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

Optimal. Leaf size=825 \[ \frac {\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac {(e+f x)^3}{6 b f}-\frac {2 \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)^2}{a d}+\frac {\left (a^2-b^2\right ) \cos (c+d x) (e+f x)^2}{a b^2 d}+\frac {\cos (c+d x) (e+f x)^2}{a d}+\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a b^3 d}-\frac {\cos (c+d x) \sin (c+d x) (e+f x)^2}{2 b d}-\frac {f \cos ^2(c+d x) (e+f x)}{2 b d^2}+\frac {2 i f \text {Li}_2\left (-e^{i (c+d x)}\right ) (e+f x)}{a d^2}-\frac {2 i f \text {Li}_2\left (e^{i (c+d x)}\right ) (e+f x)}{a d^2}+\frac {2 \left (a^2-b^2\right )^{3/2} f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)}{a b^3 d^2}-\frac {2 \left (a^2-b^2\right )^{3/2} f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)}{a b^3 d^2}-\frac {2 \left (a^2-b^2\right ) f \sin (c+d x) (e+f x)}{a b^2 d^2}-\frac {2 f \sin (c+d x) (e+f x)}{a d^2}+\frac {f^2 x}{4 b d^2}-\frac {2 \left (a^2-b^2\right ) f^2 \cos (c+d x)}{a b^2 d^3}-\frac {2 f^2 \cos (c+d x)}{a d^3}-\frac {2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac {2 i \left (a^2-b^2\right )^{3/2} f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {2 i \left (a^2-b^2\right )^{3/2} f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^3}+\frac {f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3} \]

[Out]

1/4*f^2*x/b/d^2-1/6*(f*x+e)^3/b/f+1/3*(a^2-b^2)*(f*x+e)^3/b^3/f-2*(f*x+e)^2*arctanh(exp(I*(d*x+c)))/a/d-2*f^2*
cos(d*x+c)/a/d^3-2*(a^2-b^2)*f^2*cos(d*x+c)/a/b^2/d^3+(f*x+e)^2*cos(d*x+c)/a/d+(a^2-b^2)*(f*x+e)^2*cos(d*x+c)/
a/b^2/d-1/2*f*(f*x+e)*cos(d*x+c)^2/b/d^2-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/b^3/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/b^3/d^3-2*I*f*(f*x+e)*
polylog(2,exp(I*(d*x+c)))/a/d^2+2*I*f*(f*x+e)*polylog(2,-exp(I*(d*x+c)))/a/d^2+2*(a^2-b^2)^(3/2)*f*(f*x+e)*pol
ylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a/b^3/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/b^3/d^2-2*f^2*polylog(3,-exp(I*(d*x+c)))/a/d^3+2*f^2*polylog(3,exp(I*(d*x+c)))/a/d^
3-2*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/b^3/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/b^3/d-2*f*(f*x+e)*sin(d*x+c)/a/d^2-2*(a^2-b^2)*f*(f*x+e)*si
n(d*x+c)/a/b^2/d^2+1/4*f^2*cos(d*x+c)*sin(d*x+c)/b/d^3-1/2*(f*x+e)^2*cos(d*x+c)*sin(d*x+c)/b/d

