3.236 \(\int \frac {(e+f x)^3 \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)
Optimal. Leaf size=882 \[ \frac {3 \text {Li}_3\left (e^{2 i (c+d x)}\right ) f^3}{2 a d^4}+\frac {6 i b \text {Li}_4\left (-e^{i (c+d x)}\right ) f^3}{a^2 d^4}-\frac {6 i b \text {Li}_4\left (e^{i (c+d x)}\right ) f^3}{a^2 d^4}+\frac {6 b^2 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{a^2 \sqrt {a^2-b^2} d^4}-\frac {6 b^2 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{a^2 \sqrt {a^2-b^2} d^4}-\frac {3 i (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right ) f^2}{a d^3}+\frac {6 b (e+f x) \text {Li}_3\left (-e^{i (c+d x)}\right ) f^2}{a^2 d^3}-\frac {6 b (e+f x) \text {Li}_3\left (e^{i (c+d x)}\right ) f^2}{a^2 d^3}-\frac {6 i b^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{a^2 \sqrt {a^2-b^2} d^3}+\frac {6 i b^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{a^2 \sqrt {a^2-b^2} d^3}+\frac {3 (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right ) f}{a d^2}-\frac {3 i b (e+f x)^2 \text {Li}_2\left (-e^{i (c+d x)}\right ) f}{a^2 d^2}+\frac {3 i b (e+f x)^2 \text {Li}_2\left (e^{i (c+d x)}\right ) f}{a^2 d^2}-\frac {3 b^2 (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{a^2 \sqrt {a^2-b^2} d^2}+\frac {3 b^2 (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{a^2 \sqrt {a^2-b^2} d^2}-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d} \]
[Out]
-3*I*f^2*(f*x+e)*polylog(2,exp(2*I*(d*x+c)))/a/d^3+2*b*(f*x+e)^3*arctanh(exp(I*(d*x+c)))/a^2/d-(f*x+e)^3*cot(d
*x+c)/a/d+3*f*(f*x+e)^2*ln(1-exp(2*I*(d*x+c)))/a/d^2+6*I*b*f^3*polylog(4,-exp(I*(d*x+c)))/a^2/d^4+6*I*b^2*f^2*
(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a^2/d^3/(a^2-b^2)^(1/2)-I*(f*x+e)^3/a/d+6*b*f^2*(f*x
+e)*polylog(3,-exp(I*(d*x+c)))/a^2/d^3-6*b*f^2*(f*x+e)*polylog(3,exp(I*(d*x+c)))/a^2/d^3+3/2*f^3*polylog(3,exp
(2*I*(d*x+c)))/a/d^4-I*b^2*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a^2/d/(a^2-b^2)^(1/2)+I*b^2*
(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a^2/d/(a^2-b^2)^(1/2)-3*I*b*f*(f*x+e)^2*polylog(2,-exp(
I*(d*x+c)))/a^2/d^2+3*I*b*f*(f*x+e)^2*polylog(2,exp(I*(d*x+c)))/a^2/d^2-3*b^2*f*(f*x+e)^2*polylog(2,I*b*exp(I*
(d*x+c))/(a-(a^2-b^2)^(1/2)))/a^2/d^2/(a^2-b^2)^(1/2)+3*b^2*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b
^2)^(1/2)))/a^2/d^2/(a^2-b^2)^(1/2)-6*I*b^2*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/a^2/
d^3/(a^2-b^2)^(1/2)-6*I*b*f^3*polylog(4,exp(I*(d*x+c)))/a^2/d^4+6*b^2*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a-(a^2
-b^2)^(1/2)))/a^2/d^4/(a^2-b^2)^(1/2)-6*b^2*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/a^2/d^4/(a^2
-b^2)^(1/2)
________________________________________________________________________________________
Rubi [A] time = 1.55, antiderivative size = 882, normalized size of antiderivative =
1.00, number of steps used = 29, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) =
0.393, Rules used = {4535, 4184, 3717, 2190, 2531, 2282, 6589, 4183, 6609, 3323, 2264}
\[ \frac {3 \text {PolyLog}\left (3,e^{2 i (c+d x)}\right ) f^3}{2 a d^4}+\frac {6 i b \text {PolyLog}\left (4,-e^{i (c+d x)}\right ) f^3}{a^2 d^4}-\frac {6 i b \text {PolyLog}\left (4,e^{i (c+d x)}\right ) f^3}{a^2 d^4}+\frac {6 b^2 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{a^2 \sqrt {a^2-b^2} d^4}-\frac {6 b^2 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{a^2 \sqrt {a^2-b^2} d^4}-\frac {3 i (e+f x) \text {PolyLog}\left (2,e^{2 i (c+d x)}\right ) f^2}{a d^3}+\frac {6 b (e+f x) \text {PolyLog}\left (3,-e^{i (c+d x)}\right ) f^2}{a^2 d^3}-\frac {6 b (e+f x) \text {PolyLog}\left (3,e^{i (c+d x)}\right ) f^2}{a^2 d^3}-\frac {6 i b^2 (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{a^2 \sqrt {a^2-b^2} d^3}+\frac {6 i b^2 (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{a^2 \sqrt {a^2-b^2} d^3}+\frac {3 (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right ) f}{a d^2}-\frac {3 i b (e+f x)^2 \text {PolyLog}\left (2,-e^{i (c+d x)}\right ) f}{a^2 d^2}+\frac {3 i b (e+f x)^2 \text {PolyLog}\left (2,e^{i (c+d x)}\right ) f}{a^2 d^2}-\frac {3 b^2 (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{a^2 \sqrt {a^2-b^2} d^2}+\frac {3 b^2 (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{a^2 \sqrt {a^2-b^2} d^2}-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d} \]
Antiderivative was successfully verified.
