2 + (3*((-I)*b*(-1 + E^((2*I)*c)) + a*(1 + E^((2*I)*c)))*(10*b*d^4*x^
(4/3)*Log[1 + ((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b)] + 4*a*d^5*x^(5/3)*Log[1 + ((a - I*b)*E^((2*I)*(
c + d*x^(1/3))))/(a + I*b)] - (10*I)*d^3*(2*b + a*d*x^(1/3))*x*PolyLog[2, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3)
)))/(a + I*b))] + 10*(3*b*d^2*x^(2/3) + 2*a*d^3*x)*PolyLog[3, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b
))] + (30*I)*b*d*x^(1/3)*PolyLog[4, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))] + (30*I)*a*d^2*x^(2/3)
*PolyLog[4, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))] - 15*b*PolyLog[5, -(((a - I*b)*E^((2*I)*(c + d
*x^(1/3))))/(a + I*b))] - 30*a*d*x^(1/3)*PolyLog[5, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))] - (15*
I)*a*PolyLog[6, -(((a - I*b)*E^((2*I)*(c + d*x^(1/3))))/(a + I*b))]))/((I*a + b)*d^5*E^((2*I)*c))))/(d*(b - b*
E^((2*I)*c) - I*a*(1 + E^((2*I)*c)))) + ((a + I*b)*x^2*(a*Cos[c] - b*Sin[c]))/(a*Cos[c] + b*Sin[c]) + (6*(a +
I*b)*b^2*x^(5/3)*Sin[d*x^(1/3)])/(d*(a*Cos[c] + b*Sin[c])*(a*Cos[c + d*x^(1/3)] + b*Sin[c + d*x^(1/3)])))/(2*(
a - I*b)*(a + I*b)^2)

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Maple [F]
time = 0.89, size = 0, normalized size = 0.00 \[\int \frac {x}{\left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a+b*tan(c+d*x^(1/3)))^2,x)

[Out]

int(x/(a+b*tan(c+d*x^(1/3)))^2,x)

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4345 vs. \(2 (928) = 1856\).
time = 1.60, size = 4345, normalized size = 3.76 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*tan(c+d*x^(1/3)))^2,x, algorithm="maxima")

[Out]

