3.4.26 \(\int \frac {(e+f x)^2 \cos (c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) [326]

Optimal. Leaf size=557 \[ -\frac {(e+f x)^3}{3 b f}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d}+\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac {2 \sqrt {a^2-b^2} f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d^2}+\frac {2 \sqrt {a^2-b^2} f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d^2}-\frac {2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3}-\frac {2 i \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d^3}+\frac {2 i \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a b d^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.76, antiderivative size = 557, normalized size of antiderivative = 1.00, number of steps used = 27, number of rules used = 13, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.406, Rules used = {4639, 4493, 3377, 2718, 4268, 2611, 2320, 6724, 4621, 32, 3404, 2296, 2221} \begin {gather*} -\frac {2 i f^2 \sqrt {a^2-b^2} \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d^3}+\frac {2 i f^2 \sqrt {a^2-b^2} \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b d^3}-\frac {2 f \sqrt {a^2-b^2} (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a b d^2}+\frac {2 f \sqrt {a^2-b^2} (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b d^2}-\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 f (e+f x) \text {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac {i \sqrt {a^2-b^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 d}+\frac {i \sqrt {a^2-b^2} (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{a b d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac {(e+f x)^3}{3 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]
time = 1.09, size = 917, normalized size = 1.65 \begin {gather*} -\frac {x \left (3 e^2+3 e f x+f^2 x^2\right )}{3 b}+\frac {-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 )}{a d^3}+\frac {i \sqrt {a^2-b^2} \left (2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right ) (\cos (c)+i \sin (c))-2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right ) (\cos (c)+i \sin (c))-i \left (-2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right ) (\cos (c)+i \sin (c))+2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right ) (\cos (c)+i \sin (c))+d^2 \left (\sqrt {a^2-b^2} f x (2 e+f x) \left (-\log \left (1+\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right )+\log \left (1-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right )\right ) (\cos (c)+i \sin (c))+2 e^2 \tan ^{-1}\left (\frac {b \cos (c+d x)+i (a+b \sin (c+d x))}{\sqrt {a^2-b^2}}\right ) \sqrt {\left (-a^2+b^2\right ) (\cos (2 c)+i \sin (2 c))}\right )\right )\right )}{a b d^3 \sqrt {\left (-a^2+b^2\right ) (\cos (2 c)+i \sin (2 c))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(x*(3*e^2 + 3*e*f*x + f^2*x^2))/b + (-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*d^3) + (I*Sqrt[a^2 - b^2]*(2
*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, -((b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 +
b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) - 2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, (b
*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])]*(
Cos[c] + I*Sin[c]) - I*(-2*Sqrt[a^2 - b^2]*f^2*PolyLog[3, -((b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c
] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) + 2*Sqrt[a^2 - b^2]*f^2*PolyLog
[3, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[
c])]*(Cos[c] + I*Sin[c]) + d^2*(Sqrt[a^2 - b^2]*f*x*(2*e + f*x)*(-Log[1 + (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x
]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c])] + Log[1 - (b*(Cos[2*c + d*x] + I*Sin[2
*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])])*(Cos[c] + I*Sin[c]) + 2*e^
2*ArcTan[(b*Cos[c + d*x] + I*(a + b*Sin[c + d*x]))/Sqrt[a^2 - b^2]]*Sqrt[(-a^2 + b^2)*(Cos[2*c] + I*Sin[2*c])]
))))/(a*b*d^3*Sqrt[(-a^2 + b^2)*(Cos[2*c] + I*Sin[2*c])])

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{2} \cos \left (d x +c \right ) \cot \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x+e)^2*cos(d*x+c)*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)*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. 2123 vs. \(2 (489) = 978\).
time = 0.63, size = 2123, normalized size = 3.81 \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)*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(2*a*d^3*f^2*x^3 + 6*a*d^3*f*x^2*e + 6*a*d^3*x*e^2 - 6*b*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*b*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*b*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*b*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*b*f^2*polylog(3, c
os(d*x + c) + I*sin(d*x + c)) - 6*b*f^2*polylog(3, cos(d*x + c) - I*sin(d*x + c)) + 6*b*f^2*polylog(3, -cos(d*
x + c) + I*sin(d*x + c)) + 6*b*f^2*polylog(3, -cos(d*x + c) - I*sin(d*x + c)) + 6*(-I*b*d*f^2*x - I*b*d*f*e)*s
qrt(-(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*(I*b*d*f^2*x + I*b*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*(I*b*d*f^2*x + I*b*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*(-I*b*d*f^2*x - I*b*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) - 3*(b*c^2*f^2 - 2*b
*c*d*f*e + b*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) - 3*(b*c^2*f^2 - 2*b*c*d*f*e + b*d^2*e^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) + 3*(b*c^2*f^2 - 2*b*c*d*f*e + b*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) + 3*(b*c^2*f^2 - 2*b*c*d
*f*e + b*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) + 3*(b*d^2*f^2*x^2 - b*c^2*f^2 + 2*(b*d^2*f*x + b*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) - 3*(b*d^2*f^2*x
^2 - b*c^2*f^2 + 2*(b*d^2*f*x + b*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) + 3*(b*d^2*f^2*x^2 - b*c^2*f^2 + 2*(b*d^2*f*
x + b*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) - 3*(b*d^2*f^2*x^2 - b*c^2*f^2 + 2*(b*d^2*f*x + b*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*(I*b*d*f^2*x + I*b*d*f*e)*dilog(cos(d*x + c) + I*sin(d*x + c)) + 6*(-I*b*d*f^2*x - I*b*d*f*e)*
dilog(cos(d*x + c) - I*sin(d*x + c)) + 6*(I*b*d*f^2*x + I*b*d*f*e)*dilog(-cos(d*x + c) + I*sin(d*x + c)) + 6*(
-I*b*d*f^2*x - I*b*d*f*e)*dilog(-cos(d*x + c) - I*sin(d*x + c)) + 3*(b*d^2*f^2*x^2 + 2*b*d^2*f*x*e + b*d^2*e^2
)*log(cos(d*x + c) + I*sin(d*x + c) + 1) + 3*(b*d^2*f^2*x^2 + 2*b*d^2*f*x*e + b*d^2*e^2)*log(cos(d*x + c) - I*
sin(d*x + c) + 1) - 3*(b*c^2*f^2 - 2*b*c*d*f*e + b*d^2*e^2)*log(-1/2*cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2)
- 3*(b*c^2*f^2 - 2*b*c*d*f*e + b*d^2*e^2)*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2) - 3*(b*d^2*f^2*x^2
- b*c^2*f^2 + 2*(b*d^2*f*x + b*c*d*f)*e)*log(-cos(d*x + c) + I*sin(d*x + c) + 1) - 3*(b*d^2*f^2*x^2 - b*c^2*f
^2 + 2*(b*d^2*f*x + b*c*d*f)*e)*log(-cos(d*x + c) - I*sin(d*x + c) + 1))/(a*b*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 {\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)*cot(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

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

[Out]

\text{Hanged}

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