3.56 \(\int x^{3/2} (a+b \sec (c+d \sqrt{x}))^2 \, dx\)

Optimal. Leaf size=451 \[ \frac{16 i a b x^{3/2} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{48 a b x \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{96 i a b \sqrt{x} \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 a b \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{96 a b \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{12 i b^2 x \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b^2 \sqrt{x} \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{6 i b^2 \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^2}{d} \]

[Out]

((-2*I)*b^2*x^2)/d + (2*a^2*x^(5/2))/5 - ((8*I)*a*b*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + (8*b^2*x^(3/2)*Log[
1 + E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((16*I)*a*b*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((16*I
)*a*b*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((12*I)*b^2*x*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])
/d^3 - (48*a*b*x*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))])/d^3 + (48*a*b*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/
d^3 + (12*b^2*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 - ((96*I)*a*b*Sqrt[x]*PolyLog[4, (-I)*E^(I*(
c + d*Sqrt[x]))])/d^4 + ((96*I)*a*b*Sqrt[x]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + ((6*I)*b^2*PolyLog[4, -
E^((2*I)*(c + d*Sqrt[x]))])/d^5 + (96*a*b*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (96*a*b*PolyLog[5, I*E
^(I*(c + d*Sqrt[x]))])/d^5 + (2*b^2*x^2*Tan[c + d*Sqrt[x]])/d

________________________________________________________________________________________

Rubi [A]  time = 0.537303, antiderivative size = 451, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 10, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {4204, 4190, 4181, 2531, 6609, 2282, 6589, 4184, 3719, 2190} \[ \frac{16 i a b x^{3/2} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{48 a b x \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{96 i a b \sqrt{x} \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 a b \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{96 a b \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{12 i b^2 x \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b^2 \sqrt{x} \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{6 i b^2 \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*b^2*x^2)/d + (2*a^2*x^(5/2))/5 - ((8*I)*a*b*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))])/d + (8*b^2*x^(3/2)*Log[
1 + E^((2*I)*(c + d*Sqrt[x]))])/d^2 + ((16*I)*a*b*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))])/d^2 - ((16*I
)*a*b*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))])/d^2 - ((12*I)*b^2*x*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))])
/d^3 - (48*a*b*x*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))])/d^3 + (48*a*b*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))])/
d^3 + (12*b^2*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))])/d^4 - ((96*I)*a*b*Sqrt[x]*PolyLog[4, (-I)*E^(I*(
c + d*Sqrt[x]))])/d^4 + ((96*I)*a*b*Sqrt[x]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))])/d^4 + ((6*I)*b^2*PolyLog[4, -
E^((2*I)*(c + d*Sqrt[x]))])/d^5 + (96*a*b*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))])/d^5 - (96*a*b*PolyLog[5, I*E
^(I*(c + d*Sqrt[x]))])/d^5 + (2*b^2*x^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^{3/2} \left (a+b \sec \left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^4 (a+b \sec (c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^4+2 a b x^4 \sec (c+d x)+b^2 x^4 \sec ^2(c+d x)\right ) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{5} a^2 x^{5/2}+(4 a b) \operatorname{Subst}\left (\int x^4 \sec (c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^4 \sec ^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(16 a b) \operatorname{Subst}\left (\int x^3 \log \left (1-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(16 a b) \operatorname{Subst}\left (\int x^3 \log \left (1+i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}-\frac{\left (8 b^2\right ) \operatorname{Subst}\left (\int x^3 \tan (c+d x) \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 i b^2 x^2}{d}+\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{16 i a b x^{3/2} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}-\frac{(48 i a b) \operatorname{Subst}\left (\int x^2 \text{Li}_2\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{(48 i a b) \operatorname{Subst}\left (\int x^2 \text{Li}_2\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{\left (16 i b^2\right ) \operatorname{Subst}\left (\int \frac{e^{2 i (c+d x)} x^3}{1+e^{2 i (c+d x)}} \, dx,x,\sqrt{x}\right )}{d}\\ &=-\frac{2 i b^2 x^2}{d}+\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{16 i a b x^{3/2} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{48 a b x \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(96 a b) \operatorname{Subst}\left (\int x \text{Li}_3\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}-\frac{(96 a b) \operatorname{Subst}\left (\int x \text{Li}_3\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}-\frac{\left (24 b^2\right ) \operatorname{Subst}\left (\int x^2 \log \left (1+e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=-\frac{2 i b^2 x^2}{d}+\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{16 i a b x^{3/2} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 i b^2 x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{48 a b x \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{96 i a b \sqrt{x} \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(96 i a b) \operatorname{Subst}\left (\int \text{Li}_4\left (-i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{(96 i a b) \operatorname{Subst}\left (\int \text{Li}_4\left (i e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}+\frac{\left (24 i b^2\right ) \operatorname{Subst}\left (\int x \text{Li}_2\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=-\frac{2 i b^2 x^2}{d}+\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{16 i a b x^{3/2} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 i b^2 x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{48 a b x \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b^2 \sqrt{x} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{96 i a b \sqrt{x} \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}+\frac{(96 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_4(-i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{(96 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_4(i x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{\left (12 b^2\right ) \operatorname{Subst}\left (\int \text{Li}_3\left (-e^{2 i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}\\ &=-\frac{2 i b^2 x^2}{d}+\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{16 i a b x^{3/2} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 i b^2 x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{48 a b x \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b^2 \sqrt{x} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{96 i a b \sqrt{x} \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 a b \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{96 a b \text{Li}_5\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}+\frac{\left (6 i b^2\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}\\ &=-\frac{2 i b^2 x^2}{d}+\frac{2}{5} a^2 x^{5/2}-\frac{8 i a b x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}+\frac{16 i a b x^{3/2} \text{Li}_2\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{12 i b^2 x \text{Li}_2\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{48 a b x \text{Li}_3\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b^2 \sqrt{x} \text{Li}_3\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}-\frac{96 i a b \sqrt{x} \text{Li}_4\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{Li}_4\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{6 i b^2 \text{Li}_4\left (-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{96 a b \text{Li}_5\left (-i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{96 a b \text{Li}_5\left (i e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{2 b^2 x^2 \tan \left (c+d \sqrt{x}\right )}{d}\\ \end{align*}

