\(\int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [324]

Optimal result

Integrand size = 26, antiderivative size = 753 \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 i f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2}-\frac {3 f^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}-\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^3}+\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}-\frac {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right ) d^4}+\frac {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {3 i a f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^4}+\frac {3 i a f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^4}-\frac {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))} \]

[Out]

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

Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 753, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {4507, 3405, 3404, 2296, 2221, 2611, 2320, 6724, 4615, 2317, 2438} \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\frac {3 i f^3 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^4 \left (a^2-b^2\right )}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^4 \left (a^2-b^2\right )}-\frac {3 i a f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^4 \left (a^2-b^2\right )^{3/2}}+\frac {3 i a f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^4 \left (a^2-b^2\right )^{3/2}}-\frac {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3 \left (a^2-b^2\right )^{3/2}}+\frac {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^3 \left (a^2-b^2\right )^{3/2}}-\frac {3 f^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3 \left (a^2-b^2\right )}-\frac {3 f^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^3 \left (a^2-b^2\right )}-\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{2 b d^2 \left (a^2-b^2\right )^{3/2}}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 d^2 \left (a^2-b^2\right ) (a+b \sin (c+d x))}+\frac {3 i f (e+f x)^2}{2 b d^2 \left (a^2-b^2\right )}-\frac {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2} \]

[In]

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

[Out]

(((3*I)/2)*f*(e + f*x)^2)/(b*(a^2 - b^2)*d^2) - (3*f^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 -
b^2])])/(b*(a^2 - b^2)*d^3) - (((3*I)/2)*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)^(3/2)*d^2) - (3*f^2*(e + f*x)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 -
b^2)*d^3) + (((3*I)/2)*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)^(
3/2)*d^2) + ((3*I)*f^3*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^4) - (3*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)^(3/2)*d^3) + ((3*I)*f^3*Pol
yLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)*d^4) + (3*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)^(3/2)*d^3) - ((3*I)*a*f^3*PolyLog[3, (I*b*E^(I*(c + d
*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^4) + ((3*I)*a*f^3*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + S
qrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^4) - (e + f*x)^3/(2*b*d*(a + b*Sin[c + d*x])^2) + (3*f*(e + f*x)^2*Co
s[c + d*x])/(2*(a^2 - b^2)*d^2*(a + b*Sin[c + d*x]))

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 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 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 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

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 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 3405

