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

Optimal. Leaf size=432 \[ \frac{14 i b x^3 \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{420 i b x^2 \text{PolyLog}\left (4,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{420 i b x^2 \text{PolyLog}\left (4,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{1680 b x^{3/2} \text{PolyLog}\left (5,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{1680 b x^{3/2} \text{PolyLog}\left (5,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{5040 i b x \text{PolyLog}\left (6,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{5040 i b x \text{PolyLog}\left (6,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 b \sqrt{x} \text{PolyLog}\left (7,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{10080 b \sqrt{x} \text{PolyLog}\left (7,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}-\frac{10080 i b \text{PolyLog}\left (8,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^8}+\frac{10080 i b \text{PolyLog}\left (8,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^8}+\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

[Out]

(a*x^4)/4 - (4*b*x^(7/2)*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d + ((14*I)*b*x^3*PolyLog[2, -E^(I*(c + d*Sqrt[x]))])
/d^2 - ((14*I)*b*x^3*PolyLog[2, E^(I*(c + d*Sqrt[x]))])/d^2 - (84*b*x^(5/2)*PolyLog[3, -E^(I*(c + d*Sqrt[x]))]
)/d^3 + (84*b*x^(5/2)*PolyLog[3, E^(I*(c + d*Sqrt[x]))])/d^3 - ((420*I)*b*x^2*PolyLog[4, -E^(I*(c + d*Sqrt[x])
)])/d^4 + ((420*I)*b*x^2*PolyLog[4, E^(I*(c + d*Sqrt[x]))])/d^4 + (1680*b*x^(3/2)*PolyLog[5, -E^(I*(c + d*Sqrt
[x]))])/d^5 - (1680*b*x^(3/2)*PolyLog[5, E^(I*(c + d*Sqrt[x]))])/d^5 + ((5040*I)*b*x*PolyLog[6, -E^(I*(c + d*S
qrt[x]))])/d^6 - ((5040*I)*b*x*PolyLog[6, E^(I*(c + d*Sqrt[x]))])/d^6 - (10080*b*Sqrt[x]*PolyLog[7, -E^(I*(c +
d*Sqrt[x]))])/d^7 + (10080*b*Sqrt[x]*PolyLog[7, E^(I*(c + d*Sqrt[x]))])/d^7 - ((10080*I)*b*PolyLog[8, -E^(I*(
c + d*Sqrt[x]))])/d^8 + ((10080*I)*b*PolyLog[8, E^(I*(c + d*Sqrt[x]))])/d^8

________________________________________________________________________________________

Rubi [A]  time = 0.425544, antiderivative size = 432, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {14, 4205, 4183, 2531, 6609, 2282, 6589} \[ \frac{14 i b x^3 \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{420 i b x^2 \text{PolyLog}\left (4,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{420 i b x^2 \text{PolyLog}\left (4,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{1680 b x^{3/2} \text{PolyLog}\left (5,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{1680 b x^{3/2} \text{PolyLog}\left (5,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{5040 i b x \text{PolyLog}\left (6,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{5040 i b x \text{PolyLog}\left (6,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 b \sqrt{x} \text{PolyLog}\left (7,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{10080 b \sqrt{x} \text{PolyLog}\left (7,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}-\frac{10080 i b \text{PolyLog}\left (8,-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^8}+\frac{10080 i b \text{PolyLog}\left (8,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^8}+\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*Csc[c + d*Sqrt[x]]),x]

[Out]

