3.3.29 \(\int \frac {(e+f x)^2 \sin ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [229]

Optimal. Leaf size=592 \[ -\frac {f^2 x}{4 b d^2}+\frac {a^2 (e+f x)^3}{3 b^3 f}+\frac {(e+f x)^3}{6 b f}-\frac {2 a f^2 \cos (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \cos (c+d x)}{b^2 d}+\frac {i a^3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}-\frac {i a^3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d}+\frac {2 a^3 f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}-\frac {2 a^3 f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^2}+\frac {2 i a^3 f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^3}-\frac {2 i a^3 f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 \sqrt {a^2-b^2} d^3}-\frac {2 a f (e+f x) \sin (c+d x)}{b^2 d^2}+\frac {f^2 \cos (c+d x) \sin (c+d x)}{4 b d^3}-\frac {(e+f x)^2 \cos (c+d x) \sin (c+d x)}{2 b d}+\frac {f (e+f x) \sin ^2(c+d x)}{2 b d^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.77, antiderivative size = 592, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 13, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.464, Rules used = {4611, 3392, 32, 2715, 8, 3377, 2718, 3404, 2296, 2221, 2611, 2320, 6724} \begin {gather*} \frac {2 i a^3 f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3 \sqrt {a^2-b^2}}-\frac {2 i a^3 f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^3 \sqrt {a^2-b^2}}+\frac {2 a^3 f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2 \sqrt {a^2-b^2}}-\frac {2 a^3 f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^2 \sqrt {a^2-b^2}}+\frac {a^2 (e+f x)^3}{3 b^3 f}+\frac {i a^3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d \sqrt {a^2-b^2}}-\frac {i a^3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d \sqrt {a^2-b^2}}-\frac {2 a f^2 \cos (c+d x)}{b^2 d^3}-\frac {2 a f (e+f x) \sin (c+d x)}{b^2 d^2}+\frac {a (e+f x)^2 \cos (c+d x)}{b^2 d}+\frac {f^2 \sin (c+d x) \cos (c+d x)}{4 b d^3}+\frac {f (e+f x) \sin ^2(c+d x)}{2 b d^2}-\frac {(e+f x)^2 \sin (c+d x) \cos (c+d x)}{2 b d}-\frac {f^2 x}{4 b d^2}+\frac {(e+f x)^3}{6 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 2221

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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2296

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 2320

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 2611

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 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2718

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

Rule 3377

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 3392

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

Rule 3404

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 4611

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 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

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

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Mathematica [A]
time = 2.21, size = 909, normalized size = 1.54 \begin {gather*} \frac {24 a^2 d^3 e^2 x+12 b^2 d^3 e^2 x+24 a^2 d^3 e f x^2+12 b^2 d^3 e f x^2+8 a^2 d^3 f^2 x^3+4 b^2 d^3 f^2 x^3+24 a b \left (-2 f^2+d^2 (e+f x)^2\right ) \cos (c+d x)-6 b^2 d f (e+f x) \cos (2 (c+d x))-\frac {24 i a^3 \left (-i \left (d^2 \left (\sqrt {a^2-b^2} f x (2 e+f x) \left (-\log \left (1+\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right )+\log \left (1-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right )\right ) (\cos (c)+i \sin (c))+2 e^2 \tan ^{-1}\left (\frac {b \cos (c+d x)+i (a+b \sin (c+d x))}{\sqrt {a^2-b^2}}\right ) \sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)+i \sin (c))^2\right )}\right )-2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right ) (\cos (c)+i \sin (c))+2 \sqrt {a^2-b^2} f^2 \text {Li}_3\left (\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right ) (\cos (c)+i \sin (c))\right )+2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (-\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}-a \sin (c)}\right ) (\cos (c)+i \sin (c))-2 \sqrt {a^2-b^2} d f (e+f x) \text {Li}_2\left (\frac {b (\cos (2 c+d x)+i \sin (2 c+d x))}{-i a \cos (c)+\sqrt {\left (-a^2+b^2\right ) (\cos (c)+i \sin (c))^2}+a \sin (c)}\right ) (\cos (c)+i \sin (c))\right )}{\sqrt {a^2-b^2} \sqrt {-\left (\left (a^2-b^2\right ) (\cos (c)+i \sin (c))^2\right )}}-48 a b d e f \sin (c+d x)-48 a b d f^2 x \sin (c+d x)-6 b^2 d^2 e^2 \sin (2 (c+d x))+3 b^2 f^2 \sin (2 (c+d x))-12 b^2 d^2 e f x \sin (2 (c+d x))-6 b^2 d^2 f^2 x^2 \sin (2 (c+d x))}{24 b^3 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(24*a^2*d^3*e^2*x + 12*b^2*d^3*e^2*x + 24*a^2*d^3*e*f*x^2 + 12*b^2*d^3*e*f*x^2 + 8*a^2*d^3*f^2*x^3 + 4*b^2*d^3
*f^2*x^3 + 24*a*b*(-2*f^2 + d^2*(e + f*x)^2)*Cos[c + d*x] - 6*b^2*d*f*(e + f*x)*Cos[2*(c + d*x)] - ((24*I)*a^3
*((-I)*(d^2*(Sqrt[a^2 - b^2]*f*x*(2*e + f*x)*(-Log[1 + (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + S
qrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] - a*Sin[c])] + Log[1 - (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/((-I)*a
*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])])*(Cos[c] + I*Sin[c]) + 2*e^2*ArcTan[(b*Cos[c +
d*x] + I*(a + b*Sin[c + d*x]))/Sqrt[a^2 - b^2]]*Sqrt[-((a^2 - b^2)*(Cos[c] + I*Sin[c])^2)]) - 2*Sqrt[a^2 - b^
2]*f^2*PolyLog[3, -((b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c]
)^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) + 2*Sqrt[a^2 - b^2]*f^2*PolyLog[3, (b*(Cos[2*c + d*x] + I*Sin[2*c + d*x
]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])]*(Cos[c] + I*Sin[c])) + 2*Sqrt[a^2 -
b^2]*d*f*(e + f*x)*PolyLog[2, -((b*(Cos[2*c + d*x] + I*Sin[2*c + d*x]))/(I*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[
c] + I*Sin[c])^2] - a*Sin[c]))]*(Cos[c] + I*Sin[c]) - 2*Sqrt[a^2 - b^2]*d*f*(e + f*x)*PolyLog[2, (b*(Cos[2*c +
d*x] + I*Sin[2*c + d*x]))/((-I)*a*Cos[c] + Sqrt[(-a^2 + b^2)*(Cos[c] + I*Sin[c])^2] + a*Sin[c])]*(Cos[c] + I*
Sin[c])))/(Sqrt[a^2 - b^2]*Sqrt[-((a^2 - b^2)*(Cos[c] + I*Sin[c])^2)]) - 48*a*b*d*e*f*Sin[c + d*x] - 48*a*b*d*
f^2*x*Sin[c + d*x] - 6*b^2*d^2*e^2*Sin[2*(c + d*x)] + 3*b^2*f^2*Sin[2*(c + d*x)] - 12*b^2*d^2*e*f*x*Sin[2*(c +
d*x)] - 6*b^2*d^2*f^2*x^2*Sin[2*(c + d*x)])/(24*b^3*d^3)

