3.31 \(\int x^2 (a+b \tan (c+d \sqrt{x}))^2 \, dx\)

Optimal. Leaf size=402 \[ \frac{10 i a b x^2 \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^{3/2} \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{30 i a b x \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 a b \sqrt{x} \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{15 i a b \text{PolyLog}\left (6,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{20 i b^2 x^{3/2} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 b^2 x \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 i b^2 \sqrt{x} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{15 b^2 \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a^2 x^3}{3}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{2}{3} i a b x^3+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^{5/2}}{d}-\frac{b^2 x^3}{3} \]

[Out]

((-2*I)*b^2*x^(5/2))/d + (a^2*x^3)/3 + ((2*I)/3)*a*b*x^3 - (b^2*x^3)/3 + (10*b^2*x^2*Log[1 + E^((2*I)*(c + d*S
qrt[x]))])/d^2 - (4*a*b*x^(5/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d - ((20*I)*b^2*x^(3/2)*PolyLog[2, -E^((2*
I)*(c + d*Sqrt[x]))])/d^3 + ((10*I)*a*b*x^2*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^2 + (30*b^2*x*PolyLog[3,
-E^((2*I)*(c + d*Sqrt[x]))])/d^4 - (20*a*b*x^(3/2)*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 + ((30*I)*b^2*
Sqrt[x]*PolyLog[4, -E^((2*I)*(c + d*Sqrt[x]))])/d^5 - ((30*I)*a*b*x*PolyLog[4, -E^((2*I)*(c + d*Sqrt[x]))])/d^
4 - (15*b^2*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x]))])/d^6 + (30*a*b*Sqrt[x]*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x])
)])/d^5 + ((15*I)*a*b*PolyLog[6, -E^((2*I)*(c + d*Sqrt[x]))])/d^6 + (2*b^2*x^(5/2)*Tan[c + d*Sqrt[x]])/d

________________________________________________________________________________________

Rubi [A]  time = 0.610196, antiderivative size = 402, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {3747, 3722, 3719, 2190, 2531, 6609, 2282, 6589, 3720, 30} \[ \frac{a^2 x^3}{3}+\frac{10 i a b x^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^{3/2} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{30 i a b x \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 a b \sqrt{x} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{15 i a b \text{Li}_6\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{2}{3} i a b x^3-\frac{20 i b^2 x^{3/2} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 b^2 x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 i b^2 \sqrt{x} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{15 b^2 \text{Li}_5\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^{5/2}}{d}-\frac{b^2 x^3}{3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*b^2*x^(5/2))/d + (a^2*x^3)/3 + ((2*I)/3)*a*b*x^3 - (b^2*x^3)/3 + (10*b^2*x^2*Log[1 + E^((2*I)*(c + d*S
qrt[x]))])/d^2 - (4*a*b*x^(5/2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))])/d - ((20*I)*b^2*x^(3/2)*PolyLog[2, -E^((2*
I)*(c + d*Sqrt[x]))])/d^3 + ((10*I)*a*b*x^2*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^2 + (30*b^2*x*PolyLog[3,
-E^((2*I)*(c + d*Sqrt[x]))])/d^4 - (20*a*b*x^(3/2)*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^3 + ((30*I)*b^2*
Sqrt[x]*PolyLog[4, -E^((2*I)*(c + d*Sqrt[x]))])/d^5 - ((30*I)*a*b*x*PolyLog[4, -E^((2*I)*(c + d*Sqrt[x]))])/d^
4 - (15*b^2*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x]))])/d^6 + (30*a*b*Sqrt[x]*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x])
)])/d^5 + ((15*I)*a*b*PolyLog[6, -E^((2*I)*(c + d*Sqrt[x]))])/d^6 + (2*b^2*x^(5/2)*Tan[c + d*Sqrt[x]])/d

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 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 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]