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Rubi [A]  time = 1.63, antiderivative size = 825, normalized size of antiderivative = 1.00, number of steps used = 41, number of rules used = 18, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.529, Rules used = {4543, 4408, 4405, 3310, 3296, 2638, 4183, 2531, 2282, 6589, 4525, 3311, 32, 2635, 8, 3323, 2264, 2190} \[ \frac {\left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac {(e+f x)^3}{6 b f}-\frac {2 \tanh ^{-1}\left (e^{i (c+d x)}\right ) (e+f x)^2}{a d}+\frac {\left (a^2-b^2\right ) \cos (c+d x) (e+f x)^2}{a b^2 d}+\frac {\cos (c+d x) (e+f x)^2}{a d}+\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a b^3 d}-\frac {i \left (a^2-b^2\right )^{3/2} \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)^2}{a b^3 d}-\frac {\cos (c+d x) \sin (c+d x) (e+f x)^2}{2 b d}-\frac {f \cos ^2(c+d x) (e+f x)}{2 b d^2}+\frac {2 i f \text {PolyLog}\left (2,-e^{i (c+d x)}\right ) (e+f x)}{a d^2}-\frac {2 i f \text {PolyLog}\left (2,e^{i (c+d x)}\right ) (e+f x)}{a d^2}+\frac {2 \left (a^2-b^2\right )^{3/2} f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) (e+f x)}{a b^3 d^2}-\frac {2 \left (a^2-b^2\right )^{3/2} f \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) (e+f x)}{a b^3 d^2}-\frac {2 \left (a^2-b^2\right ) f \sin (c+d x) (e+f x)}{a b^2 d^2}-\frac {2 f \sin (c+d x) (e+f x)}{a d^2}+\frac {f^2 x}{4 b d^2}-\frac {2 \left (a^2-b^2\right ) f^2 \cos (c+d x)}{a b^2 d^3}-\frac {2 f^2 \cos (c+d x)}{a d^3}-\frac {2 f^2 \text {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac {2 i \left (a^2-b^2\right )^{3/2} f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b^3 d^3}-\frac {2 i \left (a^2-b^2\right )^{3/2} f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b^3 d^3}+\frac {f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 2635

