3.232 \(\int \frac{(e+f x)^3 \csc (c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=732 \[ \frac{6 i b f^2 (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a d^3 \sqrt{a^2-b^2}}-\frac{6 i b f^2 (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a d^3 \sqrt{a^2-b^2}}+\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a d^2 \sqrt{a^2-b^2}}-\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a d^2 \sqrt{a^2-b^2}}-\frac{6 b f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a d^4 \sqrt{a^2-b^2}}+\frac{6 b f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a d^4 \sqrt{a^2-b^2}}-\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{6 i f^3 \text{PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac{6 i f^3 \text{PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}+\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a d \sqrt{a^2-b^2}}-\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a d \sqrt{a^2-b^2}}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

[Out]

(-2*(e + f*x)^3*ArcTanh[E^(I*(c + d*x))])/(a*d) + (I*b*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2
- b^2])])/(a*Sqrt[a^2 - b^2]*d) - (I*b*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*S
qrt[a^2 - b^2]*d) + ((3*I)*f*(e + f*x)^2*PolyLog[2, -E^(I*(c + d*x))])/(a*d^2) - ((3*I)*f*(e + f*x)^2*PolyLog[
2, E^(I*(c + d*x))])/(a*d^2) + (3*b*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*
Sqrt[a^2 - b^2]*d^2) - (3*b*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*Sqrt[a^2
- b^2]*d^2) - (6*f^2*(e + f*x)*PolyLog[3, -E^(I*(c + d*x))])/(a*d^3) + (6*f^2*(e + f*x)*PolyLog[3, E^(I*(c +
d*x))])/(a*d^3) + ((6*I)*b*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*Sqrt[a^2
- b^2]*d^3) - ((6*I)*b*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*Sqrt[a^2 - b^
2]*d^3) - ((6*I)*f^3*PolyLog[4, -E^(I*(c + d*x))])/(a*d^4) + ((6*I)*f^3*PolyLog[4, E^(I*(c + d*x))])/(a*d^4) -
(6*b*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*Sqrt[a^2 - b^2]*d^4) + (6*b*f^3*PolyLog[
4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*Sqrt[a^2 - b^2]*d^4)

________________________________________________________________________________________

Rubi [A]  time = 1.11721, antiderivative size = 732, normalized size of antiderivative = 1., number of steps used = 22, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {4535, 4183, 2531, 6609, 2282, 6589, 3323, 2264, 2190} \[ \frac{6 i b f^2 (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a d^3 \sqrt{a^2-b^2}}-\frac{6 i b f^2 (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a d^3 \sqrt{a^2-b^2}}+\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a d^2 \sqrt{a^2-b^2}}-\frac{3 b f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a d^2 \sqrt{a^2-b^2}}-\frac{6 b f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a d^4 \sqrt{a^2-b^2}}+\frac{6 b f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a d^4 \sqrt{a^2-b^2}}-\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{6 i f^3 \text{PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac{6 i f^3 \text{PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}+\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a d \sqrt{a^2-b^2}}-\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{a d \sqrt{a^2-b^2}}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Csc[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

(-2*(e + f*x)^3*ArcTanh[E^(I*(c + d*x))])/(a*d) + (I*b*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2
- b^2])])/(a*Sqrt[a^2 - b^2]*d) - (I*b*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*S
qrt[a^2 - b^2]*d) + ((3*I)*f*(e + f*x)^2*PolyLog[2, -E^(I*(c + d*x))])/(a*d^2) - ((3*I)*f*(e + f*x)^2*PolyLog[
2, E^(I*(c + d*x))])/(a*d^2) + (3*b*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*
Sqrt[a^2 - b^2]*d^2) - (3*b*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*Sqrt[a^2
- b^2]*d^2) - (6*f^2*(e + f*x)*PolyLog[3, -E^(I*(c + d*x))])/(a*d^3) + (6*f^2*(e + f*x)*PolyLog[3, E^(I*(c +
d*x))])/(a*d^3) + ((6*I)*b*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*Sqrt[a^2
- b^2]*d^3) - ((6*I)*b*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*Sqrt[a^2 - b^
2]*d^3) - ((6*I)*f^3*PolyLog[4, -E^(I*(c + d*x))])/(a*d^4) + ((6*I)*f^3*PolyLog[4, E^(I*(c + d*x))])/(a*d^4) -
(6*b*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a*Sqrt[a^2 - b^2]*d^4) + (6*b*f^3*PolyLog[
4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a*Sqrt[a^2 - b^2]*d^4)

