3.247 \(\int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=1512 \[ \frac {6 a \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{b \left (a^2-b^2\right ) d^4}+\frac {6 a \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{b \left (a^2-b^2\right ) d^4}+\frac {6 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{b \sqrt {a^2-b^2} d^4}-\frac {6 a^2 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {6 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{b \sqrt {a^2-b^2} d^4}+\frac {6 a^2 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {6 i a (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{b \left (a^2-b^2\right ) d^3}-\frac {6 i a (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{b \left (a^2-b^2\right ) d^3}-\frac {6 i (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{b \sqrt {a^2-b^2} d^3}+\frac {6 i a^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}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {6 i (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{b \sqrt {a^2-b^2} d^3}-\frac {6 i a^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}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {3 a (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{b \left (a^2-b^2\right ) d^2}+\frac {3 a (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{b \left (a^2-b^2\right ) d^2}-\frac {3 (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{b \sqrt {a^2-b^2} d^2}+\frac {3 a^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}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{b \sqrt {a^2-b^2} d^2}-\frac {3 a^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}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}-\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]

[Out]

-6*I*a*f^2*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^3+3*a*f*(f*x+e)^2*ln(1-I*b*
exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^2-6*I*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^
(1/2)))/b/d^3/(a^2-b^2)^(1/2)+3*a*f*(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^2-I*(
f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/d/(a^2-b^2)^(1/2)-6*I*a^2*f^2*(f*x+e)*polylog(3,I*b*ex
p(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^3+3*a^2*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2
-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2+6*I*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/d^3/(a
^2-b^2)^(1/2)-3*a^2*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^2+6*a*f^
3*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^4-6*I*a*f^2*(f*x+e)*polylog(2,I*b*exp(I*(d*x
+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)/d^3+6*a*f^3*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2
)/d^4+I*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/d/(a^2-b^2)^(1/2)-6*a^2*f^3*polylog(4,I*b*exp
(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^4+6*a^2*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/
2)))/b/(a^2-b^2)^(3/2)/d^4-a*(f*x+e)^3*cos(d*x+c)/(a^2-b^2)/d/(a+b*sin(d*x+c))-I*a^2*(f*x+e)^3*ln(1-I*b*exp(I*
(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d-I*a*(f*x+e)^3/b/(a^2-b^2)/d-3*f*(f*x+e)^2*polylog(2,I*b*exp(
I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/d^2/(a^2-b^2)^(1/2)+3*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^
(1/2)))/b/d^2/(a^2-b^2)^(1/2)+I*a^2*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d
+6*I*a^2*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(3/2)/d^3+6*f^3*polylog(4,I
*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/d^4/(a^2-b^2)^(1/2)-6*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(
1/2)))/b/d^4/(a^2-b^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 3.07, antiderivative size = 1512, normalized size of antiderivative = 1.00, number of steps used = 36, number of rules used = 10, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6742, 3324, 3323, 2264, 2190, 2531, 6609, 2282, 6589, 4519} \[ \frac {6 a \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{b \left (a^2-b^2\right ) d^4}+\frac {6 a \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{b \left (a^2-b^2\right ) d^4}+\frac {6 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{b \sqrt {a^2-b^2} d^4}-\frac {6 a^2 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^3}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {6 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{b \sqrt {a^2-b^2} d^4}+\frac {6 a^2 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^3}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {6 i a (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{b \left (a^2-b^2\right ) d^3}-\frac {6 i a (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{b \left (a^2-b^2\right ) d^3}-\frac {6 i (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f^2}{b \sqrt {a^2-b^2} d^3}+\frac {6 i a^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}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {6 i (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f^2}{b \sqrt {a^2-b^2} d^3}-\frac {6 i a^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}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {3 a (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{b \left (a^2-b^2\right ) d^2}+\frac {3 a (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{b \left (a^2-b^2\right ) d^2}-\frac {3 (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) f}{b \sqrt {a^2-b^2} d^2}+\frac {3 a^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}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) f}{b \sqrt {a^2-b^2} d^2}-\frac {3 a^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}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}-\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)^3*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

