3.204 \(\int \frac{(e+f x)^2 \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx\)
Optimal. Leaf size=327 \[ -\frac{2 i f (e+f x) \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}+\frac{2 i f (e+f x) \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{4 i f^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac{i f^2 \text{PolyLog}\left (2,e^{2 i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}-\frac{2 f^2 \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a d^2}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{(e+f x)^2 \cot (c+d x)}{a d}+\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{2 i (e+f x)^2}{a d} \]
[Out]
((-2*I)*(e + f*x)^2)/(a*d) + (2*(e + f*x)^2*ArcTanh[E^(I*(c + d*x))])/(a*d) - ((e + f*x)^2*Cot[c/2 + Pi/4 + (d
*x)/2])/(a*d) - ((e + f*x)^2*Cot[c + d*x])/(a*d) + (4*f*(e + f*x)*Log[1 - I*E^(I*(c + d*x))])/(a*d^2) + (2*f*(
e + f*x)*Log[1 - E^((2*I)*(c + d*x))])/(a*d^2) - ((2*I)*f*(e + f*x)*PolyLog[2, -E^(I*(c + d*x))])/(a*d^2) - ((
4*I)*f^2*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^3) + ((2*I)*f*(e + f*x)*PolyLog[2, E^(I*(c + d*x))])/(a*d^2) - (I
*f^2*PolyLog[2, E^((2*I)*(c + d*x))])/(a*d^3) + (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)
________________________________________________________________________________________
Rubi [A] time = 0.508825, antiderivative size = 327, normalized size of antiderivative =
1., number of steps used = 20, number of rules used = 11, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) =
0.393, Rules used = {4535, 4184, 3717, 2190, 2279, 2391, 4183, 2531, 2282, 6589, 3318}
\[ -\frac{2 i f (e+f x) \text{PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}+\frac{2 i f (e+f x) \text{PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}-\frac{4 i f^2 \text{PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac{i f^2 \text{PolyLog}\left (2,e^{2 i (c+d x)}\right )}{a d^3}+\frac{2 f^2 \text{PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}-\frac{2 f^2 \text{PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \log \left (1-e^{2 i (c+d x)}\right )}{a d^2}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{d x}{2}+\frac{\pi }{4}\right )}{a d}-\frac{(e+f x)^2 \cot (c+d x)}{a d}+\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{2 i (e+f x)^2}{a d} \]
Antiderivative was successfully verified.
[In]
Int[((e + f*x)^2*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
[Out]
((-2*I)*(e + f*x)^2)/(a*d) + (2*(e + f*x)^2*ArcTanh[E^(I*(c + d*x))])/(a*d) - ((e + f*x)^2*Cot[c/2 + Pi/4 + (d
*x)/2])/(a*d) - ((e + f*x)^2*Cot[c + d*x])/(a*d) + (4*f*(e + f*x)*Log[1 - I*E^(I*(c + d*x))])/(a*d^2) + (2*f*(
e + f*x)*Log[1 - E^((2*I)*(c + d*x))])/(a*d^2) - ((2*I)*f*(e + f*x)*PolyLog[2, -E^(I*(c + d*x))])/(a*d^2) - ((
4*I)*f^2*PolyLog[2, I*E^(I*(c + d*x))])/(a*d^3) + ((2*I)*f*(e + f*x)*PolyLog[2, E^(I*(c + d*x))])/(a*d^2) - (I
*f^2*PolyLog[2, E^((2*I)*(c + d*x))])/(a*d^3) + (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)
Rule 4535
Int[(Csc[(c_.) + (d_.)*(x_)]^(n_.)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[1/a, Int[(e + f*x)^m*Csc[c + d*x]^n, x], x] - Dist[b/a, Int[((e + f*x)^m*Csc[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]
Rule 4184
Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
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 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 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 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 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 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 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]
Rule 3318
Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])
Rubi steps
\begin{align*} \int \frac{(e+f x)^2 \csc ^2(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int (e+f x)^2 \csc ^2(c+d x) \, dx}{a}-\int \frac{(e+f x)^2 \csc (c+d x)}{a+a \sin (c+d x)} \, dx\\ &=-\frac{(e+f x)^2 \cot (c+d x)}{a d}-\frac{\int (e+f x)^2 \csc (c+d x) \, dx}{a}+\frac{(2 f) \int (e+f x) \cot (c+d x) \, dx}{a d}+\int \frac{(e+f x)^2}{a+a \sin (c+d x)} \, dx\\ &=-\frac{i (e+f x)^2}{a d}+\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x)^2 \cot (c+d x)}{a d}+\frac{\int (e+f x)^2 \csc ^2\left (\frac{1}{2} \left (c+\frac{\pi }{2}\right )+\frac{d x}{2}\right ) \, dx}{2 a}-\frac{(4 i f) \int \frac{e^{2 i (c+d x)} (e+f x)}{1-e^{2 i (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{i (e+f x)^2}{a d}+\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{(e+f x)^2 \cot (c+d x)}{a d}+\frac{2 f (e+f x) \log \left (1-e^{2 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 