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

Optimal. Leaf size=737 \[ -\frac {6 i f^3 \left (a^2-b^2\right ) \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^4}-\frac {6 i f^3 \left (a^2-b^2\right ) \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^4}-\frac {6 f^2 \left (a^2-b^2\right ) (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {6 f^2 \left (a^2-b^2\right ) (e+f x) \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^3}+\frac {3 i f \left (a^2-b^2\right ) (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {3 i f \left (a^2-b^2\right ) (e+f x)^2 \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {6 a f^3 \cos (c+d x)}{b^2 d^4}-\frac {6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac {3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac {3 f^3 \sin (c+d x) \cos (c+d x)}{8 b d^4}+\frac {3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac {3 f (e+f x)^2 \sin (c+d x) \cos (c+d x)}{4 b d^2}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac {3 f^3 x}{8 b d^3}+\frac {(e+f x)^3}{4 b d} \]

[Out]

-3/8*f^3*x/b/d^3+1/4*(f*x+e)^3/b/d+3*I*(a^2-b^2)*f*(f*x+e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))
/b^3/d^2-6*a*f^3*cos(d*x+c)/b^2/d^4+3*a*f*(f*x+e)^2*cos(d*x+c)/b^2/d^2-(a^2-b^2)*(f*x+e)^3*ln(1-I*b*exp(I*(d*x
+c))/(a-(a^2-b^2)^(1/2)))/b^3/d-(a^2-b^2)*(f*x+e)^3*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d-6*I*(a^
2-b^2)*f^3*polylog(4,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d^4-6*I*(a^2-b^2)*f^3*polylog(4,I*b*exp(I*(d*
x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d^4-6*(a^2-b^2)*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b
^3/d^3-6*(a^2-b^2)*f^2*(f*x+e)*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b^3/d^3+3*I*(a^2-b^2)*f*(f*x+
e)^2*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b^3/d^2+1/4*I*(a^2-b^2)*(f*x+e)^4/b^3/f-6*a*f^2*(f*x+e)
*sin(d*x+c)/b^2/d^3+a*(f*x+e)^3*sin(d*x+c)/b^2/d+3/8*f^3*cos(d*x+c)*sin(d*x+c)/b/d^4-3/4*f*(f*x+e)^2*cos(d*x+c
)*sin(d*x+c)/b/d^2+3/4*f^2*(f*x+e)*sin(d*x+c)^2/b/d^3-1/2*(f*x+e)^3*sin(d*x+c)^2/b/d

________________________________________________________________________________________

Rubi [A]  time = 0.88, antiderivative size = 737, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 14, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {4525, 3296, 2638, 4404, 3311, 32, 2635, 8, 4519, 2190, 2531, 6609, 2282, 6589} \[ -\frac {6 f^2 \left (a^2-b^2\right ) (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {6 f^2 \left (a^2-b^2\right ) (e+f x) \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^3}+\frac {3 i f \left (a^2-b^2\right ) (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {3 i f \left (a^2-b^2\right ) (e+f x)^2 \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^2}-\frac {6 i f^3 \left (a^2-b^2\right ) \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^4}-\frac {6 i f^3 \left (a^2-b^2\right ) \text {PolyLog}\left (4,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d^4}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac {3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac {6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac {3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac {3 f (e+f x)^2 \sin (c+d x) \cos (c+d x)}{4 b d^2}+\frac {3 f^3 \sin (c+d x) \cos (c+d x)}{8 b d^4}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac {3 f^3 x}{8 b d^3}+\frac {(e+f x)^3}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*f^3*x)/(8*b*d^3) + (e + f*x)^3/(4*b*d) + ((I/4)*(a^2 - b^2)*(e + f*x)^4)/(b^3*f) - (6*a*f^3*Cos[c + d*x])/
(b^2*d^4) + (3*a*f*(e + f*x)^2*Cos[c + d*x])/(b^2*d^2) - ((a^2 - b^2)*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x))
)/(a - Sqrt[a^2 - b^2])])/(b^3*d) - ((a^2 - b^2)*(e + f*x)^3*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2
])])/(b^3*d) + ((3*I)*(a^2 - b^2)*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*
d^2) + ((3*I)*(a^2 - b^2)*f*(e + f*x)^2*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^2) - (
6*(a^2 - b^2)*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d^3) - (6*(a^2 - b^2
)*f^2*(e + f*x)*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^3) - ((6*I)*(a^2 - b^2)*f^3*Po
lyLog[4, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b^3*d^4) - ((6*I)*(a^2 - b^2)*f^3*PolyLog[4, (I*b*E^(I
*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b^3*d^4) - (6*a*f^2*(e + f*x)*Sin[c + d*x])/(b^2*d^3) + (a*(e + f*x)^3*S
in[c + d*x])/(b^2*d) + (3*f^3*Cos[c + d*x]*Sin[c + d*x])/(8*b*d^4) - (3*f*(e + f*x)^2*Cos[c + d*x]*Sin[c + d*x
])/(4*b*d^2) + (3*f^2*(e + f*x)*Sin[c + d*x]^2)/(4*b*d^3) - ((e + f*x)^3*Sin[c + d*x]^2)/(2*b*d)