((-I)*a*(e + f*x)^3)/(b*(a^2 - b^2)*d) + (3*a*f*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2]
)])/(b*(a^2 - b^2)*d^2) + (I*a^2*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b
^2)^(3/2)*d) - (I*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d) + (3
*a*f*(e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^2) - (I*a^2*(e + f*x)^
3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d) + (I*(e + f*x)^3*Log[1 - (I*b*
E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d) - ((6*I)*a*f^2*(e + f*x)*PolyLog[2, (I*b*E^(I*(
c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^3) + (3*a^2*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))
/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^2) - (3*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sq
rt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) - ((6*I)*a*f^2*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a
^2 - b^2])])/(b*(a^2 - b^2)*d^3) - (3*a^2*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])
])/(b*(a^2 - b^2)^(3/2)*d^2) + (3*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sq
rt[a^2 - b^2]*d^2) + (6*a*f^3*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^4) + (
(6*I)*a^2*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3) - (
(6*I)*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) + (6*a*f^
3*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^4) - ((6*I)*a^2*f^2*(e + f*x)*Poly
Log[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^3) + ((6*I)*f^2*(e + f*x)*PolyLog[
3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^3) - (6*a^2*f^3*PolyLog[4, (I*b*E^(I*(c
+ d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^4) + (6*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt
[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^4) + (6*a^2*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(
b*(a^2 - b^2)^(3/2)*d^4) - (6*f^3*PolyLog[4, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*
d^4) - (a*(e + f*x)^3*Cos[c + d*x])/((a^2 - b^2)*d*(a + b*Sin[c + d*x]))

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 3324

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(b*(c + d*x)^m*Cos[
e + f*x])/(f*(a^2 - b^2)*(a + b*Sin[e + f*x])), x] + (Dist[a/(a^2 - b^2), Int[(c + d*x)^m/(a + b*Sin[e + f*x])
, x], x] - Dist[(b*d*m)/(f*(a^2 - b^2)), Int[((c + d*x)^(m - 1)*Cos[e + f*x])/(a + b*Sin[e + f*x]), x], x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 4519

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
-Simp[(I*(e + f*x)^(m + 1))/(b*f*(m + 1)), x] + (Int[((e + f*x)^m*E^(I*(c + d*x)))/(a - Rt[a^2 - b^2, 2] - I*b
*E^(I*(c + d*x))), x] + Int[((e + f*x)^m*E^(I*(c + d*x)))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x))), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

