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

Optimal. Leaf size=287 \[ \frac{12 i b x^{7/3} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac{945 b x^{2/3} \text{PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac{630 i b x \text{PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{945 i b \sqrt [3]{x} \text{PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac{945 b \text{PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9}+\frac{a x^3}{3}-\frac{3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{1}{3} i b x^3 \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.401709, antiderivative size = 287, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {14, 3747, 3719, 2190, 2531, 6609, 2282, 6589} \[ \frac{a x^3}{3}+\frac{12 i b x^{7/3} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac{945 b x^{2/3} \text{Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac{630 i b x \text{Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{945 i b \sqrt [3]{x} \text{Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac{945 b \text{Li}_9\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9}-\frac{3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{1}{3} i b x^3 \]

Antiderivative was successfully verified.

[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 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 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 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 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 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]

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 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]

Rubi steps

\begin{align*} \int x^2 \left (a+b \tan \left (c+d \sqrt [3]{x}\right )\right ) \, dx &=\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) \operatorname{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) \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 )\\ &=\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) \operatorname{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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{(84 i 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{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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}+\frac{(252 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{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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{(630 i 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{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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac{(1260 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{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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac{630 i b x \text{Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{(1890 i 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{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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac{630 i b x \text{Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{945 b x^{2/3} \text{Li}_7\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac{(1890 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{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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac{630 i b x \text{Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{945 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 \sqrt [3]{x} \text{Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac{(945 i 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{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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac{630 i b x \text{Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{945 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 \sqrt [3]{x} \text{Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac{(945 b) \operatorname{Subst}\left (\int \frac{\text{Li}_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} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{Li}_3\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}+\frac{630 i b x \text{Li}_6\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{945 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 \sqrt [3]{x} \text{Li}_8\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac{945 b \text{Li}_9\left (-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9}\\ \end{align*}

Mathematica [A]  time = 0.126838, size = 287, normalized size = 1. \[ \frac{12 i b x^{7/3} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^2}-\frac{42 b x^2 \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^3}-\frac{126 i b x^{5/3} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^4}+\frac{315 b x^{4/3} \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^5}-\frac{945 b x^{2/3} \text{PolyLog}\left (7,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^7}+\frac{630 i b x \text{PolyLog}\left (6,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^6}-\frac{945 i b \sqrt [3]{x} \text{PolyLog}\left (8,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d^8}+\frac{945 b \text{PolyLog}\left (9,-e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{2 d^9}+\frac{a x^3}{3}-\frac{3 b x^{8/3} \log \left (1+e^{2 i \left (c+d \sqrt [3]{x}\right )}\right )}{d}+\frac{1}{3} i b x^3 \]

Antiderivative was successfully verified.

[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)

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Maple [F]  time = 0.155, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b\tan \left ( c+d\sqrt [3]{x} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Maxima [B]  time = 2.37799, size = 1511, normalized size = 5.26 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)) - (5040*I*(d*x^(1/3) + c)^8*b - 23040*I*(d*x^(1/3) + c)^7*b*c + 47040*I*(d*x^(1/3) + c)^6*b*c^2 -
56448*I*(d*x^(1/3) + c)^5*b*c^3 + 44100*I*(d*x^(1/3) + c)^4*b*c^4 - 23520*I*(d*x^(1/3) + c)^3*b*c^5 + 8820*I*
(d*x^(1/3) + c)^2*b*c^6 - 2520*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) - (-20160*I*(d*x^(1/3) + c)^7*b + 80640*I*(d*x^(1/3) + c)^6*b*c - 141120*I*(d*x^(1/3) + c)^5*b*c^2 + 141
120*I*(d*x^(1/3) + c)^4*b*c^3 - 88200*I*(d*x^(1/3) + c)^3*b*c^4 + 35280*I*(d*x^(1/3) + c)^2*b*c^5 - 8820*I*(d*
x^(1/3) + c)*b*c^6 + 1260*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)) - (1587600*I*(d*x^(1/3) + c)*b - 907200*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)) - (-1058400*I*(d*x^
(1/3) + c)^3*b + 1814400*I*(d*x^(1/3) + c)^2*b*c - 1058400*I*(d*x^(1/3) + c)*b*c^2 + 211680*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)) - (211680*I*(d*x^(1/3
) + c)^5*b - 604800*I*(d*x^(1/3) + c)^4*b*c + 705600*I*(d*x^(1/3) + c)^3*b*c^2 - 423360*I*(d*x^(1/3) + c)^2*b*
c^3 + 132300*I*(d*x^(1/3) + c)*b*c^4 - 17640*I*b*c^5)*polylog(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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b x^{2} \tan \left (d x^{\frac{1}{3}} + c\right ) + a x^{2}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (d x^{\frac{1}{3}} + c\right ) + a\right )} x^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)