3.337 \(\int \frac {(e+f x)^3 \cos (c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=852 \[ \frac {i b (e+f x)^4}{4 a^2 f}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 a^2 b f}-\frac {\csc (c+d x) (e+f x)^3}{a d}-\frac {\left (a^2-b^2\right ) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^3}{a^2 b d}-\frac {\left (a^2-b^2\right ) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^3}{a^2 b d}-\frac {b \log \left (1-e^{2 i (c+d x)}\right ) (e+f x)^3}{a^2 d}-\frac {6 f \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)^2}{a d^2}+\frac {3 i \left (a^2-b^2\right ) f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a^2 b d^2}+\frac {3 i \left (a^2-b^2\right ) f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a^2 b d^2}+\frac {3 i b f \text {Li}_2\left (e^{2 i (c+d x)}\right ) (e+f x)^2}{2 a^2 d^2}+\frac {6 i f^2 \text {Li}_2\left (-e^{i (c+d x)}\right ) (e+f x)}{a d^3}-\frac {6 i f^2 \text {Li}_2\left (e^{i (c+d x)}\right ) (e+f x)}{a d^3}-\frac {6 \left (a^2-b^2\right ) f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)}{a^2 b d^3}-\frac {6 \left (a^2-b^2\right ) f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)}{a^2 b d^3}-\frac {3 b f^2 \text {Li}_3\left (e^{2 i (c+d x)}\right ) (e+f x)}{2 a^2 d^3}-\frac {6 f^3 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^4}+\frac {6 f^3 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^4}-\frac {6 i \left (a^2-b^2\right ) f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d^4}-\frac {6 i \left (a^2-b^2\right ) f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 b d^4}-\frac {3 i b f^3 \text {Li}_4\left (e^{2 i (c+d x)}\right )}{4 a^2 d^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.78, antiderivative size = 852, normalized size of antiderivative = 1.00, number of steps used = 48, number of rules used = 19, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.559, Rules used = {4543, 4408, 3296, 2638, 4410, 4183, 2531, 2282, 6589, 4404, 3311, 32, 2635, 8, 3717, 2190, 6609, 4525, 4519} \[ \frac {i b (e+f x)^4}{4 a^2 f}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 a^2 b f}-\frac {\csc (c+d x) (e+f x)^3}{a d}-\frac {\left (a^2-b^2\right ) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^3}{a^2 b d}-\frac {\left (a^2-b^2\right ) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^3}{a^2 b d}-\frac {b \log \left (1-e^{2 i (c+d x)}\right ) (e+f x)^3}{a^2 d}-\frac {6 f \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)^2}{a d^2}+\frac {3 i \left (a^2-b^2\right ) f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a^2 b d^2}+\frac {3 i \left (a^2-b^2\right ) f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a^2 b d^2}+\frac {3 i b f \text {PolyLog}\left (2,e^{2 i (c+d x)}\right ) (e+f x)^2}{2 a^2 d^2}+\frac {6 i f^2 \text {PolyLog}\left (2,-e^{i (c+d x)}\right ) (e+f x)}{a d^3}-\frac {6 i f^2 \text {PolyLog}\left (2,e^{i (c+d x)}\right ) (e+f x)}{a d^3}-\frac {6 \left (a^2-b^2\right ) f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)}{a^2 b d^3}-\frac {6 \left (a^2-b^2\right ) f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)}{a^2 b d^3}-\frac {3 b f^2 \text {PolyLog}\left (3,e^{2 i (c+d x)}\right ) (e+f x)}{2 a^2 d^3}-\frac {6 f^3 \text {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^4}+\frac {6 f^3 \text {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^4}-\frac {6 i \left (a^2-b^2\right ) f^3 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d^4}-\frac {6 i \left (a^2-b^2\right ) f^3 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 b d^4}-\frac {3 i b f^3 \text {PolyLog}\left (4,e^{2 i (c+d x)}\right )}{4 a^2 d^4} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Cos[c + d*x]*Cot[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

Rule 8

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

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 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 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 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 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 2638

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

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 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 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 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 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 4408

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 4410

Int[Cot[(a_.) + (b_.)*(x_)]^(p_.)*Csc[(a_.) + (b_.)*(x_)]^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp
[((c + d*x)^m*Csc[a + b*x]^n)/(b*n), x] + Dist[(d*m)/(b*n), Int[(c + d*x)^(m - 1)*Csc[a + b*x]^n, x], x] /; Fr
eeQ[{a, b, c, d, n}, x] && EqQ[p, 1] && GtQ[m, 0]

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 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 4543

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 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 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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [B]  time = 49.18, size = 2974, normalized size = 3.49 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)^3*Cos[c + d*x]*Cot[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

