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

Optimal. Leaf size=551 \[ \frac{20 i a b x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{80 a b x^{3/2} \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{240 i a b x \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 i a b x \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{480 a b \sqrt{x} \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{480 i a b \text{PolyLog}\left (6,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{480 i a b \text{PolyLog}\left (6,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{20 i b^2 x^{3/2} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 b^2 x \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 i b^2 \sqrt{x} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{15 b^2 \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a^2 x^3}{3}-\frac{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^{5/2}}{d} \]

[Out]

((-2*I)*b^2*x^(5/2))/d + (a^2*x^3)/3 - ((8*I)*a*b*x^(5/2)*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + (10*b^2*x^2*Log[1
+ E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((20*I)*a*b*x^2*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((20*I)*a*b
*x^2*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((20*I)*b^2*x^(3/2)*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^
3 - (80*a*b*x^(3/2)*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))])/d^3 + (80*a*b*x^(3/2)*PolyLog[3, I*E^(I*(c + d*Sqr
t[x]))])/d^3 + (30*b^2*x*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 - ((240*I)*a*b*x*PolyLog[4, (-I)*E^(I*(c
+ d*Sqrt[x]))])/d^4 + ((240*I)*a*b*x*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + ((30*I)*b^2*Sqrt[x]*PolyLog[4,
-E^((2*I)*(c + d*Sqrt[x]))])/d^5 + (480*a*b*Sqrt[x]*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (480*a*b*Sq
rt[x]*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 - (15*b^2*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x]))])/d^6 + ((480*I
)*a*b*PolyLog[6, (-I)*E^(I*(c + d*Sqrt[x]))])/d^6 - ((480*I)*a*b*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6 + (2
*b^2*x^(5/2)*Tan[c + d*Sqrt[x]])/d

________________________________________________________________________________________

Rubi [A]  time = 0.633592, antiderivative size = 551, normalized size of antiderivative = 1., number of steps used = 24, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4204, 4190, 4181, 2531, 6609, 2282, 6589, 4184, 3719, 2190} \[ \frac{20 i a b x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{80 a b x^{3/2} \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{240 i a b x \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 i a b x \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{480 a b \sqrt{x} \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{480 i a b \text{PolyLog}\left (6,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{480 i a b \text{PolyLog}\left (6,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{20 i b^2 x^{3/2} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{30 b^2 x \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{30 i b^2 \sqrt{x} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{15 b^2 \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^6}+\frac{a^2 x^3}{3}-\frac{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^{5/2} \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^{5/2}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*b^2*x^(5/2))/d + (a^2*x^3)/3 - ((8*I)*a*b*x^(5/2)*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + (10*b^2*x^2*Log[1
+ E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((20*I)*a*b*x^2*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((20*I)*a*b
*x^2*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((20*I)*b^2*x^(3/2)*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])/d^
3 - (80*a*b*x^(3/2)*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))])/d^3 + (80*a*b*x^(3/2)*PolyLog[3, I*E^(I*(c + d*Sqr
t[x]))])/d^3 + (30*b^2*x*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 - ((240*I)*a*b*x*PolyLog[4, (-I)*E^(I*(c
+ d*Sqrt[x]))])/d^4 + ((240*I)*a*b*x*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + ((30*I)*b^2*Sqrt[x]*PolyLog[4,
-E^((2*I)*(c + d*Sqrt[x]))])/d^5 + (480*a*b*Sqrt[x]*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (480*a*b*Sq
rt[x]*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))])/d^5 - (15*b^2*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x]))])/d^6 + ((480*I
)*a*b*PolyLog[6, (-I)*E^(I*(c + d*Sqrt[x]))])/d^6 - ((480*I)*a*b*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))])/d^6 + (2
*b^2*x^(5/2)*Tan[c + d*Sqrt[x]])/d

Rule 4204

Int[(x_)^(m_.)*((a_.) + (b_.)*Sec[(c_.) + (d_.)*(x_)^(n_)])^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Sec[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 4190

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

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && 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 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, 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 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]