(a*x^4)/4 - (4*b*x^(7/2)*ArcTanh[E^(I*(c + d*Sqrt[x]))])/d + ((14*I)*b*x^3*PolyLog[2, -E^(I*(c + d*Sqrt[x]))])
/d^2 - ((14*I)*b*x^3*PolyLog[2, E^(I*(c + d*Sqrt[x]))])/d^2 - (84*b*x^(5/2)*PolyLog[3, -E^(I*(c + d*Sqrt[x]))]
)/d^3 + (84*b*x^(5/2)*PolyLog[3, E^(I*(c + d*Sqrt[x]))])/d^3 - ((420*I)*b*x^2*PolyLog[4, -E^(I*(c + d*Sqrt[x])
)])/d^4 + ((420*I)*b*x^2*PolyLog[4, E^(I*(c + d*Sqrt[x]))])/d^4 + (1680*b*x^(3/2)*PolyLog[5, -E^(I*(c + d*Sqrt
[x]))])/d^5 - (1680*b*x^(3/2)*PolyLog[5, E^(I*(c + d*Sqrt[x]))])/d^5 + ((5040*I)*b*x*PolyLog[6, -E^(I*(c + d*S
qrt[x]))])/d^6 - ((5040*I)*b*x*PolyLog[6, E^(I*(c + d*Sqrt[x]))])/d^6 - (10080*b*Sqrt[x]*PolyLog[7, -E^(I*(c +
d*Sqrt[x]))])/d^7 + (10080*b*Sqrt[x]*PolyLog[7, E^(I*(c + d*Sqrt[x]))])/d^7 - ((10080*I)*b*PolyLog[8, -E^(I*(
c + d*Sqrt[x]))])/d^8 + ((10080*I)*b*PolyLog[8, E^(I*(c + d*Sqrt[x]))])/d^8

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 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 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]

Rubi steps

\begin{align*} \int x^3 \left (a+b \csc \left (c+d \sqrt{x}\right )\right ) \, dx &=\int \left (a x^3+b x^3 \csc \left (c+d \sqrt{x}\right )\right ) \, dx\\ &=\frac{a x^4}{4}+b \int x^3 \csc \left (c+d \sqrt{x}\right ) \, dx\\ &=\frac{a x^4}{4}+(2 b) \operatorname{Subst}\left (\int x^7 \csc (c+d x) \, dx,x,\sqrt{x}\right )\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}-\frac{(14 b) \operatorname{Subst}\left (\int x^6 \log \left (1-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}+\frac{(14 b) \operatorname{Subst}\left (\int x^6 \log \left (1+e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d}\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{14 i b x^3 \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{(84 i b) \operatorname{Subst}\left (\int x^5 \text{Li}_2\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}+\frac{(84 i b) \operatorname{Subst}\left (\int x^5 \text{Li}_2\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^2}\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{14 i b x^3 \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{(420 b) \operatorname{Subst}\left (\int x^4 \text{Li}_3\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}-\frac{(420 b) \operatorname{Subst}\left (\int x^4 \text{Li}_3\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^3}\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{14 i b x^3 \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{420 i b x^2 \text{Li}_4\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{420 i b x^2 \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{(1680 i b) \operatorname{Subst}\left (\int x^3 \text{Li}_4\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}-\frac{(1680 i b) \operatorname{Subst}\left (\int x^3 \text{Li}_4\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^4}\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{14 i b x^3 \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{420 i b x^2 \text{Li}_4\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{420 i b x^2 \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{1680 b x^{3/2} \text{Li}_5\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{1680 b x^{3/2} \text{Li}_5\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{(5040 b) \operatorname{Subst}\left (\int x^2 \text{Li}_5\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}+\frac{(5040 b) \operatorname{Subst}\left (\int x^2 \text{Li}_5\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^5}\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{14 i b x^3 \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{420 i b x^2 \text{Li}_4\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{420 i b x^2 \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{1680 b x^{3/2} \text{Li}_5\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{1680 b x^{3/2} \text{Li}_5\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{5040 i b x \text{Li}_6\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{5040 i b x \text{Li}_6\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{(10080 i b) \operatorname{Subst}\left (\int x \text{Li}_6\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^6}+\frac{(10080 i b) \operatorname{Subst}\left (\int x \text{Li}_6\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^6}\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{14 i b x^3 \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{420 i b x^2 \text{Li}_4\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{420 i b x^2 \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{1680 b x^{3/2} \text{Li}_5\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{1680 b x^{3/2} \text{Li}_5\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{5040 i b x \text{Li}_6\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{5040 i b x \text{Li}_6\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 b \sqrt{x} \text{Li}_7\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{10080 b \sqrt{x} \text{Li}_7\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{(10080 b) \operatorname{Subst}\left (\int \text{Li}_7\left (-e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^7}-\frac{(10080 b) \operatorname{Subst}\left (\int \text{Li}_7\left (e^{i (c+d x)}\right ) \, dx,x,\sqrt{x}\right )}{d^7}\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{14 i b x^3 \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{420 i b x^2 \text{Li}_4\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{420 i b x^2 \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{1680 b x^{3/2} \text{Li}_5\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{1680 b x^{3/2} \text{Li}_5\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{5040 i b x \text{Li}_6\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{5040 i b x \text{Li}_6\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 b \sqrt{x} \text{Li}_7\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{10080 b \sqrt{x} \text{Li}_7\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}-\frac{(10080 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_7(-x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^8}+\frac{(10080 i b) \operatorname{Subst}\left (\int \frac{\text{Li}_7(x)}{x} \, dx,x,e^{i \left (c+d \sqrt{x}\right )}\right )}{d^8}\\ &=\frac{a x^4}{4}-\frac{4 b x^{7/2} \tanh ^{-1}\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d}+\frac{14 i b x^3 \text{Li}_2\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{14 i b x^3 \text{Li}_2\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^2}-\frac{84 b x^{5/2} \text{Li}_3\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}+\frac{84 b x^{5/2} \text{Li}_3\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^3}-\frac{420 i b x^2 \text{Li}_4\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{420 i b x^2 \text{Li}_4\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^4}+\frac{1680 b x^{3/2} \text{Li}_5\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}-\frac{1680 b x^{3/2} \text{Li}_5\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^5}+\frac{5040 i b x \text{Li}_6\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{5040 i b x \text{Li}_6\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^6}-\frac{10080 b \sqrt{x} \text{Li}_7\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}+\frac{10080 b \sqrt{x} \text{Li}_7\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^7}-\frac{10080 i b \text{Li}_8\left (-e^{i \left (c+d \sqrt{x}\right )}\right )}{d^8}+\frac{10080 i b \text{Li}_8\left (e^{i \left (c+d \sqrt{x}\right )}\right )}{d^8}\\ \end{align*}