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 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 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 2635

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] && Integer
Q[2*n]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[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 3311

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 4404

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

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 4525

Int[(Cos[(c_.) + (d_.)*(x_)]^(n_)*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol
] :> Dist[a/b^2, Int[(e + f*x)^m*Cos[c + d*x]^(n - 2), x], x] + (-Dist[1/b, Int[(e + f*x)^m*Cos[c + d*x]^(n -
2)*Sin[c + d*x], x], x] - Dist[(a^2 - b^2)/b^2, Int[((e + f*x)^m*Cos[c + d*x]^(n - 2))/(a + b*Sin[c + d*x]), x
], x]) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[n, 1] && NeQ[a^2 - b^2, 0] && IGtQ[m, 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 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx &=\frac {a \int (e+f x)^3 \cos (c+d x) \, dx}{b^2}-\frac {\int (e+f x)^3 \cos (c+d x) \sin (c+d x) \, dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {(e+f x)^3 \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac {\left (a^2-b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{b^2}-\frac {\left (a^2-b^2\right ) \int \frac {e^{i (c+d x)} (e+f x)^3}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{b^2}-\frac {(3 a f) \int (e+f x)^2 \sin (c+d x) \, dx}{b^2 d}+\frac {(3 f) \int (e+f x)^2 \sin ^2(c+d x) \, dx}{2 b d}\\ &=\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}+\frac {3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}-\frac {3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac {3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d}+\frac {(3 f) \int (e+f x)^2 \, dx}{4 b d}+\frac {\left (3 \left (a^2-b^2\right ) f\right ) \int (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d}+\frac {\left (3 \left (a^2-b^2\right ) f\right ) \int (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d}-\frac {\left (6 a f^2\right ) \int (e+f x) \cos (c+d x) \, dx}{b^2 d^2}-\frac {\left (3 f^3\right ) \int \sin ^2(c+d x) \, dx}{4 b d^3}\\ &=\frac {(e+f x)^3}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}+\frac {3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {3 i \left (a^2-b^2\right ) 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^3 d^2}+\frac {3 i \left (a^2-b^2\right ) 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^3 d^2}-\frac {6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac {3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac {3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac {\left (6 i \left (a^2-b^2\right ) f^2\right ) \int (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d^2}-\frac {\left (6 i \left (a^2-b^2\right ) f^2\right ) \int (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d^2}+\frac {\left (6 a f^3\right ) \int \sin (c+d x) \, dx}{b^2 d^3}-\frac {\left (3 f^3\right ) \int 1 \, dx}{8 b d^3}\\ &=-\frac {3 f^3 x}{8 b d^3}+\frac {(e+f x)^3}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac {3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {3 i \left (a^2-b^2\right ) 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^3 d^2}+\frac {3 i \left (a^2-b^2\right ) 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^3 d^2}-\frac {6 \left (a^2-b^2\right ) 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^3 d^3}-\frac {6 \left (a^2-b^2\right ) 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^3 d^3}-\frac {6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac {3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac {3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d}+\frac {\left (6 \left (a^2-b^2\right ) f^3\right ) \int \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d^3}+\frac {\left (6 \left (a^2-b^2\right ) f^3\right ) \int \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b^3 d^3}\\ &=-\frac {3 f^3 x}{8 b d^3}+\frac {(e+f x)^3}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac {3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {3 i \left (a^2-b^2\right ) 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^3 d^2}+\frac {3 i \left (a^2-b^2\right ) 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^3 d^2}-\frac {6 \left (a^2-b^2\right ) 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^3 d^3}-\frac {6 \left (a^2-b^2\right ) 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^3 d^3}-\frac {6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac {3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac {3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d}-\frac {\left (6 i \left (a^2-b^2\right ) 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^3 d^4}-\frac {\left (6 i \left (a^2-b^2\right ) 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^3 d^4}\\ &=-\frac {3 f^3 x}{8 b d^3}+\frac {(e+f x)^3}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^4}{4 b^3 f}-\frac {6 a f^3 \cos (c+d x)}{b^2 d^4}+\frac {3 a f (e+f x)^2 \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^3 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {3 i \left (a^2-b^2\right ) 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^3 d^2}+\frac {3 i \left (a^2-b^2\right ) 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^3 d^2}-\frac {6 \left (a^2-b^2\right ) 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^3 d^3}-\frac {6 \left (a^2-b^2\right ) 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^3 d^3}-\frac {6 i \left (a^2-b^2\right ) f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^4}-\frac {6 i \left (a^2-b^2\right ) f^3 \text {Li}_4\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^4}-\frac {6 a f^2 (e+f x) \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^3 \sin (c+d x)}{b^2 d}+\frac {3 f^3 \cos (c+d x) \sin (c+d x)}{8 b d^4}-\frac {3 f (e+f x)^2 \cos (c+d x) \sin (c+d x)}{4 b d^2}+\frac {3 f^2 (e+f x) \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^3 \sin ^2(c+d x)}{2 b d}\\ \end {align*}