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 30

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

Rubi steps

\begin{align*} \int x^2 \left (a+b \tan \left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^5 (a+b \tan (c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^5+2 a b x^5 \tan (c+d x)+b^2 x^5 \tan ^2(c+d x)\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^3}{3}+(4 a b) \operatorname{Subst}\left (\int x^5 \tan (c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^5 \tan ^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^3}{3}+\frac{2}{3} i a b x^3+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}-(8 i a b) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^5}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )-\left (2 b^2\right ) \operatorname{Subst}\left (\int x^5 \, dx,x,\sqrt{x}\right )-\frac{\left (10 b^2\right ) \operatorname{Subst}\left (\int x^4 \tan (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 i b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}+\frac{2}{3} i a b x^3-\frac{b^2 x^3}{3}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(20 a b) \operatorname{Subst}\left (\int x^4 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{\left (20 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^4}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 i b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}+\frac{2}{3} i a b x^3-\frac{b^2 x^3}{3}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 i a b x^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(40 i a b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}-\frac{\left (40 b^2\right ) \operatorname{Subst}\left (\int x^3 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=-\frac{2 i b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}+\frac{2}{3} i a b x^3-\frac{b^2 x^3}{3}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{20 i b^2 x^{3/2} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{10 i a b x^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 a b x^{3/2} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(60 a b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}+\frac{\left (60 i b^2\right ) \operatorname{Subst}\left (\int x^2 \text{Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=-\frac{2 i b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}+\frac{2}{3} i a b x^3-\frac{b^2 x^3}{3}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{20 i b^2 x^{3/2} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{10 i a b x^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{30 b^2 x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{20 a b x^{3/2} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{30 i a b x \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(60 i a b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{\left (60 b^2\right ) \operatorname{Subst}\left (\int x \text{Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}\\ &=-\frac{2 i b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}+\frac{2}{3} i a b x^3-\frac{b^2 x^3}{3}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{20 i b^2 x^{3/2} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{10 i a b x^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{30 b^2 x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{20 a b x^{3/2} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 i b^2 \sqrt{x} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{30 i a b x \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 a b \sqrt{x} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(30 a b) \operatorname{Subst}\left (\int \text{Li}_5\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}-\frac{\left (30 i b^2\right ) \operatorname{Subst}\left (\int \text{Li}_4\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}\\ &=-\frac{2 i b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}+\frac{2}{3} i a b x^3-\frac{b^2 x^3}{3}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{20 i b^2 x^{3/2} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{10 i a b x^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{30 b^2 x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{20 a b x^{3/2} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 i b^2 \sqrt{x} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{30 i a b x \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 a b \sqrt{x} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(15 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{\left (15 b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_4(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}\\ &=-\frac{2 i b^2 x^{5/2}}{d}+\frac{a^2 x^3}{3}+\frac{2}{3} i a b x^3-\frac{b^2 x^3}{3}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{4 a b x^{5/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{20 i b^2 x^{3/2} \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{10 i a b x^2 \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{30 b^2 x \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{20 a b x^{3/2} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 i b^2 \sqrt{x} \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{30 i a b x \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{15 b^2 \text{Li}_5\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{30 a b \sqrt{x} \text{Li}_5\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{15 i a b \text{Li}_6\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 3.55033, size = 379, normalized size = 0.94 \[ \frac{1}{3} \left (b \left (-\frac{30 i \left (a d x^2-2 b x^{3/2}\right ) \text{PolyLog}\left (2,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{30 \left (2 a d x^{3/2}-3 b x\right ) \text{PolyLog}\left (3,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{90 i a x \text{PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{90 a \sqrt{x} \text{PolyLog}\left (5,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{45 i a \text{PolyLog}\left (6,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{90 i b \sqrt{x} \text{PolyLog}\left (4,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{45 b \text{PolyLog}\left (5,-e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{12 a x^{5/2} \log \left (1+e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{4 i a x^3}{1+e^{2 i c}}+\frac{30 b x^2 \log \left (1+e^{-2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{12 i b x^{5/2}}{d+e^{2 i c} d}\right )+x^3 \left (a^2+2 a b \tan (c)-b^2\right )+\frac{6 b^2 x^{5/2} \sec (c) \sin \left (d \sqrt{x}\right ) \sec \left (c+d \sqrt{x}\right )}{d}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(((12*I)*b*x^(5/2))/(d + d*E^((2*I)*c)) - ((4*I)*a*x^3)/(1 + E^((2*I)*c)) + (30*b*x^2*Log[1 + E^((-2*I)*(c
+ d*Sqrt[x]))])/d^2 - (12*a*x^(5/2)*Log[1 + E^((-2*I)*(c + d*Sqrt[x]))])/d - ((30*I)*(-2*b*x^(3/2) + a*d*x^2)*
PolyLog[2, -E^((-2*I)*(c + d*Sqrt[x]))])/d^3 - (30*(-3*b*x + 2*a*d*x^(3/2))*PolyLog[3, -E^((-2*I)*(c + d*Sqrt[
x]))])/d^4 - ((90*I)*b*Sqrt[x]*PolyLog[4, -E^((-2*I)*(c + d*Sqrt[x]))])/d^5 + ((90*I)*a*x*PolyLog[4, -E^((-2*I
)*(c + d*Sqrt[x]))])/d^4 - (45*b*PolyLog[5, -E^((-2*I)*(c + d*Sqrt[x]))])/d^6 + (90*a*Sqrt[x]*PolyLog[5, -E^((
-2*I)*(c + d*Sqrt[x]))])/d^5 - ((45*I)*a*PolyLog[6, -E^((-2*I)*(c + d*Sqrt[x]))])/d^6) + (6*b^2*x^(5/2)*Sec[c]
*Sec[c + d*Sqrt[x]]*Sin[d*Sqrt[x]])/d + x^3*(a^2 - b^2 + 2*a*b*Tan[c]))/3

