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

Optimal. Leaf size=421 \[ \frac{16 i a b x^{3/2} \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{48 a b x \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{96 i a b \sqrt{x} \text{PolyLog}\left (4,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{PolyLog}\left (4,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 a b \text{PolyLog}\left (5,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{96 a b \text{PolyLog}\left (5,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{12 i b^2 x \text{PolyLog}\left (2,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b^2 \sqrt{x} \text{PolyLog}\left (3,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{6 i b^2 \text{PolyLog}\left (4,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{2}{5} a^2 x^{5/2}-\frac{8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^2}{d} \]

[Out]

((-2*I)*b^2*x^2)/d + (2*a^2*x^(5/2))/5 - (8*a*b*x^2*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d - (2*b^2*x^2*Cot[c + d*S
qrt[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, -E^(I*(c
+ d*Sqrt[x]))])/d^2 - ((16*I)*a*b*x^(3/2)*PolyLog[2, 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, -E^(I*(c + d*Sqrt[x]))])/d^3 + (48*a*b*x*PolyLog[3, 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]*Poly
Log[4, -E^(I*(c + d*Sqrt[x]))])/d^4 + ((96*I)*a*b*Sqrt[x]*PolyLog[4, 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, -E^(I*(c + d*Sqrt[x]))])/d^5 - (96*a*b*PolyLog
[5, E^(I*(c + d*Sqrt[x]))])/d^5

________________________________________________________________________________________

Rubi [A]  time = 0.537797, antiderivative size = 421, 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 = {4205, 4190, 4183, 2531, 6609, 2282, 6589, 4184, 3717, 2190} \[ \frac{16 i a b x^{3/2} \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{48 a b x \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{96 i a b \sqrt{x} \text{PolyLog}\left (4,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{PolyLog}\left (4,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 a b \text{PolyLog}\left (5,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{96 a b \text{PolyLog}\left (5,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{12 i b^2 x \text{PolyLog}\left (2,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{12 b^2 \sqrt{x} \text{PolyLog}\left (3,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{6 i b^2 \text{PolyLog}\left (4,e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{2}{5} a^2 x^{5/2}-\frac{8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{8 b^2 x^{3/2} \log \left (1-e^{2 i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\right )}{d}-\frac{2 i b^2 x^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*I)*b^2*x^2)/d + (2*a^2*x^(5/2))/5 - (8*a*b*x^2*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d - (2*b^2*x^2*Cot[c + d*S
qrt[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, -E^(I*(c
+ d*Sqrt[x]))])/d^2 - ((16*I)*a*b*x^(3/2)*PolyLog[2, 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, -E^(I*(c + d*Sqrt[x]))])/d^3 + (48*a*b*x*PolyLog[3, 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]*Poly
Log[4, -E^(I*(c + d*Sqrt[x]))])/d^4 + ((96*I)*a*b*Sqrt[x]*PolyLog[4, 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, -E^(I*(c + d*Sqrt[x]))])/d^5 - (96*a*b*PolyLog
[5, E^(I*(c + d*Sqrt[x]))])/d^5

Rule 4205

Int[((a_.) + Csc[(c_.) + (d_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplif
y[(m + 1)/n] - 1)*(a + b*Csc[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 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, 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 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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && 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 \csc \left (c+d \sqrt{x}\right )\right )^2 \, dx &=2 \operatorname{Subst}\left (\int x^4 (a+b \csc (c+d x))^2 \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \left (a^2 x^4+2 a b x^4 \csc (c+d x)+b^2 x^4 \csc ^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 \csc (c+d x) \, dx,x,\sqrt{x}\right )+\left (2 b^2\right ) \operatorname{Subst}\left (\int x^4 \csc ^2(c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{2}{5} a^2 x^{5/2}-\frac{8 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\right )}{d}-\frac{(16 a b) \operatorname{Subst}\left (\int x^3 \log \left (1-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+e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{\left (8 b^2\right ) \operatorname{Subst}\left (\int x^3 \cot (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 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\right )}{d}+\frac{16 i a b x^{3/2} \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{(48 i a b) \operatorname{Subst}\left (\int x^2 \text{Li}_2\left (-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 (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 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{48 a b x \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{(96 a b) \operatorname{Subst}\left (\int x \text{Li}_3\left (-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 (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 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{96 i a b \sqrt{x} \text{Li}_4\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{(96 i a b) \operatorname{Subst}\left (\int \text{Li}_4\left (-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 (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 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{(96 a b) \operatorname{Subst}\left (\int \frac{\text{Li}_4(-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(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 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 a b \text{Li}_5\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{96 a b \text{Li}_5\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\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 a b x^2 \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{2 b^2 x^2 \cot \left (c+d \sqrt{x}\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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{16 i a b x^{3/2} \text{Li}_2\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{48 a b x \text{Li}_3\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{96 i a b \sqrt{x} \text{Li}_4\left (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 (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{96 a b \text{Li}_5\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}\\ \end{align*}

