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

Optimal. Leaf size=382 \[ \frac{12 i f^2 (e+f x) \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac{12 f^3 \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4}-\frac{6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac{3 f^2 (e+f x) \sin (c+d x) \cos (c+d x)}{4 a d^3}-\frac{6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{6 f^3 \sin (c+d x)}{a d^4}+\frac{(e+f x)^3 \cos (c+d x)}{a d}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{(e+f x)^3 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{i (e+f x)^3}{a d}+\frac{3 (e+f x)^4}{8 a f} \]

[Out]

(-3*e*f^2*x)/(4*a*d^2) - (3*f^3*x^2)/(8*a*d^2) + (I*(e + f*x)^3)/(a*d) + (3*(e + f*x)^4)/(8*a*f) - (6*f^2*(e +
f*x)*Cos[c + d*x])/(a*d^3) + ((e + f*x)^3*Cos[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) + ((12*I)*f^2*(e + f*x)*PolyLog[2, I*E^(I*(c + d*x))])
/(a*d^3) - (12*f^3*PolyLog[3, I*E^(I*(c + d*x))])/(a*d^4) + (6*f^3*Sin[c + d*x])/(a*d^4) - (3*f*(e + f*x)^2*Si
n[c + d*x])/(a*d^2) + (3*f^2*(e + f*x)*Cos[c + d*x]*Sin[c + d*x])/(4*a*d^3) - ((e + f*x)^3*Cos[c + d*x]*Sin[c
+ d*x])/(2*a*d) - (3*f^3*Sin[c + d*x]^2)/(8*a*d^4) + (3*f*(e + f*x)^2*Sin[c + d*x]^2)/(4*a*d^2)

________________________________________________________________________________________

Rubi [A]  time = 0.621256, antiderivative size = 382, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 13, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.464, Rules used = {4515, 3311, 32, 3310, 3296, 2637, 3318, 4184, 3717, 2190, 2531, 2282, 6589} \[ \frac{12 i f^2 (e+f x) \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac{12 f^3 \text{PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4}-\frac{6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac{3 f^2 (e+f x) \sin (c+d x) \cos (c+d x)}{4 a d^3}-\frac{6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{6 f^3 \sin (c+d x)}{a d^4}+\frac{(e+f x)^3 \cos (c+d x)}{a d}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{(e+f x)^3 \sin (c+d x) \cos (c+d x)}{2 a d}-\frac{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{i (e+f x)^3}{a d}+\frac{3 (e+f x)^4}{8 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*e*f^2*x)/(4*a*d^2) - (3*f^3*x^2)/(8*a*d^2) + (I*(e + f*x)^3)/(a*d) + (3*(e + f*x)^4)/(8*a*f) - (6*f^2*(e +
f*x)*Cos[c + d*x])/(a*d^3) + ((e + f*x)^3*Cos[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) + ((12*I)*f^2*(e + f*x)*PolyLog[2, I*E^(I*(c + d*x))])
/(a*d^3) - (12*f^3*PolyLog[3, I*E^(I*(c + d*x))])/(a*d^4) + (6*f^3*Sin[c + d*x])/(a*d^4) - (3*f*(e + f*x)^2*Si
n[c + d*x])/(a*d^2) + (3*f^2*(e + f*x)*Cos[c + d*x]*Sin[c + d*x])/(4*a*d^3) - ((e + f*x)^3*Cos[c + d*x]*Sin[c
+ d*x])/(2*a*d) - (3*f^3*Sin[c + d*x]^2)/(8*a*d^4) + (3*f*(e + f*x)^2*Sin[c + d*x]^2)/(4*a*d^2)

Rule 4515

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[1/b, Int[(e + f*x)^m*Sin[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sin[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 3311

Int[((c_.) + (d_.)*(x_))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*m*(c + d*x)^(m - 1)*(
b*Sin[e + f*x])^n)/(f^2*n^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)^m*(b*Sin[e + f*x])^(n - 2), x], x] - D
ist[(d^2*m*(m - 1))/(f^2*n^2), Int[(c + d*x)^(m - 2)*(b*Sin[e + f*x])^n, x], x] - Simp[(b*(c + d*x)^m*Cos[e +
f*x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 1]

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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]