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac{(2 f) \int (e+f x) \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}+\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 \log \left (1-e^{2 i (c+d x)}\right ) \, dx}{a d^2}\\ &=-\frac{2 i (e+f x)^2}{a d}+\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{(e+f x)^2 \cot (c+d x)}{a d}+\frac{2 f (e+f x) \log \left (1-e^{2 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 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}+\frac{(4 f) \int \frac{e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )} (e+f x)}{1-i e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}} \, dx}{a d}+\frac{\left (i f^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 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{\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{2 i (e+f x)^2}{a d}+\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{(e+f x)^2 \cot (c+d x)}{a d}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \log \left (1-e^{2 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 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac{i f^2 \text{Li}_2\left (e^{2 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^2 \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}-\frac{\left (4 f^2\right ) \int \log \left (1-i e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}\right ) \, dx}{a d^2}\\ &=-\frac{2 i (e+f x)^2}{a d}+\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{(e+f x)^2 \cot (c+d x)}{a d}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \log \left (1-e^{2 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 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac{i f^2 \text{Li}_2\left (e^{2 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^2 \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac{\left (4 i f^2\right ) \operatorname{Subst}\left (\int \frac{\log (1-i x)}{x} \, dx,x,e^{2 i \left (\frac{c}{2}+\frac{d x}{2}\right )}\right )}{a d^3}\\ &=-\frac{2 i (e+f x)^2}{a d}+\frac{2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}-\frac{(e+f x)^2 \cot \left (\frac{c}{2}+\frac{\pi }{4}+\frac{d x}{2}\right )}{a d}-\frac{(e+f x)^2 \cot (c+d x)}{a d}+\frac{4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac{2 f (e+f x) \log \left (1-e^{2 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{4 i f^2 \text{Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}+\frac{2 i f (e+f x) \text{Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac{i f^2 \text{Li}_2\left (e^{2 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^2 \text{Li}_3\left (e^{i (c+d x)}\right )}{a d^3}\\ \end{align*}
Mathematica [B] time = 8.33988, size = 693, normalized size = 2.12 \[ \frac{2 i f (d e+f) \text{PolyLog}\left (2,-e^{-i (c+d x)}\right )-2 i f (d e-f) \text{PolyLog}\left (2,e^{-i (c+d x)}\right )+2 f^2 \left (i d x \text{PolyLog}\left (2,-e^{-i (c+d x)}\right )+\text{PolyLog}\left (3,-e^{-i (c+d x)}\right )\right )-2 i f^2 \left (d x \text{PolyLog}\left (2,e^{-i (c+d x)}\right )-i \text{PolyLog}\left (3,e^{-i (c+d x)}\right )\right )-\frac{2 i d^2 (e+f x)^2}{-1+e^{2 i c}}-d^2 f^2 x^2 \log \left (1-e^{-i (c+d x)}\right )+d^2 f^2 x^2 \log \left (1+e^{-i (c+d x)}\right )-2 d f x (d e-f) \log \left (1-e^{-i (c+d x)}\right )+2 d f x (d e+f) \log \left (1+e^{-i (c+d x)}\right )+i d e (d e-2 f) \left (d x+i \log \left (1-e^{i (c+d x)}\right )\right )+d e (d e+2 f) \left (\log \left (1+e^{i (c+d x)}\right )-i d x\right )}{a d^3}-\frac{4 f (\cos (c)+i \sin (c)) \left (\frac{f (\cos (c)-i (\sin (c)+1)) \text{PolyLog}(2,-\sin (c+d x)-i \cos (c+d x))}{d^2}-\frac{(\sin (c)+i \cos (c)+1) (e+f x) \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac{(\cos (c)-i \sin (c)) (e+f x)^2}{2 f}\right )}{a d (\cos (c)+i (\sin (c)+1))}+\frac{2 \left (e^2 \sin \left (\frac{d x}{2}\right )+2 e f x \sin \left (\frac{d x}{2}\right )+f^2 x^2 \sin \left (\frac{d x}{2}\right )\right )}{a d \left (\sin \left (\frac{c}{2}\right )+\cos \left (\frac{c}{2}\right )\right ) \left (\sin \left (\frac{c}{2}+\frac{d x}{2}\right )+\cos \left (\frac{c}{2}+\frac{d x}{2}\right )\right )}+\frac{\csc \left (\frac{c}{2}\right ) \csc \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (e^2 \sin \left (\frac{d x}{2}\right )+2 e f x \sin \left (\frac{d x}{2}\right )+f^2 x^2 \sin \left (\frac{d x}{2}\right )\right )}{2 a d}+\frac{\sec \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}+\frac{d x}{2}\right ) \left (e^2 \sin \left (\frac{d x}{2}\right )+2 e f x \sin \left (\frac{d x}{2}\right )+f^2 x^2 \sin \left (\frac{d x}{2}\right )\right )}{2 a d} \]
Antiderivative was successfully verified.
