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

Optimal. Leaf size=597 \[ \frac {a^2 x^3}{3}+\frac {945 a b \text {Li}_9\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}-\frac {1890 i a b \sqrt [3]{x} \text {Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {1890 a b x^{2/3} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {1260 i a b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {630 a b x^{4/3} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {2}{3} i a b x^3+\frac {945 i b^2 \text {Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}+\frac {1890 b^2 \sqrt [3]{x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {1890 i b^2 x^{2/3} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {1260 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {630 i b^2 x^{4/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {3 i b^2 x^{8/3}}{d}-\frac {b^2 x^3}{3} \]

[Out]

2/3*I*a*b*x^3+1/3*a^2*x^3-1890*I*a*b*x^(1/3)*polylog(8,-exp(2*I*(c+d*x^(1/3))))/d^8-1/3*b^2*x^3+24*b^2*x^(7/3)
*ln(1+exp(2*I*(c+d*x^(1/3))))/d^2-6*a*b*x^(8/3)*ln(1+exp(2*I*(c+d*x^(1/3))))/d-252*I*a*b*x^(5/3)*polylog(4,-ex
p(2*I*(c+d*x^(1/3))))/d^4-1890*I*b^2*x^(2/3)*polylog(6,-exp(2*I*(c+d*x^(1/3))))/d^7+252*b^2*x^(5/3)*polylog(3,
-exp(2*I*(c+d*x^(1/3))))/d^4-84*a*b*x^2*polylog(3,-exp(2*I*(c+d*x^(1/3))))/d^3+630*I*b^2*x^(4/3)*polylog(4,-ex
p(2*I*(c+d*x^(1/3))))/d^5+24*I*a*b*x^(7/3)*polylog(2,-exp(2*I*(c+d*x^(1/3))))/d^2-1260*b^2*x*polylog(5,-exp(2*
I*(c+d*x^(1/3))))/d^6+630*a*b*x^(4/3)*polylog(5,-exp(2*I*(c+d*x^(1/3))))/d^5+1260*I*a*b*x*polylog(6,-exp(2*I*(
c+d*x^(1/3))))/d^6-84*I*b^2*x^2*polylog(2,-exp(2*I*(c+d*x^(1/3))))/d^3+1890*b^2*x^(1/3)*polylog(7,-exp(2*I*(c+
d*x^(1/3))))/d^8-1890*a*b*x^(2/3)*polylog(7,-exp(2*I*(c+d*x^(1/3))))/d^7+945*I*b^2*polylog(8,-exp(2*I*(c+d*x^(
1/3))))/d^9-3*I*b^2*x^(8/3)/d+945*a*b*polylog(9,-exp(2*I*(c+d*x^(1/3))))/d^9+3*b^2*x^(8/3)*tan(c+d*x^(1/3))/d

________________________________________________________________________________________

Rubi [A]  time = 0.83, antiderivative size = 597, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3747, 3722, 3719, 2190, 2531, 6609, 2282, 6589, 3720, 30} \[ \frac {a^2 x^3}{3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {630 a b x^{4/3} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {1890 a b x^{2/3} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {1260 i a b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {1890 i a b \sqrt [3]{x} \text {Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 a b \text {Li}_9\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {2}{3} i a b x^3-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {630 i b^2 x^{4/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {1890 i b^2 x^{2/3} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {1260 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {1890 b^2 \sqrt [3]{x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 i b^2 \text {Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {3 i b^2 x^{8/3}}{d}-\frac {b^2 x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-3*I)*b^2*x^(8/3))/d + (a^2*x^3)/3 + ((2*I)/3)*a*b*x^3 - (b^2*x^3)/3 + (24*b^2*x^(7/3)*Log[1 + E^((2*I)*(c +
d*x^(1/3)))])/d^2 - (6*a*b*x^(8/3)*Log[1 + E^((2*I)*(c + d*x^(1/3)))])/d - ((84*I)*b^2*x^2*PolyLog[2, -E^((2*
I)*(c + d*x^(1/3)))])/d^3 + ((24*I)*a*b*x^(7/3)*PolyLog[2, -E^((2*I)*(c + d*x^(1/3)))])/d^2 + (252*b^2*x^(5/3)
*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^4 - (84*a*b*x^2*PolyLog[3, -E^((2*I)*(c + d*x^(1/3)))])/d^3 + ((630
*I)*b^2*x^(4/3)*PolyLog[4, -E^((2*I)*(c + d*x^(1/3)))])/d^5 - ((252*I)*a*b*x^(5/3)*PolyLog[4, -E^((2*I)*(c + d
*x^(1/3)))])/d^4 - (1260*b^2*x*PolyLog[5, -E^((2*I)*(c + d*x^(1/3)))])/d^6 + (630*a*b*x^(4/3)*PolyLog[5, -E^((
2*I)*(c + d*x^(1/3)))])/d^5 - ((1890*I)*b^2*x^(2/3)*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^7 + ((1260*I)*a*
b*x*PolyLog[6, -E^((2*I)*(c + d*x^(1/3)))])/d^6 + (1890*b^2*x^(1/3)*PolyLog[7, -E^((2*I)*(c + d*x^(1/3)))])/d^
8 - (1890*a*b*x^(2/3)*PolyLog[7, -E^((2*I)*(c + d*x^(1/3)))])/d^7 + ((945*I)*b^2*PolyLog[8, -E^((2*I)*(c + d*x
^(1/3)))])/d^9 - ((1890*I)*a*b*x^(1/3)*PolyLog[8, -E^((2*I)*(c + d*x^(1/3)))])/d^8 + (945*a*b*PolyLog[9, -E^((
2*I)*(c + d*x^(1/3)))])/d^9 + (3*b^2*x^(8/3)*Tan[c + d*x^(1/3)])/d

