\(\int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx\) [303]

Optimal result

Integrand size = 28, antiderivative size = 548 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {e f x}{2 b d}+\frac {f^2 x^2}{4 b d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {2 i \left (a^2-b^2\right ) f (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {2 \left (a^2-b^2\right ) f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 \left (a^2-b^2\right ) f^2 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 a f^2 \sin (c+d x)}{b^2 d^3}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}-\frac {f (e+f x) \cos (c+d x) \sin (c+d x)}{2 b d^2}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 548, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {4621, 3377, 2717, 4489, 3391, 4615, 2221, 2611, 2320, 6724} \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {2 f^2 \left (a^2-b^2\right ) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^3}-\frac {2 f^2 \left (a^2-b^2\right ) \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^3}+\frac {2 i f \left (a^2-b^2\right ) (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d^2}+\frac {2 i f \left (a^2-b^2\right ) (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b^3 d^2}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b^3 d}-\frac {\left (a^2-b^2\right ) (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b^3 d}+\frac {i \left (a^2-b^2\right ) (e+f x)^3}{3 b^3 f}-\frac {2 a f^2 \sin (c+d x)}{b^2 d^3}+\frac {2 a f (e+f x) \cos (c+d x)}{b^2 d^2}+\frac {a (e+f x)^2 \sin (c+d x)}{b^2 d}+\frac {f^2 \sin ^2(c+d x)}{4 b d^3}-\frac {f (e+f x) \sin (c+d x) \cos (c+d x)}{2 b d^2}-\frac {(e+f x)^2 \sin ^2(c+d x)}{2 b d}+\frac {e f x}{2 b d}+\frac {f^2 x^2}{4 b d} \]

[In]

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

[Out]

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

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 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 2717

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[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 3391

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

Rule 4489

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 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 4621

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

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2283\) vs. \(2(548)=1096\).

Time = 3.29 (sec) , antiderivative size = 2283, normalized size of antiderivative = 4.17 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Result too large to show} \]

[In]

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

[Out]

((48*I)*a^2*d^3*e^2*E^((2*I)*c)*x - (48*I)*b^2*d^3*e^2*E^((2*I)*c)*x + (48*I)*a^2*d^3*e*E^((2*I)*c)*f*x^2 - (4
8*I)*b^2*d^3*e*E^((2*I)*c)*f*x^2 + (16*I)*a^2*d^3*E^((2*I)*c)*f^2*x^3 - (16*I)*b^2*d^3*E^((2*I)*c)*f^2*x^3 + (
24*I)*a*b*d^2*e^2*E^(I*c)*Cos[d*x] - (24*I)*a*b*d^2*e^2*E^((3*I)*c)*Cos[d*x] + 48*a*b*d*e*E^(I*c)*f*Cos[d*x] +
48*a*b*d*e*E^((3*I)*c)*f*Cos[d*x] - (48*I)*a*b*E^(I*c)*f^2*Cos[d*x] + (48*I)*a*b*E^((3*I)*c)*f^2*Cos[d*x] + (
48*I)*a*b*d^2*e*E^(I*c)*f*x*Cos[d*x] - (48*I)*a*b*d^2*e*E^((3*I)*c)*f*x*Cos[d*x] + 48*a*b*d*E^(I*c)*f^2*x*Cos[
d*x] + 48*a*b*d*E^((3*I)*c)*f^2*x*Cos[d*x] + (24*I)*a*b*d^2*E^(I*c)*f^2*x^2*Cos[d*x] - (24*I)*a*b*d^2*E^((3*I)
*c)*f^2*x^2*Cos[d*x] + 6*b^2*d^2*e^2*Cos[2*d*x] + 6*b^2*d^2*e^2*E^((4*I)*c)*Cos[2*d*x] - (6*I)*b^2*d*e*f*Cos[2
*d*x] + (6*I)*b^2*d*e*E^((4*I)*c)*f*Cos[2*d*x] - 3*b^2*f^2*Cos[2*d*x] - 3*b^2*E^((4*I)*c)*f^2*Cos[2*d*x] + 12*
b^2*d^2*e*f*x*Cos[2*d*x] + 12*b^2*d^2*e*E^((4*I)*c)*f*x*Cos[2*d*x] - (6*I)*b^2*d*f^2*x*Cos[2*d*x] + (6*I)*b^2*
d*E^((4*I)*c)*f^2*x*Cos[2*d*x] + 6*b^2*d^2*f^2*x^2*Cos[2*d*x] + 6*b^2*d^2*E^((4*I)*c)*f^2*x^2*Cos[2*d*x] - 48*
a^2*d^2*e^2*E^((2*I)*c)*Log[b - (2*I)*a*E^(I*(c + d*x)) - b*E^((2*I)*(c + d*x))] + 48*b^2*d^2*e^2*E^((2*I)*c)*
Log[b - (2*I)*a*E^(I*(c + d*x)) - b*E^((2*I)*(c + d*x))] - 96*a^2*d^2*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)])] + 96*b^2*d^2*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)])] - 48*a^2*d^2*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)])] + 48*b^2*d^2*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)])] - 96*a^2*d^2*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)])] + 96*b^2*d^2*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)])] - 48*a^2*d^2*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)])] + 48*b^2*d^2*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)])] + (96*I)*(a^2 - b^2)*d*E^((2*I)*c)*f*(e + f*x)*PolyLo
g[2, (I*b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])] + (96*I)*(a^2 - b^2)*d*E^((2*I)*c
)*f*(e + f*x)*PolyLog[2, -((b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))] - 96*a^2*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)])] + 96*b^2*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)])] - 96*a^2*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)]))] + 96*b^2*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)]))] + 24*a*b*d^2*e^2*E
^(I*c)*Sin[d*x] + 24*a*b*d^2*e^2*E^((3*I)*c)*Sin[d*x] - (48*I)*a*b*d*e*E^(I*c)*f*Sin[d*x] + (48*I)*a*b*d*e*E^(
(3*I)*c)*f*Sin[d*x] - 48*a*b*E^(I*c)*f^2*Sin[d*x] - 48*a*b*E^((3*I)*c)*f^2*Sin[d*x] + 48*a*b*d^2*e*E^(I*c)*f*x
*Sin[d*x] + 48*a*b*d^2*e*E^((3*I)*c)*f*x*Sin[d*x] - (48*I)*a*b*d*E^(I*c)*f^2*x*Sin[d*x] + (48*I)*a*b*d*E^((3*I
)*c)*f^2*x*Sin[d*x] + 24*a*b*d^2*E^(I*c)*f^2*x^2*Sin[d*x] + 24*a*b*d^2*E^((3*I)*c)*f^2*x^2*Sin[d*x] - (6*I)*b^
2*d^2*e^2*Sin[2*d*x] + (6*I)*b^2*d^2*e^2*E^((4*I)*c)*Sin[2*d*x] - 6*b^2*d*e*f*Sin[2*d*x] - 6*b^2*d*e*E^((4*I)*
c)*f*Sin[2*d*x] + (3*I)*b^2*f^2*Sin[2*d*x] - (3*I)*b^2*E^((4*I)*c)*f^2*Sin[2*d*x] - (12*I)*b^2*d^2*e*f*x*Sin[2
*d*x] + (12*I)*b^2*d^2*e*E^((4*I)*c)*f*x*Sin[2*d*x] - 6*b^2*d*f^2*x*Sin[2*d*x] - 6*b^2*d*E^((4*I)*c)*f^2*x*Sin
[2*d*x] - (6*I)*b^2*d^2*f^2*x^2*Sin[2*d*x] + (6*I)*b^2*d^2*E^((4*I)*c)*f^2*x^2*Sin[2*d*x])/(48*b^3*d^3*E^((2*I
)*c))