[In]
Integrate[((e + f*x)^2*Csc[c + d*x]^2)/(a + a*Sin[c + d*x]),x]
[Out]
(((-2*I)*d^2*(e + f*x)^2)/(-1 + E^((2*I)*c)) - 2*d*(d*e - f)*f*x*Log[1 - E^((-I)*(c + d*x))] - d^2*f^2*x^2*Log
[1 - E^((-I)*(c + d*x))] + 2*d*f*(d*e + f)*x*Log[1 + E^((-I)*(c + d*x))] + d^2*f^2*x^2*Log[1 + E^((-I)*(c + d*
x))] + I*d*e*(d*e - 2*f)*(d*x + I*Log[1 - E^(I*(c + d*x))]) + d*e*(d*e + 2*f)*((-I)*d*x + Log[1 + E^(I*(c + d*
x))]) + (2*I)*f*(d*e + f)*PolyLog[2, -E^((-I)*(c + d*x))] - (2*I)*(d*e - f)*f*PolyLog[2, E^((-I)*(c + d*x))] +
2*f^2*(I*d*x*PolyLog[2, -E^((-I)*(c + d*x))] + PolyLog[3, -E^((-I)*(c + d*x))]) - (2*I)*f^2*(d*x*PolyLog[2, E
^((-I)*(c + d*x))] - I*PolyLog[3, E^((-I)*(c + d*x))]))/(a*d^3) - (4*f*(Cos[c] + I*Sin[c])*(((e + f*x)^2*(Cos[
c] - I*Sin[c]))/(2*f) - ((e + f*x)*Log[1 + I*Cos[c + d*x] + Sin[c + d*x]]*(1 + I*Cos[c] + Sin[c]))/d + (f*Poly
Log[2, (-I)*Cos[c + d*x] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])))/d^2))/(a*d*(Cos[c] + I*(1 + Sin[c]))) + (C
sc[c/2]*Csc[c/2 + (d*x)/2]*(e^2*Sin[(d*x)/2] + 2*e*f*x*Sin[(d*x)/2] + f^2*x^2*Sin[(d*x)/2]))/(2*a*d) + (Sec[c/
2]*Sec[c/2 + (d*x)/2]*(e^2*Sin[(d*x)/2] + 2*e*f*x*Sin[(d*x)/2] + f^2*x^2*Sin[(d*x)/2]))/(2*a*d) + (2*(e^2*Sin[
(d*x)/2] + 2*e*f*x*Sin[(d*x)/2] + f^2*x^2*Sin[(d*x)/2]))/(a*d*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin[
c/2 + (d*x)/2]))
________________________________________________________________________________________
Maple [B] time = 0.187, size = 942, normalized size = 2.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((f*x+e)^2*csc(d*x+c)^2/(a+a*sin(d*x+c)),x)
[Out]
-2*(-2*f^2*x^2+I*exp(I*(d*x+c))*f^2*x^2-4*f*e*x+2*I*exp(I*(d*x+c))*e*f*x-2*e^2+I*exp(I*(d*x+c))*e^2+f^2*x^2*ex
p(2*I*(d*x+c))+2*e*f*x*exp(2*I*(d*x+c))+e^2*exp(2*I*(d*x+c)))/(exp(2*I*(d*x+c))-1)/(exp(I*(d*x+c))+I)/d/a-4*I*
f^2*polylog(2,I*exp(I*(d*x+c)))/a/d^3-8*f/d^2/a*ln(exp(I*(d*x+c)))*e+4*f/d^2/a*ln(exp(I*(d*x+c))+I)*e+4*f^2/d^
2/a*ln(1-I*exp(I*(d*x+c)))*x+4*f^2/d^3/a*ln(1-I*exp(I*(d*x+c)))*c+8*f^2/d^3/a*c*ln(exp(I*(d*x+c)))-4*f^2/d^3/a
*c*ln(exp(I*(d*x+c))+I)-1/a/d^3*f^2*c^2*ln(exp(I*(d*x+c))-1)-1/a/d*f^2*ln(1-exp(I*(d*x+c)))*x^2+1/a/d^3*f^2*ln
(1-exp(I*(d*x+c)))*c^2+1/a/d*f^2*ln(exp(I*(d*x+c))+1)*x^2-1/a/d*e^2*ln(exp(I*(d*x+c))-1)+1/a/d*e^2*ln(exp(I*(d
*x+c))+1)-2/a/d^2*ln(1-exp(I*(d*x+c)))*c*e*f-2/a/d*ln(1-exp(I*(d*x+c)))*e*f*x+2/a/d*ln(exp(I*(d*x+c))+1)*e*f*x
+2/a/d^2*e*f*c*ln(exp(I*(d*x+c))-1)+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*f^2*polylog(2,exp(I*(d*x+c)))/a/d^3+2*I/a/d^2*f^2*polylog(2,exp(I*(d*x+c)))*x-2*I/a/d^2*f^2*polylog(2,-
exp(I*(d*x+c)))*x-8*I/a/d^2*f^2*c*x+2*I/a/d^2*e*f*polylog(2,exp(I*(d*x+c)))-2*I/a/d^2*e*f*polylog(2,-exp(I*(d*
