\(\int \frac {(e+f x)^2 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx\) [323]

Optimal result

Integrand size = 26, antiderivative size = 357 \[ \int \frac {(e+f x)^2 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=-\frac {i a f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {i a f (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {f^2 \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^3}-\frac {a f^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^3}+\frac {a f^2 \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^3}-\frac {(e+f x)^2}{2 b d (a+b \sin (c+d x))^2}+\frac {f (e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d^2 (a+b \sin (c+d x))} \] Output:

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

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1017\) vs. \(2(357)=714\).

Time = 15.87 (sec) , antiderivative size = 1017, normalized size of antiderivative = 2.85 \[ \int \frac {(e+f x)^2 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(f^2*x*Cot[c])/(b*(-a^2 + b^2)*d^2) - ((2*I)*E^(I*c)*f*(E^(I*c)*f*x + (I*a 
*e*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*E^(I*c)*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]])/ 
Sqrt[a^2 - b^2] - ((I/2)*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^(I*c)*f*Log[b - (2*I)*a*E^(I*(c + d*x)) 
- b*E^((2*I)*(c + d*x))])/d + ((I/2)*a*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/2)*a*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/ 
2)*a*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/2)*a*E^((2*I)*c)*f*x*Lo 
g[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)] - (a*(-1 + E^((2*I)*c))*f*PolyLog[2, (I* 
b*E^(I*(2*c + d*x)))/(a*E^(I*c) + I*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/(2*d 
*Sqrt[(-a^2 + b^2)*E^((2*I)*c)]) + (a*(-1 + E^((2*I)*c))*f*PolyLog[2, -((b 
*E^(I*(2*c + d*x)))/(I*a*E^(I*c) + Sqrt[(-a^2 + b^2)*E^((2*I)*c)]))])/(2*d 
*Sqrt[(-a^2 + b^2)*E^((2*I)*c)])))/(b*(-a^2 + b^2)*d^2*(-1 + E^((2*I)*c))) 
 - (f^2*x*Csc[c/2]*Sec[c/2]*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]))/( 
2*b*(-a + b)*(a + b)*d^2) - (e + f*x)^2/(2*b*d*(a + b*Sin[c + d*x])^2) + ( 
Csc[c/2]*Sec[c/2]*(-(a*e*f*Cos[c]) - a*f^2*x*Cos[c] - b*e*f*Sin[d*x] - ...
 

Rubi [A] (verified)

Time = 1.37 (sec) , antiderivative size = 355, normalized size of antiderivative = 0.99, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {4922, 3042, 3805, 3042, 3147, 16, 3804, 2694, 27, 2620, 2715, 2838}

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.

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

\(\Big \downarrow \) 4922

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3805

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3147

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 3804

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

\(\Big \downarrow \) 2694

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

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 2715

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] 
:> Simp[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 2838

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 3042

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

rule 3147

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_.), x_Symbol] :> Simp[1/(b^p*f)   Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) 
/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p 
 - 1)/2] && NeQ[a^2 - b^2, 0]
 

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 3805

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 945 vs. \(2 (325 ) = 650\).

Time = 6.58 (sec) , antiderivative size = 946, normalized size of antiderivative = 2.65