Mathematica [A]  time = 8.35741, size = 749, normalized size = 1.78 \[ \frac{4 b \sin ^2\left (c+d \sqrt{x}\right ) \left (a+b \csc \left (c+d \sqrt{x}\right )\right )^2 \left (-2 i d^2 x \left (2 a d \sqrt{x}-3 b\right ) \text{PolyLog}\left (2,-e^{-i \left (c+d \sqrt{x}\right )}\right )+2 i d^2 x \left (2 a d \sqrt{x}+3 b\right ) \text{PolyLog}\left (2,e^{-i \left (c+d \sqrt{x}\right )}\right )-12 a d^2 x \text{PolyLog}\left (3,-e^{-i \left (c+d \sqrt{x}\right )}\right )+12 a d^2 x \text{PolyLog}\left (3,e^{-i \left (c+d \sqrt{x}\right )}\right )+24 i a d \sqrt{x} \text{PolyLog}\left (4,-e^{-i \left (c+d \sqrt{x}\right )}\right )-24 i a d \sqrt{x} \text{PolyLog}\left (4,e^{-i \left (c+d \sqrt{x}\right )}\right )+24 a \text{PolyLog}\left (5,-e^{-i \left (c+d \sqrt{x}\right )}\right )-24 a \text{PolyLog}\left (5,e^{-i \left (c+d \sqrt{x}\right )}\right )+12 b d \sqrt{x} \text{PolyLog}\left (3,-e^{-i \left (c+d \sqrt{x}\right )}\right )+12 b d \sqrt{x} \text{PolyLog}\left (3,e^{-i \left (c+d \sqrt{x}\right )}\right )-12 i b \text{PolyLog}\left (4,-e^{-i \left (c+d \sqrt{x}\right )}\right )-12 i b \text{PolyLog}\left (4,e^{-i \left (c+d \sqrt{x}\right )}\right )+a d^4 x^2 \log \left (1-e^{-i \left (c+d \sqrt{x}\right )}\right )-a d^4 x^2 \log \left (1+e^{-i \left (c+d \sqrt{x}\right )}\right )-\frac{i b d^4 x^2}{-1+e^{2 i c}}+2 b d^3 x^{3/2} \log \left (1-e^{-i \left (c+d \sqrt{x}\right )}\right )+2 b d^3 x^{3/2} \log \left (1+e^{-i \left (c+d \sqrt{x}\right )}\right )\right )}{d^5 \left (a \sin \left (c+d \sqrt{x}\right )+b\right )^2}+\frac{2 a^2 x^{5/2} \sin ^2\left (c+d \sqrt{x}\right ) \left (a+b \csc \left (c+d \sqrt{x}\right )\right )^2}{5 \left (a \sin \left (c+d \sqrt{x}\right )+b\right )^2}+\frac{b^2 x^2 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d \sqrt{x}}{2}\right ) \sin ^2\left (c+d \sqrt{x}\right ) \csc \left (\frac{c}{2}+\frac{d \sqrt{x}}{2}\right ) \left (a+b \csc \left (c+d \sqrt{x}\right )\right )^2}{d \left (a \sin \left (c+d \sqrt{x}\right )+b\right )^2}+\frac{b^2 x^2 \sec \left (\frac{c}{2}\right ) \sin \left (\frac{d \sqrt{x}}{2}\right ) \sin ^2\left (c+d \sqrt{x}\right ) \sec \left (\frac{c}{2}+\frac{d \sqrt{x}}{2}\right ) \left (a+b \csc \left (c+d \sqrt{x}\right )\right )^2}{d \left (a \sin \left (c+d \sqrt{x}\right )+b\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*x^(5/2)*(a + b*Csc[c + d*Sqrt[x]])^2*Sin[c + d*Sqrt[x]]^2)/(5*(b + a*Sin[c + d*Sqrt[x]])^2) + (4*b*(a +
b*Csc[c + d*Sqrt[x]])^2*(((-I)*b*d^4*x^2)/(-1 + E^((2*I)*c)) + 2*b*d^3*x^(3/2)*Log[1 - E^((-I)*(c + d*Sqrt[x]
))] + a*d^4*x^2*Log[1 - E^((-I)*(c + d*Sqrt[x]))] + 2*b*d^3*x^(3/2)*Log[1 + E^((-I)*(c + d*Sqrt[x]))] - a*d^4*
x^2*Log[1 + E^((-I)*(c + d*Sqrt[x]))] - (2*I)*d^2*(-3*b + 2*a*d*Sqrt[x])*x*PolyLog[2, -E^((-I)*(c + d*Sqrt[x])
)] + (2*I)*d^2*(3*b + 2*a*d*Sqrt[x])*x*PolyLog[2, E^((-I)*(c + d*Sqrt[x]))] + 12*b*d*Sqrt[x]*PolyLog[3, -E^((-
I)*(c + d*Sqrt[x]))] - 12*a*d^2*x*PolyLog[3, -E^((-I)*(c + d*Sqrt[x]))] + 12*b*d*Sqrt[x]*PolyLog[3, E^((-I)*(c
+ d*Sqrt[x]))] + 12*a*d^2*x*PolyLog[3, E^((-I)*(c + d*Sqrt[x]))] - (12*I)*b*PolyLog[4, -E^((-I)*(c + d*Sqrt[x
]))] + (24*I)*a*d*Sqrt[x]*PolyLog[4, -E^((-I)*(c + d*Sqrt[x]))] - (12*I)*b*PolyLog[4, E^((-I)*(c + d*Sqrt[x]))
] - (24*I)*a*d*Sqrt[x]*PolyLog[4, E^((-I)*(c + d*Sqrt[x]))] + 24*a*PolyLog[5, -E^((-I)*(c + d*Sqrt[x]))] - 24*
a*PolyLog[5, E^((-I)*(c + d*Sqrt[x]))])*Sin[c + d*Sqrt[x]]^2)/(d^5*(b + a*Sin[c + d*Sqrt[x]])^2) + (b^2*x^2*Cs
c[c/2]*Csc[c/2 + (d*Sqrt[x])/2]*(a + b*Csc[c + d*Sqrt[x]])^2*Sin[c + d*Sqrt[x]]^2*Sin[(d*Sqrt[x])/2])/(d*(b +
a*Sin[c + d*Sqrt[x]])^2) + (b^2*x^2*(a + b*Csc[c + d*Sqrt[x]])^2*Sec[c/2]*Sec[c/2 + (d*Sqrt[x])/2]*Sin[c + d*S
qrt[x]]^2*Sin[(d*Sqrt[x])/2])/(d*(b + a*Sin[c + d*Sqrt[x]])^2)