Rule 4535

Int[(Csc[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[1/a, Int[(e + f*x)^m*Csc[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Csc[c + d*x]^(n - 1))/(a +
b*Sin[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, 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 3323

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 2264

Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && 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 \frac{(e+f x)^3 \csc (c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{\int (e+f x)^3 \csc (c+d x) \, dx}{a}-\frac{b \int \frac{(e+f x)^3}{a+b \sin (c+d x)} \, dx}{a}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(2 b) \int \frac{e^{i (c+d x)} (e+f x)^3}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{a}-\frac{(3 f) \int (e+f x)^2 \log \left (1-e^{i (c+d x)}\right ) \, dx}{a d}+\frac{(3 f) \int (e+f x)^2 \log \left (1+e^{i (c+d x)}\right ) \, dx}{a d}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac{\left (2 i b^2\right ) \int \frac{e^{i (c+d x)} (e+f x)^3}{2 a-2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a \sqrt{a^2-b^2}}-\frac{\left (2 i b^2\right ) \int \frac{e^{i (c+d x)} (e+f x)^3}{2 a+2 \sqrt{a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a \sqrt{a^2-b^2}}-\frac{\left (6 i f^2\right ) \int (e+f x) \text{Li}_2\left (-e^{i (c+d x)}\right ) \, dx}{a d^2}+\frac{\left (6 i f^2\right ) \int (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right ) \, dx}{a d^2}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}-\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}-\frac{(3 i b f) \int (e+f x)^2 \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{a \sqrt{a^2-b^2} d}+\frac{(3 i b f) \int (e+f x)^2 \log \left (1-\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{a \sqrt{a^2-b^2} d}+\frac{\left (6 f^3\right ) \int \text{Li}_3\left (-e^{i (c+d x)}\right ) \, dx}{a d^3}-\frac{\left (6 f^3\right ) \int \text{Li}_3\left (e^{i (c+d x)}\right ) \, dx}{a d^3}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}-\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac{3 b f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^2}-\frac{3 b f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^2}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}-\frac{\left (6 b f^2\right ) \int (e+f x) \text{Li}_2\left (\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{a \sqrt{a^2-b^2} d^2}+\frac{\left (6 b f^2\right ) \int (e+f x) \text{Li}_2\left (\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{a \sqrt{a^2-b^2} d^2}-\frac{\left (6 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(-x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4}+\frac{\left (6 i f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3(x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}-\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac{3 b f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^2}-\frac{3 b f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^2}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac{6 i b f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^3}-\frac{6 i b f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^3}-\frac{6 i f^3 \text{Li}_4\left (-e^{i (c+d x)}\right )}{a d^4}+\frac{6 i f^3 \text{Li}_4\left (e^{i (c+d x)}\right )}{a d^4}-\frac{\left (6 i b f^3\right ) \int \text{Li}_3\left (\frac{2 i b e^{i (c+d x)}}{2 a-2 \sqrt{a^2-b^2}}\right ) \, dx}{a \sqrt{a^2-b^2} d^3}+\frac{\left (6 i b f^3\right ) \int \text{Li}_3\left (\frac{2 i b e^{i (c+d x)}}{2 a+2 \sqrt{a^2-b^2}}\right ) \, dx}{a \sqrt{a^2-b^2} d^3}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}-\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac{3 b f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^2}-\frac{3 b f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^2}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac{6 i b f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^3}-\frac{6 i b f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^3}-\frac{6 i f^3 \text{Li}_4\left (-e^{i (c+d x)}\right )}{a d^4}+\frac{6 i f^3 \text{Li}_4\left (e^{i (c+d x)}\right )}{a d^4}-\frac{\left (6 b f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{i b x}{a-\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a \sqrt{a^2-b^2} d^4}+\frac{\left (6 b f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_3\left (\frac{i b x}{a+\sqrt{a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a \sqrt{a^2-b^2} d^4}\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}-\frac{i b (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d}+\frac{3 i f (e+f x)^2 \text{Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac{3 b f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^2}-\frac{3 b f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^2}-\frac{6 f^2 (e+f x) \text{Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac{6 i b f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^3}-\frac{6 i b f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^3}-\frac{6 i f^3 \text{Li}_4\left (-e^{i (c+d x)}\right )}{a d^4}+\frac{6 i f^3 \text{Li}_4\left (e^{i (c+d x)}\right )}{a d^4}-\frac{6 b f^3 \text{Li}_4\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^4}+\frac{6 b f^3 \text{Li}_4\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{a \sqrt{a^2-b^2} d^4}\\ \end{align*}

