\(\int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [333]

Optimal result

Integrand size = 34, antiderivative size = 1138 \[ \int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {3 e f^2 x}{4 b d^2}+\frac {3 f^3 x^2}{8 b d^2}-\frac {(e+f x)^4}{8 b f}+\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {2 (e+f x)^3 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}-\frac {6 \left (a^2-b^2\right ) f^2 (e+f x) \cos (c+d x)}{a b^2 d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}+\frac {\left (a^2-b^2\right ) (e+f x)^3 \cos (c+d x)}{a b^2 d}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {6 i f^3 \operatorname {PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac {6 i f^3 \operatorname {PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}-\frac {6 \left (a^2-b^2\right )^{3/2} f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^4}+\frac {6 \left (a^2-b^2\right )^{3/2} f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^4}+\frac {6 f^3 \sin (c+d x)}{a d^4}+\frac {6 \left (a^2-b^2\right ) f^3 \sin (c+d x)}{a b^2 d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}-\frac {3 \left (a^2-b^2\right ) f (e+f x)^2 \sin (c+d x)}{a b^2 d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d} \]

[Out]

-3/4*f*(f*x+e)^2*cos(d*x+c)^2/b/d^2+3/8*f^3*cos(d*x+c)^2/b/d^4-6*(a^2-b^2)*f^2*(f*x+e)*cos(d*x+c)/a/b^2/d^3-3*
(a^2-b^2)*f*(f*x+e)^2*sin(d*x+c)/a/b^2/d^2+3*(a^2-b^2)^(3/2)*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-
b^2)^(1/2)))/a/b^3/d^2-3*(a^2-b^2)^(3/2)*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a/b^3/d
^2-I*(a^2-b^2)^(3/2)*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a/b^3/d-6*f^2*(f*x+e)*cos(d*x+c)/a
/d^3-3*f*(f*x+e)^2*sin(d*x+c)/a/d^2-6*(a^2-b^2)^(3/2)*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/
b^3/d^4+6*(a^2-b^2)^(3/2)*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a/b^3/d^4+6*(a^2-b^2)*f^3*sin(
d*x+c)/a/b^2/d^4+1/4*(a^2-b^2)*(f*x+e)^4/b^3/f-3*I*f*(f*x+e)^2*polylog(2,exp(I*(d*x+c)))/a/d^2+6*I*(a^2-b^2)^(
3/2)*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/b^3/d^3+I*(a^2-b^2)^(3/2)*(f*x+e)^3*ln(1-
I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/b^3/d+3/4*f^2*(f*x+e)*cos(d*x+c)*sin(d*x+c)/b/d^3+(f*x+e)^3*cos(d*x+
c)/a/d+(a^2-b^2)*(f*x+e)^3*cos(d*x+c)/a/b^2/d-6*I*(a^2-b^2)^(3/2)*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a+
(a^2-b^2)^(1/2)))/a/b^3/d^3+6*f^3*sin(d*x+c)/a/d^4-2*(f*x+e)^3*arctanh(exp(I*(d*x+c)))/a/d+3/8*f^3*x^2/b/d^2-6
*f^2*(f*x+e)*polylog(3,-exp(I*(d*x+c)))/a/d^3+6*f^2*(f*x+e)*polylog(3,exp(I*(d*x+c)))/a/d^3-6*I*f^3*polylog(4,
-exp(I*(d*x+c)))/a/d^4+3*I*f*(f*x+e)^2*polylog(2,-exp(I*(d*x+c)))/a/d^2-1/8*(f*x+e)^4/b/f+6*I*f^3*polylog(4,ex
p(I*(d*x+c)))/a/d^4+3/4*e*f^2*x/b/d^2-1/2*(f*x+e)^3*cos(d*x+c)*sin(d*x+c)/b/d

Rubi [A] (verified)

