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

Optimal. Leaf size=643 \[ \frac {6 a^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^4 \sqrt {a^2-b^2}}-\frac {6 a^2 f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^4 \sqrt {a^2-b^2}}-\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3 \sqrt {a^2-b^2}}+\frac {6 i a^2 f^2 (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^3 \sqrt {a^2-b^2}}-\frac {3 a^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2 \sqrt {a^2-b^2}}+\frac {3 a^2 f (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^2 d^2 \sqrt {a^2-b^2}}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d \sqrt {a^2-b^2}}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d \sqrt {a^2-b^2}}-\frac {a (e+f x)^4}{4 b^2 f}-\frac {6 f^3 \sin (c+d x)}{b d^4}+\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^2}-\frac {(e+f x)^3 \cos (c+d x)}{b d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.18, antiderivative size = 643, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 11, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {4515, 3296, 2637, 32, 3323, 2264, 2190, 2531, 6609, 2282, 6589} \[ -\frac {6 i a^2 f^2 (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^3 \sqrt {a^2-b^2}}+\frac {6 i a^2 f^2 (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d^3 \sqrt {a^2-b^2}}-\frac {3 a^2 f (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^2 \sqrt {a^2-b^2}}+\frac {3 a^2 f (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d^2 \sqrt {a^2-b^2}}+\frac {6 a^2 f^3 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d^4 \sqrt {a^2-b^2}}-\frac {6 a^2 f^3 \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d^4 \sqrt {a^2-b^2}}-\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^2 d \sqrt {a^2-b^2}}+\frac {i a^2 (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^2 d \sqrt {a^2-b^2}}-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^2}-\frac {6 f^3 \sin (c+d x)}{b d^4}-\frac {(e+f x)^3 \cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

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

Rule 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 4515

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sin[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbo
l] :> Dist[1/b, Int[(e + f*x)^m*Sin[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sin[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 \sin ^2(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {\int (e+f x)^3 \sin (c+d x) \, dx}{b}-\frac {a \int \frac {(e+f x)^3 \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=-\frac {(e+f x)^3 \cos (c+d x)}{b d}-\frac {a \int (e+f x)^3 \, dx}{b^2}+\frac {a^2 \int \frac {(e+f x)^3}{a+b \sin (c+d x)} \, dx}{b^2}+\frac {(3 f) \int (e+f x)^2 \cos (c+d x) \, dx}{b d}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}-\frac {(e+f x)^3 \cos (c+d x)}{b d}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^2}+\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^2}-\frac {\left (6 f^2\right ) \int (e+f x) \sin (c+d x) \, dx}{b d^2}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}-\frac {(e+f x)^3 \cos (c+d x)}{b d}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^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}{b \sqrt {a^2-b^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}{b \sqrt {a^2-b^2}}-\frac {\left (6 f^3\right ) \int \cos (c+d x) \, dx}{b d^3}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}-\frac {(e+f x)^3 \cos (c+d x)}{b 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^2 \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^2 \sqrt {a^2-b^2} d}-\frac {6 f^3 \sin (c+d x)}{b d^4}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^2}+\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^2 \sqrt {a^2-b^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^2 \sqrt {a^2-b^2} d}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}-\frac {(e+f x)^3 \cos (c+d x)}{b 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^2 \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^2 \sqrt {a^2-b^2} d}-\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^2 \sqrt {a^2-b^2} d^2}+\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^2 \sqrt {a^2-b^2} d^2}-\frac {6 f^3 \sin (c+d x)}{b d^4}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b 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^2 \sqrt {a^2-b^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^2 \sqrt {a^2-b^2} d^2}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}-\frac {(e+f x)^3 \cos (c+d x)}{b 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^2 \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^2 \sqrt {a^2-b^2} d}-\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^2 \sqrt {a^2-b^2} d^2}+\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^2 \sqrt {a^2-b^2} d^2}-\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^2 \sqrt {a^2-b^2} d^3}+\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^2 \sqrt {a^2-b^2} d^3}-\frac {6 f^3 \sin (c+d x)}{b d^4}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^2}+\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^2 \sqrt {a^2-b^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^2 \sqrt {a^2-b^2} d^3}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}-\frac {(e+f x)^3 \cos (c+d x)}{b 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^2 \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^2 \sqrt {a^2-b^2} d}-\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^2 \sqrt {a^2-b^2} d^2}+\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^2 \sqrt {a^2-b^2} d^2}-\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^2 \sqrt {a^2-b^2} d^3}+\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^2 \sqrt {a^2-b^2} d^3}-\frac {6 f^3 \sin (c+d x)}{b d^4}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^2}+\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^2 \sqrt {a^2-b^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^2 \sqrt {a^2-b^2} d^4}\\ &=-\frac {a (e+f x)^4}{4 b^2 f}+\frac {6 f^2 (e+f x) \cos (c+d x)}{b d^3}-\frac {(e+f x)^3 \cos (c+d x)}{b 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^2 \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^2 \sqrt {a^2-b^2} d}-\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^2 \sqrt {a^2-b^2} d^2}+\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^2 \sqrt {a^2-b^2} d^2}-\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^2 \sqrt {a^2-b^2} d^3}+\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^2 \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^2 \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^2 \sqrt {a^2-b^2} d^4}-\frac {6 f^3 \sin (c+d x)}{b d^4}+\frac {3 f (e+f x)^2 \sin (c+d x)}{b d^2}\\ \end {align*}