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]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \frac {(e+f x)^3 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx &=\int \left (-\frac {a (e+f x)^3}{b (a+b \sin (c+d x))^2}+\frac {(e+f x)^3}{b (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac {\int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{b}-\frac {a \int \frac {(e+f x)^3}{(a+b \sin (c+d x))^2} \, dx}{b}\\ &=-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {2 \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}{b}-\frac {a^2 \int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{b \left (a^2-b^2\right )}+\frac {(3 a f) \int \frac {(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (2 a^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}{b \left (a^2-b^2\right )}-\frac {(2 i) \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}{\sqrt {a^2-b^2}}+\frac {(2 i) \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}{\sqrt {a^2-b^2}}+\frac {(3 a f) \int \frac {e^{i (c+d x)} (e+f x)^2}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right ) d}+\frac {(3 a f) \int \frac {e^{i (c+d x)} (e+f x)^2}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\left (2 i a^2\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}{\left (a^2-b^2\right )^{3/2}}-\frac {\left (2 i a^2\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}{\left (a^2-b^2\right )^{3/2}}+\frac {(3 i f) \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}{b \sqrt {a^2-b^2} d}-\frac {(3 i f) \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}{b \sqrt {a^2-b^2} d}-\frac {\left (6 a f^2\right ) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right ) d^2}-\frac {\left (6 a f^2\right ) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right ) d^2}\\ &=-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}-\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (3 i a^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}{b \left (a^2-b^2\right )^{3/2} d}+\frac {\left (3 i a^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}{b \left (a^2-b^2\right )^{3/2} d}+\frac {\left (6 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}{b \sqrt {a^2-b^2} d^2}-\frac {\left (6 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}{b \sqrt {a^2-b^2} d^2}+\frac {\left (6 i a f^3\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right ) d^3}+\frac {\left (6 i a f^3\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b \left (a^2-b^2\right ) d^3}\\ &=-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {3 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}-\frac {3 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (6 a^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}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {\left (6 a^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}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {\left (6 a f^3\right ) \operatorname {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 )}{b \left (a^2-b^2\right ) d^4}+\frac {\left (6 a f^3\right ) \operatorname {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 )}{b \left (a^2-b^2\right ) d^4}+\frac {\left (6 i 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}{b \sqrt {a^2-b^2} d^3}-\frac {\left (6 i 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}{b \sqrt {a^2-b^2} d^3}\\ &=-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {3 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}-\frac {3 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}+\frac {6 a f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}+\frac {6 i a^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 )}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}+\frac {6 a f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}-\frac {6 i a^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 )}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}+\frac {\left (6 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 )}{b \sqrt {a^2-b^2} d^4}-\frac {\left (6 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 )}{b \sqrt {a^2-b^2} d^4}-\frac {\left (6 i a^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}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {\left (6 i a^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}{b \left (a^2-b^2\right )^{3/2} d^3}\\ &=-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {3 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}-\frac {3 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}+\frac {6 a f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}+\frac {6 i a^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 )}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}+\frac {6 a f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}-\frac {6 i a^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 )}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}+\frac {6 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}-\frac {6 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}-\frac {\left (6 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^4}+\frac {\left (6 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^4}\\ &=-\frac {i a (e+f x)^3}{b \left (a^2-b^2\right ) d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {3 a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^2}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {3 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {6 i a f^2 (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}-\frac {3 a^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 )}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}+\frac {6 a f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}+\frac {6 i a^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 )}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {6 i f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}+\frac {6 a f^3 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}-\frac {6 i a^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 )}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {6 i f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}-\frac {6 a^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^4}+\frac {6 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}+\frac {6 a^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {6 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}-\frac {a (e+f x)^3 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))}\\ \end {align*}

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Mathematica [B]  time = 22.32, size = 5446, normalized size = 3.60 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[((e + f*x)^3*Sin[c + d*x])/(a + b*Sin[c + d*x])^2,x]

[Out]