x+c)))+2/a/d^2*f^2*ln(exp(I*(d*x+c))+1)*x-2/a/d^3*f^2*c*ln(exp(I*(d*x+c))-1)+2/a/d^2*e*f*ln(exp(I*(d*x+c))-1)+
2/a/d^2*e*f*ln(exp(I*(d*x+c))+1)-4*I/a/d^3*f^2*c^2-2*I/a/d^3*f^2*polylog(2,-exp(I*(d*x+c)))-4*I/a/d*f^2*x^2+2/
a/d^2*f^2*ln(1-exp(I*(d*x+c)))*x+2/a/d^3*f^2*ln(1-exp(I*(d*x+c)))*c
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Maxima [B] time = 4.0475, size = 5003, normalized size = 15.3 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)^2*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="maxima")
[Out]
1/2*(2*c*e*f*((5*sin(d*x + c)/(cos(d*x + c) + 1) + 1)/(a*d*sin(d*x + c)/(cos(d*x + c) + 1) + a*d*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2) + 2*log(sin(d*x + c)/(cos(d*x + c) + 1))/(a*d) - sin(d*x + c)/(a*d*(cos(d*x + c) + 1))
) - e^2*((5*sin(d*x + c)/(cos(d*x + c) + 1) + 1)/(a*sin(d*x + c)/(cos(d*x + c) + 1) + a*sin(d*x + c)^2/(cos(d*
x + c) + 1)^2) + 2*log(sin(d*x + c)/(cos(d*x + c) + 1))/a - sin(d*x + c)/(a*(cos(d*x + c) + 1))) + 2*(-8*I*c^2
*f^2 + (-8*I*d*e*f + 8*I*c*f^2 + 8*(d*e*f - c*f^2)*cos(3*d*x + 3*c) + (8*I*d*e*f - 8*I*c*f^2)*cos(2*d*x + 2*c)
- 8*(d*e*f - c*f^2)*cos(d*x + c) + (8*I*d*e*f - 8*I*c*f^2)*sin(3*d*x + 3*c) - 8*(d*e*f - c*f^2)*sin(2*d*x + 2
*c) + (-8*I*d*e*f + 8*I*c*f^2)*sin(d*x + c))*arctan2(sin(d*x + c) + 1, cos(d*x + c)) - (8*(d*x + c)*f^2*cos(3*
d*x + 3*c) + 8*I*(d*x + c)*f^2*cos(2*d*x + 2*c) - 8*(d*x + c)*f^2*cos(d*x + c) + 8*I*(d*x + c)*f^2*sin(3*d*x +
3*c) - 8*(d*x + c)*f^2*sin(2*d*x + 2*c) - 8*I*(d*x + c)*f^2*sin(d*x + c) - 8*I*(d*x + c)*f^2)*arctan2(cos(d*x
+ c), sin(d*x + c) + 1) + (-2*I*(d*x + c)^2*f^2 - 4*I*d*e*f + (-2*I*c^2 + 4*I*c)*f^2 + (-4*I*d*e*f + (4*I*c -
4*I)*f^2)*(d*x + c) + 2*((d*x + c)^2*f^2 + 2*d*e*f + (c^2 - 2*c)*f^2 + 2*(d*e*f - (c - 1)*f^2)*(d*x + c))*cos
(3*d*x + 3*c) + (2*I*(d*x + c)^2*f^2 + 4*I*d*e*f + (2*I*c^2 - 4*I*c)*f^2 + (4*I*d*e*f + (-4*I*c + 4*I)*f^2)*(d
*x + c))*cos(2*d*x + 2*c) - 2*((d*x + c)^2*f^2 + 2*d*e*f + (c^2 - 2*c)*f^2 + 2*(d*e*f - (c - 1)*f^2)*(d*x + c)
)*cos(d*x + c) + (2*I*(d*x + c)^2*f^2 + 4*I*d*e*f + (2*I*c^2 - 4*I*c)*f^2 + (4*I*d*e*f + (-4*I*c + 4*I)*f^2)*(
d*x + c))*sin(3*d*x + 3*c) - 2*((d*x + c)^2*f^2 + 2*d*e*f + (c^2 - 2*c)*f^2 + 2*(d*e*f - (c - 1)*f^2)*(d*x + c
))*sin(2*d*x + 2*c) + (-2*I*(d*x + c)^2*f^2 - 4*I*d*e*f + (-2*I*c^2 + 4*I*c)*f^2 + (-4*I*d*e*f + (4*I*c - 4*I)
*f^2)*(d*x + c))*sin(d*x + c))*arctan2(sin(d*x + c), cos(d*x + c) + 1) + (-4*I*d*e*f + (2*I*c^2 + 4*I*c)*f^2 +
2*(2*d*e*f - (c^2 + 2*c)*f^2)*cos(3*d*x + 3*c) + (4*I*d*e*f + (-2*I*c^2 - 4*I*c)*f^2)*cos(2*d*x + 2*c) - 2*(2
*d*e*f - (c^2 + 2*c)*f^2)*cos(d*x + c) + (4*I*d*e*f + (-2*I*c^2 - 4*I*c)*f^2)*sin(3*d*x + 3*c) - 2*(2*d*e*f -
