\(\int \frac {(e+f x)^3 \sin (c+d x)}{a+a \sin (c+d x)} \, dx\) [179]

Optimal result

Integrand size = 26, antiderivative size = 164 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {i (e+f x)^3}{a d}+\frac {(e+f x)^4}{4 a f}+\frac {(e+f x)^3 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}-\frac {6 f (e+f x)^2 \log \left (1-i e^{i (c+d x)}\right )}{a d^2}+\frac {12 i f^2 (e+f x) \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}-\frac {12 f^3 \operatorname {PolyLog}\left (3,i e^{i (c+d x)}\right )}{a d^4} \] Output:

I*(f*x+e)^3/a/d+1/4*(f*x+e)^4/a/f+(f*x+e)^3*cot(1/2*c+1/4*Pi+1/2*d*x)/a/d- 
6*f*(f*x+e)^2*ln(1-I*exp(I*(d*x+c)))/a/d^2+12*I*f^2*(f*x+e)*polylog(2,I*ex 
p(I*(d*x+c)))/a/d^3-12*f^3*polylog(3,I*exp(I*(d*x+c)))/a/d^4
 

Mathematica [A] (verified)

Time = 2.82 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.59 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right )+\frac {24 f (\cos (c)+i \sin (c)) \left (\frac {(e+f x)^3 (\cos (c)-i \sin (c))}{3 f}-\frac {(e+f x)^2 \log (1+i \cos (c+d x)+\sin (c+d x)) (1+i \cos (c)+\sin (c))}{d}+\frac {2 f (d (e+f x) \operatorname {PolyLog}(2,-i \cos (c+d x)-\sin (c+d x))-i f \operatorname {PolyLog}(3,-i \cos (c+d x)-\sin (c+d x))) (\cos (c)-i (1+\sin (c)))}{d^3}\right )}{d (\cos (c)+i (1+\sin (c)))}-\frac {8 (e+f x)^3 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{4 a} \] Input:

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

Output:

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3) + (24*f*(Cos[c] + I*Sin[c]) 
*(((e + f*x)^3*(Cos[c] - I*Sin[c]))/(3*f) - ((e + f*x)^2*Log[1 + I*Cos[c + 
 d*x] + Sin[c + d*x]]*(1 + I*Cos[c] + Sin[c]))/d + (2*f*(d*(e + f*x)*PolyL 
og[2, (-I)*Cos[c + d*x] - Sin[c + d*x]] - I*f*PolyLog[3, (-I)*Cos[c + d*x] 
 - Sin[c + d*x]])*(Cos[c] - I*(1 + Sin[c])))/d^3))/(d*(Cos[c] + I*(1 + Sin 
[c]))) - (8*(e + f*x)^3*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[(c + d 
*x)/2] + Sin[(c + d*x)/2])))/(4*a)
 

Rubi [A] (verified)

Time = 0.98 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.18, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5026, 17, 3042, 3799, 3042, 4672, 3042, 25, 4202, 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.

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

\(\Big \downarrow \) 5026

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

\(\Big \downarrow \) 17

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\int \frac {(e+f x)^3}{\sin (c+d x) a+a}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\int \frac {(e+f x)^3}{\sin (c+d x) a+a}dx\)

\(\Big \downarrow \) 3799

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {\int (e+f x)^3 \csc ^2\left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {\int (e+f x)^3 \csc \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )^2dx}{2 a}\)

\(\Big \downarrow \) 4672

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {\frac {6 f \int (e+f x)^2 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )dx}{d}-\frac {2 (e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}}{2 a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {\frac {6 f \int -(e+f x)^2 \tan \left (\frac {c}{2}+\frac {d x}{2}+\frac {3 \pi }{4}\right )dx}{d}-\frac {2 (e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}}{2 a}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {-\frac {6 f \int (e+f x)^2 \tan \left (\frac {1}{4} (2 c+3 \pi )+\frac {d x}{2}\right )dx}{d}-\frac {2 (e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}}{2 a}\)