Mathematica [A]  time = 0.51334, size = 445, normalized size = 1.03 \[ \frac{a x^4}{4}+\frac{2 b \left (7 i d^6 x^3 \text{PolyLog}\left (2,-e^{i \left (c+d \sqrt{x}\right )}\right )-7 i d^6 x^3 \text{PolyLog}\left (2,e^{i \left (c+d \sqrt{x}\right )}\right )-42 d^5 x^{5/2} \text{PolyLog}\left (3,-e^{i \left (c+d \sqrt{x}\right )}\right )+42 d^5 x^{5/2} \text{PolyLog}\left (3,e^{i \left (c+d \sqrt{x}\right )}\right )-210 i d^4 x^2 \text{PolyLog}\left (4,-e^{i \left (c+d \sqrt{x}\right )}\right )+210 i d^4 x^2 \text{PolyLog}\left (4,e^{i \left (c+d \sqrt{x}\right )}\right )+840 d^3 x^{3/2} \text{PolyLog}\left (5,-e^{i \left (c+d \sqrt{x}\right )}\right )-840 d^3 x^{3/2} \text{PolyLog}\left (5,e^{i \left (c+d \sqrt{x}\right )}\right )+2520 i d^2 x \text{PolyLog}\left (6,-e^{i \left (c+d \sqrt{x}\right )}\right )-2520 i d^2 x \text{PolyLog}\left (6,e^{i \left (c+d \sqrt{x}\right )}\right )-5040 d \sqrt{x} \text{PolyLog}\left (7,-e^{i \left (c+d \sqrt{x}\right )}\right )+5040 d \sqrt{x} \text{PolyLog}\left (7,e^{i \left (c+d \sqrt{x}\right )}\right )-5040 i \text{PolyLog}\left (8,-e^{i \left (c+d \sqrt{x}\right )}\right )+5040 i \text{PolyLog}\left (8,e^{i \left (c+d \sqrt{x}\right )}\right )+d^7 x^{7/2} \log \left (1-e^{i \left (c+d \sqrt{x}\right )}\right )-d^7 x^{7/2} \log \left (1+e^{i \left (c+d \sqrt{x}\right )}\right )\right )}{d^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x^4)/4 + (2*b*(d^7*x^(7/2)*Log[1 - E^(I*(c + d*Sqrt[x]))] - d^7*x^(7/2)*Log[1 + E^(I*(c + d*Sqrt[x]))] + (7
*I)*d^6*x^3*PolyLog[2, -E^(I*(c + d*Sqrt[x]))] - (7*I)*d^6*x^3*PolyLog[2, E^(I*(c + d*Sqrt[x]))] - 42*d^5*x^(5
/2)*PolyLog[3, -E^(I*(c + d*Sqrt[x]))] + 42*d^5*x^(5/2)*PolyLog[3, E^(I*(c + d*Sqrt[x]))] - (210*I)*d^4*x^2*Po
lyLog[4, -E^(I*(c + d*Sqrt[x]))] + (210*I)*d^4*x^2*PolyLog[4, E^(I*(c + d*Sqrt[x]))] + 840*d^3*x^(3/2)*PolyLog
[5, -E^(I*(c + d*Sqrt[x]))] - 840*d^3*x^(3/2)*PolyLog[5, E^(I*(c + d*Sqrt[x]))] + (2520*I)*d^2*x*PolyLog[6, -E
^(I*(c + d*Sqrt[x]))] - (2520*I)*d^2*x*PolyLog[6, E^(I*(c + d*Sqrt[x]))] - 5040*d*Sqrt[x]*PolyLog[7, -E^(I*(c
+ d*Sqrt[x]))] + 5040*d*Sqrt[x]*PolyLog[7, E^(I*(c + d*Sqrt[x]))] - (5040*I)*PolyLog[8, -E^(I*(c + d*Sqrt[x]))
] + (5040*I)*PolyLog[8, E^(I*(c + d*Sqrt[x]))]))/d^8