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Mathematica [B]  time = 10.71, size = 2452, normalized size = 3.33 \[ \text {Result too large to show} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*(a^2 - b^2)*e^3*x*Cot[c] - 48*(a^2 - b^2)*e^2*f*x^2*Cot[c] - 32*(a^2 - b^2)*e*f^2*x^3*Cot[c] - 8*(a^2 - b
^2)*f^3*x^4*Cot[c] + (16*(a^2 - b^2)*((4*I)*d^4*e^3*E^((2*I)*c)*x + (6*I)*d^4*e^2*E^((2*I)*c)*f*x^2 + (4*I)*d^
4*e*E^((2*I)*c)*f^2*x^3 + I*d^4*E^((2*I)*c)*f^3*x^4 + (2*I)*d^3*e^3*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((
2*I)*(c + d*x))))] - (2*I)*d^3*e^3*E^((2*I)*c)*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))] +
d^3*e^3*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] - d^3*e^3*E^((2*I)*c)*Log[4*a^2*E^((
2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2] + 6*d^3*e^2*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c)
- Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e^2*E^((2*I)*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - S
qrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^
2)*E^((2*I)*c)])] - 6*d^3*e*E^((2*I)*c)*f^2*x^2*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)
*E^((2*I)*c)])] + 2*d^3*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])]
- 2*d^3*E^((2*I)*c)*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*
d^3*e^2*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e^2*E^((2*I)
*c)*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 6*d^3*e*f^2*x^2*Log[1
+ (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 6*d^3*e*E^((2*I)*c)*f^2*x^2*Log[1 +
(b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 2*d^3*f^3*x^3*Log[1 + (b*E^(I*(2*c + d
*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 2*d^3*E^((2*I)*c)*f^3*x^3*Log[1 + (b*E^(I*(2*c + d*x))
)/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + (6*I)*d^2*(-1 + E^((2*I)*c))*f*(e + f*x)^2*PolyLog[2, (I*b
*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + (6*I)*d^2*(-1 + E^((2*I)*c))*f*(e + f*x)
^2*PolyLog[2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 12*d*e*f^2*PolyLog[3,
(I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*d*e*E^((2*I)*c)*f^2*PolyLog[3, (
I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d*f^3*x*PolyLog[3, (I*b*E^(I*(2*c
+ d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] - 12*d*E^((2*I)*c)*f^3*x*PolyLog[3, (I*b*E^(I*(2*c +
d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + 12*d*e*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^
(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - 12*d*e*E^((2*I)*c)*f^2*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(
I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] + 12*d*f^3*x*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(
-a^2 + b^2)*E^((2*I)*c)]))] - 12*d*E^((2*I)*c)*f^3*x*PolyLog[3, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-
a^2 + b^2)*E^((2*I)*c)]))] + (12*I)*f^3*PolyLog[4, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^
((2*I)*c)])] - (12*I)*E^((2*I)*c)*f^3*PolyLog[4, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((
2*I)*c)])] + (12*I)*f^3*PolyLog[4, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] -
(12*I)*E^((2*I)*c)*f^3*PolyLog[4, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))]))/(
d^4*(-1 + E^((2*I)*c))) + (16*a*b*(-6*f^3 - (6*I)*d*f^2*(e + f*x) + 3*d^2*f*(e + f*x)^2 + I*d^3*(e + f*x)^3)*(
Cos[c + d*x] - I*Sin[c + d*x]))/d^4 + (16*a*b*(-6*f^3 + (6*I)*d*f^2*(e + f*x) + 3*d^2*f*(e + f*x)^2 - I*d^3*(e
+ f*x)^3)*(Cos[c + d*x] + I*Sin[c + d*x]))/d^4 + (b^2*((3*I)*f^3 - 6*d*f^2*(e + f*x) - (6*I)*d^2*f*(e + f*x)^
2 + 4*d^3*(e + f*x)^3)*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)]))/d^4 + (b^2*((-3*I)*f^3 - 6*d*f^2*(e + f*x) + (
6*I)*d^2*f*(e + f*x)^2 + 4*d^3*(e + f*x)^3)*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)]))/d^4)/(32*b^3)