Mathematica [A]  time = 1.32567, size = 443, normalized size = 0.98 \[ \frac{2 \left (40 i a b d^3 x^{3/2} \text{PolyLog}\left (2,-i e^{i \left (c+d \sqrt{x}\right )}\right )-40 i a b d^3 x^{3/2} \text{PolyLog}\left (2,i e^{i \left (c+d \sqrt{x}\right )}\right )-120 a b d^2 x \text{PolyLog}\left (3,-i e^{i \left (c+d \sqrt{x}\right )}\right )+120 a b d^2 x \text{PolyLog}\left (3,i e^{i \left (c+d \sqrt{x}\right )}\right )-240 i a b d \sqrt{x} \text{PolyLog}\left (4,-i e^{i \left (c+d \sqrt{x}\right )}\right )+240 i a b d \sqrt{x} \text{PolyLog}\left (4,i e^{i \left (c+d \sqrt{x}\right )}\right )+240 a b \text{PolyLog}\left (5,-i e^{i \left (c+d \sqrt{x}\right )}\right )-240 a b \text{PolyLog}\left (5,i e^{i \left (c+d \sqrt{x}\right )}\right )-30 i b^2 d^2 x \text{PolyLog}\left (2,-e^{2 i \left (c+d \sqrt{x}\right )}\right )+30 b^2 d \sqrt{x} \text{PolyLog}\left (3,-e^{2 i \left (c+d \sqrt{x}\right )}\right )+15 i b^2 \text{PolyLog}\left (4,-e^{2 i \left (c+d \sqrt{x}\right )}\right )+a^2 d^5 x^{5/2}-20 i a b d^4 x^2 \tan ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )+20 b^2 d^3 x^{3/2} \log \left (1+e^{2 i \left (c+d \sqrt{x}\right )}\right )+5 b^2 d^4 x^2 \tan \left (c+d \sqrt{x}\right )-5 i b^2 d^4 x^2\right )}{5 d^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((-5*I)*b^2*d^4*x^2 + a^2*d^5*x^(5/2) - (20*I)*a*b*d^4*x^2*ArcTan[E^(I*(c + d*Sqrt[x]))] + 20*b^2*d^3*x^(3/
2)*Log[1 + E^((2*I)*(c + d*Sqrt[x]))] + (40*I)*a*b*d^3*x^(3/2)*PolyLog[2, (-I)*E^(I*(c + d*Sqrt[x]))] - (40*I)
*a*b*d^3*x^(3/2)*PolyLog[2, I*E^(I*(c + d*Sqrt[x]))] - (30*I)*b^2*d^2*x*PolyLog[2, -E^((2*I)*(c + d*Sqrt[x]))]
- 120*a*b*d^2*x*PolyLog[3, (-I)*E^(I*(c + d*Sqrt[x]))] + 120*a*b*d^2*x*PolyLog[3, I*E^(I*(c + d*Sqrt[x]))] +
30*b^2*d*Sqrt[x]*PolyLog[3, -E^((2*I)*(c + d*Sqrt[x]))] - (240*I)*a*b*d*Sqrt[x]*PolyLog[4, (-I)*E^(I*(c + d*Sq
rt[x]))] + (240*I)*a*b*d*Sqrt[x]*PolyLog[4, I*E^(I*(c + d*Sqrt[x]))] + (15*I)*b^2*PolyLog[4, -E^((2*I)*(c + d*
Sqrt[x]))] + 240*a*b*PolyLog[5, (-I)*E^(I*(c + d*Sqrt[x]))] - 240*a*b*PolyLog[5, I*E^(I*(c + d*Sqrt[x]))] + 5*
b^2*d^4*x^2*Tan[c + d*Sqrt[x]]))/(5*d^5)