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2282

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 2531

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 3719

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 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3747

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 6589

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 6609

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 x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right )^2 \, dx &=3 \operatorname {Subst}\left (\int x^8 (a+b \tan (c+d x))^2 \, dx,x,\sqrt [3]{x}\right )\\ &=3 \operatorname {Subst}\left (\int \left (a^2 x^8+2 a b x^8 \tan (c+d x)+b^2 x^8 \tan ^2(c+d x)\right ) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {a^2 x^3}{3}+(6 a b) \operatorname {Subst}\left (\int x^8 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )+\left (3 b^2\right ) \operatorname {Subst}\left (\int x^8 \tan ^2(c+d x) \, dx,x,\sqrt [3]{x}\right )\\ &=\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-(12 i a b) \operatorname {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 )-\left (3 b^2\right ) \operatorname {Subst}\left (\int x^8 \, dx,x,\sqrt [3]{x}\right )-\frac {\left (24 b^2\right ) \operatorname {Subst}\left (\int x^7 \tan (c+d x) \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(48 a b) \operatorname {Subst}\left (\int x^7 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d}+\frac {\left (48 i b^2\right ) \operatorname {Subst}\left (\int \frac {e^{2 i (c+d x)} x^7}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt [3]{x}\right )}{d}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(168 i a b) \operatorname {Subst}\left (\int x^6 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}-\frac {\left (168 b^2\right ) \operatorname {Subst}\left (\int x^6 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^2}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(504 a b) \operatorname {Subst}\left (\int x^5 \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^3}+\frac {\left (504 i b^2\right ) \operatorname {Subst}\left (\int x^5 \text {Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^3}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(1260 i a b) \operatorname {Subst}\left (\int x^4 \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^4}-\frac {\left (1260 b^2\right ) \operatorname {Subst}\left (\int x^4 \text {Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^4}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {630 i b^2 x^{4/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac {630 a b x^{4/3} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(2520 a b) \operatorname {Subst}\left (\int x^3 \text {Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^5}-\frac {\left (2520 i b^2\right ) \operatorname {Subst}\left (\int x^3 \text {Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^5}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {630 i b^2 x^{4/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {1260 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {630 a b x^{4/3} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac {1260 i a b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}-\frac {(3780 i a b) \operatorname {Subst}\left (\int x^2 \text {Li}_6\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^6}+\frac {\left (3780 b^2\right ) \operatorname {Subst}\left (\int x^2 \text {Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^6}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {630 i b^2 x^{4/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {1260 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {630 a b x^{4/3} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {1890 i b^2 x^{2/3} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {1260 i a b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac {1890 a b x^{2/3} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(3780 a b) \operatorname {Subst}\left (\int x \text {Li}_7\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^7}+\frac {\left (3780 i b^2\right ) \operatorname {Subst}\left (\int x \text {Li}_6\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^7}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {630 i b^2 x^{4/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {1260 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {630 a b x^{4/3} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {1890 i b^2 x^{2/3} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {1260 i a b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {1890 b^2 \sqrt [3]{x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {1890 a b x^{2/3} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {1890 i a b \sqrt [3]{x} \text {Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(1890 i a b) \operatorname {Subst}\left (\int \text {Li}_8\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^8}-\frac {\left (1890 b^2\right ) \operatorname {Subst}\left (\int \text {Li}_7\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt [3]{x}\right )}{d^8}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {630 i b^2 x^{4/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {1260 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {630 a b x^{4/3} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {1890 i b^2 x^{2/3} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {1260 i a b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {1890 b^2 \sqrt [3]{x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {1890 a b x^{2/3} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}-\frac {1890 i a b \sqrt [3]{x} \text {Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}+\frac {(945 a b) \operatorname {Subst}\left (\int \frac {\text {Li}_8(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}+\frac {\left (945 i b^2\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_7(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}\\ &=-\frac {3 i b^2 x^{8/3}}{d}+\frac {a^2 x^3}{3}+\frac {2}{3} i a b x^3-\frac {b^2 x^3}{3}+\frac {24 b^2 x^{7/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac {6 a b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {84 i b^2 x^2 \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {24 i a b x^{7/3} \text {Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {252 b^2 x^{5/3} \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {84 a b x^2 \text {Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac {630 i b^2 x^{4/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {252 i a b x^{5/3} \text {Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}-\frac {1260 b^2 x \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {630 a b x^{4/3} \text {Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac {1890 i b^2 x^{2/3} \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {1260 i a b x \text {Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}+\frac {1890 b^2 \sqrt [3]{x} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}-\frac {1890 a b x^{2/3} \text {Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac {945 i b^2 \text {Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}-\frac {1890 i a b \sqrt [3]{x} \text {Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac {945 a b \text {Li}_9\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^9}+\frac {3 b^2 x^{8/3} \tan \left (c+d \sqrt [3]{x}\right )}{d}\\ \end {align*}