Rubi steps

\begin{align*} \int x^2 \left (a+b \sec \left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^5 (a+b \sec (c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^5+2 a b x^5 \sec (c+d x)+b^2 x^5 \sec ^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 \sec (c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^5 \sec ^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a^2 x^3}{3}-\frac{8 i a b x^{5/2} \tan ^{-1}\left (e^{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-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(20 a b) \operatorname{Subst}\left (\int x^4 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}-\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{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{20 i a b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{Li}_2\left (i e^{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{(80 i a b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{(80 i a b) \operatorname{Subst}\left (\int x^3 \text{Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\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{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{80 a b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (i e^{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{(240 a b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}-\frac{(240 a b) \operatorname{Subst}\left (\int x^2 \text{Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}-\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{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\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{80 a b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{240 i a b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i e^{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{(480 i a b) \operatorname{Subst}\left (\int x \text{Li}_4\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{(480 i a b) \operatorname{Subst}\left (\int x \text{Li}_4\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}+\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{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\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{80 a b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (i e^{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{240 i a b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{480 a b \sqrt{x} \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{Li}_5\left (i e^{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{(480 a b) \operatorname{Subst}\left (\int \text{Li}_5\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}+\frac{(480 a b) \operatorname{Subst}\left (\int \text{Li}_5\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}-\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{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\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{80 a b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (i e^{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{240 i a b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i e^{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{480 a b \sqrt{x} \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{Li}_5\left (i e^{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{(480 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{(480 i a b) \operatorname{Subst}\left (\int \frac{\text{Li}_5(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\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{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\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{80 a b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (i e^{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{240 i a b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i e^{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{480 a b \sqrt{x} \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{Li}_5\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{480 i a b \text{Li}_6\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{480 i a b \text{Li}_6\left (i e^{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}-\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{8 i a b x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{10 b^2 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{20 i a b x^2 \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{20 i a b x^2 \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\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{80 a b x^{3/2} \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{80 a b x^{3/2} \text{Li}_3\left (i e^{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{240 i a b x \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{240 i a b x \text{Li}_4\left (i e^{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{480 a b \sqrt{x} \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{480 a b \sqrt{x} \text{Li}_5\left (i e^{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{480 i a b \text{Li}_6\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{480 i a b \text{Li}_6\left (i e^{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 = 1.20284, size = 543, normalized size = 0.99 \[ \frac{60 i a b d^4 x^2 \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )-60 i a b d^4 x^2 \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )-240 a b d^3 x^{3/2} \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )+240 a b d^3 x^{3/2} \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )-720 i a b d^2 x \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )+720 i a b d^2 x \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )+1440 a b d \sqrt{x} \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )-1440 a b d \sqrt{x} \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )+1440 i a b \text{PolyLog}\left (6,-i e^{i \left (c+d \sqrt{x}\right )}\right )-1440 i a b \text{PolyLog}\left (6,i e^{i \left (c+d \sqrt{x}\right )}\right )-60 i b^2 d^3 x^{3/2} \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )+90 b^2 d^2 x \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )+90 i b^2 d \sqrt{x} \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )-45 b^2 \text{PolyLog}\left (5,-e^{2 i \left (c+d \sqrt{x}\right )}\right )+a^2 d^6 x^3-24 i a b d^5 x^{5/2} \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )+30 b^2 d^4 x^2 \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )+6 b^2 d^5 x^{5/2} \tan \left (c+d \sqrt{x}\right )-6 i b^2 d^5 x^{5/2}}{3 d^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-6*I)*b^2*d^5*x^(5/2) + a^2*d^6*x^3 - (24*I)*a*b*d^5*x^(5/2)*ArcTan[E^(I*(c + d*Sqrt[x]))] + 30*b^2*d^4*x^2*
Log[1 + E^((2*I)*(c + d*Sqrt[x]))] + (60*I)*a*b*d^4*x^2*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (60*I)*a*b*d^
4*x^2*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (60*I)*b^2*d^3*x^(3/2)*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))] - 24
0*a*b*d^3*x^(3/2)*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))] + 240*a*b*d^3*x^(3/2)*PolyLog[3, I*E^(I*(c + d*Sqrt[x
]))] + 90*b^2*d^2*x*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))] - (720*I)*a*b*d^2*x*PolyLog[4, (-I)*E^(I*(c + d*Sqr
t[x]))] + (720*I)*a*b*d^2*x*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))] + (90*I)*b^2*d*Sqrt[x]*PolyLog[4, -E^((2*I)*(c
+ d*Sqrt[x]))] + 1440*a*b*d*Sqrt[x]*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))] - 1440*a*b*d*Sqrt[x]*PolyLog[5, I*
E^(I*(c + d*Sqrt[x]))] - 45*b^2*PolyLog[5, -E^((2*I)*(c + d*Sqrt[x]))] + (1440*I)*a*b*PolyLog[6, (-I)*E^(I*(c
+ d*Sqrt[x]))] - (1440*I)*a*b*PolyLog[6, I*E^(I*(c + d*Sqrt[x]))] + 6*b^2*d^5*x^(5/2)*Tan[c + d*Sqrt[x]])/(3*d
^6)