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Mathematica [A]  time = 7.62, size = 1020, normalized size = 1.59 \[ \frac {-a x \left (4 e^3+6 f x e^2+4 f^2 x^2 e+f^3 x^3\right ) d^4-4 b (e+f x) \left (d^2 (e+f x)^2-6 f^2\right ) \cos (c+d x) d+\frac {4 a^2 \left (2 \sqrt {b^2-a^2} e^3 \tan ^{-1}\left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right ) d^3+\sqrt {a^2-b^2} f^3 x^3 \log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d^3+3 \sqrt {a^2-b^2} e f^2 x^2 \log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d^3+3 \sqrt {a^2-b^2} e^2 f x \log \left (1-\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d^3-\sqrt {a^2-b^2} f^3 x^3 \log \left (\frac {e^{i (c+d x)} b}{i a+\sqrt {b^2-a^2}}+1\right ) d^3-3 \sqrt {a^2-b^2} e f^2 x^2 \log \left (\frac {e^{i (c+d x)} b}{i a+\sqrt {b^2-a^2}}+1\right ) d^3-3 \sqrt {a^2-b^2} e^2 f x \log \left (\frac {e^{i (c+d x)} b}{i a+\sqrt {b^2-a^2}}+1\right ) d^3-3 i \sqrt {a^2-b^2} f (e+f x)^2 \text {Li}_2\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d^2+3 i \sqrt {a^2-b^2} f (e+f x)^2 \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right ) d^2+6 \sqrt {a^2-b^2} e f^2 \text {Li}_3\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d+6 \sqrt {a^2-b^2} f^3 x \text {Li}_3\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right ) d-6 \sqrt {a^2-b^2} e f^2 \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right ) d-6 \sqrt {a^2-b^2} f^3 x \text {Li}_3\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right ) d+6 i \sqrt {a^2-b^2} f^3 \text {Li}_4\left (\frac {b e^{i (c+d x)}}{\sqrt {b^2-a^2}-i a}\right )-6 i \sqrt {a^2-b^2} f^3 \text {Li}_4\left (-\frac {b e^{i (c+d x)}}{i a+\sqrt {b^2-a^2}}\right )\right )}{\sqrt {-\left (a^2-b^2\right )^2}}+12 b f \left (d^2 (e+f x)^2-2 f^2\right ) \sin (c+d x)}{4 b^2 d^4} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-(a*d^4*x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)) - 4*b*d*(e + f*x)*(-6*f^2 + d^2*(e + f*x)^2)*Cos[c + d
*x] + (4*a^2*(2*Sqrt[-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*Po
lyLog[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*PolyLog[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]))]))/Sqrt[-(a^2 - b^2)^2] + 12*b*f*(-2*f^2 + d^2*(e + f*x)^2)*Sin[c + d*x])/(4*b^2*d^
4)

