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

Optimal. Leaf size=352 \[ \frac{12 i f^2 (e+f x) \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{12 f^3 \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4}-\frac{6 i f^3 \text{PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac{6 i f^3 \text{PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}-\frac{6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{i (e+f x)^3}{a d} \]

[Out]

(I*(e + f*x)^3)/(a*d) - (2*(e + f*x)^3*ArcTanh[E^(I*(c + d*x))])/(a*d) + ((e + f*x)^3*Cot[c/2 + Pi/4 + (d*x)/2
])/(a*d) - (6*f*(e + f*x)^2*Log[1 - I*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) + ((12*I)*f^2*(e + f*x)*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^3) - ((3*I)*f*(e + f*x)^2*PolyLog[2
, E^(I*(c + d*x))])/(a*d^2) - (6*f^2*(e + f*x)*PolyLog[3, -E^(I*(c + d*x))])/(a*d^3) - (12*f^3*PolyLog[3, I*E^
(I*(c + d*x))])/(a*d^4) + (6*f^2*(e + f*x)*PolyLog[3, E^(I*(c + d*x))])/(a*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)

________________________________________________________________________________________

Rubi [A]  time = 0.46895, antiderivative size = 352, normalized size of antiderivative = 1., number of steps used = 17, number of rules used = 10, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.385, Rules used = {4535, 4183, 2531, 6609, 2282, 6589, 3318, 4184, 3717, 2190} \[ \frac{12 i f^2 (e+f x) \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^2 (e+f x) \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac{3 i f (e+f x)^2 \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{12 f^3 \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4}-\frac{6 i f^3 \text{PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac{6 i f^3 \text{PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}-\frac{6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{i (e+f x)^3}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(I*(e + f*x)^3)/(a*d) - (2*(e + f*x)^3*ArcTanh[E^(I*(c + d*x))])/(a*d) + ((e + f*x)^3*Cot[c/2 + Pi/4 + (d*x)/2
])/(a*d) - (6*f*(e + f*x)^2*Log[1 - I*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) + ((12*I)*f^2*(e + f*x)*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^3) - ((3*I)*f*(e + f*x)^2*PolyLog[2
, E^(I*(c + d*x))])/(a*d^2) - (6*f^2*(e + f*x)*PolyLog[3, -E^(I*(c + d*x))])/(a*d^3) - (12*f^3*PolyLog[3, I*E^
(I*(c + d*x))])/(a*d^4) + (6*f^2*(e + f*x)*PolyLog[3, E^(I*(c + d*x))])/(a*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)

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 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

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 \frac{(e+f x)^3 \csc (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x)^3 \csc (c+d x) \, dx}{a}-\int \frac{(e+f x)^3}{a+a \sin (c+d x)} \, dx\\ &=-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{\int (e+f x)^3 \csc ^2\left (\frac{1}{2} \left (c+\frac{\pi }{2}\right )+\frac{d x}{2}\right ) \, dx}{2 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{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\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{(3 f) \int (e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}-\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{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\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{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 f) \int \frac{e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )} (e+f x)^2}{1-i e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}} \, dx}{a 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{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{6 f (e+f x)^2 \log \left (1-i 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 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{\left (12 f^2\right ) \int (e+f x) \log \left (1-i e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \, dx}{a 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{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{6 f (e+f x)^2 \log \left (1-i 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{12 i f^2 (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}-\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{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 (12 i f^3\right ) \int \text{Li}_2\left (i e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \, dx}{a d^3}\\ &=\frac{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{6 f (e+f x)^2 \log \left (1-i 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{12 i f^2 (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}-\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{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 (12 f^3\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2(i x)}{x} \, dx,x,e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}\right )}{a d^4}\\ &=\frac{i (e+f x)^3}{a d}-\frac{2 (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{6 f (e+f x)^2 \log \left (1-i 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{12 i f^2 (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}-\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{12 f^3 \text{Li}_3\left (i e^{i (c+d x)}\right )}{a d^4}+\frac{6 f^2 (e+f x) \text{Li}_3\left (e^{i (c+d x)}\right )}{a 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}\\ \end{align*}