________________________________________________________________________________________

Maple [F]  time = 0.195, size = 0, normalized size = 0. \begin{align*} \int{x}^{{\frac{3}{2}}} \left ( a+b\csc \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*csc(c+d*x^(1/2)))^2,x)

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.85799, size = 3800, normalized size = 9.03 \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*csc(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(cot(d*sqrt(x) + c) + csc(d*sqrt(x) + c)) - 5*(2*b^2*c^4 -
(2*(d*sqrt(x) + c)^4*a*b + 4*b^2*c^3 - 4*(2*a*b*c + b^2)*(d*sqrt(x) + c)^3 + 12*(a*b*c^2 + b^2*c)*(d*sqrt(x) +
c)^2 - 4*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c) - 2*((d*sqrt(x) + c)^4*a*b + 2*b^2*c^3 - 2*(2*a*b*c + b^2)*(
d*sqrt(x) + c)^3 + 6*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^2 - 2*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c))*cos(2*d*
sqrt(x) + 2*c) - (2*I*(d*sqrt(x) + c)^4*a*b + 4*I*b^2*c^3 + (-8*I*a*b*c - 4*I*b^2)*(d*sqrt(x) + c)^3 + (12*I*a
*b*c^2 + 12*I*b^2*c)*(d*sqrt(x) + c)^2 + (-8*I*a*b*c^3 - 12*I*b^2*c^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c)
)*arctan2(sin(d*sqrt(x) + c), cos(d*sqrt(x) + c) + 1) + 4*(b^2*c^3*cos(2*d*sqrt(x) + 2*c) + I*b^2*c^3*sin(2*d*
sqrt(x) + 2*c) - b^2*c^3)*arctan2(sin(d*sqrt(x) + c), cos(d*sqrt(x) + c) - 1) - (2*(d*sqrt(x) + c)^4*a*b - 4*(
2*a*b*c - b^2)*(d*sqrt(x) + c)^3 + 12*(a*b*c^2 - b^2*c)*(d*sqrt(x) + c)^2 - 4*(2*a*b*c^3 - 3*b^2*c^2)*(d*sqrt(
x) + c) - 2*((d*sqrt(x) + c)^4*a*b - 2*(2*a*b*c - b^2)*(d*sqrt(x) + c)^3 + 6*(a*b*c^2 - b^2*c)*(d*sqrt(x) + c)
^2 - 2*(2*a*b*c^3 - 3*b^2*c^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) - (2*I*(d*sqrt(x) + c)^4*a*b + (-8*I*a*
b*c + 4*I*b^2)*(d*sqrt(x) + c)^3 + (12*I*a*b*c^2 - 12*I*b^2*c)*(d*sqrt(x) + c)^2 + (-8*I*a*b*c^3 + 12*I*b^2*c^
2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*arctan2(sin(d*sqrt(x) + c), -cos(d*sqrt(x) + c) + 1) + 2*((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*s
qrt(x) + 2*c) + (8*(d*sqrt(x) + c)^3*a*b - 8*a*b*c^3 - 12*b^2*c^2 - 12*(2*a*b*c + b^2)*(d*sqrt(x) + c)^2 + 24*
(a*b*c^2 + b^2*c)*(d*sqrt(x) + c) - 4*(2*(d*sqrt(x) + c)^3*a*b - 2*a*b*c^3 - 3*b^2*c^2 - 3*(2*a*b*c + b^2)*(d*