________________________________________________________________________________________

Maple [F]  time = 0.087, size = 0, normalized size = 0. \begin{align*} \int{x}^{2} \left ( a+b\sec \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*sec(c+d*x^(1/2)))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 3.03364, size = 5266, normalized size = 9.56 \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*sec(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) + tan(d*
sqrt(x) + c)) - 6*(12*b^2*c^5 + (12*(d*sqrt(x) + c)^5*a*b - 60*(d*sqrt(x) + c)^4*a*b*c + 120*(d*sqrt(x) + c)^3
*a*b*c^2 - 120*(d*sqrt(x) + c)^2*a*b*c^3 + 60*(d*sqrt(x) + c)*a*b*c^4 + 12*((d*sqrt(x) + c)^5*a*b - 5*(d*sqrt(
x) + c)^4*a*b*c + 10*(d*sqrt(x) + c)^3*a*b*c^2 - 10*(d*sqrt(x) + c)^2*a*b*c^3 + 5*(d*sqrt(x) + c)*a*b*c^4)*cos
(2*d*sqrt(x) + 2*c) - (-12*I*(d*sqrt(x) + c)^5*a*b + 60*I*(d*sqrt(x) + c)^4*a*b*c - 120*I*(d*sqrt(x) + c)^3*a*
b*c^2 + 120*I*(d*sqrt(x) + c)^2*a*b*c^3 - 60*I*(d*sqrt(x) + c)*a*b*c^4)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*
sqrt(x) + c), sin(d*sqrt(x) + c) + 1) + (12*(d*sqrt(x) + c)^5*a*b - 60*(d*sqrt(x) + c)^4*a*b*c + 120*(d*sqrt(x
) + c)^3*a*b*c^2 - 120*(d*sqrt(x) + c)^2*a*b*c^3 + 60*(d*sqrt(x) + c)*a*b*c^4 + 12*((d*sqrt(x) + c)^5*a*b - 5*
(d*sqrt(x) + c)^4*a*b*c + 10*(d*sqrt(x) + c)^3*a*b*c^2 - 10*(d*sqrt(x) + c)^2*a*b*c^3 + 5*(d*sqrt(x) + c)*a*b*
c^4)*cos(2*d*sqrt(x) + 2*c) - (-12*I*(d*sqrt(x) + c)^5*a*b + 60*I*(d*sqrt(x) + c)^4*a*b*c - 120*I*(d*sqrt(x) +
c)^3*a*b*c^2 + 120*I*(d*sqrt(x) + c)^2*a*b*c^3 - 60*I*(d*sqrt(x) + c)*a*b*c^4)*sin(2*d*sqrt(x) + 2*c))*arctan
2(cos(d*sqrt(x) + c), -sin(d*sqrt(x) + c) + 1) - (60*(d*sqrt(x) + c)^4*b^2 - 160*(d*sqrt(x) + c)^3*b^2*c + 180
*(d*sqrt(x) + c)^2*b^2*c^2 - 120*(d*sqrt(x) + c)*b^2*c^3 + 30*b^2*c^4 + 10*(6*(d*sqrt(x) + c)^4*b^2 - 16*(d*sq
rt(x) + c)^3*b^2*c + 18*(d*sqrt(x) + c)^2*b^2*c^2 - 12*(d*sqrt(x) + c)*b^2*c^3 + 3*b^2*c^4)*cos(2*d*sqrt(x) +
2*c) + (60*I*(d*sqrt(x) + c)^4*b^2 - 160*I*(d*sqrt(x) + c)^3*b^2*c + 180*I*(d*sqrt(x) + c)^2*b^2*c^2 - 120*I*(
d*sqrt(x) + c)*b^2*c^3 + 30*I*b^2*c^4)*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x)
+ 2*c) + 1) + 12*((d*sqrt(x) + c)^5*b^2 - 5*(d*sqrt(x) + c)^4*b^2*c + 10*(d*sqrt(x) + c)^3*b^2*c^2 - 10*(d*sq
