\(\int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [321]

Optimal result

Integrand size = 26, antiderivative size = 418 \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {3 i f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}+\frac {3 i f (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}+\frac {6 f^2 (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^3}-\frac {6 i f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}+\frac {6 i f^3 \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^4}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))} \] Output:

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

Mathematica [A] (warning: unable to verify)

Time = 3.15 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.07 \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {3 i f \left (-2 \sqrt {a^2-b^2} d f (e+f x) \operatorname {PolyLog}\left (2,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )+2 \sqrt {a^2-b^2} d f (e+f x) \operatorname {PolyLog}\left (2,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )-i \left (d^2 \left (2 \sqrt {-a^2+b^2} e^2 \arctan \left (\frac {i a+b e^{i (c+d x)}}{\sqrt {a^2-b^2}}\right )+\sqrt {a^2-b^2} f x (2 e+f x) \left (\log \left (1-\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-\log \left (1+\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )\right )\right )+2 \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,\frac {b e^{i (c+d x)}}{-i a+\sqrt {-a^2+b^2}}\right )-2 \sqrt {a^2-b^2} f^2 \operatorname {PolyLog}\left (3,-\frac {b e^{i (c+d x)}}{i a+\sqrt {-a^2+b^2}}\right )\right )\right )}{b \sqrt {-\left (a^2-b^2\right )^2} d^4}-\frac {(e+f x)^3}{b d (a+b \sin (c+d x))} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.52 (sec) , antiderivative size = 382, normalized size of antiderivative = 0.91, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {4922, 3042, 3804, 2694, 27, 2620, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 4922

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3804

\(\displaystyle -\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {6 f \int \frac {e^{i (c+d x)} (e+f x)^2}{2 e^{i (c+d x)} a-i b e^{2 i (c+d x)}+i b}dx}{b d}\)

\(\Big \downarrow \) 2694

\(\displaystyle -\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {6 f \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{2 \left (a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{2 \left (a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}\right )}dx}{\sqrt {a^2-b^2}}\right )}{b d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {6 f \left (\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}+\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}-\frac {i b \int \frac {e^{i (c+d x)} (e+f x)^2}{a-i b e^{i (c+d x)}-\sqrt {a^2-b^2}}dx}{2 \sqrt {a^2-b^2}}\right )}{b d}\)

\(\Big \downarrow \) 2620

\(\displaystyle -\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {6 f \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b d}\)

\(\Big \downarrow \) 3011

\(\displaystyle -\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {6 f \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )dx}{d}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b d}\)

\(\Big \downarrow \) 2720

\(\displaystyle -\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {6 f \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \int e^{-i (c+d x)} \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )de^{i (c+d x)}}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b d}\)

\(\Big \downarrow \) 7143

\(\displaystyle -\frac {(e+f x)^3}{b d (a+b \sin (c+d x))}+\frac {6 f \left (\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}-\frac {i b \left (\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2}\right )}{b d}\right )}{2 \sqrt {a^2-b^2}}\right )}{b d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2620

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] - Si 
mp[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 2694

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]}, Simp[2*(c/q)   Int 
[(f + g*x)^m*(F^u/(b - q + 2*c*F^u)), x], x] - Simp[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 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{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 3011

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] + Simp[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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3804

Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Sy 
mbol] :> Simp[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 4922

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] - Simp[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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

Maple [F]

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

Input:

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

Output:

int((f*x+e)^3*cos(d*x+c)/(a+b*sin(d*x+c))^2,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. 2284 vs. \(2 (360) = 720\).

Time = 0.28 (sec) , antiderivative size = 2284, normalized size of antiderivative = 5.46 \[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/2*(2*(a^2 - b^2)*d^3*f^3*x^3 + 6*(a^2 - b^2)*d^3*e*f^2*x^2 + 6*(a^2 - b 
^2)*d^3*e^2*f*x + 2*(a^2 - b^2)*d^3*e^3 - 6*(I*a*b*d*f^3*x + I*a*b*d*e*f^2 
 + (I*b^2*d*f^3*x + I*b^2*d*e*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*di 
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 + 1) - 6*(-I*a*b*d*f^3*x - I*a*b*d*e*f^2 
+ (-I*b^2*d*f^3*x - I*b^2*d*e*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*di 
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 + 1) - 6*(-I*a*b*d*f^3*x - I*a*b*d*e*f^2 
+ (-I*b^2*d*f^3*x - I*b^2*d*e*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*di 
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 + 1) - 6*(I*a*b*d*f^3*x + I*a*b*d*e*f^2 
+ (I*b^2*d*f^3*x + I*b^2*d*e*f^2)*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2)*dil 
og((-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*(a*b*d^2*e^2*f - 2*a*b*c*d*e*f^2 
 + a*b*c^2*f^3 + (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^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*(a*b*d^2*e^2*f - 2*a*b*c*d*e*f^2 + a 
*b*c^2*f^3 + (b^2*d^2*e^2*f - 2*b^2*c*d*e*f^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*sq 
rt(-(a^2 - b^2)/b^2) - 2*I*a) + 3*(a*b*d^2*e^2*f - 2*a*b*c*d*e*f^2 + a*...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x)^3 \cos (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {too large to display} \] Input:

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

Output:

(6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( 
c + d*x)*a**4*b*d**2*e**2*f - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)* 
a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**4*b*f**3 - 24*sqrt(a**2 - b**2)* 
atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a**3*b**2*d* 
e*f**2 + 72*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b* 
*2))*sin(c + d*x)*a**2*b**3*f**3 + 24*sqrt(a**2 - b**2)*atan((tan((c + d*x 
)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*a*b**4*d*e*f**2 - 72*sqrt(a**2 
 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(c + d*x)*b** 
5*f**3 + 6*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b** 
2))*a**5*d**2*e**2*f - 12*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/ 
sqrt(a**2 - b**2))*a**5*f**3 - 24*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2) 
*a + b)/sqrt(a**2 - b**2))*a**4*b*d*e*f**2 + 72*sqrt(a**2 - b**2)*atan((ta 
n((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**3*b**2*f**3 + 24*sqrt(a**2 - b 
**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**2*b**3*d*e*f**2 - 
 72*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a*b 
**4*f**3 + 6*cos(c + d*x)*a**5*b*d**2*e*f**2*x + 3*cos(c + d*x)*a**5*b*d** 
2*f**3*x**2 - 18*cos(c + d*x)*a**4*b**2*d*f**3*x - 18*cos(c + d*x)*a**3*b* 
*3*d**2*e*f**2*x - 9*cos(c + d*x)*a**3*b**3*d**2*f**3*x**2 + 54*cos(c + d* 
x)*a**2*b**4*d*f**3*x + 12*cos(c + d*x)*a*b**5*d**2*e*f**2*x + 6*cos(c + d 
*x)*a*b**5*d**2*f**3*x**2 - 36*cos(c + d*x)*b**6*d*f**3*x + 12*int(x**2...