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Mathematica [A]  time = 4.89, size = 599, normalized size = 1.00 \[ \frac {1}{3} \left (x^3 \left (a^2+2 a b \tan (c)-b^2\right )+b \left (\frac {9 a \left (-8 i d^7 x^{7/3} \text {Li}_2\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-28 d^6 x^2 \text {Li}_3\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+84 i d^5 x^{5/3} \text {Li}_4\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+105 \left (2 d^4 x^{4/3} \text {Li}_5\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-4 i d^3 x \text {Li}_6\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-6 d^2 x^{2/3} \text {Li}_7\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+6 i d \sqrt [3]{x} \text {Li}_8\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+3 \text {Li}_9\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )\right )\right )}{d^9}-\frac {18 a x^{8/3} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}-\frac {4 i a x^3}{1+e^{2 i c}}+\frac {72 b x^{7/3} \log \left (1+e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}+\frac {63 b \left (4 i d^6 x^2 \text {Li}_2\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+12 d^5 x^{5/3} \text {Li}_3\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-15 i \left (2 d^4 x^{4/3} \text {Li}_4\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-4 i d^3 x \text {Li}_5\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )-6 d^2 x^{2/3} \text {Li}_6\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+6 i d \sqrt [3]{x} \text {Li}_7\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )+3 \text {Li}_8\left (-e^{-2 i \left (c+d \sqrt [3]{x}\right )}\right )\right )\right )}{d^9}+\frac {18 i b x^{8/3}}{d+e^{2 i c} d}\right )+\frac {9 b^2 x^{8/3} \sec (c) \sin \left (d \sqrt [3]{x}\right ) \sec \left (c+d \sqrt [3]{x}\right )}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(((18*I)*b*x^(8/3))/(d + d*E^((2*I)*c)) - ((4*I)*a*x^3)/(1 + E^((2*I)*c)) + (72*b*x^(7/3)*Log[1 + E^((-2*I)
*(c + d*x^(1/3)))])/d^2 - (18*a*x^(8/3)*Log[1 + E^((-2*I)*(c + d*x^(1/3)))])/d + (63*b*((4*I)*d^6*x^2*PolyLog[
2, -E^((-2*I)*(c + d*x^(1/3)))] + 12*d^5*x^(5/3)*PolyLog[3, -E^((-2*I)*(c + d*x^(1/3)))] - (15*I)*(2*d^4*x^(4/
3)*PolyLog[4, -E^((-2*I)*(c + d*x^(1/3)))] - (4*I)*d^3*x*PolyLog[5, -E^((-2*I)*(c + d*x^(1/3)))] - 6*d^2*x^(2/
3)*PolyLog[6, -E^((-2*I)*(c + d*x^(1/3)))] + (6*I)*d*x^(1/3)*PolyLog[7, -E^((-2*I)*(c + d*x^(1/3)))] + 3*PolyL
og[8, -E^((-2*I)*(c + d*x^(1/3)))])))/d^9 + (9*a*((-8*I)*d^7*x^(7/3)*PolyLog[2, -E^((-2*I)*(c + d*x^(1/3)))] -
28*d^6*x^2*PolyLog[3, -E^((-2*I)*(c + d*x^(1/3)))] + (84*I)*d^5*x^(5/3)*PolyLog[4, -E^((-2*I)*(c + d*x^(1/3))
)] + 105*(2*d^4*x^(4/3)*PolyLog[5, -E^((-2*I)*(c + d*x^(1/3)))] - (4*I)*d^3*x*PolyLog[6, -E^((-2*I)*(c + d*x^(
1/3)))] - 6*d^2*x^(2/3)*PolyLog[7, -E^((-2*I)*(c + d*x^(1/3)))] + (6*I)*d*x^(1/3)*PolyLog[8, -E^((-2*I)*(c + d
*x^(1/3)))] + 3*PolyLog[9, -E^((-2*I)*(c + d*x^(1/3)))])))/d^9) + (9*b^2*x^(8/3)*Sec[c]*Sec[c + d*x^(1/3)]*Sin
[d*x^(1/3)])/d + x^3*(a^2 - b^2 + 2*a*b*Tan[c]))/3

