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

Optimal. Leaf size=1051 \[ -\frac {i \left (a^2-b^2\right )^2 (e+f x)^3}{3 a^2 b^3 f}+\frac {i b (e+f x)^3}{3 a^2 f}+\frac {b \sin ^2(c+d x) (e+f x)^2}{2 a^2 d}+\frac {\left (a^2-b^2\right ) \sin ^2(c+d x) (e+f x)^2}{2 a^2 b d}-\frac {\csc (c+d x) (e+f x)^2}{a d}+\frac {\left (a^2-b^2\right )^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^3 d}+\frac {\left (a^2-b^2\right )^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^3 d}-\frac {b \log \left (1-e^{2 i (c+d x)}\right ) (e+f x)^2}{a^2 d}-\frac {\left (a^2-b^2\right ) \sin (c+d x) (e+f x)^2}{a b^2 d}-\frac {\sin (c+d x) (e+f x)^2}{a d}-\frac {4 f \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)}{a d^2}-\frac {2 \left (a^2-b^2\right ) f \cos (c+d x) (e+f x)}{a b^2 d^2}-\frac {2 f \cos (c+d x) (e+f x)}{a d^2}-\frac {2 i \left (a^2-b^2\right )^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^3 d^2}-\frac {2 i \left (a^2-b^2\right )^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^3 d^2}+\frac {i b f \text {Li}_2\left (e^{2 i (c+d x)}\right ) (e+f x)}{a^2 d^2}+\frac {b f \cos (c+d x) \sin (c+d x) (e+f x)}{2 a^2 d^2}+\frac {\left (a^2-b^2\right ) f \cos (c+d x) \sin (c+d x) (e+f x)}{2 a^2 b d^2}-\frac {b f^2 x^2}{4 a^2 d}-\frac {\left (a^2-b^2\right ) f^2 x^2}{4 a^2 b d}-\frac {b f^2 \sin ^2(c+d x)}{4 a^2 d^3}-\frac {\left (a^2-b^2\right ) f^2 \sin ^2(c+d x)}{4 a^2 b d^3}-\frac {b e f x}{2 a^2 d}-\frac {\left (a^2-b^2\right ) e f x}{2 a^2 b 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 \left (a^2-b^2\right )^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^3 d^3}+\frac {2 \left (a^2-b^2\right )^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^3 d^3}-\frac {b f^2 \text {Li}_3\left (e^{2 i (c+d x)}\right )}{2 a^2 d^3}+\frac {2 \left (a^2-b^2\right ) f^2 \sin (c+d x)}{a b^2 d^3}+\frac {2 f^2 \sin (c+d x)}{a d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 2.24, antiderivative size = 1051, normalized size of antiderivative = 1.00, number of steps used = 60, number of rules used = 20, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4543, 4408, 3311, 3296, 2637, 2633, 4410, 4183, 2279, 2391, 4405, 3310, 4404, 3717, 2190, 2531, 2282, 6589, 4525, 4519} \[ -\frac {i \left (a^2-b^2\right )^2 (e+f x)^3}{3 a^2 b^3 f}+\frac {i b (e+f x)^3}{3 a^2 f}+\frac {b \sin ^2(c+d x) (e+f x)^2}{2 a^2 d}+\frac {\left (a^2-b^2\right ) \sin ^2(c+d x) (e+f x)^2}{2 a^2 b d}-\frac {\csc (c+d x) (e+f x)^2}{a d}+\frac {\left (a^2-b^2\right )^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^3 d}+\frac {\left (a^2-b^2\right )^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^3 d}-\frac {b \log \left (1-e^{2 i (c+d x)}\right ) (e+f x)^2}{a^2 d}-\frac {\left (a^2-b^2\right ) \sin (c+d x) (e+f x)^2}{a b^2 d}-\frac {\sin (c+d x) (e+f x)^2}{a d}-\frac {4 f \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)}{a d^2}-\frac {2 \left (a^2-b^2\right ) f \cos (c+d x) (e+f x)}{a b^2 d^2}-\frac {2 f \cos (c+d x) (e+f x)}{a d^2}-\frac {2 i \left (a^2-b^2\right )^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^3 d^2}-\frac {2 i \left (a^2-b^2\right )^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^3 d^2}+\frac {i b f \text {PolyLog}\left (2,e^{2 i (c+d x)}\right ) (e+f x)}{a^2 d^2}+\frac {b f \cos (c+d x) \sin (c+d