[In]
Int[((e + f*x)^3*Csc[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
[Out]
((-I)*(e + f*x)^3)/(a*d) + (2*b*(e + f*x)^3*ArcTanh[E^(I*(c + d*x))])/(a^2*d) - ((e + f*x)^3*Cot[c + d*x])/(a*
d) - (I*b^2*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a^2*Sqrt[a^2 - b^2]*d) + (I*b^2
*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a^2*Sqrt[a^2 - b^2]*d) + (3*f*(e + f*x)^2*
Log[1 - E^((2*I)*(c + d*x))])/(a*d^2) - ((3*I)*b*f*(e + f*x)^2*PolyLog[2, -E^(I*(c + d*x))])/(a^2*d^2) + ((3*I
)*b*f*(e + f*x)^2*PolyLog[2, E^(I*(c + d*x))])/(a^2*d^2) - (3*b^2*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)
))/(a - Sqrt[a^2 - b^2])])/(a^2*Sqrt[a^2 - b^2]*d^2) + (3*b^2*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(
a + Sqrt[a^2 - b^2])])/(a^2*Sqrt[a^2 - b^2]*d^2) - ((3*I)*f^2*(e + f*x)*PolyLog[2, E^((2*I)*(c + d*x))])/(a*d^
3) + (6*b*f^2*(e + f*x)*PolyLog[3, -E^(I*(c + d*x))])/(a^2*d^3) - (6*b*f^2*(e + f*x)*PolyLog[3, E^(I*(c + d*x)
)])/(a^2*d^3) - ((6*I)*b^2*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(a^2*Sqrt[a^
2 - b^2]*d^3) + ((6*I)*b^2*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(a^2*Sqrt[a^
2 - b^2]*d^3) + (3*f^3*PolyLog[3, E^((2*I)*(c + d*x))])/(2*a*d^4) + ((6*I)*b*f^3*PolyLog[4, -E^(I*(c + d*x))])
/(a^2*d^4) - ((6*I)*b*f^3*PolyLog[4, E^(I*(c + d*x))])/(a^2*d^4) + (6*b^2*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))
/(a - Sqrt[a^2 - b^2])])/(a^2*Sqrt[a^2 - b^2]*d^4) - (6*b^2*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2
- b^2])])/(a^2*Sqrt[a^2 - b^2]*d^4)
Rule 2190
Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
Rule 2264
Int[((F_)^(u_)*((f_.) + (g_.)*(x_))^(m_.))/((a_.) + (b_.)*(F_)^(u_) + (c_.)*(F_)^(v_)), x_Symbol] :> With[{q =
Rt[b^2 - 4*a*c, 2]}, Dist[(2*c)/q, Int[((f + g*x)^m*F^u)/(b - q + 2*c*F^u), x], x] - Dist[(2*c)/q, Int[((f +
g*x)^m*F^u)/(b + q + 2*c*F^u), x], x]] /; FreeQ[{F, a, b, c, f, g}, x] && EqQ[v, 2*u] && LinearQ[u, x] && NeQ[
b^2 - 4*a*c, 0] && IGtQ[m, 0]
Rule 2282
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]
Rule 2531
Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]
Rule 3323
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[2, Int[((c + d*x)^m*E
^(I*(e + f*x)))/(I*b + 2*a*E^(I*(e + f*x)) - I*b*E^(2*I*(e + f*x))), x], x] /; FreeQ[{a, b, c, d, e, f}, x] &&
NeQ[a^2 - b^2, 0] && IGtQ[m, 0]
Rule 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]
Rule 6609
Int[((e_.) + (f_.)*(x_))^(m_.)*PolyLog[n_, (d_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(p_.)], x_Symbol] :> Simp
[((e + f*x)^m*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p])/(b*c*p*Log[F]), x] - Dist[(f*m)/(b*c*p*Log[F]), Int[(e +
f*x)^(m - 1)*PolyLog[n + 1, d*(F^(c*(a + b*x)))^p], x], x] /; FreeQ[{F, a, b, c, d, e, f, n, p}, x] && GtQ[m,
0]
Rubi steps