Time = 1.35 (sec) , antiderivative size = 1138, normalized size of antiderivative = 1.00, number of steps used = 53, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {4639, 4493, 4490, 3392, 3377, 2717, 2713, 4268, 2611, 6744, 2320, 6724, 4621, 32, 3391, 3404, 2296, 2221} \[ \int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {(e+f x)^4}{8 b f}-\frac {2 \text {arctanh}\left (e^{i (c+d x)}\right ) (e+f x)^3}{a d}+\frac {\left (a^2-b^2\right ) \cos (c+d x) (e+f x)^3}{a b^2 d}+\frac {\cos (c+d x) (e+f x)^3}{a d}+\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^3}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^3}{a b^3 d}-\frac {\cos (c+d x) \sin (c+d x) (e+f x)^3}{2 b d}-\frac {3 f \cos ^2(c+d x) (e+f x)^2}{4 b d^2}+\frac {3 i f \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right ) (e+f x)^2}{a d^2}-\frac {3 i f \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right ) (e+f x)^2}{a d^2}+\frac {3 \left (a^2-b^2\right )^{3/2} f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a b^3 d^2}-\frac {3 \left (a^2-b^2\right )^{3/2} f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a b^3 d^2}-\frac {3 \left (a^2-b^2\right ) f \sin (c+d x) (e+f x)^2}{a b^2 d^2}-\frac {3 f \sin (c+d x) (e+f x)^2}{a d^2}-\frac {6 \left (a^2-b^2\right ) f^2 \cos (c+d x) (e+f x)}{a b^2 d^3}-\frac {6 f^2 \cos (c+d x) (e+f x)}{a d^3}-\frac {6 f^2 \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right ) (e+f x)}{a d^3}+\frac {6 f^2 \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right ) (e+f x)}{a d^3}+\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)}{a b^3 d^3}-\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)}{a b^3 d^3}+\frac {3 f^2 \cos (c+d x) \sin (c+d x) (e+f x)}{4 b d^3}+\frac {3 f^3 x^2}{8 b d^2}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}+\frac {3 e f^2 x}{4 b d^2}-\frac {6 i f^3 \operatorname {PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac {6 i f^3 \operatorname {PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}-\frac {6 \left (a^2-b^2\right )^{3/2} f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^4}+\frac {6 \left (a^2-b^2\right )^{3/2} f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^4}+\frac {6 \left (a^2-b^2\right ) f^3 \sin (c+d x)}{a b^2 d^4}+\frac {6 f^3 \sin (c+d x)}{a d^4} \]

[In]

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

[Out]

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

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2717

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

Rule 3377

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 3391

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 3392

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 3404

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 4268

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 4490

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

Rule 4493

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*Cot[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Int[
(c + d*x)^m*Cos[a + b*x]^n*Cot[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cos[a + b*x]^(n - 2)*Cot[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rule 4621