(c^2 + 2*c)*f^2)*sin(2*d*x + 2*c) + (-4*I*d*e*f + (2*I*c^2 + 4*I*c)*f^2)*sin(d*x + c))*arctan2(sin(d*x + c), c
os(d*x + c) - 1) + (-2*I*(d*x + c)^2*f^2 + (-4*I*d*e*f + (4*I*c + 4*I)*f^2)*(d*x + c) + 2*((d*x + c)^2*f^2 + 2
*(d*e*f - (c + 1)*f^2)*(d*x + c))*cos(3*d*x + 3*c) + (2*I*(d*x + c)^2*f^2 + (4*I*d*e*f + (-4*I*c - 4*I)*f^2)*(
d*x + c))*cos(2*d*x + 2*c) - 2*((d*x + c)^2*f^2 + 2*(d*e*f - (c + 1)*f^2)*(d*x + c))*cos(d*x + c) + (2*I*(d*x
+ c)^2*f^2 + (4*I*d*e*f + (-4*I*c - 4*I)*f^2)*(d*x + c))*sin(3*d*x + 3*c) - 2*((d*x + c)^2*f^2 + 2*(d*e*f - (c
+ 1)*f^2)*(d*x + c))*sin(2*d*x + 2*c) + (-2*I*(d*x + c)^2*f^2 + (-4*I*d*e*f + (4*I*c + 4*I)*f^2)*(d*x + c))*s
in(d*x + c))*arctan2(sin(d*x + c), -cos(d*x + c) + 1) - 8*((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(
3*d*x + 3*c) + (-4*I*(d*x + c)^2*f^2 + 4*I*c^2*f^2 + (-8*I*d*e*f + 8*I*c*f^2)*(d*x + c))*cos(2*d*x + 2*c) + 4*
((d*x + c)^2*f^2 - c^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(d*x + c) - (8*f^2*cos(3*d*x + 3*c) + 8*I*f^2*cos
(2*d*x + 2*c) - 8*f^2*cos(d*x + c) + 8*I*f^2*sin(3*d*x + 3*c) - 8*f^2*sin(2*d*x + 2*c) - 8*I*f^2*sin(d*x + c)
- 8*I*f^2)*dilog(I*e^(I*d*x + I*c)) + (4*I*d*e*f + 4*I*(d*x + c)*f^2 + (-4*I*c + 4*I)*f^2 - 4*(d*e*f + (d*x +
c)*f^2 - (c - 1)*f^2)*cos(3*d*x + 3*c) + (-4*I*d*e*f - 4*I*(d*x + c)*f^2 + (4*I*c - 4*I)*f^2)*cos(2*d*x + 2*c)
+ 4*(d*e*f + (d*x + c)*f^2 - (c - 1)*f^2)*cos(d*x + c) + (-4*I*d*e*f - 4*I*(d*x + c)*f^2 + (4*I*c - 4*I)*f^2)
*sin(3*d*x + 3*c) + 4*(d*e*f + (d*x + c)*f^2 - (c - 1)*f^2)*sin(2*d*x + 2*c) + (4*I*d*e*f + 4*I*(d*x + c)*f^2
+ (-4*I*c + 4*I)*f^2)*sin(d*x + c))*dilog(-e^(I*d*x + I*c)) + (-4*I*d*e*f - 4*I*(d*x + c)*f^2 + (4*I*c + 4*I)*
f^2 + 4*(d*e*f + (d*x + c)*f^2 - (c + 1)*f^2)*cos(3*d*x + 3*c) + (4*I*d*e*f + 4*I*(d*x + c)*f^2 + (-4*I*c - 4*
I)*f^2)*cos(2*d*x + 2*c) - 4*(d*e*f + (d*x + c)*f^2 - (c + 1)*f^2)*cos(d*x + c) + (4*I*d*e*f + 4*I*(d*x + c)*f
^2 + (-4*I*c - 4*I)*f^2)*sin(3*d*x + 3*c) - 4*(d*e*f + (d*x + c)*f^2 - (c + 1)*f^2)*sin(2*d*x + 2*c) + (-4*I*d
*e*f - 4*I*(d*x + c)*f^2 + (4*I*c + 4*I)*f^2)*sin(d*x + c))*dilog(e^(I*d*x + I*c)) - ((d*x + c)^2*f^2 + 2*d*e*
f + (c^2 - 2*c)*f^2 + 2*(d*e*f - (c - 1)*f^2)*(d*x + c) - (-I*(d*x + c)^2*f^2 - 2*I*d*e*f + (-I*c^2 + 2*I*c)*f
^2 + (-2*I*d*e*f + (2*I*c - 2*I)*f^2)*(d*x + c))*cos(3*d*x + 3*c) - ((d*x + c)^2*f^2 + 2*d*e*f + (c^2 - 2*c)*f
^2 + 2*(d*e*f - (c - 1)*f^2)*(d*x + c))*cos(2*d*x + 2*c) - (I*(d*x + c)^2*f^2 + 2*I*d*e*f + (I*c^2 - 2*I*c)*f^
2 + (2*I*d*e*f + (-2*I*c + 2*I)*f^2)*(d*x + c))*cos(d*x + c) - ((d*x + c)^2*f^2 + 2*d*e*f + (c^2 - 2*c)*f^2 +
2*(d*e*f - (c - 1)*f^2)*(d*x + c))*sin(3*d*x + 3*c) - (I*(d*x + c)^2*f^2 + 2*I*d*e*f + (I*c^2 - 2*I*c)*f^2 + (