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b^{2} x^{2} \tan \left (d x^{\frac {1}{3}} + c\right )^{2} + 2 \, a b x^{2} \tan \left (d x^{\frac {1}{3}} + c\right ) + a^{2} x^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.42, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a +b \tan \left (c +d \,x^{\frac {1}{3}}\right )\right )^{2}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 2.16, size = 4673, normalized size = 7.83 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((d*x^(1/3) + c)^9*a^2 - 9*(d*x^(1/3) + c)^8*a^2*c + 36*(d*x^(1/3) + c)^7*a^2*c^2 - 84*(d*x^(1/3) + c)^6*a
^2*c^3 + 126*(d*x^(1/3) + c)^5*a^2*c^4 - 126*(d*x^(1/3) + c)^4*a^2*c^5 + 84*(d*x^(1/3) + c)^3*a^2*c^6 - 36*(d*
x^(1/3) + c)^2*a^2*c^7 + 9*(d*x^(1/3) + c)*a^2*c^8 + 18*a*b*c^8*log(sec(d*x^(1/3) + c)) - 9*(-315*I*(d*x^(1/3)
+ c)*b^2*c^8 - 35*(2*a*b + I*b^2)*(d*x^(1/3) + c)^9 + 315*(2*a*b + I*b^2)*(d*x^(1/3) + c)^8*c - 1260*(2*a*b +
I*b^2)*(d*x^(1/3) + c)^7*c^2 + 2940*(2*a*b + I*b^2)*(d*x^(1/3) + c)^6*c^3 - 4410*(2*a*b + I*b^2)*(d*x^(1/3) +
c)^5*c^4 + 4410*(2*a*b + I*b^2)*(d*x^(1/3) + c)^4*c^5 - 2940*(2*a*b + I*b^2)*(d*x^(1/3) + c)^3*c^6 + 1260*(2*
a*b + I*b^2)*(d*x^(1/3) + c)^2*c^7 - 630*b^2*c^8 + (10080*(d*x^(1/3) + c)^8*a*b + 2520*b^2*c^7 - 23040*(2*a*b*
c + b^2)*(d*x^(1/3) + c)^7 + 94080*(a*b*c^2 + b^2*c)*(d*x^(1/3) + c)^6 - 56448*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1
/3) + c)^5 + 88200*(a*b*c^4 + 2*b^2*c^3)*(d*x^(1/3) + c)^4 - 23520*(2*a*b*c^5 + 5*b^2*c^4)*(d*x^(1/3) + c)^3 +
17640*(a*b*c^6 + 3*b^2*c^5)*(d*x^(1/3) + c)^2 - 2520*(2*a*b*c^7 + 7*b^2*c^6)*(d*x^(1/3) + c) + 24*(420*(d*x^(
1/3) + c)^8*a*b + 105*b^2*c^7 - 960*(2*a*b*c + b^2)*(d*x^(1/3) + c)^7 + 3920*(a*b*c^2 + b^2*c)*(d*x^(1/3) + c)
^6 - 2352*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3) + c)^5 + 3675*(a*b*c^4 + 2*b^2*c^3)*(d*x^(1/3) + c)^4 - 980*(2*a*
b*c^5 + 5*b^2*c^4)*(d*x^(1/3) + c)^3 + 735*(a*b*c^6 + 3*b^2*c^5)*(d*x^(1/3) + c)^2 - 105*(2*a*b*c^7 + 7*b^2*c^
6)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) + (10080*I*(d*x^(1/3) + c)^8*a*b + 2520*I*b^2*c^7 + (-46080*I*a*b*c