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 4507

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

Rule 4615

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 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*} \text {integral}& = -\frac {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {(3 f) \int \frac {(e+f x)^2}{(a+b \sin (c+d x))^2} \, dx}{2 b d} \\ & = -\frac {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}+\frac {(3 a f) \int \frac {(e+f x)^2}{a+b \sin (c+d x)} \, dx}{2 b \left (a^2-b^2\right ) d}-\frac {\left (3 f^2\right ) \int \frac {(e+f x) \cos (c+d x)}{a+b \sin (c+d x)} \, dx}{\left (a^2-b^2\right ) d^2} \\ & = \frac {3 i f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2}-\frac {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}+\frac {(3 a f) \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 \left (a^2-b^2\right ) d}-\frac {\left (3 f^2\right ) \int \frac {e^{i (c+d x)} (e+f x)}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right ) d^2}-\frac {\left (3 f^2\right ) \int \frac {e^{i (c+d x)} (e+f x)}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right ) d^2} \\ & = \frac {3 i f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2}-\frac {3 f^2 (e+f x) \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^3}-\frac {3 f^2 (e+f x) \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^3}-\frac {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {(3 i a f) \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}{\left (a^2-b^2\right )^{3/2} d}+\frac {(3 i a f) \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}{\left (a^2-b^2\right )^{3/2} d}+\frac {\left (3 f^3\right ) \int \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^3}+\frac {\left (3 f^3\right ) \int \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^3} \\ & = \frac {3 i f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2}-\frac {3 f^2 (e+f x) \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^3}-\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f^2 (e+f x) \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^3}+\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}+\frac {\left (3 i a f^2\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 \left (a^2-b^2\right )^{3/2} d^2}-\frac {\left (3 i a f^2\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 \left (a^2-b^2\right )^{3/2} d^2}-\frac {\left (3 i f^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\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 (3 i f^3\right ) \text {Subst}\left (\int \frac {\log \left (1-\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 {3 i f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2}-\frac {3 f^2 (e+f x) \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^3}-\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f^2 (e+f x) \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^3}+\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,\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 {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\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 {3 i f^3 \operatorname {PolyLog}\left (2,\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 {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\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 {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}+\frac {\left (3 a f^3\right ) \int \operatorname {PolyLog}\left (2,\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 (3 a f^3\right ) \int \operatorname {PolyLog}\left (2,\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 {3 i f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2}-\frac {3 f^2 (e+f x) \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^3}-\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f^2 (e+f x) \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^3}+\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,\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 {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\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 {3 i f^3 \operatorname {PolyLog}\left (2,\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 {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\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 {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))}-\frac {\left (3 i a f^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\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 (3 i a f^3\right ) \text {Subst}\left (\int \frac {\operatorname {PolyLog}\left (2,\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 {3 i f (e+f x)^2}{2 b \left (a^2-b^2\right ) d^2}-\frac {3 f^2 (e+f x) \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^3}-\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}-\frac {3 f^2 (e+f x) \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^3}+\frac {3 i a f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{2 b \left (a^2-b^2\right )^{3/2} d^2}+\frac {3 i f^3 \operatorname {PolyLog}\left (2,\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 {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\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 {3 i f^3 \operatorname {PolyLog}\left (2,\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 {3 a f^2 (e+f x) \operatorname {PolyLog}\left (2,\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 {3 i a f^3 \operatorname {PolyLog}\left (3,\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 {3 i a f^3 \operatorname {PolyLog}\left (3,\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 {(e+f x)^3}{2 b d (a+b \sin (c+d x))^2}+\frac {3 f (e+f x)^2 \cos (c+d x)}{2 \left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))} \\ \end{align*}

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2259\) vs. \(2(753)=1506\).

Time = 15.40 (sec) , antiderivative size = 2259, normalized size of antiderivative = 3.00 \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

((-3*I)*E^(I*c)*f*(2*e*E^(I*c)*f*x + E^(I*c)*f^2*x^2 + (I*a*e^2*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^
2]])/(Sqrt[a^2 - b^2]*E^(I*c)) - (I*a*e^2*E^(I*c)*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]])/Sqrt[a^2
- b^2] + (2*a*e*f*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*d*E^(I*c)) - (e*E^(I*c)*
f*ArcTan[(2*a*E^(I*(c + d*x)))/(b*(-1 + E^((2*I)*(c + d*x))))])/d + ((2*I)*a*e*f*ArcTanh[(-a + I*b*E^(I*(c + d
*x)))/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*d*E^(I*c)) - (I*e*f*Log[b - (2*I)*a*E^(I*(c + d*x)) - b*E^((2*I)*(c +
d*x))])/(d*E^(I*c)) + ((I/2)*e*E^(I*c)*f*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x)))^2])/d
+ (I*a*e*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)])])/Sqrt[(-a^2 + b^2)
*E^((2*I)*c)] - (I*a*e*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)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - (I*f^2*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^
2)*E^((2*I)*c)])])/(d*E^(I*c)) + (I*E^(I*c)*f^2*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^
2)*E^((2*I)*c)])])/d + ((I/2)*a*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)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - ((I/2)*a*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)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - (I*a*e*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)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)] + (I*a*e*E^((2*I)*c)*f*x*L
og[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)] -
(I*f^2*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/(d*E^(I*c)) + (I*E^(I
*c)*f^2*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/d - ((I/2)*a*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)])])/Sqrt[(-a^2 + b^2)*E^((2*I)*c)]
+ ((I/2)*a*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)])])/
Sqrt[(-a^2 + b^2)*E^((2*I)*c)] - ((-1 + E^((2*I)*c))*f*(-(Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*f) + a*d*E^(I*c)*(e +
f*x))*PolyLog[2, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/(d^2*E^(I*c)*Sqrt[(
-a^2 + b^2)*E^((2*I)*c)]) + ((-1 + E^((2*I)*c))*f*(Sqrt[(-a^2 + b^2)*E^((2*I)*c)]*f + a*d*E^(I*c)*(e + f*x))*P
olyLog[2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))])/(d^2*E^(I*c)*Sqrt[(-a^2 +
b^2)*E^((2*I)*c)]) + (I*a*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)]
)])/(d^2*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]) - (I*a*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)])])/(d^2*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]) - (I*a*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)]))])/(d^2*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]) + (I*a*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)]))])/(d^2*Sqrt[(-a^
2 + b^2)*E^((2*I)*c)])))/(b*(-a^2 + b^2)*d^2*(-1 + E^((2*I)*c))) - (e + f*x)^3/(2*b*d*(a + b*Sin[c + d*x])^2)
- (3*Csc[c/2]*Sec[c/2]*(a*e^2*f*Cos[c] + 2*a*e*f^2*x*Cos[c] + a*f^3*x^2*Cos[c] + b*e^2*f*Sin[d*x] + 2*b*e*f^2*
x*Sin[d*x] + b*f^3*x^2*Sin[d*x]))/(4*(a - b)*b*(a + b)*d^2*(a + b*Sin[c + d*x]))

