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

Optimal. Leaf size=840 \[ -\frac {\left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}-\frac {(e+f x)^3}{3 a f}+\frac {2 b \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)^2}{a^2 d}-\frac {b \cos (c+d x) (e+f x)^2}{a^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) (e+f x)^2}{a^2 b d}-\frac {\cot (c+d x) (e+f x)^2}{a d}-\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a^2 b^2 d}+\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a^2 b^2 d}-\frac {i (e+f x)^2}{a d}+\frac {2 f \log \left (1-e^{2 i (c+d x)}\right ) (e+f x)}{a d^2}-\frac {2 i b f \text {Li}_2\left (-e^{i (c+d x)}\right ) (e+f x)}{a^2 d^2}+\frac {2 i b f \text {Li}_2\left (e^{i (c+d x)}\right ) (e+f x)}{a^2 d^2}-\frac {2 \left (a^2-b^2\right )^{3/2} f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)}{a^2 b^2 d^2}+\frac {2 \left (a^2-b^2\right )^{3/2} f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)}{a^2 b^2 d^2}+\frac {2 b f \sin (c+d x) (e+f x)}{a^2 d^2}+\frac {2 \left (a^2-b^2\right ) f \sin (c+d x) (e+f x)}{a^2 b d^2}+\frac {2 b f^2 \cos (c+d x)}{a^2 d^3}+\frac {2 \left (a^2-b^2\right ) f^2 \cos (c+d x)}{a^2 b d^3}-\frac {i f^2 \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^3}+\frac {2 b f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a^2 d^3}-\frac {2 b f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a^2 d^3}-\frac {2 i \left (a^2-b^2\right )^{3/2} f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b^2 d^3}+\frac {2 i \left (a^2-b^2\right )^{3/2} f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 b^2 d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 2.15, antiderivative size = 840, normalized size of antiderivative = 1.00, number of steps used = 53, number of rules used = 22, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.611, Rules used = {4543, 4408, 3311, 32, 2635, 8, 3720, 3717, 2190, 2279, 2391, 4405, 3310, 3296, 2638, 4183, 2531, 2282, 6589, 4525, 3323, 2264} \[ -\frac {\left (a^2-b^2\right ) (e+f x)^3}{3 a b^2 f}-\frac {(e+f x)^3}{3 a f}+\frac {2 b \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)^2}{a^2 d}-\frac {b \cos (c+d x) (e+f x)^2}{a^2 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x) (e+f x)^2}{a^2 b d}-\frac {\cot (c+d x) (e+f x)^2}{a d}-\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a^2 b^2 d}+\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a^2 b^2 d}-\frac {i (e+f x)^2}{a d}+\frac {2 f \log \left (1-e^{2 i (c+d x)}\right ) (e+f x)}{a d^2}-\frac {2 i b f \text {PolyLog}\left (2,-e^{i (c+d x)}\right ) (e+f x)}{a^2 d^2}+\frac {2 i b f \text {PolyLog}\left (2,e^{i (c+d x)}\right ) (e+f x)}{a^2 d^2}-\frac {2 \left (a^2-b^2\right )^{3/2} f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)}{a^2 b^2 d^2}+\frac {2 \left (a^2-b^2\right )^{3/2} f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)}{a^2 b^2 d^2}+\frac {2 b f \sin (c+d x) (e+f x)}{a^2 d^2}+\frac {2 \left (a^2-b^2\right ) f \sin (c+d x) (e+f x)}{a^2 b d^2}+\frac {2 b f^2 \cos (c+d x)}{a^2 d^3}+\frac {2 \left (a^2-b^2\right ) f^2 \cos (c+d x)}{a^2 b d^3}-\frac {i f^2 \text {PolyLog}\left (2,e^{2 i (c+d x)}\right )}{a d^3}+\frac {2 b f^2 \text {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a^2 d^3}-\frac {2 b f^2 \text {PolyLog}\left (3,e^{i (c+d x)}\right )}{a^2 d^3}-\frac {2 i \left (a^2-b^2\right )^{3/2} f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b^2 d^3}+\frac {2 i \left (a^2-b^2\right )^{3/2} f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 b^2 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I)*(e + f*x)^2)/(a*d) - (e + f*x)^3/(3*a*f) - ((a^2 - b^2)*(e + f*x)^3)/(3*a*b^2*f) + (2*b*(e + f*x)^2*ArcT
anh[E^(I*(c + d*x))])/(a^2*d) + (2*b*f^2*Cos[c + d*x])/(a^2*d^3) + (2*(a^2 - b^2)*f^2*Cos[c + d*x])/(a^2*b*d^3
) - (b*(e + f*x)^2*Cos[c + d*x])/(a^2*d) - ((a^2 - b^2)*(e + f*x)^2*Cos[c + d*x])/(a^2*b*d) - ((e + f*x)^2*Cot
[c + d*x])/(a*d) - (I*(a^2 - b^2)^(3/2)*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a^2
*b^2*d) + (I*(a^2 - b^2)^(3/2)*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a^2*b^2*d) +
(2*f*(e + f*x)*Log[1 - E^((2*I)*(c + d*x))])/(a*d^2) - ((2*I)*b*f*(e + f*x)*PolyLog[2, -E^(I*(c + d*x))])/(a^
2*d^2) + ((2*I)*b*f*(e + f*x)*PolyLog[2, E^(I*(c + d*x))])/(a^2*d^2) - (2*(a^2 - b^2)^(3/2)*f*(e + f*x)*PolyLo
g[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a^2*b^2*d^2) + (2*(a^2 - b^2)^(3/2)*f*(e + f*x)*PolyLog[2,
(I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a^2*b^2*d^2) - (I*f^2*PolyLog[2, E^((2*I)*(c + d*x))])/(a*d^3)
+ (2*b*f^2*PolyLog[3, -E^(I*(c + d*x))])/(a^2*d^3) - (2*b*f^2*PolyLog[3, E^(I*(c + d*x))])/(a^2*d^3) - ((2*I)
*(a^2 - b^2)^(3/2)*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a^2*b^2*d^3) + ((2*I)*(a^2 -
b^2)^(3/2)*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a^2*b^2*d^3) + (2*b*f*(e + f*x)*Sin[c
+ d*x])/(a^2*d^2) + (2*(a^2 - b^2)*f*(e + f*x)*Sin[c + d*x])/(a^2*b*d^2)