\begin {align*} \int \frac {(e+f x)^3 \csc ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x)^3 \csc ^2(c+d x) \, dx}{a}-\frac {b \int \frac {(e+f x)^3 \csc (c+d x)}{a+b \sin (c+d x)} \, dx}{a}\\ &=-\frac {(e+f x)^3 \cot (c+d x)}{a d}-\frac {b \int (e+f x)^3 \csc (c+d x) \, dx}{a^2}+\frac {b^2 \int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{a^2}+\frac {(3 f) \int (e+f x)^2 \cot (c+d x) \, dx}{a d}\\ &=-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}+\frac {\left (2 b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{i b+2 a e^{i (c+d x)}-i b e^{2 i (c+d x)}} \, dx}{a^2}-\frac {(6 i f) \int \frac {e^{2 i (c+d x)} (e+f x)^2}{1-e^{2 i (c+d x)}} \, dx}{a d}+\frac {(3 b f) \int (e+f x)^2 \log \left (1-e^{i (c+d x)}\right ) \, dx}{a^2 d}-\frac {(3 b f) \int (e+f x)^2 \log \left (1+e^{i (c+d x)}\right ) \, dx}{a^2 d}\\ &=-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}+\frac {3 f (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d^2}-\frac {3 i b f (e+f x)^2 \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {3 i b f (e+f x)^2 \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}-\frac {\left (2 i b^3\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{2 a-2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a^2 \sqrt {a^2-b^2}}+\frac {\left (2 i b^3\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{2 a+2 \sqrt {a^2-b^2}-2 i b e^{i (c+d x)}} \, dx}{a^2 \sqrt {a^2-b^2}}-\frac {\left (6 f^2\right ) \int (e+f x) \log \left (1-e^{2 i (c+d x)}\right ) \, dx}{a d^2}+\frac {\left (6 i b f^2\right ) \int (e+f x) \text {Li}_2\left (-e^{i (c+d x)}\right ) \, dx}{a^2 d^2}-\frac {\left (6 i b f^2\right ) \int (e+f x) \text {Li}_2\left (e^{i (c+d x)}\right ) \, dx}{a^2 d^2}\\ &=-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {3 f (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d^2}-\frac {3 i b f (e+f x)^2 \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {3 i b f (e+f x)^2 \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}-\frac {3 i f^2 (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^3}+\frac {6 b f^2 (e+f x) \text {Li}_3\left (-e^{i (c+d x)}\right )}{a^2 d^3}-\frac {6 b f^2 (e+f x) \text {Li}_3\left (e^{i (c+d x)}\right )}{a^2 d^3}+\frac {\left (3 i b^2 f\right ) \int (e+f x)^2 \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{a^2 \sqrt {a^2-b^2} d}-\frac {\left (3 i b^2 f\right ) \int (e+f x)^2 \log \left (1-\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{a^2 \sqrt {a^2-b^2} d}+\frac {\left (3 i f^3\right ) \int \text {Li}_2\left (e^{2 i (c+d x)}\right ) \, dx}{a d^3}-\frac {\left (6 b f^3\right ) \int \text {Li}_3\left (-e^{i (c+d x)}\right ) \, dx}{a^2 d^3}+\frac {\left (6 b f^3\right ) \int \text {Li}_3\left (e^{i (c+d x)}\right ) \, dx}{a^2 d^3}\\ &=-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {3 f (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d^2}-\frac {3 i b f (e+f x)^2 \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {3 i b f (e+f x)^2 \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}-\frac {3 i f^2 (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^3}+\frac {6 b f^2 (e+f x) \text {Li}_3\left (-e^{i (c+d x)}\right )}{a^2 d^3}-\frac {6 b f^2 (e+f x) \text {Li}_3\left (e^{i (c+d x)}\right )}{a^2 d^3}+\frac {\left (6 b^2 f^2\right ) \int (e+f x) \text {Li}_2\left (\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{a^2 \sqrt {a^2-b^2} d^2}-\frac {\left (6 b^2 f^2\right ) \int (e+f x) \text {Li}_2\left (\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{a^2 \sqrt {a^2-b^2} d^2}+\frac {\left (3 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i (c+d x)}\right )}{2 a d^4}+\frac {\left (6 i b f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(-x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a^2 d^4}-\frac {\left (6 i b f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3(x)}{x} \, dx,x,e^{i (c+d x)}\right )}{a^2 d^4}\\ &=-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {3 f (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d^2}-\frac {3 i b f (e+f x)^2 \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {3 i b f (e+f x)^2 \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}-\frac {3 i f^2 (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^3}+\frac {6 b f^2 (e+f x) \text {Li}_3\left (-e^{i (c+d x)}\right )}{a^2 d^3}-\frac {6 b f^2 (e+f x) \text {Li}_3\left (e^{i (c+d x)}\right )}{a^2 d^3}-\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^3}+\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^3}+\frac {3 f^3 \text {Li}_3\left (e^{2 i (c+d x)}\right )}{2 a d^4}+\frac {6 i b f^3 \text {Li}_4\left (-e^{i (c+d x)}\right )}{a^2 d^4}-\frac {6 i b f^3 \text {Li}_4\left (e^{i (c+d x)}\right )}{a^2 d^4}+\frac {\left (6 i b^2 f^3\right ) \int \text {Li}_3\left (\frac {2 i b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{a^2 \sqrt {a^2-b^2} d^3}-\frac {\left (6 i b^2 f^3\right ) \int \text {Li}_3\left (\frac {2 i b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{a^2 \sqrt {a^2-b^2} d^3}\\ &=-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {3 f (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d^2}-\frac {3 i b f (e+f x)^2 \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {3 i b f (e+f x)^2 \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}-\frac {3 i f^2 (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^3}+\frac {6 b f^2 (e+f x) \text {Li}_3\left (-e^{i (c+d x)}\right )}{a^2 d^3}-\frac {6 b f^2 (e+f x) \text {Li}_3\left (e^{i (c+d x)}\right )}{a^2 d^3}-\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^3}+\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^3}+\frac {3 f^3 \text {Li}_3\left (e^{2 i (c+d x)}\right )}{2 a d^4}+\frac {6 i b f^3 \text {Li}_4\left (-e^{i (c+d x)}\right )}{a^2 d^4}-\frac {6 i b f^3 \text {Li}_4\left (e^{i (c+d x)}\right )}{a^2 d^4}+\frac {\left (6 b^2 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a^2 \sqrt {a^2-b^2} d^4}-\frac {\left (6 b^2 f^3\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{a^2 \sqrt {a^2-b^2} d^4}\\ &=-\frac {i (e+f x)^3}{a d}+\frac {2 b (e+f x)^3 \tanh ^{-1}\left (e^{i (c+d x)}\right )}{a^2 d}-\frac {(e+f x)^3 \cot (c+d x)}{a d}-\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {i b^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d}+\frac {3 f (e+f x)^2 \log \left (1-e^{2 i (c+d x)}\right )}{a d^2}-\frac {3 i b f (e+f x)^2 \text {Li}_2\left (-e^{i (c+d x)}\right )}{a^2 d^2}+\frac {3 i b f (e+f x)^2 \text {Li}_2\left (e^{i (c+d x)}\right )}{a^2 d^2}-\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}+\frac {3 b^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^2}-\frac {3 i f^2 (e+f x) \text {Li}_2\left (e^{2 i (c+d x)}\right )}{a