\(\Big \downarrow \) 4202

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {-\frac {2 (e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {6 f \left (\frac {i (e+f x)^3}{3 f}-2 i \int \frac {e^{\frac {1}{2} i (2 c+2 d x+3 \pi )} (e+f x)^2}{1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}}dx\right )}{d}}{2 a}\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {-\frac {2 (e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {6 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {2 i f \int (e+f x) \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )dx}{d}-\frac {i (e+f x)^2 \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}\right )\right )}{d}}{2 a}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {-\frac {2 (e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {6 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {2 i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}-\frac {i f \int \operatorname {PolyLog}\left (2,-e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )dx}{d}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}\right )\right )}{d}}{2 a}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {-\frac {2 (e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {6 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {2 i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}-\frac {f \int e^{-\frac {1}{2} i (2 c+2 d x+3 \pi )} \operatorname {PolyLog}\left (2,-e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )de^{\frac {1}{2} i (2 c+2 d x+3 \pi )}}{d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}\right )\right )}{d}}{2 a}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {(e+f x)^4}{4 a f}-\frac {-\frac {2 (e+f x)^3 \cot \left (\frac {c}{2}+\frac {d x}{2}+\frac {\pi }{4}\right )}{d}-\frac {6 f \left (\frac {i (e+f x)^3}{3 f}-2 i \left (\frac {2 i f \left (\frac {i (e+f x) \operatorname {PolyLog}\left (2,-e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}-\frac {f \operatorname {PolyLog}\left (3,-e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d^2}\right )}{d}-\frac {i (e+f x)^2 \log \left (1+e^{\frac {1}{2} i (2 c+2 d x+3 \pi )}\right )}{d}\right )\right )}{d}}{2 a}\)

Input:

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

Output:

(e + f*x)^4/(4*a*f) - ((-2*(e + f*x)^3*Cot[c/2 + Pi/4 + (d*x)/2])/d - (6*f 
*(((I/3)*(e + f*x)^3)/f - (2*I)*(((-I)*(e + f*x)^2*Log[1 + E^((I/2)*(2*c + 
 3*Pi + 2*d*x))])/d + ((2*I)*f*((I*(e + f*x)*PolyLog[2, -E^((I/2)*(2*c + 3 
*Pi + 2*d*x))])/d - (f*PolyLog[3, -E^((I/2)*(2*c + 3*Pi + 2*d*x))])/d^2))/ 
d)))/d)/(2*a)
 

Defintions of rubi rules used

rule 17

Int[(c_.)*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[c*((a + b*x)^(m + 1 
)/(b*(m + 1))), x] /; FreeQ[{a, b, c, m}, x] && NeQ[m, -1]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 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 3799

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) 
, x_Symbol] :> Simp[(2*a)^n   Int[(c + d*x)^m*Sin[(1/2)*(e + Pi*(a/(2*b))) 
+ f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2 - b^ 
2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])
 

rule 4202

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + (f_.)*(x_)], x_Symbol] :> Simp[I 
*((c + d*x)^(m + 1)/(d*(m + 1))), x] - Simp[2*I   Int[(c + d*x)^m*(E^(2*I*( 
e + f*x))/(1 + E^(2*I*(e + f*x)))), x], x] /; FreeQ[{c, d, e, f}, x] && IGt 
Q[m, 0]
 

rule 4672

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp 
[(-(c + d*x)^m)*(Cot[e + f*x]/f), x] + Simp[d*(m/f)   Int[(c + d*x)^(m - 1) 
*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]
 

rule 5026

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

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 [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 605 vs. \(2 (145 ) = 290\).

Time = 1.05 (sec) , antiderivative size = 606, normalized size of antiderivative = 3.70