Maple [F]

\[\int \frac {\left (f x +e \right )^{2} \left (\cos ^{3}\left (d x +c \right )\right )}{a +b \sin \left (d x +c \right )}d x\]

[In]

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

[Out]

int((f*x+e)^2*cos(d*x+c)^3/(a+b*sin(d*x+c)),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. 1779 vs. \(2 (498) = 996\).

Time = 0.47 (sec) , antiderivative size = 1779, normalized size of antiderivative = 3.25 \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \]

[In]

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

[Out]

-1/4*(b^2*d^2*f^2*x^2 + 2*b^2*d^2*e*f*x + 4*(a^2 - b^2)*f^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) + 4*(a^2 - b^2)*f^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) + 4*(a^2 - b^2)*f^2*polylo
g(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) + 4
*(a^2 - b^2)*f^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) - (2*b^2*d^2*f^2*x^2 + 4*b^2*d^2*e*f*x + 2*b^2*d^2*e^2 - b^2*f^2)*cos(d*x + c)^2 - 8*(a*b*
d*f^2*x + a*b*d*e*f)*cos(d*x + c) + 4*(-I*(a^2 - b^2)*d*f^2*x - I*(a^2 - b^2)*d*e*f)*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) + 4*(-I*(a^2 - b^2)*d
*f^2*x - I*(a^2 - b^2)*d*e*f)*dilog((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)/b + 1) + 4*(I*(a^2 - b^2)*d*f^2*x + I*(a^2 - b^2)*d*e*f)*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) + 4*(I*(a^2 - b^2)*d*
f^2*x + I*(a^2 - b^2)*d*e*f)*dilog((-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)/b + 1) + 2*((a^2 - b^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^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^2)*d^2*e^2 - 2*(a^2
- b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^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^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + (a^2 - b^2)*c^2*f^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^2)*d^2*e^2 - 2*(a^2 - b^2)*c*d*e*f + (a^2 - b
^2)*c^2*f^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^2)
*d^2*f^2*x^2 + 2*(a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^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) + 2*((a^2 - b^2)*d^2*f^2*
x^2 + 2*(a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^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) + 2*((a^2 - b^2)*d^2*f^2*x^2 + 2*(
a^2 - b^2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^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) + 2*((a^2 - b^2)*d^2*f^2*x^2 + 2*(a^2 - b^
2)*d^2*e*f*x + 2*(a^2 - b^2)*c*d*e*f - (a^2 - b^2)*c^2*f^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) - 2*(2*a*b*d^2*f^2*x^2 + 4*a*b*d^2*e*f*x + 2*a*b*d
^2*e^2 - 4*a*b*f^2 - (b^2*d*f^2*x + b^2*d*e*f)*cos(d*x + c))*sin(d*x + c))/(b^3*d^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((f*x+e)^2*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 de

Giac [F]

\[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\int { \frac {{\left (f x + e\right )}^{2} \cos \left (d x + c\right )^{3}}{b \sin \left (d x + c\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(e+f x)^2 \cos ^3(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}