Mathematica [A]  time = 2.56901, size = 443, normalized size = 1.26 \[ \frac{\frac{3 i f \left (d^2 (e+f x)^2 \text{PolyLog}(2,-\cos (c+d x)-i \sin (c+d x))+2 i d f (e+f x) \text{PolyLog}(3,-\cos (c+d x)-i \sin (c+d x))-2 f^2 \text{PolyLog}(4,-\cos (c+d x)-i \sin (c+d x))\right )}{d^3}-\frac{3 i f \left (d^2 (e+f x)^2 \text{PolyLog}(2,\cos (c+d x)+i \sin (c+d x))+2 i d f (e+f x) \text{PolyLog}(3,\cos (c+d x)+i \sin (c+d x))-2 f^2 \text{PolyLog}(4,\cos (c+d x)+i \sin (c+d x))\right )}{d^3}+\frac{6 f (\cos (c)+i \sin (c)) \left (\frac{2 f (\cos (c)-i (\sin (c)+1)) (d (e+f x) \text{PolyLog}(2,-\sin (c+d x)-i \cos (c+d x))-i f \text{PolyLog}(3,-\sin (c+d x)-i \cos (c+d x)))}{d^3}-\frac{(\sin (c)+i \cos (c)+1) (e+f x)^2 \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac{(\cos (c)-i \sin (c)) (e+f x)^3}{3 f}\right )}{\cos (c)+i (\sin (c)+1)}-\frac{2 \sin \left (\frac{d x}{2}\right ) (e+f x)^3}{\left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-2 (e+f x)^3 \tanh ^{-1}(\cos (c+d x)+i \sin (c+d x))}{a d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*(e + f*x)^3*ArcTanh[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]]))/d^3 - ((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]]))/d^3 +
(6*f*(Cos[c] + I*Sin[c])*(((e + f*x)^3*(Cos[c] - I*Sin[c]))/(3*f) - ((e + f*x)^2*Log[1 + I*Cos[c + d*x] + Sin[
c + d*x]]*(1 + I*Cos[c] + Sin[c]))/d + (2*f*(d*(e + f*x)*PolyLog[2, (-I)*Cos[c + d*x] - Sin[c + d*x]] - I*f*Po
lyLog[3, (-I)*Cos[c + d*x] - Sin[c + d*x]])*(Cos[c] - I*(1 + Sin[c])))/d^3))/(Cos[c] + I*(1 + Sin[c])) - (2*(e
+ f*x)^3*Sin[(d*x)/2])/((Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])))/(a*d)