Maple [F]

\[\int \frac {\left (f x +e \right )^{3} \cos \left (d x +c \right )}{\left (a +b \sin \left (d x +c \right )\right )^{3}}d x\]

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 4917 vs. \(2 (653) = 1306\).

Time = 0.57 (sec) , antiderivative size = 4917, normalized size of antiderivative = 6.53 \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/4*(2*(a^4 - 2*a^2*b^2 + b^4)*d^3*f^3*x^3 + 6*(a^4 - 2*a^2*b^2 + b^4)*d^3*e*f^2*x^2 + 6*(a^4 - 2*a^2*b^2 + b^
4)*d^3*e^2*f*x + 2*(a^4 - 2*a^2*b^2 + b^4)*d^3*e^3 - 6*((a^2*b^2 - b^4)*d^2*f^3*x^2 + 2*(a^2*b^2 - b^4)*d^2*e*
f^2*x + (a^2*b^2 - b^4)*d^2*e^2*f)*cos(d*x + c)*sin(d*x + c) + 6*(a*b^3*f^3*cos(d*x + c)^2 - 2*a^2*b^2*f^3*sin
(d*x + c) - (a^3*b + a*b^3)*f^3)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*co
s(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 6*(a*b^3*f^3*cos(d*x + c)^2 - 2*a^2*b^2*f^3*sin(d*
x + c) - (a^3*b + a*b^3)*f^3)*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) + 6*(a*b^3*f^3*cos(d*x + c)^2 - 2*a^2*b^2*f^3*sin(d*x +
c) - (a^3*b + a*b^3)*f^3)*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) - 6*(a*b^3*f^3*cos(d*x + c)^2 - 2*a^2*b^2*f^3*sin(d*x + c
) - (a^3*b + a*b^3)*f^3)*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) - 6*((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 + (a^3*b - a*b^3)*d^2*e^2*f)*cos(d*x + c) + 6*(I*(a^2*b^2 - b^4)*f^3*cos(d*x + c)^2 - 2*I*(a^3*b - a*b^3
)*f^3*sin(d*x + c) - I*(a^4 - b^4)*f^3 + (-I*(a^3*b + a*b^3)*d*f^3*x - I*(a^3*b + a*b^3)*d*e*f^2 + (I*a*b^3*d*
f^3*x + I*a*b^3*d*e*f^2)*cos(d*x + c)^2 + 2*(-I*a^2*b^2*d*f^3*x - I*a^2*b^2*d*e*f^2)*sin(d*x + c))*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*(I*(a^2*b^2 - b^4)*f^3*cos(d*x + c)^2 - 2*I*(a^3*b - a*b^3)*f^3*sin(d*x + c) - I*(a^4 - b
^4)*f^3 + (I*(a^3*b + a*b^3)*d*f^3*x + I*(a^3*b + a*b^3)*d*e*f^2 + (-I*a*b^3*d*f^3*x - I*a*b^3*d*e*f^2)*cos(d*
x + c)^2 + 2*(I*a^2*b^2*d*f^3*x + I*a^2*b^2*d*e*f^2)*sin(d*x + c))*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*(-I*(a^2*b^
2 - b^4)*f^3*cos(d*x + c)^2 + 2*I*(a^3*b - a*b^3)*f^3*sin(d*x + c) + I*(a^4 - b^4)*f^3 + (I*(a^3*b + a*b^3)*d*
f^3*x + I*(a^3*b + a*b^3)*d*e*f^2 + (-I*a*b^3*d*f^3*x - I*a*b^3*d*e*f^2)*cos(d*x + c)^2 + 2*(I*a^2*b^2*d*f^3*x