2*I*d*e*f + (-2*I*c + 2*I)*f^2)*(d*x + c))*sin(2*d*x + 2*c) + ((d*x + c)^2*f^2 + 2*d*e*f + (c^2 - 2*c)*f^2 + 2
*(d*e*f - (c - 1)*f^2)*(d*x + c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1) + ((
d*x + c)^2*f^2 - 2*d*e*f + (c^2 + 2*c)*f^2 + 2*(d*e*f - (c + 1)*f^2)*(d*x + c) + (I*(d*x + c)^2*f^2 - 2*I*d*e*
f + (I*c^2 + 2*I*c)*f^2 + (2*I*d*e*f + (-2*I*c - 2*I)*f^2)*(d*x + c))*cos(3*d*x + 3*c) - ((d*x + c)^2*f^2 - 2*
d*e*f + (c^2 + 2*c)*f^2 + 2*(d*e*f - (c + 1)*f^2)*(d*x + c))*cos(2*d*x + 2*c) + (-I*(d*x + c)^2*f^2 + 2*I*d*e*
f + (-I*c^2 - 2*I*c)*f^2 + (-2*I*d*e*f + (2*I*c + 2*I)*f^2)*(d*x + c))*cos(d*x + c) - ((d*x + c)^2*f^2 - 2*d*e
*f + (c^2 + 2*c)*f^2 + 2*(d*e*f - (c + 1)*f^2)*(d*x + c))*sin(3*d*x + 3*c) + (-I*(d*x + c)^2*f^2 + 2*I*d*e*f +
(-I*c^2 - 2*I*c)*f^2 + (-2*I*d*e*f + (2*I*c + 2*I)*f^2)*(d*x + c))*sin(2*d*x + 2*c) + ((d*x + c)^2*f^2 - 2*d*
e*f + (c^2 + 2*c)*f^2 + 2*(d*e*f - (c + 1)*f^2)*(d*x + c))*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 -
2*cos(d*x + c) + 1) - (4*d*e*f + 4*(d*x + c)*f^2 - 4*c*f^2 - (-4*I*d*e*f - 4*I*(d*x + c)*f^2 + 4*I*c*f^2)*cos
(3*d*x + 3*c) - 4*(d*e*f + (d*x + c)*f^2 - c*f^2)*cos(2*d*x + 2*c) - (4*I*d*e*f + 4*I*(d*x + c)*f^2 - 4*I*c*f^
2)*cos(d*x + c) - 4*(d*e*f + (d*x + c)*f^2 - c*f^2)*sin(3*d*x + 3*c) - (4*I*d*e*f + 4*I*(d*x + c)*f^2 - 4*I*c*
f^2)*sin(2*d*x + 2*c) + 4*(d*e*f + (d*x + c)*f^2 - c*f^2)*sin(d*x + c))*log(cos(d*x + c)^2 + sin(d*x + c)^2 +
2*sin(d*x + c) + 1) + (-4*I*f^2*cos(3*d*x + 3*c) + 4*f^2*cos(2*d*x + 2*c) + 4*I*f^2*cos(d*x + c) + 4*f^2*sin(3
*d*x + 3*c) + 4*I*f^2*sin(2*d*x + 2*c) - 4*f^2*sin(d*x + c) - 4*f^2)*polylog(3, -e^(I*d*x + I*c)) + (4*I*f^2*c
os(3*d*x + 3*c) - 4*f^2*cos(2*d*x + 2*c) - 4*I*f^2*cos(d*x + c) - 4*f^2*sin(3*d*x + 3*c) - 4*I*f^2*sin(2*d*x +
2*c) + 4*f^2*sin(d*x + c) + 4*f^2)*polylog(3, e^(I*d*x + I*c)) + (-8*I*(d*x + c)^2*f^2 + (-16*I*d*e*f + 16*I*
c*f^2)*(d*x + c))*sin(3*d*x + 3*c) + 4*((d*x + c)^2*f^2 - c^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*sin(2*d*x + 2
*c) + (4*I*(d*x + c)^2*f^2 - 4*I*c^2*f^2 + (8*I*d*e*f - 8*I*c*f^2)*(d*x + c))*sin(d*x + c))/(-2*I*a*d^2*cos(3*
d*x + 3*c) + 2*a*d^2*cos(2*d*x + 2*c) + 2*I*a*d^2*cos(d*x + c) + 2*a*d^2*sin(3*d*x + 3*c) + 2*I*a*d^2*sin(2*d*
x + 2*c) - 2*a*d^2*sin(d*x + c) - 2*a*d^2))/d
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Fricas [C] time = 2.81162, size = 6126, normalized size = 18.73 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)^2*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="fricas")
[Out]
-1/2*(2*d^2*f^2*x^2 + 4*d^2*e*f*x + 2*d^2*e^2 - 4*(d^2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2)*cos(d*x + c)^2 - 2*(d^
