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

Optimal. Leaf size=737 \[ -\frac{6 f^2 \left (a^2-b^2\right ) (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^3}-\frac{6 f^2 \left (a^2-b^2\right ) (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^3}+\frac{3 i f \left (a^2-b^2\right ) (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{3 i f \left (a^2-b^2\right ) (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^2}-\frac{6 i f^3 \left (a^2-b^2\right ) \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^4}-\frac{6 i f^3 \left (a^2-b^2\right ) \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^4}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d}+\frac{i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac{6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac{3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac{6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac{a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac{3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac{3 f (e+f x)^2 \sin (c+d x) \cos (c+d x)}{4 b d^2}+\frac{3 f^3 \sin (c+d x) \cos (c+d x)}{8 b d^4}-\frac{(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac{3 f^3 x}{8 b d^3}+\frac{(e+f x)^3}{4 b d} \]

[Out]

(-3*f^3*x)/(8*b*d^3) + (e + f*x)^3/(4*b*d) + ((I/4)*(a^2 - b^2)*(e + f*x)^4)/(b^3*f) - (6*a*f^3*Cos[c + d*x])/
(b^2*d^4) + (3*a*f*(e + f*x)^2*Cos[c + d*x])/(b^2*d^2) - ((a^2 - b^2)*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x))
)/(a - Sqrt[a^2 - b^2])])/(b^3*d) - ((a^2 - b^2)*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2
])])/(b^3*d) + ((3*I)*(a^2 - b^2)*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*
d^2) + ((3*I)*(a^2 - b^2)*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^2) - (
6*(a^2 - b^2)*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d^3) - (6*(a^2 - b^2
)*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^3) - ((6*I)*(a^2 - b^2)*f^3*Po
lyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d^4) - ((6*I)*(a^2 - b^2)*f^3*PolyLog[4, (I*b*E^(I
*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^4) - (6*a*f^2*(e + f*x)*Sin[c + d*x])/(b^2*d^3) + (a*(e + f*x)^3*S
in[c + d*x])/(b^2*d) + (3*f^3*Cos[c + d*x]*Sin[c + d*x])/(8*b*d^4) - (3*f*(e + f*x)^2*Cos[c + d*x]*Sin[c + d*x
])/(4*b*d^2) + (3*f^2*(e + f*x)*Sin[c + d*x]^2)/(4*b*d^3) - ((e + f*x)^3*Sin[c + d*x]^2)/(2*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.880909, antiderivative size = 737, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 14, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {4525, 3296, 2638, 4404, 3311, 32, 2635, 8, 4519, 2190, 2531, 6609, 2282, 6589} \[ -\frac{6 f^2 \left (a^2-b^2\right ) (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^3}-\frac{6 f^2 \left (a^2-b^2\right ) (e+f x) \text{PolyLog}\left (3,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^3}+\frac{3 i f \left (a^2-b^2\right ) (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{3 i f \left (a^2-b^2\right ) (e+f x)^2 \text{PolyLog}\left (2,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^2}-\frac{6 i f^3 \left (a^2-b^2\right ) \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^4}-\frac{6 i f^3 \left (a^2-b^2\right ) \text{PolyLog}\left (4,\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d^4}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{\sqrt{a^2-b^2}+a}\right )}{b^3 d}+\frac{i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac{6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac{3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac{6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac{a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac{3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac{3 f (e+f x)^2 \sin (c+d x) \cos (c+d x)}{4 b d^2}+\frac{3 f^3 \sin (c+d x) \cos (c+d x)}{8 b d^4}-\frac{(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac{3 f^3 x}{8 b d^3}+\frac{(e+f x)^3}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*f^3*x)/(8*b*d^3) + (e + f*x)^3/(4*b*d) + ((I/4)*(a^2 - b^2)*(e + f*x)^4)/(b^3*f) - (6*a*f^3*Cos[c + d*x])/
(b^2*d^4) + (3*a*f*(e + f*x)^2*Cos[c + d*x])/(b^2*d^2) - ((a^2 - b^2)*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x))
)/(a - Sqrt[a^2 - b^2])])/(b^3*d) - ((a^2 - b^2)*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2
])])/(b^3*d) + ((3*I)*(a^2 - b^2)*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*
d^2) + ((3*I)*(a^2 - b^2)*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^2) - (
6*(a^2 - b^2)*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d^3) - (6*(a^2 - b^2
)*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^3) - ((6*I)*(a^2 - b^2)*f^3*Po
lyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d^4) - ((6*I)*(a^2 - b^2)*f^3*PolyLog[4, (I*b*E^(I
*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^4) - (6*a*f^2*(e + f*x)*Sin[c + d*x])/(b^2*d^3) + (a*(e + f*x)^3*S
in[c + d*x])/(b^2*d) + (3*f^3*Cos[c + d*x]*Sin[c + d*x])/(8*b*d^4) - (3*f*(e + f*x)^2*Cos[c + d*x]*Sin[c + d*x
])/(4*b*d^2) + (3*f^2*(e + f*x)*Sin[c + d*x]^2)/(4*b*d^3) - ((e + f*x)^3*Sin[c + d*x]^2)/(2*b*d)