+ I*a^2*b^2*d*e*f^2)*sin(d*x + c))*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*(-I*(a^2*b^2 - b^4)*f^3*cos(d*x + c)^2 +
2*I*(a^3*b - a*b^3)*f^3*sin(d*x + c) + I*(a^4 - b^4)*f^3 + (-I*(a^3*b + a*b^3)*d*f^3*x - I*(a^3*b + a*b^3)*d*e
*f^2 + (I*a*b^3*d*f^3*x + I*a*b^3*d*e*f^2)*cos(d*x + c)^2 + 2*(-I*a^2*b^2*d*f^3*x - I*a^2*b^2*d*e*f^2)*sin(d*x
+ c))*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) + 3*(2*(a^4 - b^4)*d*e*f^2 - 2*(a^4 - b^4)*c*f^3 - 2*((a^2*b^2 - b^4)*d*e*
f^2 - (a^2*b^2 - b^4)*c*f^3)*cos(d*x + c)^2 + 4*((a^3*b - a*b^3)*d*e*f^2 - (a^3*b - a*b^3)*c*f^3)*sin(d*x + c)
- ((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 - (a*b^3*d^2*e^2*f - 2*a
*b^3*c*d*e*f^2 + a*b^3*c^2*f^3)*cos(d*x + c)^2 + 2*(a^2*b^2*d^2*e^2*f - 2*a^2*b^2*c*d*e*f^2 + a^2*b^2*c^2*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) + 3*(2*(a^4 - b^4)*d*e*f^2 - 2*(a^4 - b^4)*c*f^3 - 2*((a^2*b^2 - b^4)*d*e*f^2 - (a^2*b^2 - b^4)*c*f^3
)*cos(d*x + c)^2 + 4*((a^3*b - a*b^3)*d*e*f^2 - (a^3*b - a*b^3)*c*f^3)*sin(d*x + c) - ((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 - (a*b^3*d^2*e^2*f - 2*a*b^3*c*d*e*f^2 + a*b^3*c^2*
f^3)*cos(d*x + c)^2 + 2*(a^2*b^2*d^2*e^2*f - 2*a^2*b^2*c*d*e*f^2 + a^2*b^2*c^2*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) + 3*(2*(a^4 - b^4)
*d*e*f^2 - 2*(a^4 - b^4)*c*f^3 - 2*((a^2*b^2 - b^4)*d*e*f^2 - (a^2*b^2 - b^4)*c*f^3)*cos(d*x + c)^2 + 4*((a^3*
b - a*b^3)*d*e*f^2 - (a^3*b - a*b^3)*c*f^3)*sin(d*x + c) + ((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 - (a*b^3*d^2*e^2*f - 2*a*b^3*c*d*e*f^2 + a*b^3*c^2*f^3)*cos(d*x + c)^2 + 2*(a^
2*b^2*d^2*e^2*f - 2*a^2*b^2*c*d*e*f^2 + a^2*b^2*c^2*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) + 3*(2*(a^4 - b^4)*d*e*f^2 - 2*(a^4 - b^4)*c
*f^3 - 2*((a^2*b^2 - b^4)*d*e*f^2 - (a^2*b^2 - b^4)*c*f^3)*cos(d*x + c)^2 + 4*((a^3*b - a*b^3)*d*e*f^2 - (a^3*
b - a*b^3)*c*f^3)*sin(d*x + c) + ((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 - (a*b^3*d^2*e^2*f - 2*a*b^3*c*d*e*f^2 + a*b^3*c^2*f^3)*cos(d*x + c)^2 + 2*(a^2*b^2*d^2*e^2*f - 2*a^2*b^
2*c*d*e*f^2 + a^2*b^2*c^2*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) + 3*(2*(a^4 - b^4)*d*f^3*x + 2*(a^4 - b^4)*c*f^3 - 2*((a^2*b^2 - b^4)*
d*f^3*x + (a^2*b^2 - b^4)*c*f^3)*cos(d*x + c)^2 + 4*((a^3*b - a*b^3)*d*f^3*x + (a^3*b - a*b^3)*c*f^3)*sin(d*x