________________________________________________________________________________________

Maple [F]  time = 0.086, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{3}{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^(3/2)*(a+b*sec(c+d*x^(1/2)))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 2.48576, size = 3885, normalized size = 8.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*((d*sqrt(x) + c)^5*a^2 - 5*(d*sqrt(x) + c)^4*a^2*c + 10*(d*sqrt(x) + c)^3*a^2*c^2 - 10*(d*sqrt(x) + c)^2*a
^2*c^3 + 5*(d*sqrt(x) + c)*a^2*c^4 + 10*a*b*c^4*log(sec(d*sqrt(x) + c) + tan(d*sqrt(x) + c)) + 5*(6*b^2*c^4 -
(6*(d*sqrt(x) + c)^4*a*b - 24*(d*sqrt(x) + c)^3*a*b*c + 36*(d*sqrt(x) + c)^2*a*b*c^2 - 24*(d*sqrt(x) + c)*a*b*
c^3 + 6*((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)*cos(2*d*sqrt(x) + 2*c) - (-6*I*(d*sqrt(x) + c)^4*a*b + 24*I*(d*sqrt(x) + c)^3*a*b*c - 36*I*(d*sqrt(x)
+ c)^2*a*b*c^2 + 24*I*(d*sqrt(x) + c)*a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) + c), sin(d*sqrt(
x) + c) + 1) - (6*(d*sqrt(x) + c)^4*a*b - 24*(d*sqrt(x) + c)^3*a*b*c + 36*(d*sqrt(x) + c)^2*a*b*c^2 - 24*(d*sq
rt(x) + c)*a*b*c^3 + 6*((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)*cos(2*d*sqrt(x) + 2*c) - (-6*I*(d*sqrt(x) + c)^4*a*b + 24*I*(d*sqrt(x) + c)^3*a*b*c - 3
6*I*(d*sqrt(x) + c)^2*a*b*c^2 + 24*I*(d*sqrt(x) + c)*a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*arctan2(cos(d*sqrt(x) +
c), -sin(d*sqrt(x) + c) + 1) + (16*(d*sqrt(x) + c)^3*b^2 - 36*(d*sqrt(x) + c)^2*b^2*c + 36*(d*sqrt(x) + c)*b^2
*c^2 - 12*b^2*c^3 + 4*(4*(d*sqrt(x) + c)^3*b^2 - 9*(d*sqrt(x) + c)^2*b^2*c + 9*(d*sqrt(x) + c)*b^2*c^2 - 3*b^2
*c^3)*cos(2*d*sqrt(x) + 2*c) + (16*I*(d*sqrt(x) + c)^3*b^2 - 36*I*(d*sqrt(x) + c)^2*b^2*c + 36*I*(d*sqrt(x) +
c)*b^2*c^2 - 12*I*b^2*c^3)*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(2*d*sqrt(x) + 2*c), cos(2*d*sqrt(x) + 2*c) + 1)
- 6*((d*sqrt(x) + c)^4*b^2 - 4*(d*sqrt(x) + c)^3*b^2*c + 6*(d*sqrt(x) + c)^2*b^2*c^2 - 4*(d*sqrt(x) + c)*b^2*
c^3)*cos(2*d*sqrt(x) + 2*c) - (24*(d*sqrt(x) + c)^2*b^2 - 36*(d*sqrt(x) + c)*b^2*c + 18*b^2*c^2 + 6*(4*(d*sqrt
(x) + c)^2*b^2 - 6*(d*sqrt(x) + c)*b^2*c + 3*b^2*c^2)*cos(2*d*sqrt(x) + 2*c) - (-24*I*(d*sqrt(x) + c)^2*b^2 +