rt(x) + c)^2*b^2*c^3 + 5*(d*sqrt(x) + c)*b^2*c^4)*cos(2*d*sqrt(x) + 2*c) + (120*(d*sqrt(x) + c)^3*b^2 - 240*(d
*sqrt(x) + c)^2*b^2*c + 180*(d*sqrt(x) + c)*b^2*c^2 - 60*b^2*c^3 + 60*(2*(d*sqrt(x) + c)^3*b^2 - 4*(d*sqrt(x)
+ c)^2*b^2*c + 3*(d*sqrt(x) + c)*b^2*c^2 - b^2*c^3)*cos(2*d*sqrt(x) + 2*c) - (-120*I*(d*sqrt(x) + c)^3*b^2 + 2
40*I*(d*sqrt(x) + c)^2*b^2*c - 180*I*(d*sqrt(x) + c)*b^2*c^2 + 60*I*b^2*c^3)*sin(2*d*sqrt(x) + 2*c))*dilog(-e^
(2*I*d*sqrt(x) + 2*I*c)) + (60*(d*sqrt(x) + c)^4*a*b - 240*(d*sqrt(x) + c)^3*a*b*c + 360*(d*sqrt(x) + c)^2*a*b
*c^2 - 240*(d*sqrt(x) + c)*a*b*c^3 + 60*a*b*c^4 + 60*((d*sqrt(x) + c)^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d
*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3 + a*b*c^4)*cos(2*d*sqrt(x) + 2*c) - (-60*I*(d*sqrt(x) + c)
^4*a*b + 240*I*(d*sqrt(x) + c)^3*a*b*c - 360*I*(d*sqrt(x) + c)^2*a*b*c^2 + 240*I*(d*sqrt(x) + c)*a*b*c^3 - 60*
I*a*b*c^4)*sin(2*d*sqrt(x) + 2*c))*dilog(I*e^(I*d*sqrt(x) + I*c)) - (60*(d*sqrt(x) + c)^4*a*b - 240*(d*sqrt(x)
+ c)^3*a*b*c + 360*(d*sqrt(x) + c)^2*a*b*c^2 - 240*(d*sqrt(x) + c)*a*b*c^3 + 60*a*b*c^4 + 60*((d*sqrt(x) + c)
^4*a*b - 4*(d*sqrt(x) + c)^3*a*b*c + 6*(d*sqrt(x) + c)^2*a*b*c^2 - 4*(d*sqrt(x) + c)*a*b*c^3 + a*b*c^4)*cos(2*
d*sqrt(x) + 2*c) + (60*I*(d*sqrt(x) + c)^4*a*b - 240*I*(d*sqrt(x) + c)^3*a*b*c + 360*I*(d*sqrt(x) + c)^2*a*b*c
^2 - 240*I*(d*sqrt(x) + c)*a*b*c^3 + 60*I*a*b*c^4)*sin(2*d*sqrt(x) + 2*c))*dilog(-I*e^(I*d*sqrt(x) + I*c)) - (
-30*I*(d*sqrt(x) + c)^4*b^2 + 80*I*(d*sqrt(x) + c)^3*b^2*c - 90*I*(d*sqrt(x) + c)^2*b^2*c^2 + 60*I*(d*sqrt(x)
+ c)*b^2*c^3 - 15*I*b^2*c^4 + (-30*I*(d*sqrt(x) + c)^4*b^2 + 80*I*(d*sqrt(x) + c)^3*b^2*c - 90*I*(d*sqrt(x) +
c)^2*b^2*c^2 + 60*I*(d*sqrt(x) + c)*b^2*c^3 - 15*I*b^2*c^4)*cos(2*d*sqrt(x) + 2*c) + 5*(6*(d*sqrt(x) + c)^4*b^
2 - 16*(d*sqrt(x) + c)^3*b^2*c + 18*(d*sqrt(x) + c)^2*b^2*c^2 - 12*(d*sqrt(x) + c)*b^2*c^3 + 3*b^2*c^4)*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) - (-
6*I*(d*sqrt(x) + c)^5*a*b + 30*I*(d*sqrt(x) + c)^4*a*b*c - 60*I*(d*sqrt(x) + c)^3*a*b*c^2 + 60*I*(d*sqrt(x) +
c)^2*a*b*c^3 - 30*I*(d*sqrt(x) + c)*a*b*c^4 + (-6*I*(d*sqrt(x) + c)^5*a*b + 30*I*(d*sqrt(x) + c)^4*a*b*c - 60*
I*(d*sqrt(x) + c)^3*a*b*c^2 + 60*I*(d*sqrt(x) + c)^2*a*b*c^3 - 30*I*(d*sqrt(x) + c)*a*b*c^4)*cos(2*d*sqrt(x) +
