3.4.34 \(\int \frac {(e+f x)^2 \cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [334]

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

[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^2-2*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*(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*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)*polylog(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-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*f*(f*x+e)*polylog(2,exp(I*(d*x+c)))/a/d^2-2*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 = 1.05, 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 = {4639, 4493, 4490, 3391, 3377, 2718, 4268, 2611, 2320, 6724, 4621, 3392, 32, 2715, 8, 3404, 2296, 2221} \begin {gather*} \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} \end {gather*}

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 2715

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] && Integ
erQ[2*n]

Rule 2718

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

Rule 3377

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 3391

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 3392

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 3404

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 4268

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 4490

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 4493

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 4621

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 4639

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 6724

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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 = 3.11, size = 1221, normalized size = 1.48 \begin {gather*} \frac {\frac {24 \left (-2 d^2 e^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )+2 d^2 e f x \log \left (1-e^{i (c+d x)}\right )+d^2 f^2 x^2 \log \left (1-e^{i (c+d x)}\right )-2 d^2 e f x \log \left (1+e^{i (c+d x)}\right )-d^2 f^2 x^2 \log \left (1+e^{i (c+d x)}\right )+2 i d f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )-2 i d f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )-2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )+2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )\right )}{a}+\frac {24 \left (2 i \left (a^2-b^2\right )^{3/2} d^2 e^2 \sqrt {\left (a^2-b^2\right ) e^{2 i c}} \tanh ^{-1}\left (\frac {-a+i b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+2 i \left (a^2-b^2\right )^2 d^2 e e^{i c} f x \log \left (1-\frac {i b e^{i (2 c+d x)}}{a e^{i c}-\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )+i \left (a^2-b^2\right )^2 d^2 e^{i c} f^2 x^2 \log \left (1-\frac {i b e^{i (2 c+d x)}}{a e^{i c}-\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )-2 i \left (a^2-b^2\right )^2 d^2 e e^{i c} f x \log \left (1-\frac {i b e^{i (2 c+d x)}}{a e^{i c}+\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )-i \left (a^2-b^2\right )^2 d^2 e^{i c} f^2 x^2 \log \left (1-\frac {i b e^{i (2 c+d x)}}{a e^{i c}+\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )+2 \left (a^2-b^2\right )^2 d e^{i c} f (e+f x) \text {Li}_2\left (\frac {i b e^{i (2 c+d x)}}{a e^{i c}-\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )-2 \left (a^2-b^2\right )^2 d e^{i c} f (e+f x) \text {Li}_2\left (\frac {i b e^{i (2 c+d x)}}{a e^{i c}+\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )+2 i \left (a^2-b^2\right )^2 e^{i c} f^2 \text {Li}_3\left (\frac {i b e^{i (2 c+d x)}}{a e^{i c}-\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )-2 i \left (a^2-b^2\right )^2 e^{i c} f^2 \text {Li}_3\left (\frac {i b e^{i (2 c+d x)}}{a e^{i c}+\sqrt {\left (a^2-b^2\right ) e^{2 i c}}}\right )\right )}{a b^3 \sqrt {\left (a^2-b^2\right ) e^{2 i c}}}-\frac {-24 a^2 d^3 e^2 x+36 b^2 d^3 e^2 x-24 a^2 d^3 e f x^2+36 b^2 d^3 e f x^2-8 a^2 d^3 f^2 x^3+12 b^2 d^3 f^2 x^3-24 a b \left (-2 f^2+d^2 (e+f x)^2\right ) \cos (c+d x)+6 b^2 d f (e+f x) \cos (2 (c+d x))+48 a b d e f \sin (c+d x)+48 a b d f^2 x \sin (c+d x)+6 b^2 d^2 e^2 \sin (2 (c+d x))-3 b^2 f^2 \sin (2 (c+d x))+12 b^2 d^2 e f x \sin (2 (c+d x))+6 b^2 d^2 f^2 x^2 \sin (2 (c+d x))}{b^3}}{24 d^3} \end {gather*}

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]

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

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Maple [F]
time = 0.96, 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 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2802 vs. \(2 (754) = 1508\).
time = 0.73, size = 2802, normalized size = 3.40 \begin {gather*} \text {Too large to display} \end {gather*}