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 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 \frac{(e+f x)^3 \sin ^3(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x)^3 \sin ^2(c+d x) \, dx}{a}-\int \frac{(e+f x)^3 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\\ &=-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}+\frac{\int (e+f x)^3 \, dx}{2 a}-\frac{\int (e+f x)^3 \sin (c+d x) \, dx}{a}-\frac{\left (3 f^2\right ) \int (e+f x) \sin ^2(c+d x) \, dx}{2 a d^2}+\int \frac{(e+f x)^3 \sin (c+d x)}{a+a \sin (c+d x)} \, dx\\ &=\frac{(e+f x)^4}{8 a f}+\frac{(e+f x)^3 \cos (c+d x)}{a d}+\frac{3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}+\frac{\int (e+f x)^3 \, dx}{a}-\frac{(3 f) \int (e+f x)^2 \cos (c+d x) \, dx}{a d}-\frac{\left (3 f^2\right ) \int (e+f x) \, dx}{4 a d^2}-\int \frac{(e+f x)^3}{a+a \sin (c+d x)} \, dx\\ &=-\frac{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{3 (e+f x)^4}{8 a f}+\frac{(e+f x)^3 \cos (c+d x)}{a d}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac{3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\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{\left (6 f^2\right ) \int (e+f x) \sin (c+d x) \, dx}{a d^2}\\ &=-\frac{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{3 (e+f x)^4}{8 a f}-\frac{6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac{(e+f x)^3 \cos (c+d x)}{a d}+\frac{(e+f x)^3 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac{3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 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 f^3\right ) \int \cos (c+d x) \, dx}{a d^3}\\ &=-\frac{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{i (e+f x)^3}{a d}+\frac{3 (e+f x)^4}{8 a f}-\frac{6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac{(e+f x)^3 \cos (c+d x)}{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^3 \sin (c+d x)}{a d^4}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac{3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\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{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{i (e+f x)^3}{a d}+\frac{3 (e+f x)^4}{8 a f}-\frac{6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac{(e+f x)^3 \cos (c+d x)}{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{6 f^3 \sin (c+d x)}{a d^4}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac{3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}+\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{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{i (e+f x)^3}{a d}+\frac{3 (e+f x)^4}{8 a f}-\frac{6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac{(e+f x)^3 \cos (c+d x)}{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{12 i f^2 (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^3 \sin (c+d x)}{a d^4}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac{3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\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{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{i (e+f x)^3}{a d}+\frac{3 (e+f x)^4}{8 a f}-\frac{6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac{(e+f x)^3 \cos (c+d x)}{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{12 i f^2 (e+f x) \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}+\frac{6 f^3 \sin (c+d x)}{a d^4}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac{3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}-\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{3 e f^2 x}{4 a d^2}-\frac{3 f^3 x^2}{8 a d^2}+\frac{i (e+f x)^3}{a d}+\frac{3 (e+f x)^4}{8 a f}-\frac{6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac{(e+f x)^3 \cos (c+d x)}{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{12 i f^2 (e+f x) \text{Li}_2\left (i 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^3 \sin (c+d x)}{a d^4}-\frac{3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac{3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 a d^3}-\frac{(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 a d}-\frac{3 f^3 \sin ^2(c+d x)}{8 a d^4}+\frac{3 f (e+f x)^2 \sin ^2(c+d x)}{4 a d^2}\\ \end{align*}