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Maple [F]  time = 0.122, size = 0, normalized size = 0. \begin{align*} \int{x}^{3} \left ( a+b\csc \left ( c+d\sqrt{x} \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.86131, size = 2022, normalized size = 4.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((d*sqrt(x) + c)^8*a - 8*(d*sqrt(x) + c)^7*a*c + 28*(d*sqrt(x) + c)^6*a*c^2 - 56*(d*sqrt(x) + c)^5*a*c^3 +
70*(d*sqrt(x) + c)^4*a*c^4 - 56*(d*sqrt(x) + c)^3*a*c^5 + 28*(d*sqrt(x) + c)^2*a*c^6 - 8*(d*sqrt(x) + c)*a*c^
7 + 8*b*c^7*log(cot(d*sqrt(x) + c) + csc(d*sqrt(x) + c)) - 4*(2*I*(d*sqrt(x) + c)^7*b - 14*I*(d*sqrt(x) + c)^6
*b*c + 42*I*(d*sqrt(x) + c)^5*b*c^2 - 70*I*(d*sqrt(x) + c)^4*b*c^3 + 70*I*(d*sqrt(x) + c)^3*b*c^4 - 42*I*(d*sq
rt(x) + c)^2*b*c^5 + 14*I*(d*sqrt(x) + c)*b*c^6)*arctan2(sin(d*sqrt(x) + c), cos(d*sqrt(x) + c) + 1) - 4*(2*I*
(d*sqrt(x) + c)^7*b - 14*I*(d*sqrt(x) + c)^6*b*c + 42*I*(d*sqrt(x) + c)^5*b*c^2 - 70*I*(d*sqrt(x) + c)^4*b*c^3
+ 70*I*(d*sqrt(x) + c)^3*b*c^4 - 42*I*(d*sqrt(x) + c)^2*b*c^5 + 14*I*(d*sqrt(x) + c)*b*c^6)*arctan2(sin(d*sqr
t(x) + c), -cos(d*sqrt(x) + c) + 1) - 4*(-14*I*(d*sqrt(x) + c)^6*b + 84*I*(d*sqrt(x) + c)^5*b*c - 210*I*(d*sqr
t(x) + c)^4*b*c^2 + 280*I*(d*sqrt(x) + c)^3*b*c^3 - 210*I*(d*sqrt(x) + c)^2*b*c^4 + 84*I*(d*sqrt(x) + c)*b*c^5
- 14*I*b*c^6)*dilog(-e^(I*d*sqrt(x) + I*c)) - 4*(14*I*(d*sqrt(x) + c)^6*b - 84*I*(d*sqrt(x) + c)^5*b*c + 210*
I*(d*sqrt(x) + c)^4*b*c^2 - 280*I*(d*sqrt(x) + c)^3*b*c^3 + 210*I*(d*sqrt(x) + c)^2*b*c^4 - 84*I*(d*sqrt(x) +
c)*b*c^5 + 14*I*b*c^6)*dilog(e^(I*d*sqrt(x) + I*c)) - 4*((d*sqrt(x) + c)^7*b - 7*(d*sqrt(x) + c)^6*b*c + 21*(d
*sqrt(x) + c)^5*b*c^2 - 35*(d*sqrt(x) + c)^4*b*c^3 + 35*(d*sqrt(x) + c)^3*b*c^4 - 21*(d*sqrt(x) + c)^2*b*c^5 +
7*(d*sqrt(x) + c)*b*c^6)*log(cos(d*sqrt(x) + c)^2 + sin(d*sqrt(x) + c)^2 + 2*cos(d*sqrt(x) + c) + 1) + 4*((d*
sqrt(x) + c)^7*b - 7*(d*sqrt(x) + c)^6*b*c + 21*(d*sqrt(x) + c)^5*b*c^2 - 35*(d*sqrt(x) + c)^4*b*c^3 + 35*(d*s
