3.198 \(\int \frac {(e+f x)^2 \csc (c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=249 \[ \frac {4 i f^2 \text {Li}_2\left (i e^{i (c+d x)}\right )}{a d^3}-\frac {2 f^2 \text {Li}_3\left (-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {Li}_3\left (e^{i (c+d x)}\right )}{a d^3}+\frac {2 i f (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a d^2}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i (e+f x)^2}{a d} \]

[Out]

I*(f*x+e)^2/a/d-2*(f*x+e)^2*arctanh(exp(I*(d*x+c)))/a/d+(f*x+e)^2*cot(1/2*c+1/4*Pi+1/2*d*x)/a/d-4*f*(f*x+e)*ln
(1-I*exp(I*(d*x+c)))/a/d^2+2*I*f*(f*x+e)*polylog(2,-exp(I*(d*x+c)))/a/d^2+4*I*f^2*polylog(2,I*exp(I*(d*x+c)))/
a/d^3-2*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

________________________________________________________________________________________

Rubi [A]  time = 0.33, antiderivative size = 249, normalized size of antiderivative = 1.00, number of steps used = 14, number of rules used = 11, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {4535, 4183, 2531, 2282, 6589, 3318, 4184, 3717, 2190, 2279, 2391} \[ \frac {2 i f (e+f x) \text {PolyLog}\left (2,-e^{i (c+d x)}\right )}{a d^2}-\frac {2 i f (e+f x) \text {PolyLog}\left (2,e^{i (c+d x)}\right )}{a d^2}+\frac {4 i f^2 \text {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac {2 f^2 \text {PolyLog}\left (3,-e^{i (c+d x)}\right )}{a d^3}+\frac {2 f^2 \text {PolyLog}\left (3,e^{i (c+d x)}\right )}{a d^3}-\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{a d}-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a d}+\frac {i (e+f x)^2}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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) - (4*f*(e + f*x)*Log[1 - I*E^(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) - (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 2190

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

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2282

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

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2531

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

Rule 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])