method	result	size
risch	\(-\frac {3 f \,e^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a \,d^{2}}-\frac {12 f^{3} \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}+\frac {f^{2} e \,x^{3}}{a}+\frac {3 f \,e^{2} x^{2}}{2 a}+\frac {e^{3} x}{a}-\frac {3 f^{3} c^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a \,d^{4}}+\frac {6 f \,e^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}+\frac {6 f^{3} c^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}+\frac {6 f^{3} c^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}-\frac {6 f^{3} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x^{2}}{a \,d^{2}}+\frac {2 i f^{3} x^{3}}{a d}-\frac {4 i f^{3} c^{3}}{a \,d^{4}}+\frac {12 i f^{2} e c x}{a \,d^{2}}-\frac {12 i f^{2} c e \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {12 i f^{3} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{3}}+\frac {6 i f^{2} e \,x^{2}}{a d}+\frac {6 i f^{2} e \,c^{2}}{a \,d^{3}}+\frac {12 i f^{2} e \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}+\frac {6 i f^{3} c^{2} \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{4}}+\frac {6 i f \,e^{2} \arctan \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}+\frac {6 f^{2} c e \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{a \,d^{3}}-\frac {12 f^{2} e \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}-\frac {6 i f^{3} c^{2} x}{a \,d^{3}}-\frac {12 f^{2} c e \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {12 f^{2} e \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}+\frac {2 f^{3} x^{3}+6 e \,f^{2} x^{2}+6 e^{2} f x +2 e^{3}}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {f^{3} x^{4}}{4 a}+\frac {e^{4}}{4 a f}\)	\(606\)

Input:

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

Output:

2*(f^3*x^3+3*e*f^2*x^2+3*e^2*f*x+e^3)/d/a/(exp(I*(d*x+c))+I)+1/a*f^2*e*x^3 
+3/2/a*f*e^2*x^2+1/a*e^3*x+12*I/a/d^2*f^2*e*c*x-12*I/a/d^3*f^2*c*e*arctan( 
exp(I*(d*x+c)))-3/a/d^2*f*e^2*ln(exp(2*I*(d*x+c))+1)-3/a/d^4*f^3*c^2*ln(ex 
p(2*I*(d*x+c))+1)+6/a/d^2*f*e^2*ln(exp(I*(d*x+c)))+6/a/d^4*f^3*c^2*ln(exp( 
I*(d*x+c)))+6/a/d^4*f^3*c^2*ln(1-I*exp(I*(d*x+c)))-6/a/d^2*f^3*ln(1-I*exp( 
I*(d*x+c)))*x^2+2*I/a/d*f^3*x^3-4*I/a/d^4*f^3*c^3+12*I/a/d^3*f^3*polylog(2 
,I*exp(I*(d*x+c)))*x+6*I/a/d*f^2*e*x^2+6*I/a/d^3*f^2*e*c^2+12*I/a/d^3*f^2* 
e*polylog(2,I*exp(I*(d*x+c)))+6*I/a/d^4*f^3*c^2*arctan(exp(I*(d*x+c)))+6*I 
/a/d^2*f*e^2*arctan(exp(I*(d*x+c)))+6/a/d^3*f^2*c*e*ln(exp(2*I*(d*x+c))+1) 
-12/a/d^2*f^2*e*ln(1-I*exp(I*(d*x+c)))*x-6*I/a/d^3*f^3*c^2*x-12/a/d^3*f^2* 
c*e*ln(exp(I*(d*x+c)))-12/a/d^3*f^2*e*ln(1-I*exp(I*(d*x+c)))*c-12*f^3*poly 
log(3,I*exp(I*(d*x+c)))/a/d^4+1/4/a*f^3*x^4+1/4/a/f*e^4
 

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. 1044 vs. \(2 (139) = 278\).