Rule 4525

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

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 2638

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

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

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 2635

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

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 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 \frac{(e+f x)^3 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac{a \int (e+f x)^3 \cos (c+d x) \, dx}{b^2}-\frac{\int (e+f x)^3 \cos (c+d x) \sin (c+d x) \, dx}{b}-\frac{\left (a^2-b^2\right ) \int \frac{(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=\frac{i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}+\frac{a (e+f x)^3 \sin (c+d x)}{b^2 d}-\frac{(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac{\left (a^2-b^2\right ) \int \frac{e^{i (c+d x)} (e+f x)^3}{a-\sqrt{a^2-b^2}-i b e^{i (c+d x)}} \, dx}{b^2}-\frac{\left (a^2-b^2\right ) \int \frac{e^{i (c+d x)} (e+f x)^3}{a+\sqrt{a^2-b^2}-i b e^{i (c+d x)}} \, dx}{b^2}-\frac{(3 a f) \int (e+f x)^2 \sin (c+d x) \, dx}{b^2 d}+\frac{(3 f) \int (e+f x)^2 \sin ^2(c+d x) \, dx}{2 b d}\\ &=\frac{i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}+\frac{3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d}+\frac{a (e+f x)^3 \sin (c+d x)}{b^2 d}-\frac{3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac{3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac{(e+f x)^3 \sin ^2(c+d x)}{2 b d}+\frac{(3 f) \int (e+f x)^2 \, dx}{4 b d}+\frac{\left (3 \left (a^2-b^2\right ) f\right ) \int (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b^3 d}+\frac{\left (3 \left (a^2-b^2\right ) f\right ) \int (e+f x)^2 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b^3 d}-\frac{\left (6 a f^2\right ) \int (e+f x) \cos (c+d x) \, dx}{b^2 d^2}-\frac{\left (3 f^3\right ) \int \sin ^2(c+d x) \, dx}{4 b d^3}\\ &=\frac{(e+f x)^3}{4 b d}+\frac{i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}+\frac{3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d}+\frac{3 i \left (a^2-b^2\right ) f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{3 i \left (a^2-b^2\right ) f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^2}-\frac{6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac{a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac{3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac{3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac{3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac{(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac{\left (6 i \left (a^2-b^2\right ) f^2\right ) \int (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b^3 d^2}-\frac{\left (6 i \left (a^2-b^2\right ) f^2\right ) \int (e+f x) \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b^3 d^2}+\frac{\left (6 a f^3\right ) \int \sin (c+d x) \, dx}{b^2 d^3}-\frac{\left (3 f^3\right ) \int 1 \, dx}{8 b d^3}\\ &=-\frac{3 f^3 x}{8 b d^3}+\frac{(e+f x)^3}{4 b d}+\frac{i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac{6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac{3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d}+\frac{3 i \left (a^2-b^2\right ) f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{3 i \left (a^2-b^2\right ) f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^2}-\frac{6 \left (a^2-b^2\right ) f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^3}-\frac{6 \left (a^2-b^2\right ) f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^3}-\frac{6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac{a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac{3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac{3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac{3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac{(e+f x)^3 \sin ^2(c+d x)}{2 b d}+\frac{\left (6 \left (a^2-b^2\right ) f^3\right ) \int \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right ) \, dx}{b^3 d^3}+\frac{\left (6 \left (a^2-b^2\right ) f^3\right ) \int \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right ) \, dx}{b^3 d^3}\\ &=-\frac{3 f^3 x}{8 b d^3}+\frac{(e+f x)^3}{4 b d}+\frac{i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac{6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac{3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d}+\frac{3 i \left (a^2-b^2\right ) f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{3 i \left (a^2-b^2\right ) f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^2}-\frac{6 \left (a^2-b^2\right ) f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^3}-\frac{6 \left (a^2-b^2\right ) f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^3}-\frac{6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac{a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac{3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac{3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac{3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac{(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac{\left (6 i \left (a^2-b^2\right ) 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 )}{b^3 d^4}-\frac{\left (6 i \left (a^2-b^2\right ) 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 )}{b^3 d^4}\\ &=-\frac{3 f^3 x}{8 b d^3}+\frac{(e+f x)^3}{4 b d}+\frac{i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac{6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac{3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d}-\frac{\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d}+\frac{3 i \left (a^2-b^2\right ) f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^2}+\frac{3 i \left (a^2-b^2\right ) f (e+f x)^2 \text{Li}_2\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^2}-\frac{6 \left (a^2-b^2\right ) f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^3}-\frac{6 \left (a^2-b^2\right ) f^2 (e+f x) \text{Li}_3\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^3}-\frac{6 i \left (a^2-b^2\right ) f^3 \text{Li}_4\left (\frac{i b e^{i (c+d x)}}{a-\sqrt{a^2-b^2}}\right )}{b^3 d^4}-\frac{6 i \left (a^2-b^2\right ) f^3 \text{Li}_4\left (\frac{i b e^{i (c+d x)}}{a+\sqrt{a^2-b^2}}\right )}{b^3 d^4}-\frac{6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac{a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac{3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac{3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac{3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac{(e+f x)^3 \sin ^2(c+d x)}{2 b d}\\ \end{align*}