+ c) + ((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 - (a*b^3*d^2*f^3*x^2 + 2*a*b^3*d^2*e*f^2*x + 2*a*b^3*c*d*e*f^2 - a*b^3*c^2*f^3)*cos(d*x + c)^2 +
2*(a^2*b^2*d^2*f^3*x^2 + 2*a^2*b^2*d^2*e*f^2*x + 2*a^2*b^2*c*d*e*f^2 - a^2*b^2*c^2*f^3)*sin(d*x + c))*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) + 3*(2*(a^4 - b^4)*d*f^3*x + 2*(a^4 - b^4)*c*f^3 - 2*((a^2*b^2 - b^4)*d*f^3*x + (a^2*b^2 - b^4
)*c*f^3)*cos(d*x + c)^2 + 4*((a^3*b - a*b^3)*d*f^3*x + (a^3*b - a*b^3)*c*f^3)*sin(d*x + c) - ((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 - (a*b^3*d
^2*f^3*x^2 + 2*a*b^3*d^2*e*f^2*x + 2*a*b^3*c*d*e*f^2 - a*b^3*c^2*f^3)*cos(d*x + c)^2 + 2*(a^2*b^2*d^2*f^3*x^2
+ 2*a^2*b^2*d^2*e*f^2*x + 2*a^2*b^2*c*d*e*f^2 - a^2*b^2*c^2*f^3)*sin(d*x + c))*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) + 3*(2*(
a^4 - b^4)*d*f^3*x + 2*(a^4 - b^4)*c*f^3 - 2*((a^2*b^2 - b^4)*d*f^3*x + (a^2*b^2 - b^4)*c*f^3)*cos(d*x + c)^2
+ 4*((a^3*b - a*b^3)*d*f^3*x + (a^3*b - a*b^3)*c*f^3)*sin(d*x + c) + ((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 - (a*b^3*d^2*f^3*x^2 + 2*a*b^3*d^2
*e*f^2*x + 2*a*b^3*c*d*e*f^2 - a*b^3*c^2*f^3)*cos(d*x + c)^2 + 2*(a^2*b^2*d^2*f^3*x^2 + 2*a^2*b^2*d^2*e*f^2*x
+ 2*a^2*b^2*c*d*e*f^2 - a^2*b^2*c^2*f^3)*sin(d*x + c))*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) + 3*(2*(a^4 - b^4)*d*f^3*x + 2*
(a^4 - b^4)*c*f^3 - 2*((a^2*b^2 - b^4)*d*f^3*x + (a^2*b^2 - b^4)*c*f^3)*cos(d*x + c)^2 + 4*((a^3*b - a*b^3)*d*
f^3*x + (a^3*b - a*b^3)*c*f^3)*sin(d*x + c) - ((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 - (a*b^3*d^2*f^3*x^2 + 2*a*b^3*d^2*e*f^2*x + 2*a*b^3*c*d*
e*f^2 - a*b^3*c^2*f^3)*cos(d*x + c)^2 + 2*(a^2*b^2*d^2*f^3*x^2 + 2*a^2*b^2*d^2*e*f^2*x + 2*a^2*b^2*c*d*e*f^2 -
a^2*b^2*c^2*f^3)*sin(d*x + c))*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))/((a^4*b^3 - 2*a^2*b^5 + b^7)*d^4*cos(d*x + c)^2 - 2*(
a^5*b^2 - 2*a^3*b^4 + a*b^6)*d^4*sin(d*x + c) - (a^6*b - a^4*b^3 - a^2*b^5 + b^7)*d^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Exception raised: ValueError} \]

[In]

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

Giac [F]

\[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\int { \frac {{\left (f x + e\right )}^{3} \cos \left (d x + c\right )}{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}