36*I*(d*sqrt(x) + c)*b^2*c - 18*I*b^2*c^2)*sin(2*d*sqrt(x) + 2*c))*dilog(-e^(2*I*d*sqrt(x) + 2*I*c)) - (24*(d*
sqrt(x) + c)^3*a*b - 72*(d*sqrt(x) + c)^2*a*b*c + 72*(d*sqrt(x) + c)*a*b*c^2 - 24*a*b*c^3 + 24*((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)*cos(2*d*sqrt(x) + 2*c) - (-24*I*(d*
sqrt(x) + c)^3*a*b + 72*I*(d*sqrt(x) + c)^2*a*b*c - 72*I*(d*sqrt(x) + c)*a*b*c^2 + 24*I*a*b*c^3)*sin(2*d*sqrt(
x) + 2*c))*dilog(I*e^(I*d*sqrt(x) + I*c)) + (24*(d*sqrt(x) + c)^3*a*b - 72*(d*sqrt(x) + c)^2*a*b*c + 72*(d*sqr
t(x) + c)*a*b*c^2 - 24*a*b*c^3 + 24*((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)*cos(2*d*sqrt(x) + 2*c) + (24*I*(d*sqrt(x) + c)^3*a*b - 72*I*(d*sqrt(x) + c)^2*a*b*c + 72*I*(d*
sqrt(x) + c)*a*b*c^2 - 24*I*a*b*c^3)*sin(2*d*sqrt(x) + 2*c))*dilog(-I*e^(I*d*sqrt(x) + I*c)) + (-8*I*(d*sqrt(x
) + c)^3*b^2 + 18*I*(d*sqrt(x) + c)^2*b^2*c - 18*I*(d*sqrt(x) + c)*b^2*c^2 + 6*I*b^2*c^3 + (-8*I*(d*sqrt(x) +
c)^3*b^2 + 18*I*(d*sqrt(x) + c)^2*b^2*c - 18*I*(d*sqrt(x) + c)*b^2*c^2 + 6*I*b^2*c^3)*cos(2*d*sqrt(x) + 2*c) +
2*(4*(d*sqrt(x) + c)^3*b^2 - 9*(d*sqrt(x) + c)^2*b^2*c + 9*(d*sqrt(x) + c)*b^2*c^2 - 3*b^2*c^3)*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) + (-3*I*(d*
sqrt(x) + c)^4*a*b + 12*I*(d*sqrt(x) + c)^3*a*b*c - 18*I*(d*sqrt(x) + c)^2*a*b*c^2 + 12*I*(d*sqrt(x) + c)*a*b*
c^3 + (-3*I*(d*sqrt(x) + c)^4*a*b + 12*I*(d*sqrt(x) + c)^3*a*b*c - 18*I*(d*sqrt(x) + c)^2*a*b*c^2 + 12*I*(d*sq
rt(x) + c)*a*b*c^3)*cos(2*d*sqrt(x) + 2*c) + 3*((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)*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) + (3*I*(d*sqrt(x) + c)^4*a*b - 12*I*(d*sqrt(x) + c)^3*a*b*c + 18*I*(d*sqr
t(x) + c)^2*a*b*c^2 - 12*I*(d*sqrt(x) + c)*a*b*c^3 + (3*I*(d*sqrt(x) + c)^4*a*b - 12*I*(d*sqrt(x) + c)^3*a*b*c
+ 18*I*(d*sqrt(x) + c)^2*a*b*c^2 - 12*I*(d*sqrt(x) + c)*a*b*c^3)*cos(2*d*sqrt(x) + 2*c) - 3*((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)*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) + (144*I*a*b*cos(2*d*sqrt(x
) + 2*c) - 144*a*b*sin(2*d*sqrt(x) + 2*c) + 144*I*a*b)*polylog(5, I*e^(I*d*sqrt(x) + I*c)) + (-144*I*a*b*cos(2