-1/10*(30*(2*a*b*log(b*tan(d*x^(1/3) + c) + a)/(a^4 + 2*a^2*b^2 + b^4) - a*b*log(tan(d*x^(1/3) + c)^2 + 1)/(a^
4 + 2*a^2*b^2 + b^4) + (a^2 - b^2)*(d*x^(1/3) + c)/(a^4 + 2*a^2*b^2 + b^4) - b/(a^3 + a*b^2 + (a^2*b + b^3)*ta
n(d*x^(1/3) + c)))*c^5 - (5*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^6 - 30*(a^3 - I*a^2*b + a*b^2 - I*
b^3)*(d*x^(1/3) + c)^5*c + 75*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^4*c^2 - 100*(a^3 - I*a^2*b + a*b
^2 - I*b^3)*(d*x^(1/3) + c)^3*c^3 + 75*(a^3 - I*a^2*b + a*b^2 - I*b^3)*(d*x^(1/3) + c)^2*c^4 - 150*((-I*a*b^2
- b^3)*c^4*cos(2*d*x^(1/3) + 2*c) + (a*b^2 - I*b^3)*c^4*sin(2*d*x^(1/3) + 2*c) + (-I*a*b^2 + b^3)*c^4)*arctan2
(-b*cos(2*d*x^(1/3) + 2*c) + a*sin(2*d*x^(1/3) + 2*c) + b, a*cos(2*d*x^(1/3) + 2*c) + b*sin(2*d*x^(1/3) + 2*c)
+ a) - 4*(48*(I*a^2*b - a*b^2)*(d*x^(1/3) + c)^5 + 75*(I*a*b^2 - b^3 + 2*(-I*a^2*b + a*b^2)*c)*(d*x^(1/3) + c
)^4 + 200*((I*a^2*b - a*b^2)*c^2 + (-I*a*b^2 + b^3)*c)*(d*x^(1/3) + c)^3 + 75*(2*(-I*a^2*b + a*b^2)*c^3 + 3*(I
*a*b^2 - b^3)*c^2)*(d*x^(1/3) + c)^2 + 75*((I*a^2*b - a*b^2)*c^4 + 2*(-I*a*b^2 + b^3)*c^3)*(d*x^(1/3) + c) + (
48*(I*a^2*b + a*b^2)*(d*x^(1/3) + c)^5 + 75*(I*a*b^2 + b^3 + 2*(-I*a^2*b - a*b^2)*c)*(d*x^(1/3) + c)^4 + 200*(
(I*a^2*b + a*b^2)*c^2 + (-I*a*b^2 - b^3)*c)*(d*x^(1/3) + c)^3 + 75*(2*(-I*a^2*b - a*b^2)*c^3 + 3*(I*a*b^2 + b^
3)*c^2)*(d*x^(1/3) + c)^2 + 75*((I*a^2*b + a*b^2)*c^4 + 2*(-I*a*b^2 - b^3)*c^3)*(d*x^(1/3) + c))*cos(2*d*x^(1/
3) + 2*c) - (48*(a^2*b - I*a*b^2)*(d*x^(1/3) + c)^5 + 75*(a*b^2 - I*b^3 - 2*(a^2*b - I*a*b^2)*c)*(d*x^(1/3) +
c)^4 + 200*((a^2*b - I*a*b^2)*c^2 - (a*b^2 - I*b^3)*c)*(d*x^(1/3) + c)^3 - 75*(2*(a^2*b - I*a*b^2)*c^3 - 3*(a*
b^2 - I*b^3)*c^2)*(d*x^(1/3) + c)^2 + 75*((a^2*b - I*a*b^2)*c^4 - 2*(a*b^2 - I*b^3)*c^3)*(d*x^(1/3) + c))*sin(
2*d*x^(1/3) + 2*c))*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)) + 5*((a^3 - 3*I*a^
2*b - 3*a*b^2 + I*b^3)*(d*x^(1/3) + c)^6 - 6*(2*I*a*b^2 + 2*b^3 + (a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*c)*(d*x^
(1/3) + c)^5 - 60*(I*a*b^2 + b^3)*(d*x^(1/3) + c)*c^4 + 15*((a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*c^2 - 4*(-I*a*
b^2 - b^3)*c)*(d*x^(1/3) + c)^4 - 20*((a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*c^3 + 6*(I*a*b^2 + b^3)*c^2)*(d*x^(1
/3) + c)^3 + 15*((a^3 - 3*I*a^2*b - 3*a*b^2 + I*b^3)*c^4 - 8*(-I*a*b^2 - b^3)*c^3)*(d*x^(1/3) + c)^2)*cos(2*d*
x^(1/3) + 2*c) - 30*(16*(I*a^2*b - a*b^2)*(d*x^(1/3) + c)^4 + 5*(I*a^2*b - a*b^2)*c^4 + 20*(I*a*b^2 - b^3 + 2*
(-I*a^2*b + a*b^2)*c)*(d*x^(1/3) + c)^3 + 10*(-I*a*b^2 + b^3)*c^3 + 40*((I*a^2*b - a*b^2)*c^2 + (-I*a*b^2 + b^
3)*c)*(d*x^(1/3) + c)^2 + 10*(2*(-I*a^2*b + a*b^2)*c^3 + 3*(I*a*b^2 - b^3)*c^2)*(d*x^(1/3) + c) + (16*(I*a^2*b