Result too large to show

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fricas [C]  time = 1.00, size = 5200, normalized size = 3.44 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/4*(2*(6*I*b^4*f^3*sin(d*x + c) + 6*I*a*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) + 2*(-6*I*b^4*f^3*sin(d*x
+ c) - 6*I*a*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) + 2*(-6*I*b^4*f^3*sin(d*x + c) - 6*I*a*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) + 2*(6*I*b^4*f^3*sin(d*x + c) + 6*I*a*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)
- 4*((a^3*b - a*b^3)*d^3*f^3*x^3 + 3*(a^3*b - a*b^3)*d^3*e*f^2*x^2 + 3*(a^3*b - a*b^3)*d^3*e^2*f*x + (a^3*b -
a*b^3)*d^3*e^3)*cos(d*x + c) + (12*I*(a^4 - a^2*b^2)*d*f^3*x + 12*I*(a^4 - a^2*b^2)*d*e*f^2 + (12*I*(a^3*b -
a*b^3)*d*f^3*x + 12*I*(a^3*b - a*b^3)*d*e*f^2)*sin(d*x + c) + 2*(-3*I*a*b^3*d^2*f^3*x^2 - 6*I*a*b^3*d^2*e*f^2*
x - 3*I*a*b^3*d^2*e^2*f + (-3*I*b^4*d^2*f^3*x^2 - 6*I*b^4*d^2*e*f^2*x - 3*I*b^4*d^2*e^2*f)*sin(d*x + c))*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) + (12*I*(a^4 - a^2*b^2)*d*f^3*x + 12*I*(a^4 - a^2*b^2)*d*e*f^2 + (12*I*(a^
3*b - a*b^3)*d*f^3*x + 12*I*(a^3*b - a*b^3)*d*e*f^2)*sin(d*x + c) + 2*(3*I*a*b^3*d^2*f^3*x^2 + 6*I*a*b^3*d^2*e
*f^2*x + 3*I*a*b^3*d^2*e^2*f + (3*I*b^4*d^2*f^3*x^2 + 6*I*b^4*d^2*e*f^2*x + 3*I*b^4*d^2*e^2*f)*sin(d*x + c))*s
qrt(-(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) + (-12*I*(a^4 - a^2*b^2)*d*f^3*x - 12*I*(a^4 - a^2*b^2)*d*e*f^2 + (-12
*I*(a^3*b - a*b^3)*d*f^3*x - 12*I*(a^3*b - a*b^3)*d*e*f^2)*sin(d*x + c) + 2*(3*I*a*b^3*d^2*f^3*x^2 + 6*I*a*b^3
*d^2*e*f^2*x + 3*I*a*b^3*d^2*e^2*f + (3*I*b^4*d^2*f^3*x^2 + 6*I*b^4*d^2*e*f^2*x + 3*I*b^4*d^2*e^2*f)*sin(d*x +
c))*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) + (-12*I*(a^4 - a^2*b^2)*d*f^3*x - 12*I*(a^4 - a^2*b^2)*d*e*f^2
+ (-12*I*(a^3*b - a*b^3)*d*f^3*x - 12*I*(a^3*b - a*b^3)*d*e*f^2)*sin(d*x + c) + 2*(-3*I*a*b^3*d^2*f^3*x^2 - 6
*I*a*b^3*d^2*e*f^2*x - 3*I*a*b^3*d^2*e^2*f + (-3*I*b^4*d^2*f^3*x^2 - 6*I*b^4*d^2*e*f^2*x - 3*I*b^4*d^2*e^2*f)*
sin(d*x + c))*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) + 2*(3*(a^4 - a^2*b^2)*d^2*e^2*f - 6*(a^4 - a^2*b^2)*c
*d*e*f^2 + 3*(a^4 - a^2*b^2)*c^2*f^3 + 3*((a^3*b - a*b^3)*d^2*e^2*f - 2*(a^3*b - a*b^3)*c*d*e*f^2 + (a^3*b - a
*b^3)*c^2*f^3)*sin(d*x + c) - (a*b^3*d^3*e^3 - 3*a*b^3*c*d^2*e^2*f + 3*a*b^3*c^2*d*e*f^2 - a*b^3*c^3*f^3 + (b^
4*d^3*e^3 - 3*b^4*c*d^2*e^2*f + 3*b^4*c^2*d*e*f^2 - b^4*c^3*f^3)*sin(d*x + c))*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) + 2*(3*(a^4 - a^2*b^2)*d^2*e^2*f - 6*
(a^4 - a^2*b^2)*c*d*e*f^2 + 3*(a^4 - a^2*b^2)*c^2*f^3 + 3*((a^3*b - a*b^3)*d^2*e^2*f - 2*(a^3*b - a*b^3)*c*d*e