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*f*x^2*e + 3*a*b^2*d*f^2*x + 6*(2*a^3 - 3*a*b^2
)*d^3*x*e^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*si
n(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, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (
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)*po
lylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (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, -(-I*a*cos(d*x + c) + a*sin(d*x + c) + (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, -(-I
*a*cos(d*x + c) + a*sin(d*x + c) - (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*f*e)*cos(d*x + c)^2 - 12*(I*(a^2*b - b^3)*d*f^2*x + I*(a^2*b - b^3)*d*f*e)*sqrt(-(a^2 - b^2)/
b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b
)/b + 1) - 12*(-I*(a^2*b - b^3)*d*f^2*x - I*(a^2*b - b^3)*d*f*e)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c
) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 12*(-I*(a^2*b -
b^3)*d*f^2*x - I*(a^2*b - b^3)*d*f*e)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*co
s(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 12*(I*(a^2*b - b^3)*d*f^2*x + I*(a^2*b - b
^3)*d*f*e)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x +
c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 6*((a^2*b - b^3)*c^2*f^2 - 2*(a^2*b - b^3)*c*d*f*e + (a^2*b - b^3)*d^
2*e^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)*c^2*f^2 - 2*(a^2*b - b^3)*c*d*f*e + (a^2*b - b^3)*d^2*e^2)*sqrt(-(a^2 - b^2)/b^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) + 6*((a^2*b - b^3)*c^2*f^2 - 2*(a^2*b -
b^3)*c*d*f*e + (a^2*b - b^3)*d^2*e^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)*c^2*f^2 - 2*(a^2*b - b^3)*c*d*f*e + (a^2*b - b^3)*d^2*e^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 - (a^2*b - b^3)*c^2*f^2 + 2*((a^2*b - b^3)*d^2*f*x + (a^2*b - b^3)*c*d*f)*e)*sqrt(-(
a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2
)/b^2) - b)/b) - 6*((a^2*b - b^3)*d^2*f^2*x^2 - (a^2*b - b^3)*c^2*f^2 + 2*((a^2*b - b^3)*d^2*f*x + (a^2*b - b^
3)*c*d*f)*e)*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x +
c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + 6*((a^2*b - b^3)*d^2*f^2*x^2 - (a^2*b - b^3)*c^2*f^2 + 2*((a^2*b - b^3)*d
^2*f*x + (a^2*b - b^3)*c*d*f)*e)*sqrt(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x
+ c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - 6*((a^2*b - b^3)*d^2*f^2*x^2 - (a^2*b - b^3)*c^2*f^2
+ 2*((a^2*b - b^3)*d^2*f*x + (a^2*b - b^3)*c*d*f)*e)*sqrt(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x + c) - a*sin(d
*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + 12*(a^2*b*d^2*f^2*x^2 + 2*a^2*b
*d^2*f*x*e + 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 + I*b^3*d*f*e)*dilog(cos(d*x + c) +
I*sin(d*x + c)) - 12*(-I*b^3*d*f^2*x - I*b^3*d*f*e)*dilog(cos(d*x + c) - I*sin(d*x + c)) - 12*(I*b^3*d*f^2*x
+ I*b^3*d*f*e)*dilog(-cos(d*x + c) + I*sin(d*x + c)) - 12*(-I*b^3*d*f^2*x - I*b^3*d*f*e)*dilog(-cos(d*x + c) -
I*sin(d*x + c)) - 6*(b^3*d^2*f^2*x^2 + 2*b^3*d^2*f*x*e + 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*f*x*e + b^3*d^2*e^2)*log(cos(d*x + c) - I*sin(d*x + c) + 1) + 6*(b^3*c^2*f^2
- 2*b^3*c*d*f*e + b^3*d^2*e^2)*log(-1/2*cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2) + 6*(b^3*c^2*f^2 - 2*b^3*c*d*
f*e + b^3*d^2*e^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 - b^3*c^2*f^2 + 2*(b
^3*d^2*f*x + b^3*c*d*f)*e)*log(-cos(d*x + c) + I*sin(d*x + c) + 1) + 6*(b^3*d^2*f^2*x^2 - b^3*c^2*f^2 + 2*(b^3
*d^2*f*x + b^3*c*d*f)*e)*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*f*e + (2*a*b
^2*d^2*f^2*x^2 + 4*a*b^2*d^2*f*x*e + 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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

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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