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Maple [B]  time = 0.285, size = 1151, normalized size = 3.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12*f^2/d^2/a*e*ln(1-I*exp(I*(d*x+c)))*x-6*I*f^3*polylog(4,-exp(I*(d*x+c)))/a/d^4+1/d/a*e^3*ln(exp(I*(d*x+c))-
1)-1/d/a*e^3*ln(exp(I*(d*x+c))+1)+6*f^3/d^4/a*c^2*ln(exp(I*(d*x+c)))-6*f/d^2/a*ln(exp(I*(d*x+c))+I)*e^2-6*f^3/
d^4/a*c^2*ln(exp(I*(d*x+c))+I)-4*I*f^3/d^4/a*c^3+2*I*f^3/d/a*x^3+6*f/d^2/a*ln(exp(I*(d*x+c)))*e^2+2*(f^3*x^3+3
*e*f^2*x^2+3*e^2*f*x+e^3)/d/a/(exp(I*(d*x+c))+I)-3/d^2/a*e^2*f*c*ln(exp(I*(d*x+c))-1)+3/d^3/a*e*f^2*c^2*ln(exp
(I*(d*x+c))-1)-1/d/a*f^3*ln(exp(I*(d*x+c))+1)*x^3+12*I*f^2/d^2/a*e*c*x+6*I*f^3*polylog(4,exp(I*(d*x+c)))/a/d^4
-12*f^2/d^3/a*e*ln(1-I*exp(I*(d*x+c)))*c-6*f^3/d^2/a*ln(1-I*exp(I*(d*x+c)))*x^2+6*f^3/d^4/a*ln(1-I*exp(I*(d*x+
c)))*c^2-12*f^2/d^3/a*e*c*ln(exp(I*(d*x+c)))+12*f^2/d^3/a*e*c*ln(exp(I*(d*x+c))+I)-6*I*f^3/d^3/a*c^2*x+12*I*f^
3/d^3/a*polylog(2,I*exp(I*(d*x+c)))*x+12*I*f^2/d^3/a*e*polylog(2,I*exp(I*(d*x+c)))+6*I*f^2/d/a*e*x^2+6*I*f^2/d
^3/a*e*c^2+6*I/d^2/a*e*f^2*polylog(2,-exp(I*(d*x+c)))*x-6*I/d^2/a*e*f^2*polylog(2,exp(I*(d*x+c)))*x+1/d/a*f^3*
ln(1-exp(I*(d*x+c)))*x^3+1/d^4/a*f^3*ln(1-exp(I*(d*x+c)))*c^3-3/d/a*e*f^2*ln(exp(I*(d*x+c))+1)*x^2+3/d/a*ln(1-
exp(I*(d*x+c)))*e^2*f*x-3/d/a*ln(exp(I*(d*x+c))+1)*e^2*f*x+3/d/a*e*f^2*ln(1-exp(I*(d*x+c)))*x^2-3/d^3/a*e*f^2*
ln(1-exp(I*(d*x+c)))*c^2+3/d^2/a*ln(1-exp(I*(d*x+c)))*c*e^2*f+3*I/d^2/a*e^2*f*polylog(2,-exp(I*(d*x+c)))-3*I/d
^2/a*e^2*f*polylog(2,exp(I*(d*x+c)))+3*I/d^2/a*f^3*polylog(2,-exp(I*(d*x+c)))*x^2-3*I/d^2/a*f^3*polylog(2,exp(
I*(d*x+c)))*x^2-12*f^3*polylog(3,I*exp(I*(d*x+c)))/a/d^4+6/d^3/a*f^3*polylog(3,exp(I*(d*x+c)))*x-6/d^3/a*f^3*p
olylog(3,-exp(I*(d*x+c)))*x-1/d^4/a*f^3*c^3*ln(exp(I*(d*x+c))-1)+6/d^3/a*e*f^2*polylog(3,exp(I*(d*x+c)))-6/d^3
/a*e*f^2*polylog(3,-exp(I*(d*x+c)))