x) (e+f x)}{2 a^2 d^2}+\frac {\left (a^2-b^2\right ) f \cos (c+d x) \sin (c+d x) (e+f x)}{2 a^2 b d^2}-\frac {b f^2 x^2}{4 a^2 d}-\frac {\left (a^2-b^2\right ) f^2 x^2}{4 a^2 b d}-\frac {b f^2 \sin ^2(c+d x)}{4 a^2 d^3}-\frac {\left (a^2-b^2\right ) f^2 \sin ^2(c+d x)}{4 a^2 b d^3}-\frac {b e f x}{2 a^2 d}-\frac {\left (a^2-b^2\right ) e f x}{2 a^2 b d}+\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 {2 \left (a^2-b^2\right )^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^3 d^3}+\frac {2 \left (a^2-b^2\right )^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^3 d^3}-\frac {b f^2 \text {PolyLog}\left (3,e^{2 i (c+d x)}\right )}{2 a^2 d^3}+\frac {2 \left (a^2-b^2\right ) f^2 \sin (c+d x)}{a b^2 d^3}+\frac {2 f^2 \sin (c+d x)}{a d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2633

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

Rule 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 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 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 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 ^3(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x)^2 \cos ^3(c+d x) \cot ^2(c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x)^2 \cos ^4(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=-\frac {\int (e+f x)^2 \cos ^3(c+d x) \, dx}{a}+\frac {\int (e+f x)^2 \cos (c+d x) \cot ^2(c+d x) \, dx}{a}-\frac {b \int (e+f x)^2 \cos ^4(c+d x) \cot (c+d x) \, dx}{a^2}+\frac {b^2 \int \frac {(e+f x)^2 \cos ^5(c+d x)}{a+b \sin (c+d x)} \, dx}{a^2}\\ &=-\frac {2 f (e+f x) \cos ^3(c+d x)}{9 a d^2}-\frac {(e+f x)^2 \cos ^2(c+d x) \sin (c+d x)}{3 a d}-\frac {2 \int (e+f x)^2 \cos (c+d x) \, dx}{3 a}-\frac {\int (e+f x)^2 \cos (c+d x) \, dx}{a}+\frac {\int (e+f x)^2 \cos ^3(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}-\left (1-\frac {b^2}{a^2}\right ) \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx+\frac {\left (2 f^2\right ) \int \cos ^3(c+d x) \, dx}{9 a d^2}\\ &=-\frac {(e+f x)^2 \csc (c+d x)}{a d}-\frac {5 (e+f x)^2 \sin (c+d x)}{3 a d}+\frac {2 \int (e+f x)^2 \cos (c+d x) \, dx}{3 a}-\frac {b \int (e+f x)^2 \cot (c+d x) \, dx}{a^2}+\frac {b \int (e+f x)^2 \cos (c+d x) \sin (c+d x) \, dx}{a^2}-\frac {\left (a \left (1-\frac {b^2}{a^2}\right )\right ) \int (e+f x)^2 \cos (c+d x) \, dx}{b^2}-\frac {\left (-1+\frac {b^2}{a^2}\right ) \int (e+f x)^2 \cos (c+d x) \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 \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{b^2}+\frac {(4 f) \int (e+f x) \sin (c+d x) \, dx}{3 a d}+\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 {\left (2 f^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{9 a d^3}-\frac {\left (2 f^2\right ) \int \cos ^3(c+d x) \, dx}{9 a d^2}\\ &=\frac {i b (e+f x)^3}{3 a^2 f}-\frac {i \left (a^2-b^2\right )^2 (e+f x)^3}{3 a^2 b^3 f}-\frac {4 f (e+f x) \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d^2}-\frac {10 f (e+f x) \cos (c+d x)}{3 a d^2}-\frac {(e+f x)^2 \csc (c+d x)}{a d}+\frac {2 f^2 \sin (c+d x)}{9 a d^3}-\frac {(e+f x)^2 \sin (c+d x)}{a d}-\frac {a \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin (c+d x)}{b^2 d}+\frac {b (e+f x)^2 \sin ^2(c+d x)}{2 a^2 d}+\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin ^2(c+d x)}{2 b d}-\frac {2 f^2 \sin ^3(c+d x)}{27 a