sqrt(x) + c)^2 + 6*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + (-8*I*(d*sqrt(x) + c)^3*a*b + 8
*I*a*b*c^3 + 12*I*b^2*c^2 + (24*I*a*b*c + 12*I*b^2)*(d*sqrt(x) + c)^2 + (-24*I*a*b*c^2 - 24*I*b^2*c)*(d*sqrt(x
) + c))*sin(2*d*sqrt(x) + 2*c))*dilog(-e^(I*d*sqrt(x) + I*c)) - (8*(d*sqrt(x) + c)^3*a*b - 8*a*b*c^3 + 12*b^2*
c^2 - 12*(2*a*b*c - b^2)*(d*sqrt(x) + c)^2 + 24*(a*b*c^2 - b^2*c)*(d*sqrt(x) + c) - 4*(2*(d*sqrt(x) + c)^3*a*b
- 2*a*b*c^3 + 3*b^2*c^2 - 3*(2*a*b*c - b^2)*(d*sqrt(x) + c)^2 + 6*(a*b*c^2 - b^2*c)*(d*sqrt(x) + c))*cos(2*d*
sqrt(x) + 2*c) - (8*I*(d*sqrt(x) + c)^3*a*b - 8*I*a*b*c^3 + 12*I*b^2*c^2 + (-24*I*a*b*c + 12*I*b^2)*(d*sqrt(x)
+ c)^2 + (24*I*a*b*c^2 - 24*I*b^2*c)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*dilog(e^(I*d*sqrt(x) + I*c)) +
(I*(d*sqrt(x) + c)^4*a*b + 2*I*b^2*c^3 + (-4*I*a*b*c - 2*I*b^2)*(d*sqrt(x) + c)^3 + (6*I*a*b*c^2 + 6*I*b^2*c)*
(d*sqrt(x) + c)^2 + (-4*I*a*b*c^3 - 6*I*b^2*c^2)*(d*sqrt(x) + c) + (-I*(d*sqrt(x) + c)^4*a*b - 2*I*b^2*c^3 + (
4*I*a*b*c + 2*I*b^2)*(d*sqrt(x) + c)^3 + (-6*I*a*b*c^2 - 6*I*b^2*c)*(d*sqrt(x) + c)^2 + (4*I*a*b*c^3 + 6*I*b^2
*c^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + ((d*sqrt(x) + c)^4*a*b + 2*b^2*c^3 - 2*(2*a*b*c + b^2)*(d*sqrt
(x) + c)^3 + 6*(a*b*c^2 + b^2*c)*(d*sqrt(x) + c)^2 - 2*(2*a*b*c^3 + 3*b^2*c^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x
) + 2*c))*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*cos(d*sqrt(x) + c) + 1) + (-I*(d*sqrt(x) + c)^4*
a*b + 2*I*b^2*c^3 + (4*I*a*b*c - 2*I*b^2)*(d*sqrt(x) + c)^3 + (-6*I*a*b*c^2 + 6*I*b^2*c)*(d*sqrt(x) + c)^2 + (
4*I*a*b*c^3 - 6*I*b^2*c^2)*(d*sqrt(x) + c) + (I*(d*sqrt(x) + c)^4*a*b - 2*I*b^2*c^3 + (-4*I*a*b*c + 2*I*b^2)*(
d*sqrt(x) + c)^3 + (6*I*a*b*c^2 - 6*I*b^2*c)*(d*sqrt(x) + c)^2 + (-4*I*a*b*c^3 + 6*I*b^2*c^2)*(d*sqrt(x) + c))
*cos(2*d*sqrt(x) + 2*c) - ((d*sqrt(x) + c)^4*a*b - 2*b^2*c^3 - 2*(2*a*b*c - b^2)*(d*sqrt(x) + c)^3 + 6*(a*b*c^
2 - b^2*c)*(d*sqrt(x) + c)^2 - 2*(2*a*b*c^3 - 3*b^2*c^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*log(cos(d*sq
rt(x) + c)^2 + sin(d*sqrt(x) + c)^2 - 2*cos(d*sqrt(x) + c) + 1) + (48*I*a*b*cos(2*d*sqrt(x) + 2*c) - 48*a*b*si