- 23040*I*b^2)*(d*x^(1/3) + c)^7 + (94080*I*a*b*c^2 + 94080*I*b^2*c)*(d*x^(1/3) + c)^6 + (-112896*I*a*b*c^3 -
169344*I*b^2*c^2)*(d*x^(1/3) + c)^5 + (88200*I*a*b*c^4 + 176400*I*b^2*c^3)*(d*x^(1/3) + c)^4 + (-47040*I*a*b*
c^5 - 117600*I*b^2*c^4)*(d*x^(1/3) + c)^3 + (17640*I*a*b*c^6 + 52920*I*b^2*c^5)*(d*x^(1/3) + c)^2 + (-5040*I*a
*b*c^7 - 17640*I*b^2*c^6)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*c))*arctan2(sin(2*d*x^(1/3) + 2*c), cos(2*d*x^(
1/3) + 2*c) + 1) - (35*(2*a*b + I*b^2)*(d*x^(1/3) + c)^9 - 315*(2*b^2 + (2*a*b + I*b^2)*c)*(d*x^(1/3) + c)^8 +
1260*(4*b^2*c + (2*a*b + I*b^2)*c^2)*(d*x^(1/3) + c)^7 - 2940*(6*b^2*c^2 + (2*a*b + I*b^2)*c^3)*(d*x^(1/3) +
c)^6 + 4410*(8*b^2*c^3 + (2*a*b + I*b^2)*c^4)*(d*x^(1/3) + c)^5 - 4410*(10*b^2*c^4 + (2*a*b + I*b^2)*c^5)*(d*x
^(1/3) + c)^4 + 2940*(12*b^2*c^5 + (2*a*b + I*b^2)*c^6)*(d*x^(1/3) + c)^3 - 1260*(14*b^2*c^6 + (2*a*b + I*b^2)
*c^7)*(d*x^(1/3) + c)^2 - (-315*I*b^2*c^8 - 5040*b^2*c^7)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) - (40320*(d*
x^(1/3) + c)^7*a*b - 2520*a*b*c^7 - 8820*b^2*c^6 - 80640*(2*a*b*c + b^2)*(d*x^(1/3) + c)^6 + 282240*(a*b*c^2 +
b^2*c)*(d*x^(1/3) + c)^5 - 141120*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3) + c)^4 + 176400*(a*b*c^4 + 2*b^2*c^3)*(d
*x^(1/3) + c)^3 - 35280*(2*a*b*c^5 + 5*b^2*c^4)*(d*x^(1/3) + c)^2 + 17640*(a*b*c^6 + 3*b^2*c^5)*(d*x^(1/3) + c
) + 1260*(32*(d*x^(1/3) + c)^7*a*b - 2*a*b*c^7 - 7*b^2*c^6 - 64*(2*a*b*c + b^2)*(d*x^(1/3) + c)^6 + 224*(a*b*c
^2 + b^2*c)*(d*x^(1/3) + c)^5 - 112*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3) + c)^4 + 140*(a*b*c^4 + 2*b^2*c^3)*(d*x
^(1/3) + c)^3 - 28*(2*a*b*c^5 + 5*b^2*c^4)*(d*x^(1/3) + c)^2 + 14*(a*b*c^6 + 3*b^2*c^5)*(d*x^(1/3) + c))*cos(2
*d*x^(1/3) + 2*c) - (-40320*I*(d*x^(1/3) + c)^7*a*b + 2520*I*a*b*c^7 + 8820*I*b^2*c^6 + (161280*I*a*b*c + 8064
0*I*b^2)*(d*x^(1/3) + c)^6 + (-282240*I*a*b*c^2 - 282240*I*b^2*c)*(d*x^(1/3) + c)^5 + (282240*I*a*b*c^3 + 4233
60*I*b^2*c^2)*(d*x^(1/3) + c)^4 + (-176400*I*a*b*c^4 - 352800*I*b^2*c^3)*(d*x^(1/3) + c)^3 + (70560*I*a*b*c^5
+ 176400*I*b^2*c^4)*(d*x^(1/3) + c)^2 + (-17640*I*a*b*c^6 - 52920*I*b^2*c^5)*(d*x^(1/3) + c))*sin(2*d*x^(1/3)