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 4639

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

Rule 6724

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 6744

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int (e+f x)^3 \cos ^3(c+d x) \cot (c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x)^3 \cos ^4(c+d x)}{a+b \sin (c+d x)} \, dx}{a} \\ & = \frac {\int (e+f x)^3 \cos (c+d x) \cot (c+d x) \, dx}{a}-\frac {\int (e+f x)^3 \cos ^2(c+d x) \, dx}{b}+\left (\frac {a}{b}-\frac {b}{a}\right ) \int \frac {(e+f x)^3 \cos ^2(c+d x)}{a+b \sin (c+d x)} \, dx \\ & = -\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\int (e+f x)^3 \csc (c+d x) \, dx}{a}-\frac {\int (e+f x)^3 \sin (c+d x) \, dx}{a}+\left (\frac {1}{a}-\frac {a}{b^2}\right ) \int (e+f x)^3 \sin (c+d x) \, dx-\frac {\int (e+f x)^3 \, dx}{2 b}+\frac {\left (a^2-b^2\right ) \int (e+f x)^3 \, dx}{b^3}-\frac {\left (a^2-b^2\right )^2 \int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{a b^3}+\frac {\left (3 f^2\right ) \int (e+f x) \cos ^2(c+d x) \, dx}{2 b d^2} \\ & = -\frac {(e+f x)^4}{8 b f}+\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {2 (e+f x)^3 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x)^3 \cos (c+d x)}{d}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (2 \left (a^2-b^2\right )^2\right ) \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 b^3}-\frac {(3 f) \int (e+f x)^2 \cos (c+d x) \, dx}{a d}-\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 {\left (3 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f\right ) \int (e+f x)^2 \cos (c+d x) \, dx}{d}+\frac {\left (3 f^2\right ) \int (e+f x) \, dx}{4 b d^2} \\ & = \frac {3 e f^2 x}{4 b d^2}+\frac {3 f^3 x^2}{8 b d^2}-\frac {(e+f x)^4}{8 b f}+\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {2 (e+f x)^3 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x)^3 \cos (c+d x)}{d}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f (e+f x)^2 \sin (c+d x)}{d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {\left (2 i \left (a^2-b^2\right )^{3/2}\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 b^2}-\frac {\left (2 i \left (a^2-b^2\right )^{3/2}\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 b^2}-\frac {\left (6 i f^2\right ) \int (e+f x) \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right ) \, dx}{a d^2}+\frac {\left (6 i f^2\right ) \int (e+f x) \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right ) \, dx}{a d^2}+\frac {\left (6 f^2\right ) \int (e+f x) \sin (c+d x) \, dx}{a d^2}-\frac {\left (6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^2\right ) \int (e+f x) \sin (c+d x) \, dx}{d^2} \\ & = \frac {3 e f^2 x}{4 b d^2}+\frac {3 f^3 x^2}{8 b d^2}-\frac {(e+f x)^4}{8 b f}+\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {2 (e+f x)^3 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^2 (e+f x) \cos (c+d x)}{d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x)^3 \cos (c+d x)}{d}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f (e+f x)^2 \sin (c+d x)}{d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (3 i \left (a^2-b^2\right )^{3/2} f\right ) \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 b^3 d}+\frac {\left (3 i \left (a^2-b^2\right )^{3/2} f\right ) \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 b^3 d}+\frac {\left (6 f^3\right ) \int \cos (c+d x) \, dx}{a d^3}+\frac {\left (6 f^3\right ) \int \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right ) \, dx}{a d^3}-\frac {\left (6 f^3\right ) \int \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right ) \, dx}{a d^3}-\frac {\left (6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^3\right ) \int \cos (c+d x) \, dx}{d^3} \\ & = \frac {3 e f^2 x}{4 b d^2}+\frac {3 f^3 x^2}{8 b d^2}-\frac {(e+f x)^4}{8 b f}+\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {2 (e+f x)^3 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^2 (e+f x) \cos (c+d x)}{d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x)^3 \cos (c+d x)}{d}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^3 \sin (c+d x)}{d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f (e+f x)^2 \sin (c+d x)}{d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (6 \left (a^2-b^2\right )^{3/2} f^2\right ) \int (e+f x) \operatorname {PolyLog}\left (2,\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{a b^3 d^2}+\frac {\left (6 \left (a^2-b^2\right )^{3/2} f^2\right ) \int (e+f x) \operatorname {PolyLog}\left (2,\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{a b^3 d^2}-\frac {\left (6 i f^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,-x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4}+\frac {\left (6 i f^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}(3,x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a d^4} \\ & = \frac {3 e f^2 x}{4 b d^2}+\frac {3 f^3 x^2}{8 b d^2}-\frac {(e+f x)^4}{8 b f}+\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {2 (e+f x)^3 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^2 (e+f x) \cos (c+d x)}{d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x)^3 \cos (c+d x)}{d}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {6 i f^3 \operatorname {PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac {6 i f^3 \operatorname {PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^3 \sin (c+d x)}{d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f (e+f x)^2 \sin (c+d x)}{d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (6 i \left (a^2-b^2\right )^{3/2} f^3\right ) \int \operatorname {PolyLog}\left (3,\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{a b^3 d^3}+\frac {\left (6 i \left (a^2-b^2\right )^{3/2} f^3\right ) \int \operatorname {PolyLog}\left (3,\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{a b^3 d^3} \\ & = \frac {3 e f^2 x}{4 b d^2}+\frac {3 f^3 x^2}{8 b d^2}-\frac {(e+f x)^4}{8 b f}+\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {2 (e+f x)^3 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^2 (e+f x) \cos (c+d x)}{d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x)^3 \cos (c+d x)}{d}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {6 i f^3 \operatorname {PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac {6 i f^3 \operatorname {PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^3 \sin (c+d x)}{d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f (e+f x)^2 \sin (c+d x)}{d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d}-\frac {\left (6 \left (a^2-b^2\right )^{3/2} f^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a b^3 d^4}+\frac {\left (6 \left (a^2-b^2\right )^{3/2} f^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (3,\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a b^3 d^4} \\ & = \frac {3 e f^2 x}{4 b d^2}+\frac {3 f^3 x^2}{8 b d^2}-\frac {(e+f x)^4}{8 b f}+\frac {\left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {2 (e+f x)^3 \text {arctanh}\left (e^{i (c+d x)}\right )}{a d}-\frac {6 f^2 (e+f x) \cos (c+d x)}{a d^3}+\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^2 (e+f x) \cos (c+d x)}{d^3}+\frac {(e+f x)^3 \cos (c+d x)}{a d}-\frac {\left (\frac {1}{a}-\frac {a}{b^2}\right ) (e+f x)^3 \cos (c+d x)}{d}+\frac {3 f^3 \cos ^2(c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos ^2(c+d x)}{4 b d^2}+\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {3 \left (a^2-b^2\right )^{3/2} f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^2}-\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {6 i \left (a^2-b^2\right )^{3/2} f^2 (e+f x) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {6 i f^3 \operatorname {PolyLog}\left (4,-e^{i (c+d x)}\right )}{a d^4}+\frac {6 i f^3 \operatorname {PolyLog}\left (4,e^{i (c+d x)}\right )}{a d^4}-\frac {6 \left (a^2-b^2\right )^{3/2} f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^4}+\frac {6 \left (a^2-b^2\right )^{3/2} f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^4}+\frac {6 f^3 \sin (c+d x)}{a d^4}-\frac {6 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f^3 \sin (c+d x)}{d^4}-\frac {3 f (e+f x)^2 \sin (c+d x)}{a d^2}+\frac {3 \left (\frac {1}{a}-\frac {a}{b^2}\right ) f (e+f x)^2 \sin (c+d x)}{d^2}+\frac {3 f^2 (e+f x) \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^3 \cos (c+d x) \sin (c+d x)}{2 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.31 (sec) , antiderivative size = 1181, normalized size of antiderivative = 1.04 \[ \int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {8 a \left (2 a^2-3 b^2\right ) d^4 e^3 x+12 a \left (2 a^2-3 b^2\right ) d^4 e^2 f x^2+8 a \left (2 a^2-3 b^2\right ) d^4 e f^2 x^3+2 a \left (2 a^2-3 b^2\right ) d^4 f^3 x^4-32 b^3 d^3 (e+f x)^3 \text {arctanh}(\cos (c+d x)+i \sin (c+d x))-96 a^2 b d f^2 (e+f x) \cos (c+d x)+16 a^2 b d^3 (e+f x)^3 \cos (c+d x)+3 a b^2 f^3 \cos (2 (c+d x))-6 a b^2 d^2 f (e+f x)^2 \cos (2 (c+d x))+48 \left (a^2-b^2\right )^{3/2} d^2 f (e+f x)^2 \operatorname {PolyLog}\left (2,-\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )+16 i \left (a^2-b^2\right )^{3/2} \left (2 i d^3 e^3 \arctan \left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+3 d^3 e^2 f x \log \left (1+\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )+3 d^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 f^3 x^3 \log \left (1+\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )-3 d^3 e^2 f x \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )-3 d^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 f^3 x^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+3 i d^2 f (e+f x)^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+6 d f^2 (e+f x) \operatorname {PolyLog}\left (3,-\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )-6 d e f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )-6 d f^3 x \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+6 i f^3 \operatorname {PolyLog}\left (4,-\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )-6 i f^3 \operatorname {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )\right )+48 i b^3 f \left (d^2 (e+f x)^2 \operatorname {PolyLog}(2,-\cos (c+d x)-i \sin (c+d x))+2 i d f (e+f x) \operatorname {PolyLog}(3,-\cos (c+d x)-i \sin (c+d x))-2 f^2 \operatorname {PolyLog}(4,-\cos (c+d x)-i \sin (c+d x))\right )-48 i b^3 f \left (d^2 (e+f x)^2 \operatorname {PolyLog}(2,\cos (c+d x)+i \sin (c+d x))+2 i d f (e+f x) \operatorname {PolyLog}(3,\cos (c+d x)+i \sin (c+d x))-2 f^2 \operatorname {PolyLog}(4,\cos (c+d x)+i \sin (c+d x))\right )+96 a^2 b f^3 \sin (c+d x)-48 a^2 b d^2 f (e+f x)^2 \sin (c+d x)+6 a b^2 d f^2 (e+f x) \sin (2 (c+d x))-4 a b^2 d^3 (e+f x)^3 \sin (2 (c+d x))}{16 a b^3 d^4} \]