2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2)*cos(d*x + c) + (2*I*d*f^2*x + 2*I*d*e*f + (-2*I*d*f^2*x - 2*I*d*e*f + 2*I*f
^2)*cos(d*x + c)^2 - 2*I*f^2 + (2*I*d*f^2*x + 2*I*d*e*f - 2*I*f^2 + (2*I*d*f^2*x + 2*I*d*e*f - 2*I*f^2)*cos(d*
x + c))*sin(d*x + c))*dilog(cos(d*x + c) + I*sin(d*x + c)) + (-2*I*d*f^2*x - 2*I*d*e*f + (2*I*d*f^2*x + 2*I*d*
e*f - 2*I*f^2)*cos(d*x + c)^2 + 2*I*f^2 + (-2*I*d*f^2*x - 2*I*d*e*f + 2*I*f^2 + (-2*I*d*f^2*x - 2*I*d*e*f + 2*
I*f^2)*cos(d*x + c))*sin(d*x + c))*dilog(cos(d*x + c) - I*sin(d*x + c)) + (4*I*f^2*cos(d*x + c)^2 - 4*I*f^2 +
(-4*I*f^2*cos(d*x + c) - 4*I*f^2)*sin(d*x + c))*dilog(I*cos(d*x + c) - sin(d*x + c)) + (-4*I*f^2*cos(d*x + c)^
2 + 4*I*f^2 + (4*I*f^2*cos(d*x + c) + 4*I*f^2)*sin(d*x + c))*dilog(-I*cos(d*x + c) - sin(d*x + c)) + (2*I*d*f^
2*x + 2*I*d*e*f + (-2*I*d*f^2*x - 2*I*d*e*f - 2*I*f^2)*cos(d*x + c)^2 + 2*I*f^2 + (2*I*d*f^2*x + 2*I*d*e*f + 2
*I*f^2 + (2*I*d*f^2*x + 2*I*d*e*f + 2*I*f^2)*cos(d*x + c))*sin(d*x + c))*dilog(-cos(d*x + c) + I*sin(d*x + c))
+ (-2*I*d*f^2*x - 2*I*d*e*f + (2*I*d*f^2*x + 2*I*d*e*f + 2*I*f^2)*cos(d*x + c)^2 - 2*I*f^2 + (-2*I*d*f^2*x -
2*I*d*e*f - 2*I*f^2 + (-2*I*d*f^2*x - 2*I*d*e*f - 2*I*f^2)*cos(d*x + c))*sin(d*x + c))*dilog(-cos(d*x + c) - I
*sin(d*x + c)) + (d^2*f^2*x^2 + d^2*e^2 + 2*d*e*f - (d^2*f^2*x^2 + d^2*e^2 + 2*d*e*f + 2*(d^2*e*f + d*f^2)*x)*
cos(d*x + c)^2 + 2*(d^2*e*f + d*f^2)*x + (d^2*f^2*x^2 + d^2*e^2 + 2*d*e*f + 2*(d^2*e*f + d*f^2)*x + (d^2*f^2*x
^2 + d^2*e^2 + 2*d*e*f + 2*(d^2*e*f + d*f^2)*x)*cos(d*x + c))*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c)
+ 1) + 4*(d*e*f - c*f^2 - (d*e*f - c*f^2)*cos(d*x + c)^2 + (d*e*f - c*f^2 + (d*e*f - c*f^2)*cos(d*x + c))*sin(
d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) + (d^2*f^2*x^2 + d^2*e^2 + 2*d*e*f - (d^2*f^2*x^2 + d^2*e^2 +
2*d*e*f + 2*(d^2*e*f + d*f^2)*x)*cos(d*x + c)^2 + 2*(d^2*e*f + d*f^2)*x + (d^2*f^2*x^2 + d^2*e^2 + 2*d*e*f +
2*(d^2*e*f + d*f^2)*x + (d^2*f^2*x^2 + d^2*e^2 + 2*d*e*f + 2*(d^2*e*f + d*f^2)*x)*cos(d*x + c))*sin(d*x + c))*
log(cos(d*x + c) - I*sin(d*x + c) + 1) + 4*(d*f^2*x + c*f^2 - (d*f^2*x + c*f^2)*cos(d*x + c)^2 + (d*f^2*x + c*
f^2 + (d*f^2*x + c*f^2)*cos(d*x + c))*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) + 4*(d*f^2*x + c*f^
2 - (d*f^2*x + c*f^2)*cos(d*x + c)^2 + (d*f^2*x + c*f^2 + (d*f^2*x + c*f^2)*cos(d*x + c))*sin(d*x + c))*log(-I
*cos(d*x + c) + sin(d*x + c) + 1) - (d^2*e^2 - 2*(c + 1)*d*e*f + (c^2 + 2*c)*f^2 - (d^2*e^2 - 2*(c + 1)*d*e*f
+ (c^2 + 2*c)*f^2)*cos(d*x + c)^2 + (d^2*e^2 - 2*(c + 1)*d*e*f + (c^2 + 2*c)*f^2 + (d^2*e^2 - 2*(c + 1)*d*e*f
+ (c^2 + 2*c)*f^2)*cos(d*x + c))*sin(d*x + c))*log(-1/2*cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2) - (d^2*e^2 -