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

Rule 2638

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

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

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

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Mathematica [A]  time = 5.08, size = 1254, normalized size = 1.52 \[ -\frac {-24 d^2 e^2 \log \left (1-e^{i (c+d x)}\right ) b^3-24 d^2 f^2 x^2 \log \left (1-e^{i (c+d x)}\right ) b^3-48 d^2 e f x \log \left (1-e^{i (c+d x)}\right ) b^3+24 d^2 e^2 \log \left (1+e^{i (c+d x)}\right ) b^3+24 d^2 f^2 x^2 \log \left (1+e^{i (c+d x)}\right ) b^3+48 d^2 e f x \log \left (1+e^{i (c+d x)}\right ) b^3-48 i d e f \text {Li}_2\left (-e^{i (c+d x)}\right ) b^3-48 i d f^2 x \text {Li}_2\left (-e^{i (c+d x)}\right ) b^3+48 i d e f \text {Li}_2\left (e^{i (c+d x)}\right ) b^3+48 i d f^2 x \text {Li}_2\left (e^{i (c+d x)}\right ) b^3+48 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right ) b^3-48 f^2 \text {Li}_3\left (e^{i (c+d x)}\right ) b^3+12 a d^3 f^2 x^3 b^2+36 a d^3 e f x^2 b^2+36 a d^3 e^2 x b^2+6 a d e f \cos (2 (c+d x)) b^2+6 a d f^2 x \cos (2 (c+d x)) b^2+6 a d^2 e^2 \sin (2 (c+d x)) b^2-3 a f^2 \sin (2 (c+d x)) b^2+6 a d^2 f^2 x^2 \sin (2 (c+d x)) b^2+12 a d^2 e f x \sin (2 (c+d x)) b^2-24 a^2 d^2 e^2 \cos (c+d x) b+48 a^2 f^2 \cos (c+d x) b-24 a^2 d^2 f^2 x^2 \cos (c+d x) b-48 a^2 d^2 e f x \cos (c+d x) b+48 a^2 d e f \sin (c+d x) b+48 a^2 d f^2 x \sin (c+d x) b-8 a^3 d^3 f^2 x^3-24 a^3 d^3 e f x^2-24 a^3 d^3 e^2 x+48 \left (a^2-b^2\right )^{3/2} d^2 e^2 \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )-24 i \left (a^2-b^2\right )^{3/2} d^2 f^2 x^2 \log \left (\frac {i e^{i (c+d x)} b}{\sqrt {a^2-b^2}-a}+1\right )-48 i \left (a^2-b^2\right )^{3/2} d^2 e f x \log \left (\frac {i e^{i (c+d x)} b}{\sqrt {a^2-b^2}-a}+1\right )+24 i \left (a^2-b^2\right )^{3/2} d^2 f^2 x^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+48 i \left (a^2-b^2\right )^{3/2} d^2 e f x \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )-48 \left (a^2-b^2\right )^{3/2} d e f \text {Li}_2\left (-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )-48 \left (a^2-b^2\right )^{3/2} d f^2 x \text {Li}_2\left (-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )+48 \left (a^2-b^2\right )^{3/2} d e f \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+48 \left (a^2-b^2\right )^{3/2} d f^2 x \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )-48 i \left (a^2-b^2\right )^{3/2} f^2 \text {Li}_3\left (-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}-a}\right )+48 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 )}{24 a b^3 d^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(-24*a^3*d^3*e^2*x + 36*a*b^2*d^3*e^2*x - 24*a^3*d^3*e*f*x^2 + 36*a*b^2*d^3*e*f*x^2 - 8*a^3*d^3*f^2*x^3
+ 12*a*b^2*d^3*f^2*x^3 + 48*(a^2 - b^2)^(3/2)*d^2*e^2*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] - 24*a
^2*b*d^2*e^2*Cos[c + d*x] + 48*a^2*b*f^2*Cos[c + d*x] - 48*a^2*b*d^2*e*f*x*Cos[c + d*x] - 24*a^2*b*d^2*f^2*x^2
*Cos[c + d*x] + 6*a*b^2*d*e*f*Cos[2*(c + d*x)] + 6*a*b^2*d*f^2*x*Cos[2*(c + d*x)] - 24*b^3*d^2*e^2*Log[1 - E^(
I*(c + d*x))] - 48*b^3*d^2*e*f*x*Log[1 - E^(I*(c + d*x))] - 24*b^3*d^2*f^2*x^2*Log[1 - E^(I*(c + d*x))] + 24*b
^3*d^2*e^2*Log[1 + E^(I*(c + d*x))] + 48*b^3*d^2*e*f*x*Log[1 + E^(I*(c + d*x))] + 24*b^3*d^2*f^2*x^2*Log[1 + E
^(I*(c + d*x))] - (48*I)*(a^2 - b^2)^(3/2)*d^2*e*f*x*Log[1 + (I*b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])] - (
24*I)*(a^2 - b^2)^(3/2)*d^2*f^2*x^2*Log[1 + (I*b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])] + (48*I)*(a^2 - b^2)
^(3/2)*d^2*e*f*x*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + (24*I)*(a^2 - b^2)^(3/2)*d^2*f^2*x^2*L
og[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] - (48*I)*b^3*d*e*f*PolyLog[2, -E^(I*(c + d*x))] - (48*I)*b
^3*d*f^2*x*PolyLog[2, -E^(I*(c + d*x))] + (48*I)*b^3*d*e*f*PolyLog[2, E^(I*(c + d*x))] + (48*I)*b^3*d*f^2*x*Po
lyLog[2, E^(I*(c + d*x))] - 48*(a^2 - b^2)^(3/2)*d*e*f*PolyLog[2, ((-I)*b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^
2])] - 48*(a^2 - b^2)^(3/2)*d*f^2*x*PolyLog[2, ((-I)*b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])] + 48*(a^2 - b^
2)^(3/2)*d*e*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + 48*(a^2 - b^2)^(3/2)*d*f^2*x*PolyLog[
2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + 48*b^3*f^2*PolyLog[3, -E^(I*(c + d*x))] - 48*b^3*f^2*PolyLog
[3, E^(I*(c + d*x))] - (48*I)*(a^2 - b^2)^(3/2)*f^2*PolyLog[3, ((-I)*b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])
] + (48*I)*(a^2 - b^2)^(3/2)*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + 48*a^2*b*d*e*f*Sin[
c + d*x] + 48*a^2*b*d*f^2*x*Sin[c + d*x] + 6*a*b^2*d^2*e^2*Sin[2*(c + d*x)] - 3*a*b^2*f^2*Sin[2*(c + d*x)] + 1
2*a*b^2*d^2*e*f*x*Sin[2*(c + d*x)] + 6*a*b^2*d^2*f^2*x^2*Sin[2*(c + d*x)])/(a*b^3*d^3)