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fricas [C]  time = 0.82, size = 2704, normalized size = 3.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(2*b^2*d^3*f^3*x^3 + 6*b^2*d^3*e*f^2*x^2 + 24*I*(a^2 - b^2)*f^3*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) + 24*I*(a^2 - b^2)*f^3*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) - 24*I*(a^2 - b^2)*f^3*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*si
n(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 24*I*(a^2 - b^2)*f^3*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*(2*b^2*d^3*f^3*x^3 + 6*b^2*d^3*e
*f^2*x^2 + 2*b^2*d^3*e^3 - 3*b^2*d*e*f^2 + 3*(2*b^2*d^3*e^2*f - b^2*d*f^3)*x)*cos(d*x + c)^2 + 3*(2*b^2*d^3*e^
2*f - b^2*d*f^3)*x - 24*(a*b*d^2*f^3*x^2 + 2*a*b*d^2*e*f^2*x + a*b*d^2*e^2*f - 2*a*b*f^3)*cos(d*x + c) - (-12*
I*(a^2 - b^2)*d^2*f^3*x^2 - 24*I*(a^2 - b^2)*d^2*e*f^2*x - 12*I*(a^2 - b^2)*d^2*e^2*f)*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^2 - b^2)*d^2*f^3*x^2 - 24*I*(a^2 - b^2)*d^2*e*f^2*x - 12*I*(a^2 - b^2)*d^2*e^2*f)*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^2 - b^2)*d^2*f^3*x^2 + 24*I*(a^2 - b^2)*d^2*e*f^2*x + 12*I*(a^2 - b^2)*d^2*e^2*f)*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) - (1
2*I*(a^2 - b^2)*d^2*f^3*x^2 + 24*I*(a^2 - b^2)*d^2*e*f^2*x + 12*I*(a^2 - b^2)*d^2*e^2*f)*dilog(-1/2*(-2*I*a*co
s(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) + 4
*((a^2 - b^2)*d^3*e^3 - 3*(a^2 - b^2)*c*d^2*e^2*f + 3*(a^2 - b^2)*c^2*d*e*f^2 - (a^2 - b^2)*c^3*f^3)*log(2*b*c
os(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + 4*((a^2 - b^2)*d^3*e^3 - 3*(a^2 - b^2
)*c*d^2*e^2*f + 3*(a^2 - b^2)*c^2*d*e*f^2 - (a^2 - b^2)*c^3*f^3)*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) + 4*((a^2 - b^2)*d^3*e^3 - 3*(a^2 - b^2)*c*d^2*e^2*f + 3*(a^2 - b^2)*c^2*d*
e*f^2 - (a^2 - b^2)*c^3*f^3)*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)
+ 4*((a^2 - b^2)*d^3*e^3 - 3*(a^2 - b^2)*c*d^2*e^2*f + 3*(a^2 - b^2)*c^2*d*e*f^2 - (a^2 - b^2)*c^3*f^3)*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) + 4*((a^2 - b^2)*d^3*f^3*x^3 + 3*(a