+ a*b^2)*(d*x^(1/3) + c)^4 + 5*(I*a^2*b + a*b^2)*c^4 + 20*(I*a*b^2 + b^3 + 2*(-I*a^2*b - a*b^2)*c)*(d*x^(1/3)
+ c)^3 + 10*(-I*a*b^2 - b^3)*c^3 + 40*((I*a^2*b + a*b^2)*c^2 + (-I*a*b^2 - b^3)*c)*(d*x^(1/3) + c)^2 + 10*(2*
(-I*a^2*b - a*b^2)*c^3 + 3*(I*a*b^2 + b^3)*c^2)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) - (16*(a^2*b - I*a*b^2
)*(d*x^(1/3) + c)^4 + 5*(a^2*b - I*a*b^2)*c^4 + 20*(a*b^2 - I*b^3 - 2*(a^2*b - I*a*b^2)*c)*(d*x^(1/3) + c)^3 -
10*(a*b^2 - I*b^3)*c^3 + 40*((a^2*b - I*a*b^2)*c^2 - (a*b^2 - I*b^3)*c)*(d*x^(1/3) + c)^2 - 10*(2*(a^2*b - I*
a*b^2)*c^3 - 3*(a*b^2 - I*b^3)*c^2)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*c))*dilog((I*a + b)*e^(2*I*d*x^(1/3)
+ 2*I*c)/(-I*a + b)) + 75*((a*b^2 - I*b^3)*c^4*cos(2*d*x^(1/3) + 2*c) - (-I*a*b^2 - b^3)*c^4*sin(2*d*x^(1/3) +
2*c) + (a*b^2 + I*b^3)*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)) + 2*(48*(a^2*b + I*a*b^2)*(d
*x^(1/3) + c)^5 + 75*(a*b^2 + I*b^3 - 2*(a^2*b + I*a*b^2)*c)*(d*x^(1/3) + c)^4 + 200*((a^2*b + I*a*b^2)*c^2 -
(a*b^2 + I*b^3)*c)*(d*x^(1/3) + c)^3 - 75*(2*(a^2*b + I*a*b^2)*c^3 - 3*(a*b^2 + I*b^3)*c^2)*(d*x^(1/3) + c)^2
+ 75*((a^2*b + I*a*b^2)*c^4 - 2*(a*b^2 + I*b^3)*c^3)*(d*x^(1/3) + c) + (48*(a^2*b - I*a*b^2)*(d*x^(1/3) + c)^5
+ 75*(a*b^2 - I*b^3 - 2*(a^2*b - I*a*b^2)*c)*(d*x^(1/3) + c)^4 + 200*((a^2*b - I*a*b^2)*c^2 - (a*b^2 - I*b^3)
*c)*(d*x^(1/3) + c)^3 - 75*(2*(a^2*b - I*a*b^2)*c^3 - 3*(a*b^2 - I*b^3)*c^2)*(d*x^(1/3) + c)^2 + 75*((a^2*b -
I*a*b^2)*c^4 - 2*(a*b^2 - I*b^3)*c^3)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) - (48*(-I*a^2*b - a*b^2)*(d*x^(1
/3) + c)^5 + 75*(-I*a*b^2 - b^3 + 2*(I*a^2*b + a*b^2)*c)*(d*x^(1/3) + c)^4 + 200*((-I*a^2*b - a*b^2)*c^2 + (I*
a*b^2 + b^3)*c)*(d*x^(1/3) + c)^3 + 75*(2*(I*a^2*b + a*b^2)*c^3 + 3*(-I*a*b^2 - b^3)*c^2)*(d*x^(1/3) + c)^2 +
75*((-I*a^2*b - a*b^2)*c^4 + 2*(I*a*b^2 + b^3)*c^3)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*c))*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)) - 720*(I*a^2*b - a*b^2 + (I*a^2*b + a*b^2)*cos(2*d*x^(1/3) +
2*c) - (a^2*b - I*a*b^2)*sin(2*d*x^(1/3) + 2*c...

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*tan(c+d*x^(1/3)))^2,x, algorithm="fricas")

[Out]

integral(x/(b^2*tan(d*x^(1/3) + c)^2 + 2*a*b*tan(d*x^(1/3) + c) + a^2), x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*tan(c+d*x**(1/3)))**2,x)

[Out]

Integral(x/(a + b*tan(c + d*x**(1/3)))**2, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(a+b*tan(c+d*x^(1/3)))^2,x, algorithm="giac")

[Out]

integrate(x/(b*tan(d*x^(1/3) + c) + a)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x}{{\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a + b*tan(c + d*x^(1/3)))^2,x)

[Out]

int(x/(a + b*tan(c + d*x^(1/3)))^2, x)

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