[In]

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

[Out]

(8*a*(2*a^2 - 3*b^2)*d^4*e^3*x + 12*a*(2*a^2 - 3*b^2)*d^4*e^2*f*x^2 + 8*a*(2*a^2 - 3*b^2)*d^4*e*f^2*x^3 + 2*a*
(2*a^2 - 3*b^2)*d^4*f^3*x^4 - 32*b^3*d^3*(e + f*x)^3*ArcTanh[Cos[c + d*x] + I*Sin[c + d*x]] - 96*a^2*b*d*f^2*(
e + f*x)*Cos[c + d*x] + 16*a^2*b*d^3*(e + f*x)^3*Cos[c + d*x] + 3*a*b^2*f^3*Cos[2*(c + d*x)] - 6*a*b^2*d^2*f*(
e + f*x)^2*Cos[2*(c + d*x)] + 48*(a^2 - b^2)^(3/2)*d^2*f*(e + f*x)^2*PolyLog[2, ((-I)*b*E^(I*(c + d*x)))/(-a +
Sqrt[a^2 - b^2])] + (16*I)*(a^2 - b^2)^(3/2)*((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*PolyLog[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])]) + (48*I)*b^3*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]]) - (48*I)*b^3*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]]) + 96*a^2*b*f^3*Sin[c + d*x] - 48*a^2*b*d^2*f*(e + f*x)^2*Sin[c + d*x] + 6*a*b^2*d*
f^2*(e + f*x)*Sin[2*(c + d*x)] - 4*a*b^2*d^3*(e + f*x)^3*Sin[2*(c + d*x)])/(16*a*b^3*d^4)

Maple [F]

\[\int \frac {\left (f x +e \right )^{3} \left (\cos ^{3}\left (d x +c \right )\right ) \cot \left (d x +c \right )}{a +b \sin \left (d x +c \right )}d x\]

[In]

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

[Out]

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

Fricas [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   Too many variables

Sympy [F]

\[ \int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\left (e + f x\right )^{3} \cos ^{3}{\left (c + d x \right )} \cot {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}