Mathematica [B]  time = 10.1211, size = 2452, normalized size = 3.33 \[ \text{Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*(a^2 - b^2)*e^3*x*Cot[c] - 48*(a^2 - b^2)*e^2*f*x^2*Cot[c] - 32*(a^2 - b^2)*e*f^2*x^3*Cot[c] - 8*(a^2 - b
^2)*f^3*x^4*Cot[c] + (16*(a^2 - b^2)*((4*I)*d^4*e^3*E^((2*I)*c)*x + (6*I)*d^4*e^2*E^((2*I)*c)*f*x^2 + (4*I)*d^
4*e*E^((2*I)*c)*f^2*x^3 + I*d^4*E^((2*I)*c)*f^3*x^4 + (2*I)*d^3*e^3*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((
2*I)*(c + d*x))))] - (2*I)*d^3*e^3*E^((2*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] +
d^3*e^3*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] - d^3*e^3*E^((2*I)*c)*Log[4*a^2*E^((
2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] + 6*d^3*e^2*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c)
- Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e^2*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - S
qrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^
2)*E^((2*I)*c)])] - 6*d^3*e*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)
*E^((2*I)*c)])] + 2*d^3*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])]
- 2*d^3*E^((2*I)*c)*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*
d^3*e^2*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e^2*E^((2*I)
*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e*f^2*x^2*Log[1
+ (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e*E^((2*I)*c)*f^2*x^2*Log[1 +
(b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 2*d^3*f^3*x^3*Log[1 + (b*E^(I*(2*c + d
*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 2*d^3*E^((2*I)*c)*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x))
)/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + (6*I)*d^2*(-1 + E^((2*I)*c))*f*(e + f*x)^2*PolyLog[2, (I*b
*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + (6*I)*d^2*(-1 + E^((2*I)*c))*f*(e + f*x)
^2*PolyLog[2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 12*d*e*f^2*PolyLog[3,
(I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*d*e*E^((2*I)*c)*f^2*PolyLog[3, (
I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d*f^3*x*PolyLog[3, (I*b*E^(I*(2*c
+ d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*d*E^((2*I)*c)*f^3*x*PolyLog[3, (I*b*E^(I*(2*c +
d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d*e*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^
(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - 12*d*e*E^((2*I)*c)*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(
I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 12*d*f^3*x*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(
-a^2 + b^2)*E^((2*I)*c)]))] - 12*d*E^((2*I)*c)*f^3*x*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-
a^2 + b^2)*E^((2*I)*c)]))] + (12*I)*f^3*PolyLog[4, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^
((2*I)*c)])] - (12*I)*E^((2*I)*c)*f^3*PolyLog[4, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((
2*I)*c)])] + (12*I)*f^3*PolyLog[4, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] -
(12*I)*E^((2*I)*c)*f^3*PolyLog[4, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))]))/(
d^4*(-1 + E^((2*I)*c))) + (16*a*b*(-6*f^3 - (6*I)*d*f^2*(e + f*x) + 3*d^2*f*(e + f*x)^2 + I*d^3*(e + f*x)^3)*(
Cos[c + d*x] - I*Sin[c + d*x]))/d^4 + (16*a*b*(-6*f^3 + (6*I)*d*f^2*(e + f*x) + 3*d^2*f*(e + f*x)^2 - I*d^3*(e
+ f*x)^3)*(Cos[c + d*x] + I*Sin[c + d*x]))/d^4 + (b^2*((3*I)*f^3 - 6*d*f^2*(e + f*x) - (6*I)*d^2*f*(e + f*x)^
2 + 4*d^3*(e + f*x)^3)*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)]))/d^4 + (b^2*((-3*I)*f^3 - 6*d*f^2*(e + f*x) + (
6*I)*d^2*f*(e + f*x)^2 + 4*d^3*(e + f*x)^3)*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]))/d^4)/(32*b^3)