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Maxima [B]  time = 3.40972, size = 3750, normalized size = 10.65 \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+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-(3*c*e^2*f*(2/(a*d + a*d*sin(d*x + c)/(cos(d*x + c) + 1)) + log(sin(d*x + c)/(cos(d*x + c) + 1))/(a*d)) - e^3
*(log(sin(d*x + c)/(cos(d*x + c) + 1))/a + 2/(a + a*sin(d*x + c)/(cos(d*x + c) + 1))) + (12*I*c^2*d*e*f^2 - 4*
I*c^3*f^3 + (12*I*d^2*e^2*f - 24*I*c*d*e*f^2 + 12*I*c^2*f^3 + 12*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*cos(d*x +
c) + (12*I*d^2*e^2*f - 24*I*c*d*e*f^2 + 12*I*c^2*f^3)*sin(d*x + c))*arctan2(sin(d*x + c) + 1, cos(d*x + c)) +
(-12*I*(d*x + c)^2*f^3 + (-24*I*d*e*f^2 + 24*I*c*f^3)*(d*x + c) - 12*((d*x + c)^2*f^3 + 2*(d*e*f^2 - c*f^3)*(
d*x + c))*cos(d*x + c) + (-12*I*(d*x + c)^2*f^3 + (-24*I*d*e*f^2 + 24*I*c*f^3)*(d*x + c))*sin(d*x + c))*arctan
2(cos(d*x + c), sin(d*x + c) + 1) + (6*I*c^2*d*e*f^2 + 2*I*(d*x + c)^3*f^3 - 2*I*c^3*f^3 + (6*I*d*e*f^2 - 6*I*
c*f^3)*(d*x + c)^2 + (6*I*d^2*e^2*f - 12*I*c*d*e*f^2 + 6*I*c^2*f^3)*(d*x + c) + 2*(3*c^2*d*e*f^2 + (d*x + c)^3
*f^3 - c^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x + c))*cos(d*x +
c) + (6*I*c^2*d*e*f^2 + 2*I*(d*x + c)^3*f^3 - 2*I*c^3*f^3 + (6*I*d*e*f^2 - 6*I*c*f^3)*(d*x + c)^2 + (6*I*d^2*e
^2*f - 12*I*c*d*e*f^2 + 6*I*c^2*f^3)*(d*x + c))*sin(d*x + c))*arctan2(sin(d*x + c), cos(d*x + c) + 1) + (-6*I*
c^2*d*e*f^2 + 2*I*c^3*f^3 - 2*(3*c^2*d*e*f^2 - c^3*f^3)*cos(d*x + c) + (-6*I*c^2*d*e*f^2 + 2*I*c^3*f^3)*sin(d*
x + c))*arctan2(sin(d*x + c), cos(d*x + c) - 1) + (2*I*(d*x + c)^3*f^3 + (6*I*d*e*f^2 - 6*I*c*f^3)*(d*x + c)^2
+ (6*I*d^2*e^2*f - 12*I*c*d*e*f^2 + 6*I*c^2*f^3)*(d*x + c) + 2*((d*x + c)^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x +
c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x + c))*cos(d*x + c) + (2*I*(d*x + c)^3*f^3 + (6*I*d*e*f^2 - 6
*I*c*f^3)*(d*x + c)^2 + (6*I*d^2*e^2*f - 12*I*c*d*e*f^2 + 6*I*c^2*f^3)*(d*x + c))*sin(d*x + c))*arctan2(sin(d*
x + c), -cos(d*x + c) + 1) - 4*((d*x + c)^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2
+ c^2*f^3)*(d*x + c))*cos(d*x + c) + (-24*I*d*e*f^2 - 24*I*(d*x + c)*f^3 + 24*I*c*f^3 - 24*(d*e*f^2 + (d*x +
c)*f^3 - c*f^3)*cos(d*x + c) + (-24*I*d*e*f^2 - 24*I*(d*x + c)*f^3 + 24*I*c*f^3)*sin(d*x + c))*dilog(I*e^(I*d*
x + I*c)) + (-6*I*d^2*e^2*f + 12*I*c*d*e*f^2 - 6*I*(d*x + c)^2*f^3 - 6*I*c^2*f^3 + (-12*I*d*e*f^2 + 12*I*c*f^3
)*(d*x + c) - 6*(d^2*e^2*f - 2*c*d*e*f^2 + (d*x + c)^2*f^3 + c^2*f^3 + 2*(d*e*f^2 - c*f^3)*(d*x + c))*cos(d*x
+ c) + (-6*I*d^2*e^2*f + 12*I*c*d*e*f^2 - 6*I*(d*x + c)^2*f^3 - 6*I*c^2*f^3 + (-12*I*d*e*f^2 + 12*I*c*f^3)*(d*
x + c))*sin(d*x + c))*dilog(-e^(I*d*x + I*c)) + (6*I*d^2*e^2*f - 12*I*c*d*e*f^2 + 6*I*(d*x + c)^2*f^3 + 6*I*c^
2*f^3 + (12*I*d*e*f^2 - 12*I*c*f^3)*(d*x + c) + 6*(d^2*e^2*f - 2*c*d*e*f^2 + (d*x + c)^2*f^3 + c^2*f^3 + 2*(d*
e*f^2 - c*f^3)*(d*x + c))*cos(d*x + c) + (6*I*d^2*e^2*f - 12*I*c*d*e*f^2 + 6*I*(d*x + c)^2*f^3 + 6*I*c^2*f^3 +
(12*I*d*e*f^2 - 12*I*c*f^3)*(d*x + c))*sin(d*x + c))*dilog(e^(I*d*x + I*c)) + (3*c^2*d*e*f^2 + (d*x + c)^3*f^