*f^2 + (a^3*b - a*b^3)*c^2*f^3)*sin(d*x + c) - (a*b^3*d^3*e^3 - 3*a*b^3*c*d^2*e^2*f + 3*a*b^3*c^2*d*e*f^2 - a*
b^3*c^3*f^3 + (b^4*d^3*e^3 - 3*b^4*c*d^2*e^2*f + 3*b^4*c^2*d*e*f^2 - b^4*c^3*f^3)*sin(d*x + c))*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) + 2*(3*(a^4 - a^2*b^
2)*d^2*e^2*f - 6*(a^4 - a^2*b^2)*c*d*e*f^2 + 3*(a^4 - a^2*b^2)*c^2*f^3 + 3*((a^3*b - a*b^3)*d^2*e^2*f - 2*(a^3
*b - a*b^3)*c*d*e*f^2 + (a^3*b - a*b^3)*c^2*f^3)*sin(d*x + c) + (a*b^3*d^3*e^3 - 3*a*b^3*c*d^2*e^2*f + 3*a*b^3
*c^2*d*e*f^2 - a*b^3*c^3*f^3 + (b^4*d^3*e^3 - 3*b^4*c*d^2*e^2*f + 3*b^4*c^2*d*e*f^2 - b^4*c^3*f^3)*sin(d*x + c
))*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) +
2*(3*(a^4 - a^2*b^2)*d^2*e^2*f - 6*(a^4 - a^2*b^2)*c*d*e*f^2 + 3*(a^4 - a^2*b^2)*c^2*f^3 + 3*((a^3*b - a*b^3)*
d^2*e^2*f - 2*(a^3*b - a*b^3)*c*d*e*f^2 + (a^3*b - a*b^3)*c^2*f^3)*sin(d*x + c) + (a*b^3*d^3*e^3 - 3*a*b^3*c*d
^2*e^2*f + 3*a*b^3*c^2*d*e*f^2 - a*b^3*c^3*f^3 + (b^4*d^3*e^3 - 3*b^4*c*d^2*e^2*f + 3*b^4*c^2*d*e*f^2 - b^4*c^
3*f^3)*sin(d*x + c))*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) + 2*(3*(a^4 - a^2*b^2)*d^2*f^3*x^2 + 6*(a^4 - a^2*b^2)*d^2*e*f^2*x + 6*(a^4 - a^2*b^2)*c*d*e*f
^2 - 3*(a^4 - a^2*b^2)*c^2*f^3 + 3*((a^3*b - a*b^3)*d^2*f^3*x^2 + 2*(a^3*b - a*b^3)*d^2*e*f^2*x + 2*(a^3*b - a
*b^3)*c*d*e*f^2 - (a^3*b - a*b^3)*c^2*f^3)*sin(d*x + c) - (a*b^3*d^3*f^3*x^3 + 3*a*b^3*d^3*e*f^2*x^2 + 3*a*b^3
*d^3*e^2*f*x + 3*a*b^3*c*d^2*e^2*f - 3*a*b^3*c^2*d*e*f^2 + a*b^3*c^3*f^3 + (b^4*d^3*f^3*x^3 + 3*b^4*d^3*e*f^2*
x^2 + 3*b^4*d^3*e^2*f*x + 3*b^4*c*d^2*e^2*f - 3*b^4*c^2*d*e*f^2 + b^4*c^3*f^3)*sin(d*x + c))*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) + 2*(3*(a^4 - a^2*b^2)*d^2*f^3*x^2 + 6*(a^4 - a^2*b^2)*d^2*e*f^2*x + 6*(a^4 - a^2*b^2)*c*d*e*
f^2 - 3*(a^4 - a^2*b^2)*c^2*f^3 + 3*((a^3*b - a*b^3)*d^2*f^3*x^2 + 2*(a^3*b - a*b^3)*d^2*e*f^2*x + 2*(a^3*b -
a*b^3)*c*d*e*f^2 - (a^3*b - a*b^3)*c^2*f^3)*sin(d*x + c) + (a*b^3*d^3*f^3*x^3 + 3*a*b^3*d^3*e*f^2*x^2 + 3*a*b^
3*d^3*e^2*f*x + 3*a*b^3*c*d^2*e^2*f - 3*a*b^3*c^2*d*e*f^2 + a*b^3*c^3*f^3 + (b^4*d^3*f^3*x^3 + 3*b^4*d^3*e*f^2
*x^2 + 3*b^4*d^3*e^2*f*x + 3*b^4*c*d^2*e^2*f - 3*b^4*c^2*d*e*f^2 + b^4*c^3*f^3)*sin(d*x + c))*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) + 2*(3*(a^4 - a^2*b^2)*d^2*f^3*x^2 + 6*(a^4 - a^2*b^2)*d^2*e*f^2*x + 6*(a^4 - a^2*b^2)*c*d*e
*f^2 - 3*(a^4 - a^2*b^2)*c^2*f^3 + 3*((a^3*b - a*b^3)*d^2*f^3*x^2 + 2*(a^3*b - a*b^3)*d^2*e*f^2*x + 2*(a^3*b -
a*b^3)*c*d*e*f^2 - (a^3*b - a*b^3)*c^2*f^3)*sin(d*x + c) - (a*b^3*d^3*f^3*x^3 + 3*a*b^3*d^3*e*f^2*x^2 + 3*a*b
^3*d^3*e^2*f*x + 3*a*b^3*c*d^2*e^2*f - 3*a*b^3*c^2*d*e*f^2 + a*b^3*c^3*f^3 + (b^4*d^3*f^3*x^3 + 3*b^4*d^3*e*f^
