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

Optimal. Leaf size=616 \[ -\frac {2 f^2 \left (a^2-b^2\right ) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d^3}-\frac {2 f^2 \left (a^2-b^2\right ) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 b d^3}+\frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d^2}+\frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 b d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a^2 b f}-\frac {b f^2 \text {Li}_3\left (e^{2 i (c+d x)}\right )}{2 a^2 d^3}+\frac {i b f (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a^2 d^2}-\frac {b (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a^2 d}+\frac {i b (e+f x)^3}{3 a^2 f}+\frac {2 i f^2 \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^3}-\frac {2 i f^2 \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^3}-\frac {4 f (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d^2}-\frac {(e+f x)^2 \csc (c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.39, antiderivative size = 616, normalized size of antiderivative = 1.00, number of steps used = 37, number of rules used = 17, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4543, 4408, 3296, 2637, 4410, 4183, 2279, 2391, 4404, 3310, 3717, 2190, 2531, 2282, 6589, 4525, 4519} \[ \frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d^2}+\frac {2 i f \left (a^2-b^2\right ) (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 b d^2}-\frac {2 f^2 \left (a^2-b^2\right ) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d^3}-\frac {2 f^2 \left (a^2-b^2\right ) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 b d^3}+\frac {i b f (e+f x) \text {PolyLog}\left (2,e^{2 i (c+d x)}\right )}{a^2 d^2}-\frac {b f^2 \text {PolyLog}\left (3,e^{2 i (c+d x)}\right )}{2 a^2 d^3}+\frac {2 i f^2 \text {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^3}-\frac {2 i f^2 \text {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^3}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 b d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a^2 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 a^2 b f}-\frac {b (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a^2 d}+\frac {i b (e+f x)^3}{3 a^2 f}-\frac {4 f (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d^2}-\frac {(e+f x)^2 \csc (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 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]

Rubi steps

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

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Mathematica [B]  time = 14.23, size = 1833, normalized size = 2.98 \[ \text {result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((2*I)*b*(e + f*x)^3)/((-1 + E^((2*I)*c))*f) + (6*f*(-(b*d*e) + a*f)*x*Log[1 - E^((-I)*(c + d*x))])/d^2 - (3*
b*f^2*x^2*Log[1 - E^((-I)*(c + d*x))])/d - (6*f*(b*d*e + a*f)*x*Log[1 + E^((-I)*(c + d*x))])/d^2 - (3*b*f^2*x^
2*Log[1 + E^((-I)*(c + d*x))])/d + ((3*I)*e*(b*d*e - 2*a*f)*(d*x + I*Log[1 - E^(I*(c + d*x))]))/d^2 + ((3*I)*e
*(b*d*e + 2*a*f)*(d*x + I*Log[1 + E^(I*(c + d*x))]))/d^2 - ((6*I)*f*(b*d*e + a*f)*PolyLog[2, -E^((-I)*(c + d*x
))])/d^3 + ((6*I)*f*(-(b*d*e) + a*f)*PolyLog[2, E^((-I)*(c + d*x))])/d^3 - ((6*I)*b*f^2*(d*x*PolyLog[2, -E^((-
I)*(c + d*x))] - I*PolyLog[3, -E^((-I)*(c + d*x))]))/d^3 - ((6*I)*b*f^2*(d*x*PolyLog[2, E^((-I)*(c + d*x))] -
I*PolyLog[3, E^((-I)*(c + d*x))]))/d^3)/(3*a^2) + ((a^2 - b^2)*((12*I)*d^3*e^2*E^((2*I)*c)*x + (12*I)*d^3*e*E^
((2*I)*c)*f*x^2 + (4*I)*d^3*E^((2*I)*c)*f^2*x^3 + (6*I)*d^2*e^2*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)
*(c + d*x))))] - (6*I)*d^2*e^2*E^((2*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] + 3*d^
2*e^2*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] - 3*d^2*e^2*E^((2*I)*c)*Log[4*a^2*E^((
2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] + 12*d^2*e*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)])] - 12*d^2*e*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqr
t[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^2*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^2*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)])] + 12*d^2*e*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)])] - 12*d^
2*e*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^2*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^2*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)])] + (12*I)*d*(-1 + E^((2*I)*c))*f*
(e + f*x)*PolyLog[2, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + (12*I)*d*(-1 +
E^((2*I)*c))*f*(e + f*x)*PolyLog[2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] +
12*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*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*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*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)]))]))/(6*a^2*b*d^3*(-1 + E^((2*I)*c))) + ((-3*b*e^2
- 6*b*e*f*x - 3*b*f^2*x^2 - 3*a*d*e^2*x*Cos[c] - 3*a*d*e*f*x^2*Cos[c] - a*d*f^2*x^3*Cos[c])*Csc[c/2]*Sec[c/2]
)/(6*a*b*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]*(-(e^2*Sin[(d*x)/2]) - 2*e*f*x*Sin[(d*x)/2] - f^2*x^2*Sin[(d*x)/2])
)/(2*a*d) + (Csc[c/2]*Csc[c/2 + (d*x)/2]*(e^2*Sin[(d*x)/2] + 2*e*f*x*Sin[(d*x)/2] + f^2*x^2*Sin[(d*x)/2]))/(2*
a*d)