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Maple [F]  time = 1.418, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{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*cos(d*x+c)^3/(a+b*sin(d*x+c)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [C]  time = 4.34691, size = 6086, normalized size = 8.26 \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*cos(d*x+c)^3/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/8*(2*b^2*d^3*f^3*x^3 + 6*b^2*d^3*e*f^2*x^2 + 24*I*(a^2 - b^2)*f^3*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) + 24*I*(a^2 - b^2)*f^3*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) - 24*I*(a^2 - b^2)*f^3*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) - 24*I*(a^2 - b^2)*f^3*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) - 2*(2*b^2*d^3*f^3*x^3 + 6*b^2*d^3*e*f^2*x^2 + 2*b^2*d^3
*e^3 - 3*b^2*d*e*f^2 + 3*(2*b^2*d^3*e^2*f - b^2*d*f^3)*x)*cos(d*x + c)^2 + 3*(2*b^2*d^3*e^2*f - b^2*d*f^3)*x -
24*(a*b*d^2*f^3*x^2 + 2*a*b*d^2*e*f^2*x + a*b*d^2*e^2*f - 2*a*b*f^3)*cos(d*x + c) - (-12*I*(a^2 - b^2)*d^2*f^
3*x^2 - 24*I*(a^2 - b^2)*d^2*e*f^2*x - 12*I*(a^2 - b^2)*d^2*e^2*f)*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) - (-12*I*(a^2 - b^2)*d^2*f
^3*x^2 - 24*I*(a^2 - b^2)*d^2*e*f^2*x - 12*I*(a^2 - b^2)*d^2*e^2*f)*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) - (12*I*(a^2 - b^2)*d^2*f
^3*x^2 + 24*I*(a^2 - b^2)*d^2*e*f^2*x + 12*I*(a^2 - b^2)*d^2*e^2*f)*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) - (12*I*(a^2 - b^2)*d^2*
f^3*x^2 + 24*I*(a^2 - b^2)*d^2*e*f^2*x + 12*I*(a^2 - b^2)*d^2*e^2*f)*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) + 4*((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(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) + 4*((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(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) + 4*((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(-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) + 4*((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(-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) + 4*((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(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) + 4*((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(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) + 4*((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(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) + 4*((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(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) + 24*((a^
2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e*f^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) + 24*((a^2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e*f^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)
+ 24*((a^2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e*f^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) + 24*((a^2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e*f^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) - (8*a*
b*d^3*f^3*x^3 + 24*a*b*d^3*e*f^2*x^2 + 8*a*b*d^3*e^3 - 48*a*b*d*e*f^2 + 24*(a*b*d^3*e^2*f - 2*a*b*d*f^3)*x - 3
*(2*b^2*d^2*f^3*x^2 + 4*b^2*d^2*e*f^2*x + 2*b^2*d^2*e^2*f - b^2*f^3)*cos(d*x + c))*sin(d*x + c))/(b^3*d^4)

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Sympy [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*cos(d*x+c)**3/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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

[Out]

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