qrt(x) + c)^3*b*c^4 - 21*(d*sqrt(x) + c)^2*b*c^5 + 7*(d*sqrt(x) + c)*b*c^6)*log(cos(d*sqrt(x) + c)^2 + sin(d*s
qrt(x) + c)^2 - 2*cos(d*sqrt(x) + c) + 1) - 40320*I*b*polylog(8, -e^(I*d*sqrt(x) + I*c)) + 40320*I*b*polylog(8
, e^(I*d*sqrt(x) + I*c)) - 40320*((d*sqrt(x) + c)*b - b*c)*polylog(7, -e^(I*d*sqrt(x) + I*c)) + 40320*((d*sqrt
(x) + c)*b - b*c)*polylog(7, e^(I*d*sqrt(x) + I*c)) - 4*(-5040*I*(d*sqrt(x) + c)^2*b + 10080*I*(d*sqrt(x) + c)
*b*c - 5040*I*b*c^2)*polylog(6, -e^(I*d*sqrt(x) + I*c)) - 4*(5040*I*(d*sqrt(x) + c)^2*b - 10080*I*(d*sqrt(x) +
c)*b*c + 5040*I*b*c^2)*polylog(6, e^(I*d*sqrt(x) + I*c)) + 6720*((d*sqrt(x) + c)^3*b - 3*(d*sqrt(x) + c)^2*b*
c + 3*(d*sqrt(x) + c)*b*c^2 - b*c^3)*polylog(5, -e^(I*d*sqrt(x) + I*c)) - 6720*((d*sqrt(x) + c)^3*b - 3*(d*sqr
t(x) + c)^2*b*c + 3*(d*sqrt(x) + c)*b*c^2 - b*c^3)*polylog(5, e^(I*d*sqrt(x) + I*c)) - 4*(420*I*(d*sqrt(x) + c
)^4*b - 1680*I*(d*sqrt(x) + c)^3*b*c + 2520*I*(d*sqrt(x) + c)^2*b*c^2 - 1680*I*(d*sqrt(x) + c)*b*c^3 + 420*I*b
*c^4)*polylog(4, -e^(I*d*sqrt(x) + I*c)) - 4*(-420*I*(d*sqrt(x) + c)^4*b + 1680*I*(d*sqrt(x) + c)^3*b*c - 2520
*I*(d*sqrt(x) + c)^2*b*c^2 + 1680*I*(d*sqrt(x) + c)*b*c^3 - 420*I*b*c^4)*polylog(4, e^(I*d*sqrt(x) + I*c)) - 3
36*((d*sqrt(x) + c)^5*b - 5*(d*sqrt(x) + c)^4*b*c + 10*(d*sqrt(x) + c)^3*b*c^2 - 10*(d*sqrt(x) + c)^2*b*c^3 +
5*(d*sqrt(x) + c)*b*c^4 - b*c^5)*polylog(3, -e^(I*d*sqrt(x) + I*c)) + 336*((d*sqrt(x) + c)^5*b - 5*(d*sqrt(x)
+ c)^4*b*c + 10*(d*sqrt(x) + c)^3*b*c^2 - 10*(d*sqrt(x) + c)^2*b*c^3 + 5*(d*sqrt(x) + c)*b*c^4 - b*c^5)*polylo
g(3, e^(I*d*sqrt(x) + I*c)))/d^8

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(b*x^3*csc(d*sqrt(x) + c) + a*x^3, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*(a + b*csc(c + d*sqrt(x))), 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 )} x^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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