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fricas [C]  time = 0.87, size = 2797, normalized size = 3.39 \[ \text {result too large to display} \]

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)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/12*(2*(2*a^3 - 3*a*b^2)*d^3*f^2*x^3 + 6*(2*a^3 - 3*a*b^2)*d^3*e*f*x^2 + 12*b^3*f^2*polylog(3, cos(d*x + c) +
I*sin(d*x + c)) + 12*b^3*f^2*polylog(3, cos(d*x + c) - I*sin(d*x + c)) - 12*b^3*f^2*polylog(3, -cos(d*x + c)
+ I*sin(d*x + c)) - 12*b^3*f^2*polylog(3, -cos(d*x + c) - I*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) - 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*si
n(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 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))*s
qrt(-(a^2 - b^2)/b^2))/b) - 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) - 6*(a*b^2*d*f^2*x + a*b^2
*d*e*f)*cos(d*x + c)^2 + 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) + 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)
+ 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) + 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*s
in(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) - 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*s
in(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) - 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) + 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) + 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) - 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) + 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)
- 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) + 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) + 3*(2*(2*a^3 - 3*a*b^2)*d^3*e^2 + a*b^2*
d*f^2)*x + 12*(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) + (-12*I*b^3*
d*f^2*x - 12*I*b^3*d*e*f)*dilog(cos(d*x + c) + I*sin(d*x + c)) + (12*I*b^3*d*f^2*x + 12*I*b^3*d*e*f)*dilog(cos
(d*x + c) - I*sin(d*x + c)) + (-12*I*b^3*d*f^2*x - 12*I*b^3*d*e*f)*dilog(-cos(d*x + c) + I*sin(d*x + c)) + (12
*I*b^3*d*f^2*x + 12*I*b^3*d*e*f)*dilog(-cos(d*x + c) - I*sin(d*x + c)) - 6*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*e*f*x
+ b^3*d^2*e^2)*log(cos(d*x + c) + I*sin(d*x + c) + 1) - 6*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*e*f*x + b^3*d^2*e^2)*lo
g(cos(d*x + c) - I*sin(d*x + c) + 1) + 6*(b^3*d^2*e^2 - 2*b^3*c*d*e*f + b^3*c^2*f^2)*log(-1/2*cos(d*x + c) + 1
/2*I*sin(d*x + c) + 1/2) + 6*(b^3*d^2*e^2 - 2*b^3*c*d*e*f + b^3*c^2*f^2)*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x
+ c) + 1/2) + 6*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*e*f*x + 2*b^3*c*d*e*f - b^3*c^2*f^2)*log(-cos(d*x + c) + I*sin(d
*x + c) + 1) + 6*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*e*f*x + 2*b^3*c*d*e*f - b^3*c^2*f^2)*log(-cos(d*x + c) - I*sin(d
*x + c) + 1) - 3*(8*a^2*b*d*f^2*x + 8*a^2*b*d*e*f + (2*a*b^2*d^2*f^2*x^2 + 4*a*b^2*d^2*e*f*x + 2*a*b^2*d^2*e^2
- a*b^2*f^2)*cos(d*x + c))*sin(d*x + c))/(a*b^3*d^3)

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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)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

Timed out

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maple [F]  time = 6.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{2} \left (\cos ^{3}\left (d x +c \right )\right ) \cot \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx \]

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)/(a+b*sin(d*x+c)),x)

[Out]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more 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)^3*cot(c + d*x)*(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 \[ \int \frac {\left (e + f x\right )^{2} \cos ^{3}{\left (c + d x \right )} \cot {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

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)/(a+b*sin(d*x+c)),x)

[Out]

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

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