d^3}+\frac {6 b f^2 (e+f x) \text {Li}_3\left (-e^{i (c+d x)}\right )}{a^2 d^3}-\frac {6 b f^2 (e+f x) \text {Li}_3\left (e^{i (c+d x)}\right )}{a^2 d^3}-\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^3}+\frac {6 i b^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^3}+\frac {3 f^3 \text {Li}_3\left (e^{2 i (c+d x)}\right )}{2 a d^4}+\frac {6 i b f^3 \text {Li}_4\left (-e^{i (c+d x)}\right )}{a^2 d^4}-\frac {6 i b f^3 \text {Li}_4\left (e^{i (c+d x)}\right )}{a^2 d^4}+\frac {6 b^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^4}-\frac {6 b^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2} d^4}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 48.21, size = 1680, normalized size = 1.90 \[ \text {result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[((e + f*x)^3*Csc[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
[Out]
(((-2*I)*a*d^3*(e + f*x)^3)/(-1 + E^((2*I)*c)) - 3*d^2*e*f*(b*d*e - 2*a*f)*x*Log[1 - E^((-I)*(c + d*x))] - 3*d
^2*f^2*(b*d*e - a*f)*x^2*Log[1 - E^((-I)*(c + d*x))] - b*d^3*f^3*x^3*Log[1 - E^((-I)*(c + d*x))] + 3*d^2*e*f*(
b*d*e + 2*a*f)*x*Log[1 + E^((-I)*(c + d*x))] + 3*d^2*f^2*(b*d*e + a*f)*x^2*Log[1 + E^((-I)*(c + d*x))] + b*d^3
*f^3*x^3*Log[1 + E^((-I)*(c + d*x))] + I*d^2*e^2*(b*d*e - 3*a*f)*(d*x + I*Log[1 - E^(I*(c + d*x))]) + d^2*e^2*
(b*d*e + 3*a*f)*((-I)*d*x + Log[1 + E^(I*(c + d*x))]) + (3*I)*d*e*f*(b*d*e + 2*a*f)*PolyLog[2, -E^((-I)*(c + d
*x))] - (3*I)*d*e*f*(b*d*e - 2*a*f)*PolyLog[2, E^((-I)*(c + d*x))] + 6*f^2*(b*d*e + a*f)*(I*d*x*PolyLog[2, -E^
((-I)*(c + d*x))] + PolyLog[3, -E^((-I)*(c + d*x))]) + 6*f^2*(-(b*d*e) + a*f)*(I*d*x*PolyLog[2, E^((-I)*(c + d
*x))] + PolyLog[3, E^((-I)*(c + d*x))]) + 3*b*f^3*(I*d^2*x^2*PolyLog[2, -E^((-I)*(c + d*x))] + 2*d*x*PolyLog[3
, -E^((-I)*(c + d*x))] - (2*I)*PolyLog[4, -E^((-I)*(c + d*x))]) - (3*I)*b*f^3*(d^2*x^2*PolyLog[2, E^((-I)*(c +
d*x))] - (2*I)*d*x*PolyLog[3, E^((-I)*(c + d*x))] - 2*PolyLog[4, E^((-I)*(c + d*x))]))/(a^2*d^4) + (b^2*(2*Sq
rt[-a^2 + b^2]*d^3*e^3*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]] + 3*Sqrt[a^2 - b^2]*d^3*e^2*f*x*Log[1
- (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + 3*Sqrt[a^2 - b^2]*d^3*e*f^2*x^2*Log[1 - (b*E^(I*(c + d*x
)))/((-I)*a + Sqrt[-a^2 + b^2])] + Sqrt[a^2 - b^2]*d^3*f^3*x^3*Log[1 - (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2
+ b^2])] - 3*Sqrt[a^2 - b^2]*d^3*e^2*f*x*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] - 3*Sqrt[a^2 -
b^2]*d^3*e*f^2*x^2*Log[1 + (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] - Sqrt[a^2 - b^2]*d^3*f^3*x^3*Log[1
+ (b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2])] - (3*I)*Sqrt[a^2 - b^2]*d^2*f*(e + f*x)^2*PolyLog[2, (b*E^(I*(
c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] + (3*I)*Sqrt[a^2 - b^2]*d^2*f*(e + f*x)^2*PolyLog[2, -((b*E^(I*(c + d*
x)))/(I*a + Sqrt[-a^2 + b^2]))] + 6*Sqrt[a^2 - b^2]*d*e*f^2*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2
+ b^2])] + 6*Sqrt[a^2 - b^2]*d*f^3*x*PolyLog[3, (b*E^(I*(c + d*x)))/((-I)*a + Sqrt[-a^2 + b^2])] - 6*Sqrt[a^2
- b^2]*d*e*f^2*PolyLog[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] - 6*Sqrt[a^2 - b^2]*d*f^3*x*PolyLo