Mathematica [A]  time = 2.62818, size = 538, normalized size = 1.41 \[ \frac{\frac{192 f (\cos (c)+i \sin (c)) \left (\frac{2 f (\cos (c)-i (\sin (c)+1)) (d (e+f x) \text{PolyLog}(2,-\sin (c+d x)-i \cos (c+d x))-i f \text{PolyLog}(3,-\sin (c+d x)-i \cos (c+d x)))}{d^3}-\frac{(\sin (c)+i \cos (c)+1) (e+f x)^2 \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac{(\cos (c)-i \sin (c)) (e+f x)^3}{3 f}\right )}{d (\cos (c)+i (\sin (c)+1))}+\frac{16 \left (-3 i d^2 f (e+f x)^2+d^3 (e+f x)^3-6 d f^2 (e+f x)+6 i f^3\right ) (\cos (c+d x)-i \sin (c+d x))}{d^4}+\frac{16 \left (3 i d^2 f (e+f x)^2+d^3 (e+f x)^3-6 d f^2 (e+f x)-6 i f^3\right ) (\cos (c+d x)+i \sin (c+d x))}{d^4}+\frac{\left (-6 d^2 f (e+f x)^2-4 i d^3 (e+f x)^3+6 i d f^2 (e+f x)+3 f^3\right ) (\cos (2 (c+d x))-i \sin (2 (c+d x)))}{d^4}+\frac{\left (-6 d^2 f (e+f x)^2+4 i d^3 (e+f x)^3-6 i d f^2 (e+f x)+3 f^3\right ) (\cos (2 (c+d x))+i \sin (2 (c+d x)))}{d^4}-\frac{64 \sin \left (\frac{d x}{2}\right ) (e+f x)^3}{d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}+72 e^2 f x^2+48 e^3 x+48 e f^2 x^3+12 f^3 x^4}{32 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(48*e^3*x + 72*e^2*f*x^2 + 48*e*f^2*x^3 + 12*f^3*x^4 + (192*f*(Cos[c] + I*Sin[c])*(((e + f*x)^3*(Cos[c] - I*Si
n[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*PolyLog[3, (-I)*Cos[c + d*x] - Sin[c + d*x]])*(Cos[c] -
I*(1 + Sin[c])))/d^3))/(d*(Cos[c] + I*(1 + Sin[c]))) - (64*(e + f*x)^3*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])
*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (16*((6*I)*f^3 - 6*d*f^2*(e + f*x) - (3*I)*d^2*f*(e + f*x)^2 + d^3*(
e + f*x)^3)*(Cos[c + d*x] - I*Sin[c + d*x]))/d^4 + (16*((-6*I)*f^3 - 6*d*f^2*(e + f*x) + (3*I)*d^2*f*(e + f*x)
^2 + d^3*(e + f*x)^3)*(Cos[c + d*x] + I*Sin[c + d*x]))/d^4 + ((3*f^3 + (6*I)*d*f^2*(e + f*x) - 6*d^2*f*(e + f*
x)^2 - (4*I)*d^3*(e + f*x)^3)*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)]))/d^4 + ((3*f^3 - (6*I)*d*f^2*(e + f*x) -
6*d^2*f*(e + f*x)^2 + (4*I)*d^3*(e + f*x)^3)*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]))/d^4)/(32*a)

________________________________________________________________________________________

Maple [B]  time = 0.18, size = 974, normalized size = 2.6 \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*sin(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

-12*f^2/d^2/a*e*ln(1-I*exp(I*(d*x+c)))*x+3/8/a*f^3*x^4+3/2/a*e^3*x+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+3/2/a*e*f^2*x^3+9/4/a*e^2*f*x^2+1/32*I*(4*f^3*x^3*d^3+6*I*d^2*f^3*x^2+12*d^3*e*f^2*x^2+12
*I*d^2*e*f^2*x+12*d^3*e^2*f*x+6*I*d^2*e^2*f+4*d^3*e^3-6*d*f^3*x-3*I*f^3-6*f^2*e*d)/a/d^4*exp(2*I*(d*x+c))-1/32
*I*(4*f^3*x^3*d^3-6*I*d^2*f^3*x^2+12*d^3*e*f^2*x^2-12*I*d^2*e*f^2*x+12*d^3*e^2*f*x-6*I*d^2*e^2*f+4*d^3*e^3-6*d
*f^3*x+3*I*f^3-6*f^2*e*d)/a/d^4*exp(-2*I*(d*x+c))+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)
+12*I*f^2/d^2/a*e*c*x-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+1/2*(f^3*x^3*d^3-3*I*d^2*f^3*x^2+3*d^3*e*f^2*x^2-6*I*d^2*e*f^2*x+3*d^3*e^2*f*x
-3*I*d^2*e^2*f+d^3*e^3-6*d*f^3*x+6*I*f^3-6*f^2*e*d)/a/d^4*exp(-I*(d*x+c))+1/2*(f^3*x^3*d^3+3*I*d^2*f^3*x^2+3*d
^3*e*f^2*x^2+6*I*d^2*e*f^2*x+3*d^3*e^2*f*x+3*I*d^2*e^2*f+d^3*e^3-6*d*f^3*x-6*I*f^3-6*f^2*e*d)/a/d^4*exp(I*(d*x
+c))-12*f^3*polylog(3,I*exp(I*(d*x+c)))/a/d^4