-1/2*(((-I)*b*(e + f*x)^4)/((-1 + E^((2*I)*c))*f) + (6*e*f*(b*d*e - 2*a*f)*x*Log[1 - E^((-I)*(c + d*x))])/d^2
+ (6*f^2*(b*d*e - a*f)*x^2*Log[1 - E^((-I)*(c + d*x))])/d^2 + (2*b*f^3*x^3*Log[1 - E^((-I)*(c + d*x))])/d + (6
*e*f*(b*d*e + 2*a*f)*x*Log[1 + E^((-I)*(c + d*x))])/d^2 + (6*f^2*(b*d*e + a*f)*x^2*Log[1 + E^((-I)*(c + d*x))]
)/d^2 + (2*b*f^3*x^3*Log[1 + E^((-I)*(c + d*x))])/d + (2*e^2*(b*d*e - 3*a*f)*((-I)*d*x + Log[1 - E^(I*(c + d*x
))]))/d^2 + (2*e^2*(b*d*e + 3*a*f)*((-I)*d*x + Log[1 + E^(I*(c + d*x))]))/d^2 + ((6*I)*e*f*(b*d*e + 2*a*f)*Pol
yLog[2, -E^((-I)*(c + d*x))])/d^3 + ((6*I)*e*f*(b*d*e - 2*a*f)*PolyLog[2, E^((-I)*(c + d*x))])/d^3 + (12*f^2*(
b*d*e + a*f)*(I*d*x*PolyLog[2, -E^((-I)*(c + d*x))] + PolyLog[3, -E^((-I)*(c + d*x))]))/d^4 + (12*f^2*(b*d*e -
a*f)*(I*d*x*PolyLog[2, E^((-I)*(c + d*x))] + PolyLog[3, E^((-I)*(c + d*x))]))/d^4 + (6*b*f^3*(I*d^2*x^2*PolyL
og[2, -E^((-I)*(c + d*x))] + 2*d*x*PolyLog[3, -E^((-I)*(c + d*x))] - (2*I)*PolyLog[4, -E^((-I)*(c + d*x))]))/d
^4 + (6*b*f^3*(I*d^2*x^2*PolyLog[2, E^((-I)*(c + d*x))] + 2*d*x*PolyLog[3, E^((-I)*(c + d*x))] - (2*I)*PolyLog
[4, E^((-I)*(c + d*x))]))/d^4)/a^2 + ((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)*L
og[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) - 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) - Sq
rt[(-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*Po
lyLog[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)]))]))/(2*a^2*b*d^4*(-1 + E^((2*I)*c))) + ((-4*b*e^3 - 12*b*e^2*f*x - 12*b*e*f^2*x^2 - 4*b*f^3*x^3 - 4*a*
d*e^3*x*Cos[c] - 6*a*d*e^2*f*x^2*Cos[c] - 4*a*d*e*f^2*x^3*Cos[c] - a*d*f^3*x^4*Cos[c])*Csc[c/2]*Sec[c/2])/(8*a
*b*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]*(-(e^3*Sin[(d*x)/2]) - 3*e^2*f*x*Sin[(d*x)/2] - 3*e*f^2*x^2*Sin[(d*x)/2]
- f^3*x^3*Sin[(d*x)/2]))/(2*a*d) + (Csc[c/2]*Csc[c/2 + (d*x)/2]*(e^3*Sin[(d*x)/2] + 3*e^2*f*x*Sin[(d*x)/2] + 3
*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2]))/(2*a*d)