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 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 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 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 \csc (c+d x)}{a+a \sin (c+d x)} \, dx &=\frac {\int (e+f x)^2 \csc (c+d x) \, dx}{a}-\int \frac {(e+f x)^2}{a+a \sin (c+d x)} \, dx\\ &=-\frac {2 (e+f x)^2 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{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 {(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 {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 {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 {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 {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 (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 {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 {4 f (e+f x) \log \left (1-i 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 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a 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 (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 {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 {4 f (e+f x) \log \left (1-i 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 i f (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )}{a 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 (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 {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 {4 f (e+f x) \log \left (1-i 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 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 {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 [A]  time = 2.20, size = 330, normalized size = 1.33 \[ \frac {\frac {2 i f \left (d (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right )+i f \text {Li}_3\left (-e^{i (c+d x)}\right )\right )}{d^2}+\frac {2 f \left (f \text {Li}_3\left (e^{i (c+d x)}\right )-i d (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right )\right )}{d^2}+\frac {4 f (\cos (c)+i \sin (c)) \left (\frac {f (\cos (c)-i (\sin (c)+1)) \text {Li}_2(-i \cos (c+d x)-\sin (c+d x))}{d^2}-\frac {(\sin (c)+i \cos (c)+1) (e+f x) \log (\sin (c+d x)+i \cos (c+d x)+1)}{d}+\frac {(\cos (c)-i \sin (c)) (e+f x)^2}{2 f}\right )}{\cos (c)+i (\sin (c)+1)}+(e+f x)^2 \log \left (1-e^{i (c+d x)}\right )-(e+f x)^2 \log \left (1+e^{i (c+d x)}\right )-\frac {2 \sin \left (\frac {d x}{2}\right ) (e+f x)^2}{\left (\sin \left (\frac {c}{2}\right )+\cos \left (\frac {c}{2}\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [C]  time = 0.58, size = 1636, normalized size = 6.57 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*csc(d*x+c)/(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 + 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)*cos(d*x + c) + (-2*I*d*f^2*x - 2*I*d*e*f)*sin(d*x + c))*dilog(c
os(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)*cos(d*x + c) + (2*I*d*f^2
*x + 2*I*d*e*f)*sin(d*x + c))*dilog(cos(d*x + c) - I*sin(d*x + c)) + (4*I*f^2*cos(d*x + c) + 4*I*f^2*sin(d*x +
c) + 4*I*f^2)*dilog(I*cos(d*x + c) - sin(d*x + c)) + (-4*I*f^2*cos(d*x + c) - 4*I*f^2*sin(d*x + c) - 4*I*f^2)
*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)*cos(d*x + c) +
(-2*I*d*f^2*x - 2*I*d*e*f)*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)*cos(d*x + c) + (2*I*d*f^2*x + 2*I*d*e*f)*sin(d*x + c))*dilog(-cos(d*x + c) - I*sin(d*
x + c)) - (d^2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2)*cos(d*x + c) + (d^2*f^2
*x^2 + 2*d^2*e*f*x + d^2*e^2)*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) + (d*e*f - c*f^2)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) - (d^2*f^2*x^2 +
2*d^2*e*f*x + d^2*e^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + d^2*e^2)*cos(d*x + c) + (d^2*f^2*x^2 + 2*d^2*e*f*x + d^2
*e^2)*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) + (d*f^2*x + c*f^2)*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) + (d*f^2*x + c*f^2)*sin(d*x + c))*log(-I*cos(d*x + c) + sin(d*x + c) + 1) + (d^2*e^2 - 2*c
*d*e*f + c^2*f^2 + (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*cos(d*x + c) + (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*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*d*e*f + c^2*f^2 + (d^2*e^2 - 2*c*d*e*f +
c^2*f^2)*cos(d*x + c) + (d^2*e^2 - 2*c*d*e*f + c^2*f^2)*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*d^2*e*f*x + 2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^
2)*cos(d*x + c) + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*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) + (d*e*f - c*f^2)*sin(d*x + c))*log(-cos(d*x + c
) + I*sin(d*x + c) + I) + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2 + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*
d*e*f - c^2*f^2)*cos(d*x + c) + (d^2*f^2*x^2 + 2*d^2*e*f*x + 2*c*d*e*f - c^2*f^2)*sin(d*x + c))*log(-cos(d*x +
c) - I*sin(d*x + c) + 1) + 2*(f^2*cos(d*x + c) + f^2*sin(d*x + c) + f^2)*polylog(3, cos(d*x + c) + I*sin(d*x
+ c)) + 2*(f^2*cos(d*x + c) + f^2*sin(d*x + c) + f^2)*polylog(3, cos(d*x + c) - I*sin(d*x + c)) - 2*(f^2*cos(d
*x + c) + f^2*sin(d*x + c) + f^2)*polylog(3, -cos(d*x + c) + I*sin(d*x + c)) - 2*(f^2*cos(d*x + c) + f^2*sin(d
*x + c) + f^2)*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)*sin(d*x +
c))/(a*d^3*cos(d*x + c) + a*d^3*sin(d*x + c) + a*d^3)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{2} \csc \left (d x + c\right )}{a \sin \left (d x + c\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*csc(d*x+c)/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

integrate((f*x + e)^2*csc(d*x + c)/(a*sin(d*x + c) + a), x)