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 2264

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 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 3310

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

Rule 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 3323

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 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 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[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 4405

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 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 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]

Rubi steps

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

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Mathematica [A]  time = 10.85, size = 951, normalized size = 1.13 \[ \frac {12 \left (-b d^2 x^2 \log \left (1-e^{-i (c+d x)}\right ) f^2+b d^2 x^2 \log \left (1+e^{-i (c+d x)}\right ) f^2+2 b \left (i d x \text {Li}_2\left (-e^{-i (c+d x)}\right )+\text {Li}_3\left (-e^{-i (c+d x)}\right )\right ) f^2-2 i b \left (d x \text {Li}_2\left (e^{-i (c+d x)}\right )-i \text {Li}_3\left (e^{-i (c+d x)}\right )\right ) f^2-2 d (b d e-a f) x \log \left (1-e^{-i (c+d x)}\right ) f+2 d (b d e+a f) x \log \left (1+e^{-i (c+d x)}\right ) f+2 i (b d e+a f) \text {Li}_2\left (-e^{-i (c+d x)}\right ) f+2 i (a f-b d e) \text {Li}_2\left (e^{-i (c+d x)}\right ) f-\frac {2 i a d^2 (e+f x)^2}{-1+e^{2 i c}}+i d e (b d e-2 a f) \left (d x+i \log \left (1-e^{i (c+d x)}\right )\right )+d e (b d e+2 a f) \left (\log \left (1+e^{i (c+d x)}\right )-i d x\right )\right )-\frac {12 i \sqrt {-\left (a^2-b^2\right )^2} \left (-2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )+2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )-i \left (\left (2 \sqrt {b^2-a^2} \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right ) e^2+\sqrt {a^2-b^2} f x (2 e+f x) \left (\log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-\log \left (\frac {e^{i (c+d x)} b}{i a+\sqrt {b^2-a^2}}+1\right )\right )\right ) d^2+2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )\right )\right )}{b^2}+\frac {a \csc (c) \csc (c+d x) \left (-2 a^2 x \left (3 e^2+3 f x e+f^2 x^2\right ) \cos (d x) d^3+2 a^2 x \left (3 e^2+3 f x e+f^2 x^2\right ) \cos (2 c+d x) d^3+3 b \left (-a \left (d^2 (e+f x)^2-2 f^2\right ) \cos (c+2 d x)+a \left (d^2 (e+f x)^2-2 f^2\right ) \cos (3 c+2 d x)+2 d (e+f x) \left (4 a f \sin (c) \sin ^2(c+d x)+2 b d (e+f x) \sin (d x)\right )\right )\right )}{b^2}}{12 a^2 d^3} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(12*(((-2*I)*a*d^2*(e + f*x)^2)/(-1 + E^((2*I)*c)) - 2*d*f*(b*d*e - a*f)*x*Log[1 - E^((-I)*(c + d*x))] - b*d^2
*f^2*x^2*Log[1 - E^((-I)*(c + d*x))] + 2*d*f*(b*d*e + a*f)*x*Log[1 + E^((-I)*(c + d*x))] + b*d^2*f^2*x^2*Log[1
+ E^((-I)*(c + d*x))] + I*d*e*(b*d*e - 2*a*f)*(d*x + I*Log[1 - E^(I*(c + d*x))]) + d*e*(b*d*e + 2*a*f)*((-I)*
d*x + Log[1 + E^(I*(c + d*x))]) + (2*I)*f*(b*d*e + a*f)*PolyLog[2, -E^((-I)*(c + d*x))] + (2*I)*f*(-(b*d*e) +
a*f)*PolyLog[2, E^((-I)*(c + d*x))] + 2*b*f^2*(I*d*x*PolyLog[2, -E^((-I)*(c + d*x))] + PolyLog[3, -E^((-I)*(c
+ d*x))]) - (2*I)*b*f^2*(d*x*PolyLog[2, E^((-I)*(c + d*x))] - I*PolyLog[3, E^((-I)*(c + d*x))])) - ((12*I)*Sqr
t[-(a^2 - b^2)^2]*(-2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])
] + 2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] - I*(d^2*(2*Sq
rt[-a^2 + b^2]*e^2*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + Sqrt[a^2 - b^2]*f*x*(2*e + f*x)*(Log[1
- (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])])) +
2*Sqrt[a^2 - b^2]*f^2*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 2*Sqrt[a^2 - b^2]*f^2*Pol
yLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))])))/b^2 + (a*Csc[c]*Csc[c + d*x]*(-2*a^2*d^3*x*(3*e^2
+ 3*e*f*x + f^2*x^2)*Cos[d*x] + 2*a^2*d^3*x*(3*e^2 + 3*e*f*x + f^2*x^2)*Cos[2*c + d*x] + 3*b*(-(a*(-2*f^2 + d^
2*(e + f*x)^2)*Cos[c + 2*d*x]) + a*(-2*f^2 + d^2*(e + f*x)^2)*Cos[3*c + 2*d*x] + 2*d*(e + f*x)*(2*b*d*(e + f*x
)*Sin[d*x] + 4*a*f*Sin[c]*Sin[c + d*x]^2))))/b^2)/(12*a^2*d^3)