g[3, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2 + b^2]))] + (6*I)*Sqrt[a^2 - b^2]*f^3*PolyLog[4, (b*E^(I*(c + d*x)
))/((-I)*a + Sqrt[-a^2 + b^2])] - (6*I)*Sqrt[a^2 - b^2]*f^3*PolyLog[4, -((b*E^(I*(c + d*x)))/(I*a + Sqrt[-a^2
+ b^2]))]))/(a^2*Sqrt[-(a^2 - b^2)^2]*d^4) + (Csc[c/2]*Csc[c/2 + (d*x)/2]*(e^3*Sin[(d*x)/2] + 3*e^2*f*x*Sin[(d
*x)/2] + 3*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2]))/(2*a*d) + (Sec[c/2]*Sec[c/2 + (d*x)/2]*(e^3*Sin[(d*
x)/2] + 3*e^2*f*x*Sin[(d*x)/2] + 3*e*f^2*x^2*Sin[(d*x)/2] + f^3*x^3*Sin[(d*x)/2]))/(2*a*d)
________________________________________________________________________________________
fricas [C] time = 1.01, size = 4584, normalized size = 5.20 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)^3*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")
[Out]
-1/4*(12*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(b*cos(d*x
+ c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b)*sin(d*x + c) - 12*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*polylo
g(4, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)
)/b)*sin(d*x + c) - 12*I*b^3*f^3*sqrt(-(a^2 - b^2)/b^2)*polylog(4, 1/2*(-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c)
+ 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b)*sin(d*x + c) + 12*I*b^3*f^3*sqrt(-(a^2 - b
^2)/b^2)*polylog(4, 1/2*(-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-
(a^2 - b^2)/b^2))/b)*sin(d*x + c) + 12*I*(a^2*b - b^3)*f^3*polylog(4, cos(d*x + c) + I*sin(d*x + c))*sin(d*x +
c) - 12*I*(a^2*b - b^3)*f^3*polylog(4, cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + 12*I*(a^2*b - b^3)*f^3*p
olylog(4, -cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) - 12*I*(a^2*b - b^3)*f^3*polylog(4, -cos(d*x + c) - I*s
in(d*x + c))*sin(d*x + c) + 2*(-3*I*b^3*d^2*f^3*x^2 - 6*I*b^3*d^2*e*f^2*x - 3*I*b^3*d^2*e^2*f)*sqrt(-(a^2 - b^
2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 -
b^2)/b^2) + 2*b)/b + 1)*sin(d*x + c) + 2*(3*I*b^3*d^2*f^3*x^2 + 6*I*b^3*d^2*e*f^2*x + 3*I*b^3*d^2*e^2*f)*sqrt
(-(a^2 - b^2)/b^2)*dilog(-1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*s
qrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1)*sin(d*x + c) + 2*(3*I*b^3*d^2*f^3*x^2 + 6*I*b^3*d^2*e*f^2*x + 3*I*b^3*d^2*
e^2*f)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin
(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1)*sin(d*x + c) + 2*(-3*I*b^3*d^2*f^3*x^2 - 6*I*b^3*d^2*e*f^2*x -
3*I*b^3*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*dilog(-1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x +
c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b + 1)*sin(d*x + c) - 2*(b^3*d^3*e^3 - 3*b^3*c*d^2*e^2*f
+ 3*b^3*c^2*d*e*f^2 - b^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(2*b*cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqr
t(-(a^2 - b^2)/b^2) + 2*I*a)*sin(d*x + c) - 2*(b^3*d^3*e^3 - 3*b^3*c*d^2*e^2*f + 3*b^3*c^2*d*e*f^2 - b^3*c^3*f
^3)*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)*sin
(d*x + c) + 2*(b^3*d^3*e^3 - 3*b^3*c*d^2*e^2*f + 3*b^3*c^2*d*e*f^2 - b^3*c^3*f^3)*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)*sin(d*x + c) + 2*(b^3*d^3*e^3 - 3*