________________________________________________________________________________________

maple [B]  time = 0.24, size = 643, normalized size = 2.58 \[ -\frac {f^{2} \ln \left (1-{\mathrm e}^{i \left (d x +c \right )}\right ) c^{2}}{a \,d^{3}}-\frac {f^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) x^{2}}{a d}+\frac {f^{2} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a \,d^{3}}+\frac {f^{2} \ln \left (1-{\mathrm e}^{i \left (d x +c \right )}\right ) x^{2}}{a d}-\frac {4 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e}{a \,d^{2}}+\frac {4 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right ) e}{a \,d^{2}}+\frac {2 i f^{2} x^{2}}{a d}+\frac {2 i f^{2} c^{2}}{a \,d^{3}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}-\frac {e^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a d}+\frac {e^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a d}-\frac {2 f^{2} \polylog \left (3, -{\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {2 f^{2} \polylog \left (3, {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {4 i f^{2} \polylog \left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a \,d^{3}}-\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {2 i f^{2} \polylog \left (2, -{\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {2 e f c \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a \,d^{2}}+\frac {2 \ln \left (1-{\mathrm e}^{i \left (d x +c \right )}\right ) e f x}{a d}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) e f x}{a d}+\frac {2 \ln \left (1-{\mathrm e}^{i \left (d x +c \right )}\right ) c e f}{a \,d^{2}}-\frac {2 i e f \polylog \left (2, {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}+\frac {2 i e f \polylog \left (2, -{\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}-\frac {2 i f^{2} \polylog \left (2, {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}+\frac {2 f^{2} x^{2}+4 f e x +2 e^{2}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {4 i f^{2} c x}{a \,d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [B]  time = 1.39, size = 1410, normalized size = 5.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*c*e*f*(2/(a*d + a*d*sin(d*x + c)/(cos(d*x + c) + 1)) + log(sin(d*x + c)/(cos(d*x + c) + 1))/(a*d)) - e^2*(
log(sin(d*x + c)/(cos(d*x + c) + 1))/a + 2/(a + a*sin(d*x + c)/(cos(d*x + c) + 1))) + (4*I*c^2*f^2 + (8*I*d*e*
f - 8*I*c*f^2 + 8*(d*e*f - c*f^2)*cos(d*x + 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(d*x + 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 + 2*I*c^2*f^2 + (4*I*d*e*f - 4*I*c*f^2)*(d*x + c) + 2*
((d*x + c)^2*f^2 + c^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(d*x + c) + (2*I*(d*x + c)^2*f^2 + 2*I*c^2*f^2 +
(4*I*d*e*f - 4*I*c*f^2)*(d*x + c))*sin(d*x + c))*arctan2(sin(d*x + c), cos(d*x + c) + 1) - (2*c^2*f^2*cos(d*x
+ c) + 2*I*c^2*f^2*sin(d*x + c) + 2*I*c^2*f^2)*arctan2(sin(d*x + c), cos(d*x + c) - 1) + (2*I*(d*x + c)^2*f^2
+ (4*I*d*e*f - 4*I*c*f^2)*(d*x + c) + 2*((d*x + c)^2*f^2 + 2*(d*e*f - c*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*f^2)*(d*x + c))*sin(d*x + c))*arctan2(sin(d*x + c), -cos(d*x + c) + 1) - 4*
((d*x + c)^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c))*cos(d*x + c) - (8*f^2*cos(d*x + 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*f^2 - 4*(d*e*f + (d*x + c)*f^2 - c*f
^2)*cos(d*x + c) + (-4*I*d*e*f - 4*I*(d*x + c)*f^2 + 4*I*c*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*f^2 + 4*(d*e*f + (d*x + c)*f^2 - c*f^2)*cos(d*x + c) + (4*I*d*e*f + 4*I*(d*x
+ c)*f^2 - 4*I*c*f^2)*sin(d*x + c))*dilog(e^(I*d*x + I*c)) + ((d*x + c)^2*f^2 + c^2*f^2 + 2*(d*e*f - c*f^2)*(d
*x + c) + (-I*(d*x + c)^2*f^2 - I*c^2*f^2 + (-2*I*d*e*f + 2*I*c*f^2)*(d*x + c))*cos(d*x + c) + ((d*x + c)^2*f^
2 + c^2*f^2 + 2*(d*e*f - c*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 + c^2*f^2 + 2*(d*e*f - c*f^2)*(d*x + c) - (I*(d*x + c)^2*f^2 + I*c^2*f^2 + (2*I*d*e*f
- 2*I*c*f^2)*(d*x + c))*cos(d*x + c) + ((d*x + c)^2*f^2 + c^2*f^2 + 2*(d*e*f - c*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(d*x + 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(d*x + c) - f^2*sin(d*x + c) - f^2)*polylog(3, -e^
(I*d*x + I*c)) - 4*(-I*f^2*cos(d*x + c) + f^2*sin(d*x + c) + f^2)*polylog(3, e^(I*d*x + I*c)) + (-4*I*(d*x + 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(d*x + c) + 2*a*d^2*sin(d*x + c) +
2*a*d^2))/d

________________________________________________________________________________________

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((e + f*x)^2/(sin(c + d*x)*(a + a*sin(c + d*x))),x)

[Out]

\text{Hanged}

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {e^{2} \csc {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{2} x^{2} \csc {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {2 e f x \csc {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*csc(d*x+c)/(a+a*sin(d*x+c)),x)

[Out]

(Integral(e**2*csc(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**2*x**2*csc(c + d*x)/(sin(c + d*x) + 1), x) +
Integral(2*e*f*x*csc(c + d*x)/(sin(c + d*x) + 1), x))/a

________________________________________________________________________________________