\(\int x^2 (a+b \tan (c+d \sqrt [3]{x})) \, dx\) [47]

Optimal result

Integrand size = 18, antiderivative size = 287 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9} \]

[Out]

1/3*a*x^3+1/3*I*b*x^3-3*b*x^(8/3)*ln(1+exp(2*I*(c+d*x^(1/3))))/d+12*I*b*x^(7/3)*polylog(2,-exp(2*I*(c+d*x^(1/3
))))/d^2-42*b*x^2*polylog(3,-exp(2*I*(c+d*x^(1/3))))/d^3-126*I*b*x^(5/3)*polylog(4,-exp(2*I*(c+d*x^(1/3))))/d^
4+315*b*x^(4/3)*polylog(5,-exp(2*I*(c+d*x^(1/3))))/d^5+630*I*b*x*polylog(6,-exp(2*I*(c+d*x^(1/3))))/d^6-945*b*
x^(2/3)*polylog(7,-exp(2*I*(c+d*x^(1/3))))/d^7-945*I*b*x^(1/3)*polylog(8,-exp(2*I*(c+d*x^(1/3))))/d^8+945/2*b*
polylog(9,-exp(2*I*(c+d*x^(1/3))))/d^9

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {14, 3832, 3800, 2221, 2611, 6744, 2320, 6724} \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {945 b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {1}{3} i b x^3 \]

[In]

Int[x^2*(a + b*Tan[c + d*x^(1/3)]),x]

[Out]

(a*x^3)/3 + (I/3)*b*x^3 - (3*b*x^(8/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d + ((12*I)*b*x^(7/3)*PolyLog[2, -E
^((2*I)*(c + d*x^(1/3)))])/d^2 - (42*b*x^2*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 - ((126*I)*b*x^(5/3)*Po
lyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 + (315*b*x^(4/3)*PolyLog[5, -E^((2*I)*(c + d*x^(1/3)))])/d^5 + ((630
*I)*b*x*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^6 - (945*b*x^(2/3)*PolyLog[7, -E^((2*I)*(c + d*x^(1/3)))])/d
^7 - ((945*I)*b*x^(1/3)*PolyLog[8, -E^((2*I)*(c + d*x^(1/3)))])/d^8 + (945*b*PolyLog[9, -E^((2*I)*(c + d*x^(1/
3)))])/(2*d^9)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I*((c + d*x)^(m + 1)/(d*(m + 1))), x
] - Dist[2*I, Int[(c + d*x)^m*(E^(2*I*(e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] &&
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}& = \int \left (a x^2+b x^2 \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx \\ & = \frac {a x^3}{3}+b \int x^2 \tan \left (c+d \sqrt [3]{x}\right ) \, dx \\ & = \frac {a x^3}{3}+(3 b) \text {Subst}\left (\int x^8 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-(6 i b) \text {Subst}\left (\int \frac {e^{2 i (c+d x)} x^8}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right ) \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {(24 b) \text {Subst}\left (\int x^7 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {(84 i b) \text {Subst}\left (\int x^6 \operatorname {PolyLog}\left (2,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {(252 b) \text {Subst}\left (\int x^5 \operatorname {PolyLog}\left (3,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^3} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {(630 i b) \text {Subst}\left (\int x^4 \operatorname {PolyLog}\left (4,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^4} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {(1260 b) \text {Subst}\left (\int x^3 \operatorname {PolyLog}\left (5,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^5} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {(1890 i b) \text {Subst}\left (\int x^2 \operatorname {PolyLog}\left (6,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^6} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {(1890 b) \text {Subst}\left (\int x \operatorname {PolyLog}\left (7,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^7} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {(945 i b) \text {Subst}\left (\int \operatorname {PolyLog}\left (8,-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^8} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {(945 b) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(8,-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9} \\ & = \frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.00 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\frac {a x^3}{3}+\frac {1}{3} i b x^3-\frac {3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {12 i b x^{7/3} \operatorname {PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {42 b x^2 \operatorname {PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {126 i b x^{5/3} \operatorname {PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {315 b x^{4/3} \operatorname {PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {630 i b x \operatorname {PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {945 b x^{2/3} \operatorname {PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {945 i b \sqrt [3]{x} \operatorname {PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 b \operatorname {PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9} \]

[In]

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

[Out]

(a*x^3)/3 + (I/3)*b*x^3 - (3*b*x^(8/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d + ((12*I)*b*x^(7/3)*PolyLog[2, -E
^((2*I)*(c + d*x^(1/3)))])/d^2 - (42*b*x^2*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 - ((126*I)*b*x^(5/3)*Po
lyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^4 + (315*b*x^(4/3)*PolyLog[5, -E^((2*I)*(c + d*x^(1/3)))])/d^5 + ((630
*I)*b*x*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^6 - (945*b*x^(2/3)*PolyLog[7, -E^((2*I)*(c + d*x^(1/3)))])/d
^7 - ((945*I)*b*x^(1/3)*PolyLog[8, -E^((2*I)*(c + d*x^(1/3)))])/d^8 + (945*b*PolyLog[9, -E^((2*I)*(c + d*x^(1/
3)))])/(2*d^9)

Maple [F]

\[\int x^{2} \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x^{2} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x^{2} \left (a + b \tan {\left (c + d \sqrt [3]{x} \right )}\right )\, dx \]

[In]

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

[Out]

Integral(x**2*(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. 1119 vs. \(2 (222) = 444\).