2*c) + 6*((d*sqrt(x) + c)^5*a*b - 5*(d*sqrt(x) + c)^4*a*b*c + 10*(d*sqrt(x) + c)^3*a*b*c^2 - 10*(d*sqrt(x) +
c)^2*a*b*c^3 + 5*(d*sqrt(x) + c)*a*b*c^4)*sin(2*d*sqrt(x) + 2*c))*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c
)^2 + 2*sin(d*sqrt(x) + c) + 1) - (6*I*(d*sqrt(x) + c)^5*a*b - 30*I*(d*sqrt(x) + c)^4*a*b*c + 60*I*(d*sqrt(x)
+ c)^3*a*b*c^2 - 60*I*(d*sqrt(x) + c)^2*a*b*c^3 + 30*I*(d*sqrt(x) + c)*a*b*c^4 + (6*I*(d*sqrt(x) + c)^5*a*b -
30*I*(d*sqrt(x) + c)^4*a*b*c + 60*I*(d*sqrt(x) + c)^3*a*b*c^2 - 60*I*(d*sqrt(x) + c)^2*a*b*c^3 + 30*I*(d*sqrt(
x) + c)*a*b*c^4)*cos(2*d*sqrt(x) + 2*c) - 6*((d*sqrt(x) + c)^5*a*b - 5*(d*sqrt(x) + c)^4*a*b*c + 10*(d*sqrt(x)
+ c)^3*a*b*c^2 - 10*(d*sqrt(x) + c)^2*a*b*c^3 + 5*(d*sqrt(x) + c)*a*b*c^4)*sin(2*d*sqrt(x) + 2*c))*log(cos(d*
sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*sin(d*sqrt(x) + c) + 1) + (1440*a*b*cos(2*d*sqrt(x) + 2*c) + 1440*I*
a*b*sin(2*d*sqrt(x) + 2*c) + 1440*a*b)*polylog(6, I*e^(I*d*sqrt(x) + I*c)) - (1440*a*b*cos(2*d*sqrt(x) + 2*c)
+ 1440*I*a*b*sin(2*d*sqrt(x) + 2*c) + 1440*a*b)*polylog(6, -I*e^(I*d*sqrt(x) + I*c)) - (90*I*b^2*cos(2*d*sqrt(
x) + 2*c) - 90*b^2*sin(2*d*sqrt(x) + 2*c) + 90*I*b^2)*polylog(5, -e^(2*I*d*sqrt(x) + 2*I*c)) - (1440*I*(d*sqrt
(x) + c)*a*b - 1440*I*a*b*c + (1440*I*(d*sqrt(x) + c)*a*b - 1440*I*a*b*c)*cos(2*d*sqrt(x) + 2*c) - 1440*((d*sq
rt(x) + c)*a*b - a*b*c)*sin(2*d*sqrt(x) + 2*c))*polylog(5, I*e^(I*d*sqrt(x) + I*c)) - (-1440*I*(d*sqrt(x) + c)
*a*b + 1440*I*a*b*c + (-1440*I*(d*sqrt(x) + c)*a*b + 1440*I*a*b*c)*cos(2*d*sqrt(x) + 2*c) + 1440*((d*sqrt(x) +
c)*a*b - a*b*c)*sin(2*d*sqrt(x) + 2*c))*polylog(5, -I*e^(I*d*sqrt(x) + I*c)) - (180*(d*sqrt(x) + c)*b^2 - 120
*b^2*c + 60*(3*(d*sqrt(x) + c)*b^2 - 2*b^2*c)*cos(2*d*sqrt(x) + 2*c) + (180*I*(d*sqrt(x) + c)*b^2 - 120*I*b^2*
c)*sin(2*d*sqrt(x) + 2*c))*polylog(4, -e^(2*I*d*sqrt(x) + 2*I*c)) - (720*(d*sqrt(x) + c)^2*a*b - 1440*(d*sqrt(
x) + c)*a*b*c + 720*a*b*c^2 + 720*((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c + a*b*c^2)*cos(2*d*sqrt(x)
+ 2*c) + (720*I*(d*sqrt(x) + c)^2*a*b - 1440*I*(d*sqrt(x) + c)*a*b*c + 720*I*a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*
polylog(4, I*e^(I*d*sqrt(x) + I*c)) + (720*(d*sqrt(x) + c)^2*a*b - 1440*(d*sqrt(x) + c)*a*b*c + 720*a*b*c^2 +