*d*sqrt(x) + 2*c) + 144*a*b*sin(2*d*sqrt(x) + 2*c) - 144*I*a*b)*polylog(5, -I*e^(I*d*sqrt(x) + I*c)) + (12*b^2
*cos(2*d*sqrt(x) + 2*c) + 12*I*b^2*sin(2*d*sqrt(x) + 2*c) + 12*b^2)*polylog(4, -e^(2*I*d*sqrt(x) + 2*I*c)) + (
144*(d*sqrt(x) + c)*a*b - 144*a*b*c + 144*((d*sqrt(x) + c)*a*b - a*b*c)*cos(2*d*sqrt(x) + 2*c) + (144*I*(d*sqr
t(x) + c)*a*b - 144*I*a*b*c)*sin(2*d*sqrt(x) + 2*c))*polylog(4, I*e^(I*d*sqrt(x) + I*c)) - (144*(d*sqrt(x) + c
)*a*b - 144*a*b*c + 144*((d*sqrt(x) + c)*a*b - a*b*c)*cos(2*d*sqrt(x) + 2*c) - (-144*I*(d*sqrt(x) + c)*a*b + 1
44*I*a*b*c)*sin(2*d*sqrt(x) + 2*c))*polylog(4, -I*e^(I*d*sqrt(x) + I*c)) + (-24*I*(d*sqrt(x) + c)*b^2 + 18*I*b
^2*c + (-24*I*(d*sqrt(x) + c)*b^2 + 18*I*b^2*c)*cos(2*d*sqrt(x) + 2*c) + 6*(4*(d*sqrt(x) + c)*b^2 - 3*b^2*c)*s
in(2*d*sqrt(x) + 2*c))*polylog(3, -e^(2*I*d*sqrt(x) + 2*I*c)) + (-72*I*(d*sqrt(x) + c)^2*a*b + 144*I*(d*sqrt(x
) + c)*a*b*c - 72*I*a*b*c^2 + (-72*I*(d*sqrt(x) + c)^2*a*b + 144*I*(d*sqrt(x) + c)*a*b*c - 72*I*a*b*c^2)*cos(2
*d*sqrt(x) + 2*c) + 72*((d*sqrt(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c + a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*pol
ylog(3, I*e^(I*d*sqrt(x) + I*c)) + (72*I*(d*sqrt(x) + c)^2*a*b - 144*I*(d*sqrt(x) + c)*a*b*c + 72*I*a*b*c^2 +
(72*I*(d*sqrt(x) + c)^2*a*b - 144*I*(d*sqrt(x) + c)*a*b*c + 72*I*a*b*c^2)*cos(2*d*sqrt(x) + 2*c) - 72*((d*sqrt
(x) + c)^2*a*b - 2*(d*sqrt(x) + c)*a*b*c + a*b*c^2)*sin(2*d*sqrt(x) + 2*c))*polylog(3, -I*e^(I*d*sqrt(x) + I*c
)) + (-6*I*(d*sqrt(x) + c)^4*b^2 + 24*I*(d*sqrt(x) + c)^3*b^2*c - 36*I*(d*sqrt(x) + c)^2*b^2*c^2 + 24*I*(d*sqr
t(x) + c)*b^2*c^3)*sin(2*d*sqrt(x) + 2*c))/(-3*I*cos(2*d*sqrt(x) + 2*c) + 3*sin(2*d*sqrt(x) + 2*c) - 3*I))/d^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b^2*x^(3/2)*sec(d*sqrt(x) + c)^2 + 2*a*b*x^(3/2)*sec(d*sqrt(x) + c) + a^2*x^(3/2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{\frac{3}{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**(3/2)*(a+b*sec(c+d*x**(1/2)))**2,x)

[Out]

Integral(x**(3/2)*(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^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sec(d*sqrt(x) + c) + a)^2*x^(3/2), x)