+ 2*c))*dilog(-e^(2*I*d*x^(1/3) + 2*I*c)) + (-5040*I*(d*x^(1/3) + c)^8*a*b - 1260*I*b^2*c^7 + (23040*I*a*b*c +
11520*I*b^2)*(d*x^(1/3) + c)^7 + (-47040*I*a*b*c^2 - 47040*I*b^2*c)*(d*x^(1/3) + c)^6 + (56448*I*a*b*c^3 + 84
672*I*b^2*c^2)*(d*x^(1/3) + c)^5 + (-44100*I*a*b*c^4 - 88200*I*b^2*c^3)*(d*x^(1/3) + c)^4 + (23520*I*a*b*c^5 +
58800*I*b^2*c^4)*(d*x^(1/3) + c)^3 + (-8820*I*a*b*c^6 - 26460*I*b^2*c^5)*(d*x^(1/3) + c)^2 + (2520*I*a*b*c^7
+ 8820*I*b^2*c^6)*(d*x^(1/3) + c) + (-5040*I*(d*x^(1/3) + c)^8*a*b - 1260*I*b^2*c^7 + (23040*I*a*b*c + 11520*I
*b^2)*(d*x^(1/3) + c)^7 + (-47040*I*a*b*c^2 - 47040*I*b^2*c)*(d*x^(1/3) + c)^6 + (56448*I*a*b*c^3 + 84672*I*b^
2*c^2)*(d*x^(1/3) + c)^5 + (-44100*I*a*b*c^4 - 88200*I*b^2*c^3)*(d*x^(1/3) + c)^4 + (23520*I*a*b*c^5 + 58800*I
*b^2*c^4)*(d*x^(1/3) + c)^3 + (-8820*I*a*b*c^6 - 26460*I*b^2*c^5)*(d*x^(1/3) + c)^2 + (2520*I*a*b*c^7 + 8820*I
*b^2*c^6)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) + 12*(420*(d*x^(1/3) + c)^8*a*b + 105*b^2*c^7 - 960*(2*a*b*c
+ b^2)*(d*x^(1/3) + c)^7 + 3920*(a*b*c^2 + b^2*c)*(d*x^(1/3) + c)^6 - 2352*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3)
+ c)^5 + 3675*(a*b*c^4 + 2*b^2*c^3)*(d*x^(1/3) + c)^4 - 980*(2*a*b*c^5 + 5*b^2*c^4)*(d*x^(1/3) + c)^3 + 735*(
a*b*c^6 + 3*b^2*c^5)*(d*x^(1/3) + c)^2 - 105*(2*a*b*c^7 + 7*b^2*c^6)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*c))*
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) + (1587600*I*a*b*cos(2
*d*x^(1/3) + 2*c) - 1587600*a*b*sin(2*d*x^(1/3) + 2*c) + 1587600*I*a*b)*polylog(9, -e^(2*I*d*x^(1/3) + 2*I*c))
+ (3175200*(d*x^(1/3) + c)*a*b - 1814400*a*b*c - 907200*b^2 + 453600*(7*(d*x^(1/3) + c)*a*b - 4*a*b*c - 2*b^2
)*cos(2*d*x^(1/3) + 2*c) + (3175200*I*(d*x^(1/3) + c)*a*b - 1814400*I*a*b*c - 907200*I*b^2)*sin(2*d*x^(1/3) +
2*c))*polylog(8, -e^(2*I*d*x^(1/3) + 2*I*c)) + (-3175200*I*(d*x^(1/3) + c)^2*a*b - 1058400*I*a*b*c^2 - 1058400
*I*b^2*c + (3628800*I*a*b*c + 1814400*I*b^2)*(d*x^(1/3) + c) + (-3175200*I*(d*x^(1/3) + c)^2*a*b - 1058400*I*a
*b*c^2 - 1058400*I*b^2*c + (3628800*I*a*b*c + 1814400*I*b^2)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) + 151200*