3 - c^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x + c) + (-3*I*c^2*d*
e*f^2 - I*(d*x + c)^3*f^3 + I*c^3*f^3 + (-3*I*d*e*f^2 + 3*I*c*f^3)*(d*x + c)^2 + (-3*I*d^2*e^2*f + 6*I*c*d*e*f
^2 - 3*I*c^2*f^3)*(d*x + c))*cos(d*x + c) + (3*c^2*d*e*f^2 + (d*x + c)^3*f^3 - c^3*f^3 + 3*(d*e*f^2 - c*f^3)*(
d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x + c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^
2 + 2*cos(d*x + c) + 1) - (3*c^2*d*e*f^2 + (d*x + c)^3*f^3 - c^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^
2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x + c) - (3*I*c^2*d*e*f^2 + I*(d*x + c)^3*f^3 - I*c^3*f^3 + (3*I*d*e*f^2 -
3*I*c*f^3)*(d*x + c)^2 + (3*I*d^2*e^2*f - 6*I*c*d*e*f^2 + 3*I*c^2*f^3)*(d*x + c))*cos(d*x + c) + (3*c^2*d*e*f
^2 + (d*x + c)^3*f^3 - c^3*f^3 + 3*(d*e*f^2 - c*f^3)*(d*x + c)^2 + 3*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*(d*x
+ c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*cos(d*x + c) + 1) + (6*d^2*e^2*f - 12*c*d*e*f^2 +
6*(d*x + c)^2*f^3 + 6*c^2*f^3 + 12*(d*e*f^2 - c*f^3)*(d*x + c) + (-6*I*d^2*e^2*f + 12*I*c*d*e*f^2 - 6*I*(d*x +
c)^2*f^3 - 6*I*c^2*f^3 + (-12*I*d*e*f^2 + 12*I*c*f^3)*(d*x + c))*cos(d*x + c) + 6*(d^2*e^2*f - 2*c*d*e*f^2 +
(d*x + c)^2*f^3 + c^2*f^3 + 2*(d*e*f^2 - c*f^3)*(d*x + c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 +
2*sin(d*x + c) + 1) + (12*f^3*cos(d*x + c) + 12*I*f^3*sin(d*x + c) + 12*I*f^3)*polylog(4, -e^(I*d*x + I*c)) -
(12*f^3*cos(d*x + c) + 12*I*f^3*sin(d*x + c) + 12*I*f^3)*polylog(4, e^(I*d*x + I*c)) - 24*(I*f^3*cos(d*x + c)
- f^3*sin(d*x + c) - f^3)*polylog(3, I*e^(I*d*x + I*c)) + (12*d*e*f^2 + 12*(d*x + c)*f^3 - 12*c*f^3 + (-12*I*
d*e*f^2 - 12*I*(d*x + c)*f^3 + 12*I*c*f^3)*cos(d*x + c) + 12*(d*e*f^2 + (d*x + c)*f^3 - c*f^3)*sin(d*x + c))*p
olylog(3, -e^(I*d*x + I*c)) - (12*d*e*f^2 + 12*(d*x + c)*f^3 - 12*c*f^3 - (12*I*d*e*f^2 + 12*I*(d*x + c)*f^3 -
12*I*c*f^3)*cos(d*x + c) + 12*(d*e*f^2 + (d*x + c)*f^3 - c*f^3)*sin(d*x + c))*polylog(3, e^(I*d*x + I*c)) + (
-4*I*(d*x + c)^3*f^3 + (-12*I*d*e*f^2 + 12*I*c*f^3)*(d*x + c)^2 + (-12*I*d^2*e^2*f + 24*I*c*d*e*f^2 - 12*I*c^2
*f^3)*(d*x + c))*sin(d*x + c))/(-2*I*a*d^3*cos(d*x + c) + 2*a*d^3*sin(d*x + c) + 2*a*d^3))/d

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Fricas [C]  time = 3.22015, size = 6904, normalized size = 19.61 \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+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{e^{3} \csc{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{f^{3} x^{3} \csc{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{3 e f^{2} x^{2} \csc{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{3 e^{2} f x \csc{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(e**3*csc(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**3*x**3*csc(c + d*x)/(sin(c + d*x) + 1), x) +
Integral(3*e*f**2*x**2*csc(c + d*x)/(sin(c + d*x) + 1), x) + Integral(3*e**2*f*x*csc(c + d*x)/(sin(c + d*x) +
1), x))/a

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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