Time = 0.11 (sec) , antiderivative size = 1044, normalized size of antiderivative = 6.37 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/4*(d^4*f^3*x^4 + 4*d^3*e^3 + 4*(d^4*e*f^2 + d^3*f^3)*x^3 + 6*(d^4*e^2*f 
+ 2*d^3*e*f^2)*x^2 + 4*(d^4*e^3 + 3*d^3*e^2*f)*x + (d^4*f^3*x^4 + 4*d^3*e^ 
3 + 4*(d^4*e*f^2 + d^3*f^3)*x^3 + 6*(d^4*e^2*f + 2*d^3*e*f^2)*x^2 + 4*(d^4 
*e^3 + 3*d^3*e^2*f)*x)*cos(d*x + c) - 24*(-I*d*f^3*x - I*d*e*f^2 + (-I*d*f 
^3*x - I*d*e*f^2)*cos(d*x + c) + (-I*d*f^3*x - I*d*e*f^2)*sin(d*x + c))*di 
log(I*cos(d*x + c) - sin(d*x + c)) - 24*(I*d*f^3*x + I*d*e*f^2 + (I*d*f^3* 
x + I*d*e*f^2)*cos(d*x + c) + (I*d*f^3*x + I*d*e*f^2)*sin(d*x + c))*dilog( 
-I*cos(d*x + c) - sin(d*x + c)) - 12*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3 + 
(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3)*cos(d*x + c) + (d^2*e^2*f - 2*c*d*e*f^ 
2 + c^2*f^3)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x + c) + I) - 12*(d^ 
2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3 + (d^2*f^3*x^2 + 2*d^2*e 
*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c) + (d^2*f^3*x^2 + 2*d^2*e*f^2* 
x + 2*c*d*e*f^2 - c^2*f^3)*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) 
 + 1) - 12*(d^2*f^3*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3 + (d^2*f^3 
*x^2 + 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*cos(d*x + c) + (d^2*f^3*x^2 
+ 2*d^2*e*f^2*x + 2*c*d*e*f^2 - c^2*f^3)*sin(d*x + c))*log(-I*cos(d*x + c) 
 + sin(d*x + c) + 1) - 12*(d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3 + (d^2*e^2*f 
- 2*c*d*e*f^2 + c^2*f^3)*cos(d*x + c) + (d^2*e^2*f - 2*c*d*e*f^2 + c^2*f^3 
)*sin(d*x + c))*log(-cos(d*x + c) + I*sin(d*x + c) + I) - 24*(f^3*cos(d*x 
+ c) + f^3*sin(d*x + c) + f^3)*polylog(3, I*cos(d*x + c) - sin(d*x + c)...
 

Sympy [F]

\[ \int \frac {(e+f x)^3 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\frac {\int \frac {e^{3} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {f^{3} x^{3} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e f^{2} x^{2} \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx + \int \frac {3 e^{2} f x \sin {\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

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

Output:

(Integral(e**3*sin(c + d*x)/(sin(c + d*x) + 1), x) + Integral(f**3*x**3*si 
n(c + d*x)/(sin(c + d*x) + 1), x) + Integral(3*e*f**2*x**2*sin(c + d*x)/(s 
in(c + d*x) + 1), x) + Integral(3*e**2*f*x*sin(c + d*x)/(sin(c + d*x) + 1) 
, x))/a
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1303 vs. \(2 (139) = 278\).