d^3}+\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}-\frac {\left (\left (a^2-b^2\right ) \left (-1+\frac {b^2}{a^2}\right )\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}{b^2}-\frac {\left (\left (a^2-b^2\right ) \left (-1+\frac {b^2}{a^2}\right )\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}{b^2}-\frac {(4 f) \int (e+f x) \sin (c+d x) \, dx}{3 a d}-\frac {(b f) \int (e+f x) \sin ^2(c+d x) \, dx}{a^2 d}+\frac {\left (2 a \left (1-\frac {b^2}{a^2}\right ) f\right ) \int (e+f x) \sin (c+d x) \, dx}{b^2 d}-\frac {\left (\left (1-\frac {b^2}{a^2}\right ) f\right ) \int (e+f x) \sin ^2(c+d x) \, dx}{b d}+\frac {\left (2 f^2\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{9 a d^3}+\frac {\left (4 f^2\right ) \int \cos (c+d x) \, dx}{3 a d^2}+\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 (a^2-b^2\right )^2 (e+f x)^3}{3 a^2 b^3 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 {2 a \left (1-\frac {b^2}{a^2}\right ) f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {(e+f x)^2 \csc (c+d x)}{a d}+\frac {\left (a^2-b^2\right )^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^3 d}+\frac {\left (a^2-b^2\right )^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^3 d}-\frac {b (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a^2 d}+\frac {10 f^2 \sin (c+d x)}{3 a d^3}-\frac {(e+f x)^2 \sin (c+d x)}{a d}-\frac {a \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin (c+d x)}{b^2 d}+\frac {b f (e+f x) \cos (c+d x) \sin (c+d x)}{2 a^2 d^2}+\frac {\left (1-\frac {b^2}{a^2}\right ) f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}-\frac {b f^2 \sin ^2(c+d x)}{4 a^2 d^3}-\frac {\left (1-\frac {b^2}{a^2}\right ) f^2 \sin ^2(c+d x)}{4 b d^3}+\frac {b (e+f x)^2 \sin ^2(c+d x)}{2 a^2 d}+\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin ^2(c+d x)}{2 b d}-\frac {(b f) \int (e+f x) \, dx}{2 a^2 d}+\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 (a^2-b^2\right )^2 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}{a^2 b^3 d}-\frac {\left (2 \left (a^2-b^2\right )^2 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}{a^2 b^3 d}-\frac {\left (\left (1-\frac {b^2}{a^2}\right ) f\right ) \int (e+f x) \, dx}{2 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 (4 f^2\right ) \int \cos (c+d x) \, dx}{3 a d^2}+\frac {\left (2 a \left (1-\frac {b^2}{a^2}\right ) f^2\right ) \int \cos (c+d x) \, dx}{b^2 d^2}\\ &=-\frac {b e f x}{2 a^2 d}-\frac {\left (1-\frac {b^2}{a^2}\right ) e f x}{2 b d}-\frac {b f^2 x^2}{4 a^2 d}-\frac {\left (1-\frac {b^2}{a^2}\right ) f^2 x^2}{4 b d}+\frac {i b (e+f x)^3}{3 a^2 f}-\frac {i \left (a^2-b^2\right )^2 (e+f x)^3}{3 a^2 b^3 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 {2 a \left (1-\frac {b^2}{a^2}\right ) f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {(e+f x)^2 \csc (c+d x)}{a d}+\frac {\left (a^2-b^2\right )^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^3 d}+\frac {\left (a^2-b^2\right )^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^3 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 (a^2-b^2\right )^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^3 d^2}-\frac {2 i \left (a^2-b^2\right )^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^3 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 f^2 \sin (c+d x)}{a