method	result	size
risch	\(\frac {2 a^{2} d \,f^{2} x^{2} {\mathrm e}^{2 i \left (d x +c \right )}-2 b^{2} d \,f^{2} x^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 i a^{2} f^{2} x \,{\mathrm e}^{2 i \left (d x +c \right )}-2 i b^{2} f^{2} x +4 a^{2} d e f x \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b a \,f^{2} x \,{\mathrm e}^{3 i \left (d x +c \right )}-4 b^{2} d e f x \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i b^{2} e f \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i b^{2} f^{2} x \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a^{2} d \,e^{2} {\mathrm e}^{2 i \left (d x +c \right )}+2 b a e f \,{\mathrm e}^{3 i \left (d x +c \right )}-2 b^{2} d \,e^{2} {\mathrm e}^{2 i \left (d x +c \right )}+4 i a^{2} e f \,{\mathrm e}^{2 i \left (d x +c \right )}-6 a b \,f^{2} x \,{\mathrm e}^{i \left (d x +c \right )}-2 i b^{2} e f -6 a b e f \,{\mathrm e}^{i \left (d x +c \right )}}{\left (2 i a \,{\mathrm e}^{i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )^{2} d^{2} \left (a^{2}-b^{2}\right ) b}+\frac {f^{2} \ln \left (i b \,{\mathrm e}^{2 i \left (d x +c \right )}-2 a \,{\mathrm e}^{i \left (d x +c \right )}-i b \right )}{b \left (-a^{2}+b^{2}\right ) d^{3}}-\frac {2 f^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{b \left (-a^{2}+b^{2}\right ) d^{3}}+\frac {2 i f^{2} a c \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}-\frac {2 i f a e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2}}-\frac {f^{2} a \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b -\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) x}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2}}+\frac {f^{2} a \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) x}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{2}}-\frac {f^{2} a \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b -\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right ) c}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}+\frac {f^{2} a \ln \left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right ) c}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}+\frac {i f^{2} a \operatorname {dilog}\left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b -\sqrt {-a^{2}+b^{2}}}{i a -\sqrt {-a^{2}+b^{2}}}\right )}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}-\frac {i f^{2} a \operatorname {dilog}\left (\frac {i a +{\mathrm e}^{i \left (d x +c \right )} b +\sqrt {-a^{2}+b^{2}}}{i a +\sqrt {-a^{2}+b^{2}}}\right )}{b \left (-a^{2}+b^{2}\right )^{\frac {3}{2}} d^{3}}\)	\(946\)

Input:

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

Output:

2*(a^2*d*f^2*x^2*exp(2*I*(d*x+c))-b^2*d*f^2*x^2*exp(2*I*(d*x+c))+2*I*a^2*f 
^2*x*exp(2*I*(d*x+c))-I*b^2*f^2*x+2*a^2*d*e*f*x*exp(2*I*(d*x+c))+b*a*f^2*x 
*exp(3*I*(d*x+c))-2*b^2*d*e*f*x*exp(2*I*(d*x+c))+I*b^2*e*f*exp(2*I*(d*x+c) 
)+I*b^2*f^2*x*exp(2*I*(d*x+c))+a^2*d*e^2*exp(2*I*(d*x+c))+b*a*e*f*exp(3*I* 
(d*x+c))-b^2*d*e^2*exp(2*I*(d*x+c))+2*I*a^2*e*f*exp(2*I*(d*x+c))-3*a*b*f^2 
*x*exp(I*(d*x+c))-I*b^2*e*f-3*a*b*e*f*exp(I*(d*x+c)))/(2*I*a*exp(I*(d*x+c) 
)+b*exp(2*I*(d*x+c))-b)^2/d^2/(a^2-b^2)/b+1/b/(-a^2+b^2)/d^3*f^2*ln(I*b*ex 
p(2*I*(d*x+c))-2*a*exp(I*(d*x+c))-I*b)-2/b/(-a^2+b^2)/d^3*f^2*ln(exp(I*(d* 
x+c)))+2*I/b/(-a^2+b^2)^(3/2)/d^3*f^2*a*c*arctan(1/2*(2*I*b*exp(I*(d*x+c)) 
-2*a)/(-a^2+b^2)^(1/2))-2*I/b/(-a^2+b^2)^(3/2)/d^2*f*a*e*arctan(1/2*(2*I*b 
*exp(I*(d*x+c))-2*a)/(-a^2+b^2)^(1/2))-1/b/(-a^2+b^2)^(3/2)/d^2*f^2*a*ln(( 
I*a+exp(I*(d*x+c))*b-(-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+1/b/(-a^2 
+b^2)^(3/2)/d^2*f^2*a*ln((I*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^ 
2+b^2)^(1/2)))*x-1/b/(-a^2+b^2)^(3/2)/d^3*f^2*a*ln((I*a+exp(I*(d*x+c))*b-( 
-a^2+b^2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*c+1/b/(-a^2+b^2)^(3/2)/d^3*f^2*a* 
ln((I*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*c+I/b/( 
-a^2+b^2)^(3/2)/d^3*f^2*a*dilog((I*a+exp(I*(d*x+c))*b-(-a^2+b^2)^(1/2))/(I 
*a-(-a^2+b^2)^(1/2)))-I/b/(-a^2+b^2)^(3/2)/d^3*f^2*a*dilog((I*a+exp(I*(d*x 
+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))
 

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. 2375 vs. \(2 (317) = 634\).