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

________________________________________________________________________________________

Fricas [C]  time = 2.78325, size = 3534, normalized size = 9.25 \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*sin(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(6*d^4*f^3*x^4 + 16*d^3*e^3 - 42*d^2*e^2*f + 8*(3*d^4*e*f^2 + 2*d^3*f^3)*x^3 + 2*(4*d^3*f^3*x^3 + 4*d^3*e
^3 - 6*d^2*e^2*f - 6*d*e*f^2 + 3*f^3 + 6*(2*d^3*e*f^2 - d^2*f^3)*x^2 + 6*(2*d^3*e^2*f - 2*d^2*e*f^2 - d*f^3)*x
)*cos(d*x + c)^3 + 93*f^3 + 6*(6*d^4*e^2*f + 8*d^3*e*f^2 - 7*d^2*f^3)*x^2 + 2*(8*d^3*f^3*x^3 + 8*d^3*e^3 + 18*
d^2*e^2*f - 48*d*e*f^2 - 45*f^3 + 6*(4*d^3*e*f^2 + 3*d^2*f^3)*x^2 + 12*(2*d^3*e^2*f + 3*d^2*e*f^2 - 4*d*f^3)*x
)*cos(d*x + c)^2 + 12*(2*d^4*e^3 + 4*d^3*e^2*f - 7*d^2*e*f^2)*x + 3*(2*d^4*f^3*x^4 + 8*d^3*e^3 + 2*d^2*e^2*f -
28*d*e*f^2 + 8*(d^4*e*f^2 + d^3*f^3)*x^3 - f^3 + 2*(6*d^4*e^2*f + 12*d^3*e*f^2 + d^2*f^3)*x^2 + 4*(2*d^4*e^3
+ 6*d^3*e^2*f + d^2*e*f^2 - 7*d*f^3)*x)*cos(d*x + c) + (96*I*d*f^3*x + 96*I*d*e*f^2 + (96*I*d*f^3*x + 96*I*d*e
*f^2)*cos(d*x + c) + (96*I*d*f^3*x + 96*I*d*e*f^2)*sin(d*x + c))*dilog(I*cos(d*x + c) - sin(d*x + c)) + (-96*I
*d*f^3*x - 96*I*d*e*f^2 + (-96*I*d*f^3*x - 96*I*d*e*f^2)*cos(d*x + c) + (-96*I*d*f^3*x - 96*I*d*e*f^2)*sin(d*x
+ c))*dilog(-I*cos(d*x + c) - sin(d*x + c)) - 48*(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) - 48*(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) - 48*(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) - 48*(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) - 96*(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)) - 96*(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^4*f^3*x^4 - 16*d^3*e^3 - 42
*d^2*e^2*f + 8*(3*d^4*e*f^2 - 2*d^3*f^3)*x^3 + 93*f^3 + 6*(6*d^4*e^2*f - 8*d^3*e*f^2 - 7*d^2*f^3)*x^2 - 2*(4*d
^3*f^3*x^3 + 4*d^3*e^3 + 6*d^2*e^2*f - 6*d*e*f^2 - 3*f^3 + 6*(2*d^3*e*f^2 + d^2*f^3)*x^2 + 6*(2*d^3*e^2*f + 2*
d^2*e*f^2 - d*f^3)*x)*cos(d*x + c)^2 + 12*(2*d^4*e^3 - 4*d^3*e^2*f - 7*d^2*e*f^2)*x + 4*(2*d^3*f^3*x^3 + 2*d^3
*e^3 - 12*d^2*e^2*f - 21*d*e*f^2 + 24*f^3 + 6*(d^3*e*f^2 - 2*d^2*f^3)*x^2 + 3*(2*d^3*e^2*f - 8*d^2*e*f^2 - 7*d
*f^3)*x)*cos(d*x + c))*sin(d*x + c))/(a*d^4*cos(d*x + c) + a*d^4*sin(d*x + c) + a*d^4)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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