Mathematica [A]  time = 2.60291, size = 894, normalized size = 1.22 \[ \frac{-2 d^3 \tanh ^{-1}(\cos (c+d x)+i \sin (c+d x)) (e+f x)^3+\frac{b \left (3 d^2 f \text{PolyLog}\left (2,-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}-a}\right ) (e+f x)^2+i \left (2 i e^3 \tan ^{-1}\left (\frac{i a+b e^{i (c+d x)}}{\sqrt{a^2-b^2}}\right ) d^3+f^3 x^3 \log \left (\frac{i e^{i (c+d x)} b}{\sqrt{a^2-b^2}-a}+1\right ) d^3+3 e f^2 x^2 \log \left (\frac{i e^{i (c+d x)} b}{\sqrt{a^2-b^2}-a}+1\right ) d^3+3 e^2 f x \log \left (\frac{i e^{i (c+d x)} b}{\sqrt{a^2-b^2}-a}+1\right ) d^3-f^3 x^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) d^3-3 e f^2 x^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) d^3-3 e^2 f x \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) d^3+3 i f (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) d^2+6 f^2 (e+f x) \text{PolyLog}\left (3,-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}-a}\right ) d-6 e f^2 \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) d-6 f^3 x \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) d+6 i f^3 \text{PolyLog}\left (4,-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}-a}\right )-6 i f^3 \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )\right )\right )}{\sqrt{a^2-b^2}}+3 i f \left (-2 \text{PolyLog}(4,-\cos (c+d x)-i \sin (c+d x)) f^2+2 i d (e+f x) \text{PolyLog}(3,-\cos (c+d x)-i \sin (c+d x)) f+d^2 (e+f x)^2 \text{PolyLog}(2,-\cos (c+d x)-i \sin (c+d x))\right )-3 i f \left (-2 \text{PolyLog}(4,\cos (c+d x)+i \sin (c+d x)) f^2+2 i d (e+f x) \text{PolyLog}(3,\cos (c+d x)+i \sin (c+d x)) f+d^2 (e+f x)^2 \text{PolyLog}(2,\cos (c+d x)+i \sin (c+d x))\right )}{a d^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)^3*Csc[c + d*x])/(a + b*Sin[c + d*x]),x]

[Out]

(-2*d^3*(e + f*x)^3*ArcTanh[Cos[c + d*x] + I*Sin[c + d*x]] + (b*(3*d^2*f*(e + f*x)^2*PolyLog[2, ((-I)*b*E^(I*(
c + d*x)))/(-a + Sqrt[a^2 - b^2])] + I*((2*I)*d^3*e^3*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + 3*d^
3*e^2*f*x*Log[1 + (I*b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])] + 3*d^3*e*f^2*x^2*Log[1 + (I*b*E^(I*(c + d*x))
)/(-a + Sqrt[a^2 - b^2])] + d^3*f^3*x^3*Log[1 + (I*b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])] - 3*d^3*e^2*f*x*
Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] - 3*d^3*e*f^2*x^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt
[a^2 - b^2])] - d^3*f^3*x^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + (3*I)*d^2*f*(e + f*x)^2*Pol
yLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + 6*d*f^2*(e + f*x)*PolyLog[3, ((-I)*b*E^(I*(c + d*x)))/(
-a + Sqrt[a^2 - b^2])] - 6*d*e*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] - 6*d*f^3*x*PolyLog
[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + (6*I)*f^3*PolyLog[4, ((-I)*b*E^(I*(c + d*x)))/(-a + Sqrt[a^
2 - b^2])] - (6*I)*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])))/Sqrt[a^2 - b^2] + (3*I)*f*(d
^2*(e + f*x)^2*PolyLog[2, -Cos[c + d*x] - I*Sin[c + d*x]] + (2*I)*d*f*(e + f*x)*PolyLog[3, -Cos[c + d*x] - I*S
in[c + d*x]] - 2*f^2*PolyLog[4, -Cos[c + d*x] - I*Sin[c + d*x]]) - (3*I)*f*(d^2*(e + f*x)^2*PolyLog[2, Cos[c +
d*x] + I*Sin[c + d*x]] + (2*I)*d*f*(e + f*x)*PolyLog[3, Cos[c + d*x] + I*Sin[c + d*x]] - 2*f^2*PolyLog[4, Cos
[c + d*x] + I*Sin[c + d*x]]))/(a*d^4)