(21*(d*x^(1/3) + c)^2*a*b + 7*a*b*c^2 + 7*b^2*c - 12*(2*a*b*c + b^2)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*c))*
polylog(7, -e^(2*I*d*x^(1/3) + 2*I*c)) - (2116800*(d*x^(1/3) + c)^3*a*b - 423360*a*b*c^3 - 635040*b^2*c^2 - 18
14400*(2*a*b*c + b^2)*(d*x^(1/3) + c)^2 + 2116800*(a*b*c^2 + b^2*c)*(d*x^(1/3) + c) + 30240*(70*(d*x^(1/3) + c
)^3*a*b - 14*a*b*c^3 - 21*b^2*c^2 - 60*(2*a*b*c + b^2)*(d*x^(1/3) + c)^2 + 70*(a*b*c^2 + b^2*c)*(d*x^(1/3) + c
))*cos(2*d*x^(1/3) + 2*c) - (-2116800*I*(d*x^(1/3) + c)^3*a*b + 423360*I*a*b*c^3 + 635040*I*b^2*c^2 + (3628800
*I*a*b*c + 1814400*I*b^2)*(d*x^(1/3) + c)^2 + (-2116800*I*a*b*c^2 - 2116800*I*b^2*c)*(d*x^(1/3) + c))*sin(2*d*
x^(1/3) + 2*c))*polylog(6, -e^(2*I*d*x^(1/3) + 2*I*c)) + (1058400*I*(d*x^(1/3) + c)^4*a*b + 132300*I*a*b*c^4 +
264600*I*b^2*c^3 + (-2419200*I*a*b*c - 1209600*I*b^2)*(d*x^(1/3) + c)^3 + (2116800*I*a*b*c^2 + 2116800*I*b^2*
c)*(d*x^(1/3) + c)^2 + (-846720*I*a*b*c^3 - 1270080*I*b^2*c^2)*(d*x^(1/3) + c) + (1058400*I*(d*x^(1/3) + c)^4*
a*b + 132300*I*a*b*c^4 + 264600*I*b^2*c^3 + (-2419200*I*a*b*c - 1209600*I*b^2)*(d*x^(1/3) + c)^3 + (2116800*I*
a*b*c^2 + 2116800*I*b^2*c)*(d*x^(1/3) + c)^2 + (-846720*I*a*b*c^3 - 1270080*I*b^2*c^2)*(d*x^(1/3) + c))*cos(2*
d*x^(1/3) + 2*c) - 3780*(280*(d*x^(1/3) + c)^4*a*b + 35*a*b*c^4 + 70*b^2*c^3 - 320*(2*a*b*c + b^2)*(d*x^(1/3)
+ c)^3 + 560*(a*b*c^2 + b^2*c)*(d*x^(1/3) + c)^2 - 112*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3) + c))*sin(2*d*x^(1/3
) + 2*c))*polylog(5, -e^(2*I*d*x^(1/3) + 2*I*c)) + (423360*(d*x^(1/3) + c)^5*a*b - 35280*a*b*c^5 - 88200*b^2*c
^4 - 604800*(2*a*b*c + b^2)*(d*x^(1/3) + c)^4 + 1411200*(a*b*c^2 + b^2*c)*(d*x^(1/3) + c)^3 - 423360*(2*a*b*c^
3 + 3*b^2*c^2)*(d*x^(1/3) + c)^2 + 264600*(a*b*c^4 + 2*b^2*c^3)*(d*x^(1/3) + c) + 2520*(168*(d*x^(1/3) + c)^5*
a*b - 14*a*b*c^5 - 35*b^2*c^4 - 240*(2*a*b*c + b^2)*(d*x^(1/3) + c)^4 + 560*(a*b*c^2 + b^2*c)*(d*x^(1/3) + c)^
3 - 168*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3) + c)^2 + 105*(a*b*c^4 + 2*b^2*c^3)*(d*x^(1/3) + c))*cos(2*d*x^(1/3)
+ 2*c) + (423360*I*(d*x^(1/3) + c)^5*a*b - 35280*I*a*b*c^5 - 88200*I*b^2*c^4 + (-1209600*I*a*b*c - 604800*I*b