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Maple [F]  time = 0.28, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b\tan \left ( c+d\sqrt{x} \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 2.65109, size = 3244, normalized size = 8.07 \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^(1/2)))^2,x, algorithm="maxima")

[Out]

1/3*((d*sqrt(x) + c)^6*a^2 - 6*(d*sqrt(x) + c)^5*a^2*c + 15*(d*sqrt(x) + c)^4*a^2*c^2 - 20*(d*sqrt(x) + c)^3*a
^2*c^3 + 15*(d*sqrt(x) + c)^2*a^2*c^4 - 6*(d*sqrt(x) + c)*a^2*c^5 - 12*a*b*c^5*log(sec(d*sqrt(x) + c)) - 6*(30
*I*(d*sqrt(x) + c)*b^2*c^5 - (10*a*b + 5*I*b^2)*(d*sqrt(x) + c)^6 + (60*a*b + 30*I*b^2)*(d*sqrt(x) + c)^5*c -
(150*a*b + 75*I*b^2)*(d*sqrt(x) + c)^4*c^2 + (200*a*b + 100*I*b^2)*(d*sqrt(x) + c)^3*c^3 - (150*a*b + 75*I*b^2
)*(d*sqrt(x) + c)^2*c^4 + 60*b^2*c^5 + (192*(d*sqrt(x) + c)^5*a*b - 150*b^2*c^4 - 300*(2*a*b*c + b^2)*(d*sqrt(
x) + c)^4 + 800*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^3 - 300*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c)^2 + 300*(a*b
*c^4 + 2*b^2*c^3)*(d*sqrt(x) + c) + 2*(96*(d*sqrt(x) + c)^5*a*b - 75*b^2*c^4 - 150*(2*a*b*c + b^2)*(d*sqrt(x)
+ c)^4 + 400*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^3 - 150*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c)^2 + 150*(a*b*c^
4 + 2*b^2*c^3)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + (192*I*(d*sqrt(x) + c)^5*a*b - 150*I*b^2*c^4 + (-600*
I*a*b*c - 300*I*b^2)*(d*sqrt(x) + c)^4 + (800*I*a*b*c^2 + 800*I*b^2*c)*(d*sqrt(x) + c)^3 + (-600*I*a*b*c^3 - 9
00*I*b^2*c^2)*(d*sqrt(x) + c)^2 + (300*I*a*b*c^4 + 600*I*b^2*c^3)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arc
tan2(sin(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c) + 1) - ((10*a*b + 5*I*b^2)*(d*sqrt(x) + c)^6 - (60*b^2 + (
60*a*b + 30*I*b^2)*c)*(d*sqrt(x) + c)^5 + (300*b^2*c + (150*a*b + 75*I*b^2)*c^2)*(d*sqrt(x) + c)^4 - (600*b^2*
c^2 + (200*a*b + 100*I*b^2)*c^3)*(d*sqrt(x) + c)^3 + (600*b^2*c^3 + (150*a*b + 75*I*b^2)*c^4)*(d*sqrt(x) + c)^
2 - (30*I*b^2*c^5 + 300*b^2*c^4)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (480*(d*sqrt(x) + c)^4*a*b + 150*a*
b*c^4 + 300*b^2*c^3 - 600*(2*a*b*c + b^2)*(d*sqrt(x) + c)^3 + 1200*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^2 - 300*(
2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c) + 30*(16*(d*sqrt(x) + c)^4*a*b + 5*a*b*c^4 + 10*b^2*c^3 - 20*(2*a*b*c +
b^2)*(d*sqrt(x) + c)^3 + 40*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^2 - 10*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c))
*cos(2*d*sqrt(x) + 2*c) - (-480*I*(d*sqrt(x) + c)^4*a*b - 150*I*a*b*c^4 - 300*I*b^2*c^3 + (1200*I*a*b*c + 600*
I*b^2)*(d*sqrt(x) + c)^3 + (-1200*I*a*b*c^2 - 1200*I*b^2*c)*(d*sqrt(x) + c)^2 + (600*I*a*b*c^3 + 900*I*b^2*c^2
)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) + (-96*I*(d*sqrt(x) + c)^5*a*b +
75*I*b^2*c^4 + (300*I*a*b*c + 150*I*b^2)*(d*sqrt(x) + c)^4 + (-400*I*a*b*c^2 - 400*I*b^2*c)*(d*sqrt(x) + c)^3