^2 - b^2)*d^3*e*f^2*x^2 + 3*(a^2 - b^2)*d^3*e^2*f*x + 3*(a^2 - b^2)*c*d^2*e^2*f - 3*(a^2 - b^2)*c^2*d*e*f^2 +
(a^2 - b^2)*c^3*f^3)*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))*sq
rt(-(a^2 - b^2)/b^2) + 2*b)/b) + 4*((a^2 - b^2)*d^3*f^3*x^3 + 3*(a^2 - b^2)*d^3*e*f^2*x^2 + 3*(a^2 - b^2)*d^3*
e^2*f*x + 3*(a^2 - b^2)*c*d^2*e^2*f - 3*(a^2 - b^2)*c^2*d*e*f^2 + (a^2 - b^2)*c^3*f^3)*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) + 4*((a^2 - b
^2)*d^3*f^3*x^3 + 3*(a^2 - b^2)*d^3*e*f^2*x^2 + 3*(a^2 - b^2)*d^3*e^2*f*x + 3*(a^2 - b^2)*c*d^2*e^2*f - 3*(a^2
- b^2)*c^2*d*e*f^2 + (a^2 - b^2)*c^3*f^3)*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) + 4*((a^2 - b^2)*d^3*f^3*x^3 + 3*(a^2 - b^2)*d^3*e*f^2*x
^2 + 3*(a^2 - b^2)*d^3*e^2*f*x + 3*(a^2 - b^2)*c*d^2*e^2*f - 3*(a^2 - b^2)*c^2*d*e*f^2 + (a^2 - b^2)*c^3*f^3)*
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) + 24*((a^2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e*f^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) + 24*((a^2 - b^2)*d*f^3*x + (a^2 - b^2)
*d*e*f^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) + 24*((a^2 - b^2)*d*f^3*x + (a^2 - b^2)*d*e*f^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) + 24*((a^2 - b^2)*d*f^3*x +
(a^2 - b^2)*d*e*f^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) - (8*a*b*d^3*f^3*x^3 + 24*a*b*d^3*e*f^2*x^2 + 8*a*b*d^3*e^3 - 48*a*b*d*e*f^2
+ 24*(a*b*d^3*e^2*f - 2*a*b*d*f^3)*x - 3*(2*b^2*d^2*f^3*x^2 + 4*b^2*d^2*e*f^2*x + 2*b^2*d^2*e^2*f - b^2*f^3)*c
os(d*x + c))*sin(d*x + c))/(b^3*d^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 2.84, size = 0, normalized size = 0.00 \[ \int \frac {\left (f x +e \right )^{3} \left (\cos ^{3}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((f*x+e)^3*cos(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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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