d^3}+\frac {2 a \left (1-\frac {b^2}{a^2}\right ) f^2 \sin (c+d x)}{b^2 d^3}-\frac {(e+f x)^2 \sin (c+d x)}{a d}-\frac {a \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin (c+d x)}{b^2 d}+\frac {b f (e+f x) \cos (c+d x) \sin (c+d x)}{2 a^2 d^2}+\frac {\left (1-\frac {b^2}{a^2}\right ) f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}-\frac {b f^2 \sin ^2(c+d x)}{4 a^2 d^3}-\frac {\left (1-\frac {b^2}{a^2}\right ) f^2 \sin ^2(c+d x)}{4 b d^3}+\frac {b (e+f x)^2 \sin ^2(c+d x)}{2 a^2 d}+\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin ^2(c+d x)}{2 b d}-\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 (a^2-b^2\right )^2 f^2\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{a^2 b^3 d^2}+\frac {\left (2 i \left (a^2-b^2\right )^2 f^2\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{a^2 b^3 d^2}\\ &=-\frac {b e f x}{2 a^2 d}-\frac {\left (1-\frac {b^2}{a^2}\right ) e f x}{2 b d}-\frac {b f^2 x^2}{4 a^2 d}-\frac {\left (1-\frac {b^2}{a^2}\right ) f^2 x^2}{4 b d}+\frac {i b (e+f x)^3}{3 a^2 f}-\frac {i \left (a^2-b^2\right )^2 (e+f x)^3}{3 a^2 b^3 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 {2 a \left (1-\frac {b^2}{a^2}\right ) f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {(e+f x)^2 \csc (c+d x)}{a d}+\frac {\left (a^2-b^2\right )^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^3 d}+\frac {\left (a^2-b^2\right )^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^3 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 (a^2-b^2\right )^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^3 d^2}-\frac {2 i \left (a^2-b^2\right )^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^3 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 f^2 \sin (c+d x)}{a d^3}+\frac {2 a \left (1-\frac {b^2}{a^2}\right ) f^2 \sin (c+d x)}{b^2 d^3}-\frac {(e+f x)^2 \sin (c+d x)}{a d}-\frac {a \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin (c+d x)}{b^2 d}+\frac {b f (e+f x) \cos (c+d x) \sin (c+d x)}{2 a^2 d^2}+\frac {\left (1-\frac {b^2}{a^2}\right ) f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}-\frac {b f^2 \sin ^2(c+d x)}{4 a^2 d^3}-\frac {\left (1-\frac {b^2}{a^2}\right ) f^2 \sin ^2(c+d x)}{4 b d^3}+\frac {b (e+f x)^2 \sin ^2(c+d x)}{2 a^2 d}+\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin ^2(c+d x)}{2 b d}-\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 (a^2-b^2\right )^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^3 d^3}+\frac {\left (2 \left (a^2-b^2\right )^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^3 d^3}\\ &=-\frac {b e f x}{2 a^2 d}-\frac {\left (1-\frac {b^2}{a^2}\right ) e f x}{2 b d}-\frac {b f^2 x^2}{4 a^2 d}-\frac {\left (1-\frac {b^2}{a^2}\right ) f^2 x^2}{4 b d}+\frac {i b (e+f x)^3}{3 a^2 f}-\frac {i \left (a^2-b^2\right )^2 (e+f x)^3}{3 a^2 b^3 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 {2 a \left (1-\frac {b^2}{a^2}\right ) f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {(e+f x)^2 \csc (c+d x)}{a d}+\frac {\left (a^2-b^2\right )^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^3 d}+\frac {\left (a^2-b^2\right )^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^3 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 (a^2-b^2\right )^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^3 d^2}-\frac {2 i \left (a^2-b^2\right )^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^3 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 (a^2-b^2\right )^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^3 d^3}+\frac {2 \left (a^2-b^2\right )^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^3 d^3}-\frac {b f^2 \text {Li}_3\left (e^{2 i (c+d x)}\right )}{2 a^2 d^3}+\frac {2 f^2 \sin (c+d x)}{a d^3}+\frac {2 a \left (1-\frac {b^2}{a^2}\right ) f^2 \sin (c+d x)}{b^2 d^3}-\frac {(e+f x)^2 \sin (c+d x)}{a d}-\frac {a \left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin (c+d x)}{b^2 d}+\frac {b f (e+f x) \cos (c+d x) \sin (c+d x)}{2 a^2 d^2}+\frac {\left (1-\frac {b^2}{a^2}\right ) f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}-\frac {b f^2 \sin ^2(c+d x)}{4 a^2 d^3}-\frac {\left (1-\frac {b^2}{a^2}\right ) f^2 \sin ^2(c+d x)}{4 b d^3}+\frac {b (e+f x)^2 \sin ^2(c+d x)}{2 a^2 d}+\frac {\left (1-\frac {b^2}{a^2}\right ) (e+f x)^2 \sin ^2(c+d x)}{2 b d}\\ \end {align*}

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Mathematica [B]  time = 13.98, size = 5156, normalized size = 4.91 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

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fricas [C]  time = 0.99, size = 3145, normalized size = 2.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(8*b^4*f^2*polylog(3, cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + 8*b^4*f^2*polylog(3, cos(d*x + c) - I
*sin(d*x + c))*sin(d*x + c) + 8*b^4*f^2*polylog(3, -cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + 8*b^4*f^2*po
lylog(3, -cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + 8*(a^3*b + a*b^3)*d^2*f^2*x^2 - 16*a^3*b*f^2 + 16*(a^3
*b + a*b^3)*d^2*e*f*x + 8*(a^3*b + a*b^3)*d^2*e^2 - 8*(a^4 - 2*a^2*b^2 + b^4)*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) - 8*
(a^4 - 2*a^2*b^2 + b^4)*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*si
n(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b)*sin(d*x + c) - 8*(a^4 - 2*a^2*b^2 + b^4)*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) -
8*(a^4 - 2*a^2*b^2 + b^4)*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) + 4*(a^2*b^2*d*f^2*x + a^2*b^2*d*e*f)*cos(d*x + c)^3 -
8*(a^3*b*d^2*f^2*x^2 + 2*a^3*b*d^2*e*f*x + a^3*b*d^2*e^2 - 2*a^3*b*f^2)*cos(d*x + c)^2 - (8*I*(a^4 - 2*a^2*b^
2 + b^4)*d*f^2*x + 8*I*(a^4 - 2*a^2*b^2 + b^4)*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) - (8*I*(a^4 - 2*a^2*b^2 +
b^4)*d*f^2*x + 8*I*(a^4 - 2*a^2*b^2 + b^4)*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) - (-8*I*(a^4 - 2*a^2*b^2 + b^4
)*d*f^2*x - 8*I*(a^4 - 2*a^2*b^2 + b^4)*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) - (-8*I*(a^4 - 2*a^2*b^2 + b^4)*
d*f^2*x - 8*I*(a^4 - 2*a^2*b^2 + b^4)*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) - (8*I*b^4*d*f^2*x + 8*I*b^4*d*e*f
- 8*I*a*b^3*f^2)*dilog(cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - (-8*I*b^4*d*f^2*x - 8*I*b^4*d*e*f + 8*I*
a*b^3*f^2)*dilog(cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) - (-8*I*b^4*d*f^2*x - 8*I*b^4*d*e*f - 8*I*a*b^3*f
^2)*dilog(-cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - (8*I*b^4*d*f^2*x + 8*I*b^4*d*e*f + 8*I*a*b^3*f^2)*dil
og(-cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) - 4*((a^4 - 2*a^2*b^2 + b^4)*d^2*e^2 - 2*(a^4 - 2*a^2*b^2 + b^