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fricas [C]  time = 0.79, size = 2691, normalized size = 4.19 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((a^3 - a*b^2)*d^4*f^3*x^4 + 4*(a^3 - a*b^2)*d^4*e*f^2*x^3 + 6*(a^3 - a*b^2)*d^4*e^2*f*x^2 + 4*(a^3 - a*b
^2)*d^4*e^3*x + 12*I*a^2*b*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) - 12*I*a^2*b*f^3*sqrt(-(a^2 - b^2)/b^2)*polyl
og(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) - 12*I*a^2*b*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*co
s(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*I*a^2*b*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*(-3*I*a^2*b*d^2*f^3*x^2 - 6*I*a^2*b*d^2*e*f^2*x - 3*I*a^2*b*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) + 2*(3*I*a^2*b*d^2*f^3*x^2 + 6*I*a^2*b*d^2*e*f^2*x + 3*I*a^2*b*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) + 2*(3*I*a^2*b*d^2*f^3*x^2 + 6*I*a^2*b*d^2*e*f^2*x + 3*I*a^2*b*d^2*e^2*f)*sqrt(-(a^2 - b^2)/b^2)*dilo
g(-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*I*a^2*b*d^2*f^3*x^2 - 6*I*a^2*b*d^2*e*f^2*x - 3*I*a^2*b*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) - 2*(a^2*b*d^3*e^3 - 3*a^2*b*c*d^2*e^2*f + 3*a^2*b*c^2*d*e*f^2 - a^2*b*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) - 2*(a^2*b*d^3*e^
3 - 3*a^2*b*c*d^2*e^2*f + 3*a^2*b*c^2*d*e*f^2 - a^2*b*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) + 2*(a^2*b*d^3*e^3 - 3*a^2*b*c*d^2*e^2*f + 3*a^2*b*c^2
*d*e*f^2 - a^2*b*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) + 2*(a^2*b*d^3*e^3 - 3*a^2*b*c*d^2*e^2*f + 3*a^2*b*c^2*d*e*f^2 - a^2*b*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) - 2*(a^2*b*d^3*f
^3*x^3 + 3*a^2*b*d^3*e*f^2*x^2 + 3*a^2*b*d^3*e^2*f*x + 3*a^2*b*c*d^2*e^2*f - 3*a^2*b*c^2*d*e*f^2 + a^2*b*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) + 2*(a^2*b*d^3*f^3*x^3 + 3*a^2*b*d^3*e*f^2*x^2 + 3*a^2*b*d^3*e^2*f*x + 3*
a^2*b*c*d^2*e^2*f - 3*a^2*b*c^2*d*e*f^2 + a^2*b*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) - 2*(a^2*b*d^3*f^3*x
^3 + 3*a^2*b*d^3*e*f^2*x^2 + 3*a^2*b*d^3*e^2*f*x + 3*a^2*b*c*d^2*e^2*f - 3*a^2*b*c^2*d*e*f^2 + a^2*b*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) + 2*(a^2*b*d^3*f^3*x^3 + 3*a^2*b*d^3*e*f^2*x^2 + 3*a^2*b*d^3*e^2*f*x + 3*a^2
*b*c*d^2*e^2*f - 3*a^2*b*c^2*d*e*f^2 + a^2*b*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) + 12*(a^2*b*d*f^3*x +
a^2*b*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) - 12*(a^2*b*d*f^3*x + a^2*b*d*e*f^2)*sqrt(-(a^2 - b^2)/b^2)*p
olylog(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^2*b*d*f^3*x + a^2*b*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) - 12*(a^2*b*d*f^3*x + a^2*b*
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) + 4*((a^2*b - b^3)*d^3*f^3*x^3 + 3*(a^2*b - b^3)*d^3*e*f^2*x^2 + (
a^2*b - b^3)*d^3*e^3 - 6*(a^2*b - b^3)*d*e*f^2 + 3*((a^2*b - b^3)*d^3*e^2*f - 2*(a^2*b - b^3)*d*f^3)*x)*cos(d*
x + c) - 12*((a^2*b - b^3)*d^2*f^3*x^2 + 2*(a^2*b - b^3)*d^2*e*f^2*x + (a^2*b - b^3)*d^2*e^2*f - 2*(a^2*b - b^
3)*f^3)*sin(d*x + c))/((a^2*b^2 - b^4)*d^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.53, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \left (\sin ^{2}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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