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Maple [F]  time = 0.864, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3}\csc \left ( dx+c \right ) }{a+b\sin \left ( dx+c \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*csc(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

int((f*x+e)^3*csc(d*x+c)/(a+b*sin(d*x+c)),x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*csc(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [C]  time = 4.77588, size = 8263, normalized size = 11.29 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*csc(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/4*(-12*I*b^2*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(b*cos(d*
x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*I*b^2*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, 1/2*(2*
I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*I*
b^2*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x
+ c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*I*b^2*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, -(I*a*cos(d*x + c) + a*sin(d
*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*I*(a^2 - b^2)*f^3*polylog(4, cos
(d*x + c) + I*sin(d*x + c)) + 12*I*(a^2 - b^2)*f^3*polylog(4, cos(d*x + c) - I*sin(d*x + c)) - 12*I*(a^2 - b^2
)*f^3*polylog(4, -cos(d*x + c) + I*sin(d*x + c)) + 12*I*(a^2 - b^2)*f^3*polylog(4, -cos(d*x + c) - I*sin(d*x +
c)) + 2*(3*I*b^2*d^2*f^3*x^2 + 6*I*b^2*d^2*e*f^2*x + 3*I*b^2*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*
I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b +
1) + 2*(-3*I*b^2*d^2*f^3*x^2 - 6*I*b^2*d^2*e*f^2*x - 3*I*b^2*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I
*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1
) + 2*(-3*I*b^2*d^2*f^3*x^2 - 6*I*b^2*d^2*e*f^2*x - 3*I*b^2*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I
*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1
) + 2*(3*I*b^2*d^2*f^3*x^2 + 6*I*b^2*d^2*e*f^2*x + 3*I*b^2*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*
a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1)
+ 2*(b^2*d^3*e^3 - 3*b^2*c*d^2*e^2*f + 3*b^2*c^2*d*e*f^2 - b^2*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*
x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 2*(b^2*d^3*e^3 - 3*b^2*c*d^2*e^2*f + 3*b^2
*c^2*d*e*f^2 - b^2*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2
- b^2)/b^2) - 2*I*a) - 2*(b^2*d^3*e^3 - 3*b^2*c*d^2*e^2*f + 3*b^2*c^2*d*e*f^2 - b^2*c^3*f^3)*sqrt(-(a^2 - b^2)
/b^2)*log(-2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - 2*(b^2*d^3*e^3 - 3*b^
2*c*d^2*e^2*f + 3*b^2*c^2*d*e*f^2 - b^2*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) - 2*I*b*sin(d*x
+ c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + 2*(b^2*d^3*f^3*x^3 + 3*b^2*d^3*e*f^2*x^2 + 3*b^2*d^3*e^2*f*x + 3*
b^2*c*d^2*e^2*f - 3*b^2*c^2*d*e*f^2 + b^2*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*si
n(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(b^2*d^3*f^3*x^3 + 3*b
^2*d^3*e*f^2*x^2 + 3*b^2*d^3*e^2*f*x + 3*b^2*c*d^2*e^2*f - 3*b^2*c^2*d*e*f^2 + b^2*c^3*f^3)*sqrt(-(a^2 - b^2)/
b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/
b^2) + 2*b)/b) + 2*(b^2*d^3*f^3*x^3 + 3*b^2*d^3*e*f^2*x^2 + 3*b^2*d^3*e^2*f*x + 3*b^2*c*d^2*e^2*f - 3*b^2*c^2*