b^3*c*d^2*e^2*f + 3*b^3*c^2*d*e*f^2 - b^3*c^3*f^3)*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)*sin(d*x + c) - 2*(b^3*d^3*f^3*x^3 + 3*b^3*d^3*e*f^2*x^2 + 3*b^3*d
^3*e^2*f*x + 3*b^3*c*d^2*e^2*f - 3*b^3*c^2*d*e*f^2 + b^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*
x + c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b)*sin(d*x + c
) + 2*(b^3*d^3*f^3*x^3 + 3*b^3*d^3*e*f^2*x^2 + 3*b^3*d^3*e^2*f*x + 3*b^3*c*d^2*e^2*f - 3*b^3*c^2*d*e*f^2 + b^3
*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(
d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b)*sin(d*x + c) - 2*(b^3*d^3*f^3*x^3 + 3*b^3*d^3*e*f^2*x^2 + 3*b^3*d^3
*e^2*f*x + 3*b^3*c*d^2*e^2*f - 3*b^3*c^2*d*e*f^2 + b^3*c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x
+ c) + 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b)*sin(d*x + c)
+ 2*(b^3*d^3*f^3*x^3 + 3*b^3*d^3*e*f^2*x^2 + 3*b^3*d^3*e^2*f*x + 3*b^3*c*d^2*e^2*f - 3*b^3*c^2*d*e*f^2 + b^3*
c^3*f^3)*sqrt(-(a^2 - b^2)/b^2)*log(1/2*(-2*I*a*cos(d*x + c) + 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(
d*x + c))*sqrt(-(a^2 - b^2)/b^2) + 2*b)/b)*sin(d*x + c) + 12*(b^3*d*f^3*x + b^3*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2
)*polylog(3, 1/2*(2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b
^2)/b^2))/b)*sin(d*x + c) - 12*(b^3*d*f^3*x + b^3*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, 1/2*(2*I*a*cos(d*
x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b)*sin(d*x + c) + 12
*(b^3*d*f^3*x + b^3*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, 1/2*(-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) + 2
*(b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b)*sin(d*x + c) - 12*(b^3*d*f^3*x + b^3*d*e*f^2)*
sqrt(-(a^2 - b^2)/b^2)*polylog(3, 1/2*(-2*I*a*cos(d*x + c) - 2*a*sin(d*x + c) - 2*(b*cos(d*x + c) - I*b*sin(d*
x + c))*sqrt(-(a^2 - b^2)/b^2))/b)*sin(d*x + c) + (-6*I*(a^2*b - b^3)*d^2*f^3*x^2 - 6*I*(a^2*b - b^3)*d^2*e^2*
f + 12*I*(a^3 - a*b^2)*d*e*f^2 - 12*I*((a^2*b - b^3)*d^2*e*f^2 - (a^3 - a*b^2)*d*f^3)*x)*dilog(cos(d*x + c) +
I*sin(d*x + c))*sin(d*x + c) + (6*I*(a^2*b - b^3)*d^2*f^3*x^2 + 6*I*(a^2*b - b^3)*d^2*e^2*f - 12*I*(a^3 - a*b^
2)*d*e*f^2 + 12*I*((a^2*b - b^3)*d^2*e*f^2 - (a^3 - a*b^2)*d*f^3)*x)*dilog(cos(d*x + c) - I*sin(d*x + c))*sin(
d*x + c) + (-6*I*(a^2*b - b^3)*d^2*f^3*x^2 - 6*I*(a^2*b - b^3)*d^2*e^2*f - 12*I*(a^3 - a*b^2)*d*e*f^2 - 12*I*(
(a^2*b - b^3)*d^2*e*f^2 + (a^3 - a*b^2)*d*f^3)*x)*dilog(-cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + (6*I*(a
^2*b - b^3)*d^2*f^3*x^2 + 6*I*(a^2*b - b^3)*d^2*e^2*f + 12*I*(a^3 - a*b^2)*d*e*f^2 + 12*I*((a^2*b - b^3)*d^2*e
*f^2 + (a^3 - a*b^2)*d*f^3)*x)*dilog(-cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) - 2*((a^2*b - b^3)*d^3*f^3*x
^3 + (a^2*b - b^3)*d^3*e^3 + 3*(a^3 - a*b^2)*d^2*e^2*f + 3*((a^2*b - b^3)*d^3*e*f^2 + (a^3 - a*b^2)*d^2*f^3)*x
^2 + 3*((a^2*b - b^3)*d^3*e^2*f + 2*(a^3 - a*b^2)*d^2*e*f^2)*x)*log(cos(d*x + c) + I*sin(d*x + c) + 1)*sin(d*x
+ c) - 2*((a^2*b - b^3)*d^3*f^3*x^3 + (a^2*b - b^3)*d^3*e^3 + 3*(a^3 - a*b^2)*d^2*e^2*f + 3*((a^2*b - b^3)*d^
3*e*f^2 + (a^3 - a*b^2)*d^2*f^3)*x^2 + 3*((a^2*b - b^3)*d^3*e^2*f + 2*(a^3 - a*b^2)*d^2*e*f^2)*x)*log(cos(d*x