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fricas [C]  time = 0.71, size = 2543, normalized size = 4.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a*b*d^2*f^2*x^2 + 4*a*b*d^2*e*f*x + 2*a*b*d^2*e^2 + 2*b^2*f^2*polylog(3, cos(d*x + c) + I*sin(d*x + c)
)*sin(d*x + c) + 2*b^2*f^2*polylog(3, cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + 2*b^2*f^2*polylog(3, -cos(
d*x + c) + I*sin(d*x + c))*sin(d*x + c) + 2*b^2*f^2*polylog(3, -cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) +
2*(a^2 - b^2)*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) + 2*(a^2 - b^2)*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) + 2*(a^2 - b^2)*f^2*poly
log(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*(a^2 - b^2)*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) - (-2*I*(a^2 - b^2)*d*f^2*x - 2*I*(a^2 - b^2
)*d*e*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) - (-2*I*(a^2 - b^2)*d*f^2*x - 2*I*(a^2 - b^2)*d*e*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)*s
in(d*x + c) - (2*I*(a^2 - b^2)*d*f^2*x + 2*I*(a^2 - b^2)*d*e*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) - (2*I*(a^2 - b
^2)*d*f^2*x + 2*I*(a^2 - b^2)*d*e*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) - (2*I*b^2*d*f^2*x + 2*I*b^2*d*e*f - 2*I*a
*b*f^2)*dilog(cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - (-2*I*b^2*d*f^2*x - 2*I*b^2*d*e*f + 2*I*a*b*f^2)*d
ilog(cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) - (-2*I*b^2*d*f^2*x - 2*I*b^2*d*e*f - 2*I*a*b*f^2)*dilog(-cos
(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - (2*I*b^2*d*f^2*x + 2*I*b^2*d*e*f + 2*I*a*b*f^2)*dilog(-cos(d*x + c)
- I*sin(d*x + c))*sin(d*x + c) + ((a^2 - b^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^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) + ((a^2 - b^2)*d^2*e^2 -
2*(a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^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) + ((a^2 - b^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^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) + ((a^2 - b^2)*d^2*e^2 - 2*(
a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^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) + ((a^2 - b^2)*d^2*f^2*x^2 + 2*(a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2
- b^2)*c^2*f^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) + ((a^2 - b^2)*d^2*f^2*x^2 + 2*(a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*
d*e*f - (a^2 - b^2)*c^2*f^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) + ((a^2 - b^2)*d^2*f^2*x^2 + 2*(a^2 - b^2)*d^2*e*f*x + 2*(
a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^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) + ((a^2 - b^2)*d^2*f^2*x^2 + 2*(a^2 - b^2)*d
^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^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) + (b^2*d^2*f^2*x^2 + b^2*d^2*e
^2 + 2*a*b*d*e*f + 2*(b^2*d^2*e*f + a*b*d*f^2)*x)*log(cos(d*x + c) + I*sin(d*x + c) + 1)*sin(d*x + c) + (b^2*d
^2*f^2*x^2 + b^2*d^2*e^2 + 2*a*b*d*e*f + 2*(b^2*d^2*e*f + a*b*d*f^2)*x)*log(cos(d*x + c) - I*sin(d*x + c) + 1)
*sin(d*x + c) + (b^2*d^2*e^2 - 2*(b^2*c + a*b)*d*e*f + (b^2*c^2 + 2*a*b*c)*f^2)*log(-1/2*cos(d*x + c) + 1/2*I*
sin(d*x + c) + 1/2)*sin(d*x + c) + (b^2*d^2*e^2 - 2*(b^2*c + a*b)*d*e*f + (b^2*c^2 + 2*a*b*c)*f^2)*log(-1/2*co
s(d*x + c) - 1/2*I*sin(d*x + c) + 1/2)*sin(d*x + c) + (b^2*d^2*f^2*x^2 + 2*b^2*c*d*e*f - (b^2*c^2 + 2*a*b*c)*f
^2 + 2*(b^2*d^2*e*f - a*b*d*f^2)*x)*log(-cos(d*x + c) + I*sin(d*x + c) + 1)*sin(d*x + c) + (b^2*d^2*f^2*x^2 +
2*b^2*c*d*e*f - (b^2*c^2 + 2*a*b*c)*f^2 + 2*(b^2*d^2*e*f - a*b*d*f^2)*x)*log(-cos(d*x + c) - I*sin(d*x + c) +
1)*sin(d*x + c))/(a^2*b*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)*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 5.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \cos \left (d x +c \right ) \left (\cot ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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