Time = 0.25 (sec) , antiderivative size = 1303, normalized size of antiderivative = 7.95 \[ \int \frac {(e+f x)^3 \sin (c+d x)}{a+a \sin (c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(12*c^2*e*f^2*(1/(a*d^2 + a*d^2*sin(d*x + c)/(cos(d*x + c) + 1)) + arc 
tan(sin(d*x + c)/(cos(d*x + c) + 1))/(a*d^2)) - 12*c*e^2*f*(1/(a*d + a*d*s 
in(d*x + c)/(cos(d*x + c) + 1)) + arctan(sin(d*x + c)/(cos(d*x + c) + 1))/ 
(a*d)) - 6*((d*x + c)^2*cos(d*x + c)^2 + (d*x + c)^2*sin(d*x + c)^2 + 2*(d 
*x + c)^2*sin(d*x + c) + (d*x + c)^2 + 4*(d*x + c)*cos(d*x + c) - 2*(cos(d 
*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1)*log(cos(d*x + c)^2 + sin( 
d*x + c)^2 + 2*sin(d*x + c) + 1))*c*e*f^2/(a*d^2*cos(d*x + c)^2 + a*d^2*si 
n(d*x + c)^2 + 2*a*d^2*sin(d*x + c) + a*d^2) + 4*e^3*(arctan(sin(d*x + c)/ 
(cos(d*x + c) + 1))/a + 1/(a + a*sin(d*x + c)/(cos(d*x + c) + 1))) + 3*((d 
*x + c)^2*cos(d*x + c)^2 + (d*x + c)^2*sin(d*x + c)^2 + 2*(d*x + c)^2*sin( 
d*x + c) + (d*x + c)^2 + 4*(d*x + c)*cos(d*x + c) - 2*(cos(d*x + c)^2 + si 
n(d*x + c)^2 + 2*sin(d*x + c) + 1)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2 
*sin(d*x + c) + 1))*e^2*f/(a*d*cos(d*x + c)^2 + a*d*sin(d*x + c)^2 + 2*a*d 
*sin(d*x + c) + a*d) + 2*((d*x + c)^4*f^3 + 6*(d*x + c)^2*c^2*f^3 - 4*(d*x 
 + c)*c^3*f^3 + 8*I*c^3*f^3 + 4*(d*e*f^2 - c*f^3)*(d*x + c)^3 - 24*(c^2*f^ 
3*cos(d*x + c) + I*c^2*f^3*sin(d*x + c) + I*c^2*f^3)*arctan2(sin(d*x + c) 
+ 1, cos(d*x + c)) + 24*(I*(d*x + c)^2*f^3 + 2*(I*d*e*f^2 - I*c*f^3)*(d*x 
+ c) + ((d*x + c)^2*f^3 + 2*(d*e*f^2 - c*f^3)*(d*x + c))*cos(d*x + c) + (I 
*(d*x + c)^2*f^3 + 2*(I*d*e*f^2 - I*c*f^3)*(d*x + c))*sin(d*x + c))*arctan 
2(cos(d*x + c), sin(d*x + c) + 1) - (I*(d*x + c)^4*f^3 - 4*(I*c^3 + 6*c...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 24*int(x**2/(tan((c + d*x)/2)**2 + 2*tan((c + d*x)/2) + 1),x)*tan((c + 
 d*x)/2)*d**3*f**3 - 24*int(x**2/(tan((c + d*x)/2)**2 + 2*tan((c + d*x)/2) 
 + 1),x)*d**3*f**3 - 48*int(x/(tan((c + d*x)/2)**2 + 2*tan((c + d*x)/2) + 
1),x)*tan((c + d*x)/2)*d**3*e*f**2 + 48*int(x/(tan((c + d*x)/2)**2 + 2*tan 
((c + d*x)/2) + 1),x)*tan((c + d*x)/2)*d**2*f**3 - 48*int(x/(tan((c + d*x) 
/2)**2 + 2*tan((c + d*x)/2) + 1),x)*d**3*e*f**2 + 48*int(x/(tan((c + d*x)/ 
2)**2 + 2*tan((c + d*x)/2) + 1),x)*d**2*f**3 + 12*log(tan((c + d*x)/2)**2 
+ 1)*tan((c + d*x)/2)*d**2*e**2*f - 24*log(tan((c + d*x)/2)**2 + 1)*tan((c 
 + d*x)/2)*d*e*f**2 + 24*log(tan((c + d*x)/2)**2 + 1)*tan((c + d*x)/2)*f** 
3 + 12*log(tan((c + d*x)/2)**2 + 1)*d**2*e**2*f - 24*log(tan((c + d*x)/2)* 
*2 + 1)*d*e*f**2 + 24*log(tan((c + d*x)/2)**2 + 1)*f**3 - 24*log(tan((c + 
d*x)/2) + 1)*tan((c + d*x)/2)*d**2*e**2*f + 48*log(tan((c + d*x)/2) + 1)*t 
an((c + d*x)/2)*d*e*f**2 - 48*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)*f 
**3 - 24*log(tan((c + d*x)/2) + 1)*d**2*e**2*f + 48*log(tan((c + d*x)/2) + 
 1)*d*e*f**2 - 48*log(tan((c + d*x)/2) + 1)*f**3 + 4*tan((c + d*x)/2)*d**4 
*e**3*x + 6*tan((c + d*x)/2)*d**4*e**2*f*x**2 + 4*tan((c + d*x)/2)*d**4*e* 
f**2*x**3 + tan((c + d*x)/2)*d**4*f**3*x**4 - 8*tan((c + d*x)/2)*d**3*e**3 
 - 12*tan((c + d*x)/2)*d**3*e**2*f*x - 12*tan((c + d*x)/2)*d**3*e*f**2*x** 
2 - 4*tan((c + d*x)/2)*d**3*f**3*x**3 + 24*tan((c + d*x)/2)*d**2*e*f**2*x 
+ 12*tan((c + d*x)/2)*d**2*f**3*x**2 - 24*tan((c + d*x)/2)*d*f**3*x + 4...