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Maple [F]
time = 0.32, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)^2*sin(d*x+c)^3/(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 de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2059 vs. \(2 (532) = 1064\).
time = 0.59, size = 2059, normalized size = 3.48 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(2*(2*a^4 - a^2*b^2 - b^4)*d^3*f^2*x^3 + 6*(2*a^4 - a^2*b^2 - b^4)*d^3*f*x^2*e - 12*a^3*b*f^2*sqrt(-(a^2
- b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 -
b^2)/b^2))/b) + 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*
x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 12*a^3*b*f^2*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*
cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) + 12*a^3*b*f^2*
sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*s
qrt(-(a^2 - b^2)/b^2))/b) + 6*(2*a^4 - a^2*b^2 - b^4)*d^3*x*e^2 + 3*(a^2*b^2 - b^4)*d*f^2*x - 6*((a^2*b^2 - b^
4)*d*f^2*x + (a^2*b^2 - b^4)*d*f*e)*cos(d*x + c)^2 + 12*(-I*a^3*b*d*f^2*x - I*a^3*b*d*f*e)*sqrt(-(a^2 - b^2)/b
^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)
/b + 1) + 12*(I*a^3*b*d*f^2*x + I*a^3*b*d*f*e)*sqrt(-(a^2 - b^2)/b^2)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c)
- (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + 12*(I*a^3*b*d*f^2*x + I*a^3*b*d*f*
e)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt
(-(a^2 - b^2)/b^2) - b)/b + 1) + 12*(-I*a^3*b*d*f^2*x - I*a^3*b*d*f*e)*sqrt(-(a^2 - b^2)/b^2)*dilog((-I*a*cos(
d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 6*(a^3*b*
c^2*f^2 - 2*a^3*b*c*d*f*e + a^3*b*d^2*e^2)*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) - 6*(a^3*b*c^2*f^2 - 2*a^3*b*c*d*f*e + a^3*b*d^2*e^2)*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) + 6*(a^3*b*c^2*f^2 - 2*a^3*
b*c*d*f*e + a^3*b*d^2*e^2)*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) + 6*(a^3*b*c^2*f^2 - 2*a^3*b*c*d*f*e + a^3*b*d^2*e^2)*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) + 6*(a^3*b*d^2*f^2*x^2 - a^3*b*c^2*f^2 +
2*(a^3*b*d^2*f*x + a^3*b*c*d*f)*e)*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x
+ c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - 6*(a^3*b*d^2*f^2*x^2 - a^3*b*c^2*f^2 + 2*(a^3*b*d^2
*f*x + a^3*b*c*d*f)*e)*sqrt(-(a^2 - b^2)/b^2)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*
sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + 6*(a^3*b*d^2*f^2*x^2 - a^3*b*c^2*f^2 + 2*(a^3*b*d^2*f*x + a^3*b
*c*d*f)*e)*sqrt(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c
))*sqrt(-(a^2 - b^2)/b^2) - b)/b) - 6*(a^3*b*d^2*f^2*x^2 - a^3*b*c^2*f^2 + 2*(a^3*b*d^2*f*x + a^3*b*c*d*f)*e)*
sqrt(-(a^2 - b^2)/b^2)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a
^2 - b^2)/b^2) - b)/b) + 12*((a^3*b - a*b^3)*d^2*f^2*x^2 + 2*(a^3*b - a*b^3)*d^2*f*x*e + (a^3*b - a*b^3)*d^2*e
^2 - 2*(a^3*b - a*b^3)*f^2)*cos(d*x + c) - 3*(8*(a^3*b - a*b^3)*d*f^2*x + 8*(a^3*b - a*b^3)*d*f*e + (2*(a^2*b^
2 - b^4)*d^2*f^2*x^2 + 4*(a^2*b^2 - b^4)*d^2*f*x*e + 2*(a^2*b^2 - b^4)*d^2*e^2 - (a^2*b^2 - b^4)*f^2)*cos(d*x
+ c))*sin(d*x + c))/((a^2*b^3 - b^5)*d^3)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F(-1)]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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