^2)*(d*x^(1/3) + c)^4 + (1411200*I*a*b*c^2 + 1411200*I*b^2*c)*(d*x^(1/3) + c)^3 + (-846720*I*a*b*c^3 - 1270080
*I*b^2*c^2)*(d*x^(1/3) + c)^2 + (264600*I*a*b*c^4 + 529200*I*b^2*c^3)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*c))
*polylog(4, -e^(2*I*d*x^(1/3) + 2*I*c)) + (-141120*I*(d*x^(1/3) + c)^6*a*b - 8820*I*a*b*c^6 - 26460*I*b^2*c^5
+ (483840*I*a*b*c + 241920*I*b^2)*(d*x^(1/3) + c)^5 + (-705600*I*a*b*c^2 - 705600*I*b^2*c)*(d*x^(1/3) + c)^4 +
(564480*I*a*b*c^3 + 846720*I*b^2*c^2)*(d*x^(1/3) + c)^3 + (-264600*I*a*b*c^4 - 529200*I*b^2*c^3)*(d*x^(1/3) +
c)^2 + (70560*I*a*b*c^5 + 176400*I*b^2*c^4)*(d*x^(1/3) + c) + (-141120*I*(d*x^(1/3) + c)^6*a*b - 8820*I*a*b*c
^6 - 26460*I*b^2*c^5 + (483840*I*a*b*c + 241920*I*b^2)*(d*x^(1/3) + c)^5 + (-705600*I*a*b*c^2 - 705600*I*b^2*c
)*(d*x^(1/3) + c)^4 + (564480*I*a*b*c^3 + 846720*I*b^2*c^2)*(d*x^(1/3) + c)^3 + (-264600*I*a*b*c^4 - 529200*I*
b^2*c^3)*(d*x^(1/3) + c)^2 + (70560*I*a*b*c^5 + 176400*I*b^2*c^4)*(d*x^(1/3) + c))*cos(2*d*x^(1/3) + 2*c) + 12
60*(112*(d*x^(1/3) + c)^6*a*b + 7*a*b*c^6 + 21*b^2*c^5 - 192*(2*a*b*c + b^2)*(d*x^(1/3) + c)^5 + 560*(a*b*c^2
+ b^2*c)*(d*x^(1/3) + c)^4 - 224*(2*a*b*c^3 + 3*b^2*c^2)*(d*x^(1/3) + c)^3 + 210*(a*b*c^4 + 2*b^2*c^3)*(d*x^(1
/3) + c)^2 - 28*(2*a*b*c^5 + 5*b^2*c^4)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*c))*polylog(3, -e^(2*I*d*x^(1/3)
+ 2*I*c)) + ((-70*I*a*b + 35*b^2)*(d*x^(1/3) + c)^9 + (630*I*b^2 + (630*I*a*b - 315*b^2)*c)*(d*x^(1/3) + c)^8
+ (-5040*I*b^2*c + (-2520*I*a*b + 1260*b^2)*c^2)*(d*x^(1/3) + c)^7 + (17640*I*b^2*c^2 + (5880*I*a*b - 2940*b^2
)*c^3)*(d*x^(1/3) + c)^6 + (-35280*I*b^2*c^3 + (-8820*I*a*b + 4410*b^2)*c^4)*(d*x^(1/3) + c)^5 + (44100*I*b^2*
c^4 + (8820*I*a*b - 4410*b^2)*c^5)*(d*x^(1/3) + c)^4 + (-35280*I*b^2*c^5 + (-5880*I*a*b + 2940*b^2)*c^6)*(d*x^
(1/3) + c)^3 + (17640*I*b^2*c^6 + (2520*I*a*b - 1260*b^2)*c^7)*(d*x^(1/3) + c)^2 + 315*(b^2*c^8 - 16*I*b^2*c^7
)*(d*x^(1/3) + c))*sin(2*d*x^(1/3) + 2*c))/(-315*I*cos(2*d*x^(1/3) + 2*c) + 315*sin(2*d*x^(1/3) + 2*c) - 315*I
))/d^9

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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