+ (300*I*a*b*c^3 + 450*I*b^2*c^2)*(d*sqrt(x) + c)^2 + (-150*I*a*b*c^4 - 300*I*b^2*c^3)*(d*sqrt(x) + c) + (-96*
I*(d*sqrt(x) + c)^5*a*b + 75*I*b^2*c^4 + (300*I*a*b*c + 150*I*b^2)*(d*sqrt(x) + c)^4 + (-400*I*a*b*c^2 - 400*I
*b^2*c)*(d*sqrt(x) + c)^3 + (300*I*a*b*c^3 + 450*I*b^2*c^2)*(d*sqrt(x) + c)^2 + (-150*I*a*b*c^4 - 300*I*b^2*c^
3)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + (96*(d*sqrt(x) + c)^5*a*b - 75*b^2*c^4 - 150*(2*a*b*c + b^2)*(d*s
qrt(x) + c)^4 + 400*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^3 - 150*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c)^2 + 150*
(a*b*c^4 + 2*b^2*c^3)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*log(cos(2*d*sqrt(x) + 2*c)^2 + sin(2*d*sqrt(x)
+ 2*c)^2 + 2*cos(2*d*sqrt(x) + 2*c) + 1) - (720*a*b*cos(2*d*sqrt(x) + 2*c) + 720*I*a*b*sin(2*d*sqrt(x) + 2*c)
+ 720*a*b)*polylog(6, -e^(2*I*d*sqrt(x) + 2*I*c)) + (1440*I*(d*sqrt(x) + c)*a*b - 900*I*a*b*c - 450*I*b^2 + (1
440*I*(d*sqrt(x) + c)*a*b - 900*I*a*b*c - 450*I*b^2)*cos(2*d*sqrt(x) + 2*c) - 90*(16*(d*sqrt(x) + c)*a*b - 10*
a*b*c - 5*b^2)*sin(2*d*sqrt(x) + 2*c))*polylog(5, -e^(2*I*d*sqrt(x) + 2*I*c)) + (1440*(d*sqrt(x) + c)^2*a*b +
600*a*b*c^2 + 600*b^2*c - 900*(2*a*b*c + b^2)*(d*sqrt(x) + c) + 60*(24*(d*sqrt(x) + c)^2*a*b + 10*a*b*c^2 + 10
*b^2*c - 15*(2*a*b*c + b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + (1440*I*(d*sqrt(x) + c)^2*a*b + 600*I*a*
b*c^2 + 600*I*b^2*c + (-1800*I*a*b*c - 900*I*b^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*polylog(4, -e^(2*I*
d*sqrt(x) + 2*I*c)) + (-960*I*(d*sqrt(x) + c)^3*a*b + 300*I*a*b*c^3 + 450*I*b^2*c^2 + (1800*I*a*b*c + 900*I*b^
2)*(d*sqrt(x) + c)^2 + (-1200*I*a*b*c^2 - 1200*I*b^2*c)*(d*sqrt(x) + c) + (-960*I*(d*sqrt(x) + c)^3*a*b + 300*
I*a*b*c^3 + 450*I*b^2*c^2 + (1800*I*a*b*c + 900*I*b^2)*(d*sqrt(x) + c)^2 + (-1200*I*a*b*c^2 - 1200*I*b^2*c)*(d
*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + 30*(32*(d*sqrt(x) + c)^3*a*b - 10*a*b*c^3 - 15*b^2*c^2 - 30*(2*a*b*c +
b^2)*(d*sqrt(x) + c)^2 + 40*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*polylog(3, -e^(2*I*d*s
qrt(x) + 2*I*c)) - (5*(2*I*a*b - b^2)*(d*sqrt(x) + c)^6 - (60*I*b^2 - 30*(-2*I*a*b + b^2)*c)*(d*sqrt(x) + c)^5
- (-300*I*b^2*c - 75*(2*I*a*b - b^2)*c^2)*(d*sqrt(x) + c)^4 - (600*I*b^2*c^2 - 100*(-2*I*a*b + b^2)*c^3)*(d*s
qrt(x) + c)^3 - (-600*I*b^2*c^3 - 75*(2*I*a*b - b^2)*c^4)*(d*sqrt(x) + c)^2 + 30*(b^2*c^5 - 10*I*b^2*c^4)*(d*s
qrt(x) + c))*sin(2*d*sqrt(x) + 2*c))/(-30*I*cos(2*d*sqrt(x) + 2*c) + 30*sin(2*d*sqrt(x) + 2*c) - 30*I))/d^6

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

[Out]

integral(b^2*x^2*tan(d*sqrt(x) + c)^2 + 2*a*b*x^2*tan(d*sqrt(x) + c) + a^2*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{x} \right )}\right )^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*tan(c + d*sqrt(x)))**2, x)

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

[Out]

integrate((b*tan(d*sqrt(x) + c) + a)^2*x^2, x)