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fricas [C]  time = 0.90, size = 3923, normalized size = 4.60 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cos(d*x+c)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*a*b*d^3*f^3*x^3 + 6*a*b*d^3*e*f^2*x^2 + 6*a*b*d^3*e^2*f*x + 2*a*b*d^3*e^3 + 6*I*b^2*f^3*polylog(4, cos
(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - 6*I*b^2*f^3*polylog(4, cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c)
- 6*I*b^2*f^3*polylog(4, -cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + 6*I*b^2*f^3*polylog(4, -cos(d*x + c) -
I*sin(d*x + c))*sin(d*x + c) + 6*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)*sin(d*x + c) + 6*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)*s
in(d*x + c) - 6*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)*sin(d*x + c) - 6*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)*sin(d*x + c) - (-
3*I*(a^2 - b^2)*d^2*f^3*x^2 - 6*I*(a^2 - b^2)*d^2*e*f^2*x - 3*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)*sin(d*
x + c) - (-3*I*(a^2 - b^2)*d^2*f^3*x^2 - 6*I*(a^2 - b^2)*d^2*e*f^2*x - 3*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)*sin(d*x + c) - (3*I*(a^2 - b^2)*d^2*f^3*x^2 + 6*I*(a^2 - b^2)*d^2*e*f^2*x + 3*I*(a^2 - b^2)*d^2*e^2*f)*di
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 + 1)*sin(d*x + c) - (3*I*(a^2 - b^2)*d^2*f^3*x^2 + 6*I*(a^2 - b^2)*d^2*e*f^2*x + 3*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)*sin(d*x + c) - (3*I*b^2*d^2*f^3*x^2 + 3*I*b^2*d^2*e^2*f - 6*I*a*b*d*e*f^2 + 6*I*(b^
2*d^2*e*f^2 - a*b*d*f^3)*x)*dilog(cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - (-3*I*b^2*d^2*f^3*x^2 - 3*I*b^
2*d^2*e^2*f + 6*I*a*b*d*e*f^2 - 6*I*(b^2*d^2*e*f^2 - a*b*d*f^3)*x)*dilog(cos(d*x + c) - I*sin(d*x + c))*sin(d*
x + c) - (-3*I*b^2*d^2*f^3*x^2 - 3*I*b^2*d^2*e^2*f - 6*I*a*b*d*e*f^2 - 6*I*(b^2*d^2*e*f^2 + a*b*d*f^3)*x)*dilo
g(-cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - (3*I*b^2*d^2*f^3*x^2 + 3*I*b^2*d^2*e^2*f + 6*I*a*b*d*e*f^2 +
6*I*(b^2*d^2*e*f^2 + a*b*d*f^3)*x)*dilog(-cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + ((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)*sin(d*x + c) + ((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)*sin(d*x + c) + ((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)*s
in(d*x + c) + ((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)*sin(d*x + c) + ((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)*sin(d*x + c) + ((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)*sin(d*x + c) + ((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)*sin(d*x + c)
+ ((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*c
os(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b)*sin(d*x + c) + (b^2*d^3*f^3*x^3 + b^2*d^3*e^3
+ 3*a*b*d^2*e^2*f + 3*(b^2*d^3*e*f^2 + a*b*d^2*f^3)*x^2 + 3*(b^2*d^3*e^2*f + 2*a*b*d^2*e*f^2)*x)*log(cos(d*x
+ c) + I*sin(d*x + c) + 1)*sin(d*x + c) + (b^2*d^3*f^3*x^3 + b^2*d^3*e^3 + 3*a*b*d^2*e^2*f + 3*(b^2*d^3*e*f^2
+ a*b*d^2*f^3)*x^2 + 3*(b^2*d^3*e^2*f + 2*a*b*d^2*e*f^2)*x)*log(cos(d*x + c) - I*sin(d*x + c) + 1)*sin(d*x + c
) + (b^2*d^3*e^3 - 3*(b^2*c + a*b)*d^2*e^2*f + 3*(b^2*c^2 + 2*a*b*c)*d*e*f^2 - (b^2*c^3 + 3*a*b*c^2)*f^3)*log(
-1/2*cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2)*sin(d*x + c) + (b^2*d^3*e^3 - 3*(b^2*c + a*b)*d^2*e^2*f + 3*(b^2
*c^2 + 2*a*b*c)*d*e*f^2 - (b^2*c^3 + 3*a*b*c^2)*f^3)*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2)*sin(d*x
+ c) + (b^2*d^3*f^3*x^3 + 3*b^2*c*d^2*e^2*f - 3*(b^2*c^2 + 2*a*b*c)*d*e*f^2 + (b^2*c^3 + 3*a*b*c^2)*f^3 + 3*(
b^2*d^3*e*f^2 - a*b*d^2*f^3)*x^2 + 3*(b^2*d^3*e^2*f - 2*a*b*d^2*e*f^2)*x)*log(-cos(d*x + c) + I*sin(d*x + c) +
1)*sin(d*x + c) + (b^2*d^3*f^3*x^3 + 3*b^2*c*d^2*e^2*f - 3*(b^2*c^2 + 2*a*b*c)*d*e*f^2 + (b^2*c^3 + 3*a*b*c^2
)*f^3 + 3*(b^2*d^3*e*f^2 - a*b*d^2*f^3)*x^2 + 3*(b^2*d^3*e^2*f - 2*a*b*d^2*e*f^2)*x)*log(-cos(d*x + c) - I*sin
(d*x + c) + 1)*sin(d*x + c) + 6*((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)*sin(d*x + c) + 6*((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)*sin(d*x + c) + 6*((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)*sin(d*x + c) + 6*((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)*sin(d*x + c) + 6*(b^2*d*f
^3*x + b^2*d*e*f^2 - a*b*f^3)*polylog(3, cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + 6*(b^2*d*f^3*x + b^2*d*
e*f^2 - a*b*f^3)*polylog(3, cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + 6*(b^2*d*f^3*x + b^2*d*e*f^2 + a*b*f
^3)*polylog(3, -cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + 6*(b^2*d*f^3*x + b^2*d*e*f^2 + a*b*f^3)*polylog(
3, -cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c))/(a^2*b*d^4*sin(d*x + c))

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cos(d*x+c)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 7.30, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \cos \left (d x +c \right ) \left (\cot ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*cos(d*x+c)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

int((f*x+e)^3*cos(d*x+c)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*cos(d*x+c)*cot(d*x+c)^2/(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 details)Is 4*b^2-4*a^2 positive or negative?

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)*cot(c + d*x)^2*(e + f*x)^3)/(a + b*sin(c + d*x)),x)

[Out]

\text{Hanged}

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right )^{3} \cos {\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**3*cos(d*x+c)*cot(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

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

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