Time = 0.49 (sec) , antiderivative size = 1119, normalized size of antiderivative = 3.90 \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/105*(35*(d*x^(1/3) + c)^9*a + 35*I*(d*x^(1/3) + c)^9*b - 315*(d*x^(1/3) + c)^8*a*c - 315*I*(d*x^(1/3) + c)^8
*b*c + 1260*(d*x^(1/3) + c)^7*a*c^2 + 1260*I*(d*x^(1/3) + c)^7*b*c^2 - 2940*(d*x^(1/3) + c)^6*a*c^3 - 2940*I*(
d*x^(1/3) + c)^6*b*c^3 + 4410*(d*x^(1/3) + c)^5*a*c^4 + 4410*I*(d*x^(1/3) + c)^5*b*c^4 - 4410*(d*x^(1/3) + c)^
4*a*c^5 - 4410*I*(d*x^(1/3) + c)^4*b*c^5 + 2940*(d*x^(1/3) + c)^3*a*c^6 + 2940*I*(d*x^(1/3) + c)^3*b*c^6 - 126
0*(d*x^(1/3) + c)^2*a*c^7 - 1260*I*(d*x^(1/3) + c)^2*b*c^7 + 315*(d*x^(1/3) + c)*a*c^8 + 315*b*c^8*log(sec(d*x
^(1/3) + c)) + 12*(-420*I*(d*x^(1/3) + c)^8*b + 1920*I*(d*x^(1/3) + c)^7*b*c - 3920*I*(d*x^(1/3) + c)^6*b*c^2
+ 4704*I*(d*x^(1/3) + c)^5*b*c^3 - 3675*I*(d*x^(1/3) + c)^4*b*c^4 + 1960*I*(d*x^(1/3) + c)^3*b*c^5 - 735*I*(d*
x^(1/3) + c)^2*b*c^6 + 210*I*(d*x^(1/3) + c)*b*c^7)*arctan2(sin(2*d*x^(1/3) + 2*c), cos(2*d*x^(1/3) + 2*c) + 1
) + 1260*(16*I*(d*x^(1/3) + c)^7*b - 64*I*(d*x^(1/3) + c)^6*b*c + 112*I*(d*x^(1/3) + c)^5*b*c^2 - 112*I*(d*x^(
1/3) + c)^4*b*c^3 + 70*I*(d*x^(1/3) + c)^3*b*c^4 - 28*I*(d*x^(1/3) + c)^2*b*c^5 + 7*I*(d*x^(1/3) + c)*b*c^6 -
I*b*c^7)*dilog(-e^(2*I*d*x^(1/3) + 2*I*c)) - 6*(420*(d*x^(1/3) + c)^8*b - 1920*(d*x^(1/3) + c)^7*b*c + 3920*(d
*x^(1/3) + c)^6*b*c^2 - 4704*(d*x^(1/3) + c)^5*b*c^3 + 3675*(d*x^(1/3) + c)^4*b*c^4 - 1960*(d*x^(1/3) + c)^3*b
*c^5 + 735*(d*x^(1/3) + c)^2*b*c^6 - 210*(d*x^(1/3) + c)*b*c^7)*log(cos(2*d*x^(1/3) + 2*c)^2 + sin(2*d*x^(1/3)
+ 2*c)^2 + 2*cos(2*d*x^(1/3) + 2*c) + 1) + 793800*b*polylog(9, -e^(2*I*d*x^(1/3) + 2*I*c)) + 226800*(-7*I*(d*
x^(1/3) + c)*b + 4*I*b*c)*polylog(8, -e^(2*I*d*x^(1/3) + 2*I*c)) - 75600*(21*(d*x^(1/3) + c)^2*b - 24*(d*x^(1/
3) + c)*b*c + 7*b*c^2)*polylog(7, -e^(2*I*d*x^(1/3) + 2*I*c)) + 30240*(35*I*(d*x^(1/3) + c)^3*b - 60*I*(d*x^(1
/3) + c)^2*b*c + 35*I*(d*x^(1/3) + c)*b*c^2 - 7*I*b*c^3)*polylog(6, -e^(2*I*d*x^(1/3) + 2*I*c)) + 1890*(280*(d
*x^(1/3) + c)^4*b - 640*(d*x^(1/3) + c)^3*b*c + 560*(d*x^(1/3) + c)^2*b*c^2 - 224*(d*x^(1/3) + c)*b*c^3 + 35*b
*c^4)*polylog(5, -e^(2*I*d*x^(1/3) + 2*I*c)) + 1260*(-168*I*(d*x^(1/3) + c)^5*b + 480*I*(d*x^(1/3) + c)^4*b*c
- 560*I*(d*x^(1/3) + c)^3*b*c^2 + 336*I*(d*x^(1/3) + c)^2*b*c^3 - 105*I*(d*x^(1/3) + c)*b*c^4 + 14*I*b*c^5)*po
lylog(4, -e^(2*I*d*x^(1/3) + 2*I*c)) - 630*(112*(d*x^(1/3) + c)^6*b - 384*(d*x^(1/3) + c)^5*b*c + 560*(d*x^(1/
3) + c)^4*b*c^2 - 448*(d*x^(1/3) + c)^3*b*c^3 + 210*(d*x^(1/3) + c)^2*b*c^4 - 56*(d*x^(1/3) + c)*b*c^5 + 7*b*c
^6)*polylog(3, -e^(2*I*d*x^(1/3) + 2*I*c)))/d^9

Giac [F]

\[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int { {\left (b \tan \left (d x^{\frac {1}{3}} + c\right ) + a\right )} x^{2} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx=\int x^2\,\left (a+b\,\mathrm {tan}\left (c+d\,x^{1/3}\right )\right ) \,d x \]

[In]

int(x^2*(a + b*tan(c + d*x^(1/3))),x)

[Out]

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