4)*c*d*e*f + (a^4 - 2*a^2*b^2 + b^4)*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) - 4*((a^4 - 2*a^2*b^2 + b^4)*d^2*e^2 - 2*(a^4 - 2*a^2*b^2 + b^4)*c*d*e*f + (a^4 -
2*a^2*b^2 + b^4)*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) - 4*((a^4 - 2*a^2*b^2 + b^4)*d^2*e^2 - 2*(a^4 - 2*a^2*b^2 + b^4)*c*d*e*f + (a^4 - 2*a^2*b^2 + b^4)*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) - 4*((a^
4 - 2*a^2*b^2 + b^4)*d^2*e^2 - 2*(a^4 - 2*a^2*b^2 + b^4)*c*d*e*f + (a^4 - 2*a^2*b^2 + b^4)*c^2*f^2)*log(-2*b*c
os(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) - 4*((a^4 - 2*a^2*b^2 + b^
4)*d^2*f^2*x^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*d^2*e*f*x + 2*(a^4 - 2*a^2*b^2 + b^4)*c*d*e*f - (a^4 - 2*a^2*b^2 +
b^4)*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) - 4*((a^4 - 2*a^2*b^2 + b^4)*d^2*f^2*x^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*d^2*
e*f*x + 2*(a^4 - 2*a^2*b^2 + b^4)*c*d*e*f - (a^4 - 2*a^2*b^2 + b^4)*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) - 4*((a^4
- 2*a^2*b^2 + b^4)*d^2*f^2*x^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*d^2*e*f*x + 2*(a^4 - 2*a^2*b^2 + b^4)*c*d*e*f - (a^
4 - 2*a^2*b^2 + b^4)*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) - 4*((a^4 - 2*a^2*b^2 + b^4)*d^2*f^2*x^2 + 2*(a^4 - 2*a^
2*b^2 + b^4)*d^2*e*f*x + 2*(a^4 - 2*a^2*b^2 + b^4)*c*d*e*f - (a^4 - 2*a^2*b^2 + b^4)*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) + 4*(b^4*d^2*f^2*x^2 + b^4*d^2*e^2 + 2*a*b^3*d*e*f + 2*(b^4*d^2*e*f + a*b^3*d*f^2)*x)*log(cos(d*x + c)
+ I*sin(d*x + c) + 1)*sin(d*x + c) + 4*(b^4*d^2*f^2*x^2 + b^4*d^2*e^2 + 2*a*b^3*d*e*f + 2*(b^4*d^2*e*f + a*b^
3*d*f^2)*x)*log(cos(d*x + c) - I*sin(d*x + c) + 1)*sin(d*x + c) + 4*(b^4*d^2*e^2 - 2*(b^4*c + a*b^3)*d*e*f + (
b^4*c^2 + 2*a*b^3*c)*f^2)*log(-1/2*cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2)*sin(d*x + c) + 4*(b^4*d^2*e^2 - 2*
(b^4*c + a*b^3)*d*e*f + (b^4*c^2 + 2*a*b^3*c)*f^2)*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2)*sin(d*x +
c) + 4*(b^4*d^2*f^2*x^2 + 2*b^4*c*d*e*f - (b^4*c^2 + 2*a*b^3*c)*f^2 + 2*(b^4*d^2*e*f - a*b^3*d*f^2)*x)*log(-c
os(d*x + c) + I*sin(d*x + c) + 1)*sin(d*x + c) + 4*(b^4*d^2*f^2*x^2 + 2*b^4*c*d*e*f - (b^4*c^2 + 2*a*b^3*c)*f^
2 + 2*(b^4*d^2*e*f - a*b^3*d*f^2)*x)*log(-cos(d*x + c) - I*sin(d*x + c) + 1)*sin(d*x + c) - 4*(a^2*b^2*d*f^2*x
+ a^2*b^2*d*e*f)*cos(d*x + c) - (2*a^2*b^2*d^2*f^2*x^2 + 4*a^2*b^2*d^2*e*f*x + 2*a^2*b^2*d^2*e^2 - a^2*b^2*f^
2 - 2*(2*a^2*b^2*d^2*f^2*x^2 + 4*a^2*b^2*d^2*e*f*x + 2*a^2*b^2*d^2*e^2 - a^2*b^2*f^2)*cos(d*x + c)^2 - 16*(a^3
*b*d*f^2*x + a^3*b*d*e*f)*cos(d*x + c))*sin(d*x + c))/(a^2*b^3*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)^3*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 10.58, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \left (\cot ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x+e)^2*cos(d*x+c)^3*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 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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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)^3*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 \[ \int \frac {\left (e + f x\right )^{2} \cos ^{3}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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