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fricas [C]  time = 0.95, size = 3085, normalized size = 3.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(24*a^2*b*d*f^2*x - 12*b^3*f^2*polylog(3, cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - 12*b^3*f^2*polylo
g(3, cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + 12*b^3*f^2*polylog(3, -cos(d*x + c) + I*sin(d*x + c))*sin(d
*x + c) + 12*b^3*f^2*polylog(3, -cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + 24*a^2*b*d*e*f - 12*(a^2*b - b^
3)*f^2*sqrt(-(a^2 - b^2)/b^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) + 12*(a^2*b - b^3)*f^2*sqrt(-(a^2 - b^2)/b^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) - 12*(a^2*b - b^3)*f^2*sqrt(-(a^2 - b^2)/b^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) + 12*(a^2*b - b^3)*f^2*sq
rt(-(a^2 - b^2)/b^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) + 2*(6*I*(a^2*b - b^3)*d*f^2*x + 6*I*(a^2*b - b^3)*d*e*f)*sqrt(-
(a^2 - b^2)/b^2)*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))*sqr
t(-(a^2 - b^2)/b^2) + 2*b)/b + 1)*sin(d*x + c) + 2*(-6*I*(a^2*b - b^3)*d*f^2*x - 6*I*(a^2*b - b^3)*d*e*f)*sqrt
(-(a^2 - b^2)/b^2)*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))*s
qrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1)*sin(d*x + c) + 2*(-6*I*(a^2*b - b^3)*d*f^2*x - 6*I*(a^2*b - b^3)*d*e*f)*sq
rt(-(a^2 - b^2)/b^2)*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) + 2*(6*I*(a^2*b - b^3)*d*f^2*x + 6*I*(a^2*b - b^3)*d*e*f)*
sqrt(-(a^2 - b^2)/b^2)*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) + 6*((a^2*b - b^3)*d^2*e^2 - 2*(a^2*b - b^3)*c*d*e*f + (
a^2*b - b^3)*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*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) + 6*((a^2*b - b^3)*d^2*e^2 - 2*(a^2*b - b^3)*c*d*e*f + (a^2*b - b^3)*c^2*f^2)*sqrt
(-(a^2 - b^2)/b^2)*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
) - 6*((a^2*b - b^3)*d^2*e^2 - 2*(a^2*b - b^3)*c*d*e*f + (a^2*b - b^3)*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*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) - 6*((a^2*b - b^3)*d^2*
e^2 - 2*(a^2*b - b^3)*c*d*e*f + (a^2*b - b^3)*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*log(-2*b*cos(d*x + c) - 2*I*b*si
n(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a)*sin(d*x + c) + 6*((a^2*b - b^3)*d^2*f^2*x^2 + 2*(a^2*b - b^3)
*d^2*e*f*x + 2*(a^2*b - b^3)*c*d*e*f - (a^2*b - b^3)*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*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) -
6*((a^2*b - b^3)*d^2*f^2*x^2 + 2*(a^2*b - b^3)*d^2*e*f*x + 2*(a^2*b - b^3)*c*d*e*f - (a^2*b - b^3)*c^2*f^2)*sq
rt(-(a^2 - b^2)/b^2)*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))*sq