2*x^2 + 3*b^4*d^3*e^2*f*x + 3*b^4*c*d^2*e^2*f - 3*b^4*c^2*d*e*f^2 + b^4*c^3*f^3)*sin(d*x + c))*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) + 2*(3*(a^4 - a^2*b^2)*d^2*f^3*x^2 + 6*(a^4 - a^2*b^2)*d^2*e*f^2*x + 6*(a^4 - a^2*b^2)*c*d
*e*f^2 - 3*(a^4 - a^2*b^2)*c^2*f^3 + 3*((a^3*b - a*b^3)*d^2*f^3*x^2 + 2*(a^3*b - a*b^3)*d^2*e*f^2*x + 2*(a^3*b
- a*b^3)*c*d*e*f^2 - (a^3*b - a*b^3)*c^2*f^3)*sin(d*x + c) + (a*b^3*d^3*f^3*x^3 + 3*a*b^3*d^3*e*f^2*x^2 + 3*a
*b^3*d^3*e^2*f*x + 3*a*b^3*c*d^2*e^2*f - 3*a*b^3*c^2*d*e*f^2 + a*b^3*c^3*f^3 + (b^4*d^3*f^3*x^3 + 3*b^4*d^3*e*
f^2*x^2 + 3*b^4*d^3*e^2*f*x + 3*b^4*c*d^2*e^2*f - 3*b^4*c^2*d*e*f^2 + b^4*c^3*f^3)*sin(d*x + c))*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) + 12*((a^3*b - a*b^3)*f^3*sin(d*x + c) + (a^4 - a^2*b^2)*f^3 + (a*b^3*d*f^3*x + a*b^3*d*
e*f^2 + (b^4*d*f^3*x + b^4*d*e*f^2)*sin(d*x + c))*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) + 12*((a^3*b - a*b^3)*f^3
*sin(d*x + c) + (a^4 - a^2*b^2)*f^3 - (a*b^3*d*f^3*x + a*b^3*d*e*f^2 + (b^4*d*f^3*x + b^4*d*e*f^2)*sin(d*x + c
))*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) + 12*((a^3*b - a*b^3)*f^3*sin(d*x + c) + (a^4 - a^2*b^2)*f^3 + (a*b^3*d*
f^3*x + a*b^3*d*e*f^2 + (b^4*d*f^3*x + b^4*d*e*f^2)*sin(d*x + c))*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) + 12*((a
^3*b - a*b^3)*f^3*sin(d*x + c) + (a^4 - a^2*b^2)*f^3 - (a*b^3*d*f^3*x + a*b^3*d*e*f^2 + (b^4*d*f^3*x + b^4*d*e
*f^2)*sin(d*x + c))*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))/((a^4*b^2 - 2*a^2*b^4 + b^6)*d^4*sin(d*x + c) + (a^5*
b - 2*a^3*b^3 + a*b^5)*d^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (f x + e\right )}^{3} \sin \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((f*x + e)^3*sin(d*x + c)/(b*sin(d*x + c) + a)^2, x)

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maple [F]  time = 3.01, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \sin \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^2,x)

[Out]

int((f*x+e)^3*sin(d*x+c)/(a+b*sin(d*x+c))^2,x)

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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*sin(d*x+c)/(a+b*sin(d*x+c))^2,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?

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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((sin(c + d*x)*(e + f*x)^3)/(a + b*sin(c + d*x))^2,x)

[Out]

\text{Hanged}

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sympy [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*sin(d*x+c)/(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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