Time = 0.27 (sec) , antiderivative size = 2375, normalized size of antiderivative = 6.65 \[ \int \frac {(e+f x)^2 \cos (c+d x)}{(a+b \sin (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*((a^4 - 2*a^2*b^2 + b^4)*d^2*f^2*x^2 + 2*(a^4 - 2*a^2*b^2 + b^4)*d^2*e 
*f*x + (a^4 - 2*a^2*b^2 + b^4)*d^2*e^2 - 2*((a^2*b^2 - b^4)*d*f^2*x + (a^2 
*b^2 - b^4)*d*e*f)*cos(d*x + c)*sin(d*x + c) - (-I*a*b^3*f^2*cos(d*x + c)^ 
2 + 2*I*a^2*b^2*f^2*sin(d*x + c) + I*(a^3*b + a*b^3)*f^2)*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*si 
n(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - (I*a*b^3*f^2*cos(d*x + c) 
^2 - 2*I*a^2*b^2*f^2*sin(d*x + c) - I*(a^3*b + a*b^3)*f^2)*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*s 
in(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - (I*a*b^3*f^2*cos(d*x + c 
)^2 - 2*I*a^2*b^2*f^2*sin(d*x + c) - I*(a^3*b + a*b^3)*f^2)*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) - (-I*a*b^3*f^2*cos(d*x 
+ c)^2 + 2*I*a^2*b^2*f^2*sin(d*x + c) + I*(a^3*b + a*b^3)*f^2)*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) + ((a^3*b + a*b^3)*d* 
f^2*x + (a^3*b + a*b^3)*c*f^2 - (a*b^3*d*f^2*x + a*b^3*c*f^2)*cos(d*x + c) 
^2 + 2*(a^2*b^2*d*f^2*x + a^2*b^2*c*f^2)*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^3*b + a*b^3)*d*f^2*x + (a^3*b 
 + a*b^3)*c*f^2 - (a*b^3*d*f^2*x + a*b^3*c*f^2)*cos(d*x + c)^2 + 2*(a^2...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

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

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

Output:

(4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( 
c + d*x)**2*a**2*b**2*e*f + 8*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*e*f + 4*sqrt(a**2 - b**2)*atan( 
(tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a**4*e*f + 2*cos(c + d*x)*sin( 
c + d*x)*a**3*b**2*e*f - 2*cos(c + d*x)*sin(c + d*x)*a*b**4*e*f + 2*cos(c 
+ d*x)*a**4*b*e*f - 2*cos(c + d*x)*a**2*b**3*e*f + 2*int((cos(c + d*x)*x** 
2)/(sin(c + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2* 
b + a**3),x)*sin(c + d*x)**2*a**5*b**3*d**2*f**2 - 4*int((cos(c + d*x)*x** 
2)/(sin(c + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2* 
b + a**3),x)*sin(c + d*x)**2*a**3*b**5*d**2*f**2 + 2*int((cos(c + d*x)*x** 
2)/(sin(c + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2* 
b + a**3),x)*sin(c + d*x)**2*a*b**7*d**2*f**2 + 4*int((cos(c + d*x)*x**2)/ 
(sin(c + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b + 
 a**3),x)*sin(c + d*x)*a**6*b**2*d**2*f**2 - 8*int((cos(c + d*x)*x**2)/(si 
n(c + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b + a* 
*3),x)*sin(c + d*x)*a**4*b**4*d**2*f**2 + 4*int((cos(c + d*x)*x**2)/(sin(c 
 + d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b + a**3) 
,x)*sin(c + d*x)*a**2*b**6*d**2*f**2 + 2*int((cos(c + d*x)*x**2)/(sin(c + 
d*x)**3*b**3 + 3*sin(c + d*x)**2*a*b**2 + 3*sin(c + d*x)*a**2*b + a**3),x) 
*a**7*b*d**2*f**2 - 4*int((cos(c + d*x)*x**2)/(sin(c + d*x)**3*b**3 + 3...