rt(-(a^2 - b^2)/b^2) + 2*b)/b)*sin(d*x + c) + 6*((a^2*b - b^3)*d^2*f^2*x^2 + 2*(a^2*b - b^3)*d^2*e*f*x + 2*(a^
2*b - b^3)*c*d*e*f - (a^2*b - b^3)*c^2*f^2)*sqrt(-(a^2 - b^2)/b^2)*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) - 6*((a^2*b - b^3)*
d^2*f^2*x^2 + 2*(a^2*b - b^3)*d^2*e*f*x + 2*(a^2*b - b^3)*c*d*e*f - (a^2*b - b^3)*c^2*f^2)*sqrt(-(a^2 - b^2)/b
^2)*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) - 24*(a^2*b*d*f^2*x + a^2*b*d*e*f)*cos(d*x + c)^2 + (12*I*b^3*d*f^2*x + 12*I*b^3*d
*e*f - 12*I*a*b^2*f^2)*dilog(cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + (-12*I*b^3*d*f^2*x - 12*I*b^3*d*e*f
+ 12*I*a*b^2*f^2)*dilog(cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + (12*I*b^3*d*f^2*x + 12*I*b^3*d*e*f + 12
*I*a*b^2*f^2)*dilog(-cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + (-12*I*b^3*d*f^2*x - 12*I*b^3*d*e*f - 12*I*
a*b^2*f^2)*dilog(-cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + 6*(b^3*d^2*f^2*x^2 + b^3*d^2*e^2 + 2*a*b^2*d*e
*f + 2*(b^3*d^2*e*f + a*b^2*d*f^2)*x)*log(cos(d*x + c) + I*sin(d*x + c) + 1)*sin(d*x + c) + 6*(b^3*d^2*f^2*x^2
+ b^3*d^2*e^2 + 2*a*b^2*d*e*f + 2*(b^3*d^2*e*f + a*b^2*d*f^2)*x)*log(cos(d*x + c) - I*sin(d*x + c) + 1)*sin(d
*x + c) - 6*(b^3*d^2*e^2 - 2*(b^3*c + a*b^2)*d*e*f + (b^3*c^2 + 2*a*b^2*c)*f^2)*log(-1/2*cos(d*x + c) + 1/2*I*
sin(d*x + c) + 1/2)*sin(d*x + c) - 6*(b^3*d^2*e^2 - 2*(b^3*c + a*b^2)*d*e*f + (b^3*c^2 + 2*a*b^2*c)*f^2)*log(-
1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2)*sin(d*x + c) - 6*(b^3*d^2*f^2*x^2 + 2*b^3*c*d*e*f - (b^3*c^2 + 2*
a*b^2*c)*f^2 + 2*(b^3*d^2*e*f - a*b^2*d*f^2)*x)*log(-cos(d*x + c) + I*sin(d*x + c) + 1)*sin(d*x + c) - 6*(b^3*
d^2*f^2*x^2 + 2*b^3*c*d*e*f - (b^3*c^2 + 2*a*b^2*c)*f^2 + 2*(b^3*d^2*e*f - a*b^2*d*f^2)*x)*log(-cos(d*x + c) -
I*sin(d*x + c) + 1)*sin(d*x + c) - 12*(a*b^2*d^2*f^2*x^2 + 2*a*b^2*d^2*e*f*x + a*b^2*d^2*e^2)*cos(d*x + c) -
4*(a^3*d^3*f^2*x^3 + 3*a^3*d^3*e*f*x^2 + 3*a^3*d^3*e^2*x + 3*(a^2*b*d^2*f^2*x^2 + 2*a^2*b*d^2*e*f*x + a^2*b*d^
2*e^2 - 2*a^2*b*f^2)*cos(d*x + c))*sin(d*x + c))/(a^2*b^2*d^3*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)^2*cos(d*x+c)^2*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 6.58, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \left (\cos ^{2}\left (d x +c \right )\right ) \left (\cot ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x+e)^2*cos(d*x+c)^2*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)^2*cos(d*x+c)^2*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)^2*cot(c + d*x)^2*(e + f*x)^2)/(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 \[ \int \frac {\left (e + f x\right )^{2} \cos ^{2}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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