d*e*f^2 + b^2*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x +
c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 2*(b^2*d^3*f^3*x^3 + 3*b^2*d^3*e*f^2*x^2 + 3*b^2*d^
3*e^2*f*x + 3*b^2*c*d^2*e^2*f - 3*b^2*c^2*d*e*f^2 + b^2*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*
x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b) - 12*(b^2*d
*f^3*x + b^2*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(b*cos(
d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*(b^2*d*f^3*x + b^2*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2
)*polylog(3, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b
^2)/b^2))/b) + 12*(b^2*d*f^3*x + b^2*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x
+ c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*(b^2*d*f^3*x + b^2*d*e*f^2)*sqrt(-
(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a
^2 - b^2)/b^2))/b) + (6*I*(a^2 - b^2)*d^2*f^3*x^2 + 12*I*(a^2 - b^2)*d^2*e*f^2*x + 6*I*(a^2 - b^2)*d^2*e^2*f)*
dilog(cos(d*x + c) + I*sin(d*x + c)) + (-6*I*(a^2 - b^2)*d^2*f^3*x^2 - 12*I*(a^2 - b^2)*d^2*e*f^2*x - 6*I*(a^2
- b^2)*d^2*e^2*f)*dilog(cos(d*x + c) - I*sin(d*x + c)) + (6*I*(a^2 - b^2)*d^2*f^3*x^2 + 12*I*(a^2 - b^2)*d^2*
e*f^2*x + 6*I*(a^2 - b^2)*d^2*e^2*f)*dilog(-cos(d*x + c) + I*sin(d*x + c)) + (-6*I*(a^2 - b^2)*d^2*f^3*x^2 - 1
2*I*(a^2 - b^2)*d^2*e*f^2*x - 6*I*(a^2 - b^2)*d^2*e^2*f)*dilog(-cos(d*x + c) - I*sin(d*x + c)) + 2*((a^2 - b^2
)*d^3*f^3*x^3 + 3*(a^2 - b^2)*d^3*e*f^2*x^2 + 3*(a^2 - b^2)*d^3*e^2*f*x + (a^2 - b^2)*d^3*e^3)*log(cos(d*x + c
) + I*sin(d*x + c) + 1) + 2*((a^2 - b^2)*d^3*f^3*x^3 + 3*(a^2 - b^2)*d^3*e*f^2*x^2 + 3*(a^2 - b^2)*d^3*e^2*f*x
+ (a^2 - b^2)*d^3*e^3)*log(cos(d*x + c) - I*sin(d*x + c) + 1) - 2*((a^2 - b^2)*d^3*e^3 - 3*(a^2 - b^2)*c*d^2*
e^2*f + 3*(a^2 - b^2)*c^2*d*e*f^2 - (a^2 - b^2)*c^3*f^3)*log(-1/2*cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2) - 2
*((a^2 - b^2)*d^3*e^3 - 3*(a^2 - b^2)*c*d^2*e^2*f + 3*(a^2 - b^2)*c^2*d*e*f^2 - (a^2 - b^2)*c^3*f^3)*log(-1/2*
cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2) - 2*((a^2 - b^2)*d^3*f^3*x^3 + 3*(a^2 - b^2)*d^3*e*f^2*x^2 + 3*(a^2 -
b^2)*d^3*e^2*f*x + 3*(a^2 - b^2)*c*d^2*e^2*f - 3*(a^2 - b^2)*c^2*d*e*f^2 + (a^2 - b^2)*c^3*f^3)*log(-cos(d*x
+ c) + I*sin(d*x + c) + 1) - 2*((a^2 - b^2)*d^3*f^3*x^3 + 3*(a^2 - b^2)*d^3*e*f^2*x^2 + 3*(a^2 - b^2)*d^3*e^2*
f*x + 3*(a^2 - b^2)*c*d^2*e^2*f - 3*(a^2 - b^2)*c^2*d*e*f^2 + (a^2 - b^2)*c^3*f^3)*log(-cos(d*x + c) - I*sin(d
*x + c) + 1) - 12*((a^2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e*f^2)*polylog(3, cos(d*x + c) + I*sin(d*x + c)) - 12*(
(a^2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e*f^2)*polylog(3, cos(d*x + c) - I*sin(d*x + c)) + 12*((a^2 - b^2)*d*f^3*x
+ (a^2 - b^2)*d*e*f^2)*polylog(3, -cos(d*x + c) + I*sin(d*x + c)) + 12*((a^2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e
*f^2)*polylog(3, -cos(d*x + c) - I*sin(d*x + c)))/((a^3 - a*b^2)*d^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*csc(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)**3*csc(c + d*x)/(a + b*sin(c + d*x)), x)

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Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*csc(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out