2*(c + 1)*d*e*f + (c^2 + 2*c)*f^2 - (d^2*e^2 - 2*(c + 1)*d*e*f + (c^2 + 2*c)*f^2)*cos(d*x + c)^2 + (d^2*e^2 -
2*(c + 1)*d*e*f + (c^2 + 2*c)*f^2 + (d^2*e^2 - 2*(c + 1)*d*e*f + (c^2 + 2*c)*f^2)*cos(d*x + c))*sin(d*x + c))*
log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2) - (d^2*f^2*x^2 + 2*c*d*e*f - (c^2 + 2*c)*f^2 - (d^2*f^2*x^2
+ 2*c*d*e*f - (c^2 + 2*c)*f^2 + 2*(d^2*e*f - d*f^2)*x)*cos(d*x + c)^2 + 2*(d^2*e*f - d*f^2)*x + (d^2*f^2*x^2 +
2*c*d*e*f - (c^2 + 2*c)*f^2 + 2*(d^2*e*f - d*f^2)*x + (d^2*f^2*x^2 + 2*c*d*e*f - (c^2 + 2*c)*f^2 + 2*(d^2*e*f
- d*f^2)*x)*cos(d*x + c))*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + 1) + 4*(d*e*f - c*f^2 - (d*e*f -
c*f^2)*cos(d*x + c)^2 + (d*e*f - c*f^2 + (d*e*f - c*f^2)*cos(d*x + c))*sin(d*x + c))*log(-cos(d*x + c) + I*si
n(d*x + c) + I) - (d^2*f^2*x^2 + 2*c*d*e*f - (c^2 + 2*c)*f^2 - (d^2*f^2*x^2 + 2*c*d*e*f - (c^2 + 2*c)*f^2 + 2*
(d^2*e*f - d*f^2)*x)*cos(d*x + c)^2 + 2*(d^2*e*f - d*f^2)*x + (d^2*f^2*x^2 + 2*c*d*e*f - (c^2 + 2*c)*f^2 + 2*(
d^2*e*f - d*f^2)*x + (d^2*f^2*x^2 + 2*c*d*e*f - (c^2 + 2*c)*f^2 + 2*(d^2*e*f - d*f^2)*x)*cos(d*x + c))*sin(d*x
+ c))*log(-cos(d*x + c) - I*sin(d*x + c) + 1) + 2*(f^2*cos(d*x + c)^2 - f^2 - (f^2*cos(d*x + c) + f^2)*sin(d*
x + c))*polylog(3, cos(d*x + c) + I*sin(d*x + c)) + 2*(f^2*cos(d*x + c)^2 - f^2 - (f^2*cos(d*x + c) + f^2)*sin
(d*x + c))*polylog(3, cos(d*x + c) - I*sin(d*x + c)) - 2*(f^2*cos(d*x + c)^2 - f^2 - (f^2*cos(d*x + c) + f^2)*
sin(d*x + c))*polylog(3, -cos(d*x + c) + I*sin(d*x + c)) - 2*(f^2*cos(d*x + c)^2 - f^2 - (f^2*cos(d*x + c) + f
^2)*sin(d*x + c))*polylog(3, -cos(d*x + c) - I*sin(d*x + c)) - 2*(d^2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2 + 2*(d^2
*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2)*cos(d*x + c))*sin(d*x + c))/(a*d^3*cos(d*x + c)^2 - a*d^3 - (a*d^3*cos(d*x +
c) + a*d^3)*sin(d*x + c))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{e^{2} \csc ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{f^{2} x^{2} \csc ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx + \int \frac{2 e f x \csc ^{2}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)**2*csc(d*x+c)**2/(a+a*sin(d*x+c)),x)
[Out]
(Integral(e**2*csc(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(f**2*x**2*csc(c + d*x)**2/(sin(c + d*x) + 1),
x) + Integral(2*e*f*x*csc(c + d*x)**2/(sin(c + d*x) + 1), x))/a
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{2} \csc \left (d x + c\right )^{2}}{a \sin \left (d x + c\right ) + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)^2*csc(d*x+c)^2/(a+a*sin(d*x+c)),x, algorithm="giac")
[Out]
integrate((f*x + e)^2*csc(d*x + c)^2/(a*sin(d*x + c) + a), x)