+ c) - I*sin(d*x + c) + 1)*sin(d*x + c) + 2*((a^2*b - b^3)*d^3*e^3 - 3*(a^3 - a*b^2 + (a^2*b - b^3)*c)*d^2*e^2
*f + 3*((a^2*b - b^3)*c^2 + 2*(a^3 - a*b^2)*c)*d*e*f^2 - ((a^2*b - b^3)*c^3 + 3*(a^3 - a*b^2)*c^2)*f^3)*log(-1
/2*cos(d*x + c) + 1/2*I*sin(d*x + c) + 1/2)*sin(d*x + c) + 2*((a^2*b - b^3)*d^3*e^3 - 3*(a^3 - a*b^2 + (a^2*b
- b^3)*c)*d^2*e^2*f + 3*((a^2*b - b^3)*c^2 + 2*(a^3 - a*b^2)*c)*d*e*f^2 - ((a^2*b - b^3)*c^3 + 3*(a^3 - a*b^2)
*c^2)*f^3)*log(-1/2*cos(d*x + c) - 1/2*I*sin(d*x + c) + 1/2)*sin(d*x + c) + 2*((a^2*b - b^3)*d^3*f^3*x^3 + 3*(
a^2*b - b^3)*c*d^2*e^2*f - 3*((a^2*b - b^3)*c^2 + 2*(a^3 - a*b^2)*c)*d*e*f^2 + ((a^2*b - b^3)*c^3 + 3*(a^3 - a
*b^2)*c^2)*f^3 + 3*((a^2*b - b^3)*d^3*e*f^2 - (a^3 - a*b^2)*d^2*f^3)*x^2 + 3*((a^2*b - b^3)*d^3*e^2*f - 2*(a^3
- a*b^2)*d^2*e*f^2)*x)*log(-cos(d*x + c) + I*sin(d*x + c) + 1)*sin(d*x + c) + 2*((a^2*b - b^3)*d^3*f^3*x^3 +
3*(a^2*b - b^3)*c*d^2*e^2*f - 3*((a^2*b - b^3)*c^2 + 2*(a^3 - a*b^2)*c)*d*e*f^2 + ((a^2*b - b^3)*c^3 + 3*(a^3
- a*b^2)*c^2)*f^3 + 3*((a^2*b - b^3)*d^3*e*f^2 - (a^3 - a*b^2)*d^2*f^3)*x^2 + 3*((a^2*b - b^3)*d^3*e^2*f - 2*(
a^3 - a*b^2)*d^2*e*f^2)*x)*log(-cos(d*x + c) - I*sin(d*x + c) + 1)*sin(d*x + c) + 12*((a^2*b - b^3)*d*f^3*x +
(a^2*b - b^3)*d*e*f^2 - (a^3 - a*b^2)*f^3)*polylog(3, cos(d*x + c) + I*sin(d*x + c))*sin(d*x + c) + 12*((a^2*b
- b^3)*d*f^3*x + (a^2*b - b^3)*d*e*f^2 - (a^3 - a*b^2)*f^3)*polylog(3, cos(d*x + c) - I*sin(d*x + c))*sin(d*x
+ c) - 12*((a^2*b - b^3)*d*f^3*x + (a^2*b - b^3)*d*e*f^2 + (a^3 - a*b^2)*f^3)*polylog(3, -cos(d*x + c) + I*si
n(d*x + c))*sin(d*x + c) - 12*((a^2*b - b^3)*d*f^3*x + (a^2*b - b^3)*d*e*f^2 + (a^3 - a*b^2)*f^3)*polylog(3, -
cos(d*x + c) - I*sin(d*x + c))*sin(d*x + c) + 4*((a^3 - a*b^2)*d^3*f^3*x^3 + 3*(a^3 - a*b^2)*d^3*e*f^2*x^2 + 3
*(a^3 - a*b^2)*d^3*e^2*f*x + (a^3 - a*b^2)*d^3*e^3)*cos(d*x + c))/((a^4 - a^2*b^2)*d^4*sin(d*x + c))
________________________________________________________________________________________
giac [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)^3*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")
[Out]
Timed out
________________________________________________________________________________________
maple [F] time = 4.66, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \left (\csc ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((f*x+e)^3*csc(d*x+c)^2/(a+b*sin(d*x+c)),x)
[Out]
int((f*x+e)^3*csc(d*x+c)^2/(a+b*sin(d*x+c)),x)
________________________________________________________________________________________
maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)^3*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?
________________________________________________________________________________________
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)^3/(sin(c + d*x)^2*(a + b*sin(c + d*x))),x)
[Out]
\text{Hanged}
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (e + f x\right )^{3} \csc ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((f*x+e)**3*csc(d*x+c)**2/(a+b*sin(d*x+c)),x)
[Out]
Integral((e + f*x)**3*csc(c + d*x)**2/(a + b*sin(c + d*x)), x)
________________________________________________________________________________________