n(2*d*sqrt(x) + 2*c) - 48*I*a*b)*polylog(5, -e^(I*d*sqrt(x) + I*c)) + (-48*I*a*b*cos(2*d*sqrt(x) + 2*c) + 48*a
*b*sin(2*d*sqrt(x) + 2*c) + 48*I*a*b)*polylog(5, e^(I*d*sqrt(x) + I*c)) - (48*(d*sqrt(x) + c)*a*b - 48*a*b*c -
24*b^2 - 24*(2*(d*sqrt(x) + c)*a*b - 2*a*b*c - b^2)*cos(2*d*sqrt(x) + 2*c) - (48*I*(d*sqrt(x) + c)*a*b - 48*I
*a*b*c - 24*I*b^2)*sin(2*d*sqrt(x) + 2*c))*polylog(4, -e^(I*d*sqrt(x) + I*c)) + (48*(d*sqrt(x) + c)*a*b - 48*a
*b*c + 24*b^2 - 24*(2*(d*sqrt(x) + c)*a*b - 2*a*b*c + b^2)*cos(2*d*sqrt(x) + 2*c) + (-48*I*(d*sqrt(x) + c)*a*b
+ 48*I*a*b*c - 24*I*b^2)*sin(2*d*sqrt(x) + 2*c))*polylog(4, e^(I*d*sqrt(x) + I*c)) + (24*I*(d*sqrt(x) + c)^2*
a*b + 24*I*a*b*c^2 + 24*I*b^2*c + (-48*I*a*b*c - 24*I*b^2)*(d*sqrt(x) + c) + (-24*I*(d*sqrt(x) + c)^2*a*b - 24
*I*a*b*c^2 - 24*I*b^2*c + (48*I*a*b*c + 24*I*b^2)*(d*sqrt(x) + c))*cos(2*d*sqrt(x) + 2*c) + 24*((d*sqrt(x) + c
)^2*a*b + a*b*c^2 + b^2*c - (2*a*b*c + b^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(x) + 2*c))*polylog(3, -e^(I*d*sqrt(x
) + I*c)) + (-24*I*(d*sqrt(x) + c)^2*a*b - 24*I*a*b*c^2 + 24*I*b^2*c + (48*I*a*b*c - 24*I*b^2)*(d*sqrt(x) + c)
+ (24*I*(d*sqrt(x) + c)^2*a*b + 24*I*a*b*c^2 - 24*I*b^2*c + (-48*I*a*b*c + 24*I*b^2)*(d*sqrt(x) + c))*cos(2*d
*sqrt(x) + 2*c) - 24*((d*sqrt(x) + c)^2*a*b + a*b*c^2 - b^2*c - (2*a*b*c - b^2)*(d*sqrt(x) + c))*sin(2*d*sqrt(
x) + 2*c))*polylog(3, e^(I*d*sqrt(x) + I*c)) + (2*I*(d*sqrt(x) + c)^4*b^2 - 8*I*(d*sqrt(x) + c)^3*b^2*c + 12*I
*(d*sqrt(x) + c)^2*b^2*c^2 - 8*I*(d*sqrt(x) + c)*b^2*c^3)*sin(2*d*sqrt(x) + 2*c))/(-I*cos(2*d*sqrt(x) + 2*c) +
sin(2*d*sqrt(x) + 2*c) + 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}} \csc \left (d \sqrt{x} + c\right )^{2} + 2 \, a b x^{\frac{3}{2}} \csc \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*csc(c+d*x^(1/2)))^2,x, algorithm="fricas")

[Out]

integral(b^2*x^(3/2)*csc(d*sqrt(x) + c)^2 + 2*a*b*x^(3/2)*csc(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 \csc{\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*csc(c+d*x**(1/2)))**2,x)

[Out]

Integral(x**(3/2)*(a + b*csc(c + d*sqrt(x)))**2, x)

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

[Out]

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