720*((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c + a*b*c^2)*cos(2*d*sqrt(x) + 2*c) - (-720*I*(d*sqrt(x) +
c)^2*a*b + 1440*I*(d*sqrt(x) + c)*a*b*c - 720*I*a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*polylog(4, -I*e^(I*d*sqrt(x)
+ I*c)) - (-180*I*(d*sqrt(x) + c)^2*b^2 + 240*I*(d*sqrt(x) + c)*b^2*c - 90*I*b^2*c^2 + (-180*I*(d*sqrt(x) + c)
^2*b^2 + 240*I*(d*sqrt(x) + c)*b^2*c - 90*I*b^2*c^2)*cos(2*d*sqrt(x) + 2*c) + 30*(6*(d*sqrt(x) + c)^2*b^2 - 8*
(d*sqrt(x) + c)*b^2*c + 3*b^2*c^2)*sin(2*d*sqrt(x) + 2*c))*polylog(3, -e^(2*I*d*sqrt(x) + 2*I*c)) - (-240*I*(d
*sqrt(x) + c)^3*a*b + 720*I*(d*sqrt(x) + c)^2*a*b*c - 720*I*(d*sqrt(x) + c)*a*b*c^2 + 240*I*a*b*c^3 + (-240*I*
(d*sqrt(x) + c)^3*a*b + 720*I*(d*sqrt(x) + c)^2*a*b*c - 720*I*(d*sqrt(x) + c)*a*b*c^2 + 240*I*a*b*c^3)*cos(2*d
*sqrt(x) + 2*c) + 240*((d*sqrt(x) + c)^3*a*b - 3*(d*sqrt(x) + c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2 - a*b*c^3
)*sin(2*d*sqrt(x) + 2*c))*polylog(3, I*e^(I*d*sqrt(x) + I*c)) - (240*I*(d*sqrt(x) + c)^3*a*b - 720*I*(d*sqrt(x
) + c)^2*a*b*c + 720*I*(d*sqrt(x) + c)*a*b*c^2 - 240*I*a*b*c^3 + (240*I*(d*sqrt(x) + c)^3*a*b - 720*I*(d*sqrt(
x) + c)^2*a*b*c + 720*I*(d*sqrt(x) + c)*a*b*c^2 - 240*I*a*b*c^3)*cos(2*d*sqrt(x) + 2*c) - 240*((d*sqrt(x) + c)
^3*a*b - 3*(d*sqrt(x) + c)^2*a*b*c + 3*(d*sqrt(x) + c)*a*b*c^2 - a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*polylog(3, -
I*e^(I*d*sqrt(x) + I*c)) - (-12*I*(d*sqrt(x) + c)^5*b^2 + 60*I*(d*sqrt(x) + c)^4*b^2*c - 120*I*(d*sqrt(x) + c)
^3*b^2*c^2 + 120*I*(d*sqrt(x) + c)^2*b^2*c^3 - 60*I*(d*sqrt(x) + c)*b^2*c^4)*sin(2*d*sqrt(x) + 2*c))/(-6*I*cos
(2*d*sqrt(x) + 2*c) + 6*sin(2*d*sqrt(x) + 2*c) - 6*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} \sec \left (d \sqrt{x} + c\right )^{2} + 2 \, a b x^{2} \sec \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*sec(c+d*x^(1/2)))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^2*sec(d*sqrt(x) + c)^2 + 2*a*b*x^2*sec(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 \sec{\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*sec(c+d*x**(1/2)))**2,x)

[Out]

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

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

[Out]

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