\(\int \frac {(c+d x)^3}{a-a \sin (e+f x)} \, dx\) [117]

Optimal result

Integrand size = 21, antiderivative size = 147 \[ \int \frac {(c+d x)^3}{a-a \sin (e+f x)} \, dx=-\frac {i (c+d x)^3}{a f}+\frac {6 d (c+d x)^2 \log \left (1+i e^{i (e+f x)}\right )}{a f^2}-\frac {12 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{a f^3}+\frac {12 d^3 \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{a f^4}+\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f} \] Output:

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

Mathematica [A] (verified)

Time = 1.28 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.84 \[ \int \frac {(c+d x)^3}{a-a \sin (e+f x)} \, dx=\frac {-12 i d^2 f (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )+12 d^3 \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )+f^2 (c+d x)^2 \left (-i f (c+d x)+6 d \log \left (1+i e^{i (e+f x)}\right )+f (c+d x) \tan \left (\frac {1}{4} (2 e+\pi +2 f x)\right )\right )}{a f^4} \] Input:

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

Output:

((-12*I)*d^2*f*(c + d*x)*PolyLog[2, (-I)*E^(I*(e + f*x))] + 12*d^3*PolyLog 
[3, (-I)*E^(I*(e + f*x))] + f^2*(c + d*x)^2*((-I)*f*(c + d*x) + 6*d*Log[1 
+ I*E^(I*(e + f*x))] + f*(c + d*x)*Tan[(2*e + Pi + 2*f*x)/4]))/(a*f^4)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.04, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3799, 3042, 4672, 25, 3042, 4200, 26, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3799

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4672

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4200

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

\(\Big \downarrow \) 26

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

rule 26

Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a])   I 
nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
 

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 4200

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (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*k*Pi)*(E^(2*I*(e + f*x))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))), x] 
, x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[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 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. 483 vs. \(2 (130 ) = 260\).

Time = 0.94 (sec) , antiderivative size = 484, normalized size of antiderivative = 3.29

method	result	size
risch	\(\frac {2 d^{3} x^{3}+6 c \,d^{2} x^{2}+6 c^{2} d x +2 c^{3}}{f a \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}-\frac {12 i c \,d^{2} e x}{a \,f^{2}}+\frac {6 \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c^{2} d}{a \,f^{2}}-\frac {12 i d^{3} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{3}}+\frac {12 e c \,d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {12 i c \,d^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {6 i c \,d^{2} x^{2}}{a f}-\frac {2 i d^{3} x^{3}}{a f}+\frac {12 c \,d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{2}}-\frac {12 e c \,d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a \,f^{3}}+\frac {6 e^{2} d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a \,f^{4}}-\frac {6 e^{2} d^{3} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}+\frac {12 c \,d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{a \,f^{3}}-\frac {6 e^{2} d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}+\frac {12 d^{3} \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}-\frac {6 i c \,d^{2} e^{2}}{a \,f^{3}}+\frac {6 d^{3} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{a \,f^{2}}-\frac {6 \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right ) c^{2} d}{a \,f^{2}}+\frac {6 i d^{3} e^{2} x}{a \,f^{3}}+\frac {4 i d^{3} e^{3}}{a \,f^{4}}\)	\(484\)

Input:

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

Output:

2*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/f/a/(exp(I*(f*x+e))-I)-12*I/a/f^2*c* 
d^2*e*x+6/a/f^2*ln(exp(I*(f*x+e))-I)*c^2*d-12*I/a/f^3*d^3*polylog(2,-I*exp 
(I*(f*x+e)))*x+12/a/f^3*e*c*d^2*ln(exp(I*(f*x+e)))-12*I/a/f^3*c*d^2*polylo 
g(2,-I*exp(I*(f*x+e)))-6*I/a/f*c*d^2*x^2-2*I/a/f*d^3*x^3+12/a/f^2*c*d^2*ln 
(1+I*exp(I*(f*x+e)))*x-12/a/f^3*e*c*d^2*ln(exp(I*(f*x+e))-I)+6/a/f^4*e^2*d 
^3*ln(exp(I*(f*x+e))-I)-6/a/f^4*e^2*d^3*ln(1+I*exp(I*(f*x+e)))+12/a/f^3*c* 
d^2*ln(1+I*exp(I*(f*x+e)))*e-6/a/f^4*e^2*d^3*ln(exp(I*(f*x+e)))+12*d^3*pol 
ylog(3,-I*exp(I*(f*x+e)))/a/f^4-6*I/a/f^3*c*d^2*e^2+6/a/f^2*d^3*ln(1+I*exp 
(I*(f*x+e)))*x^2-6/a/f^2*ln(exp(I*(f*x+e)))*c^2*d+6*I/a/f^3*d^3*e^2*x+4*I/ 
a/f^4*d^3*e^3
 

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. 916 vs. \(2 (124) = 248\).

Time = 0.10 (sec) , antiderivative size = 916, normalized size of antiderivative = 6.23 \[ \int \frac {(c+d x)^3}{a-a \sin (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d*f^3*x + c^3*f^3 + (d^3*f^3*x^3 + 
3*c*d^2*f^3*x^2 + 3*c^2*d*f^3*x + c^3*f^3)*cos(f*x + e) + 6*(I*d^3*f*x + I 
*c*d^2*f + (I*d^3*f*x + I*c*d^2*f)*cos(f*x + e) + (-I*d^3*f*x - I*c*d^2*f) 
*sin(f*x + e))*dilog(I*cos(f*x + e) + sin(f*x + e)) + 6*(-I*d^3*f*x - I*c* 
d^2*f + (-I*d^3*f*x - I*c*d^2*f)*cos(f*x + e) + (I*d^3*f*x + I*c*d^2*f)*si 
n(f*x + e))*dilog(-I*cos(f*x + e) + sin(f*x + e)) + 3*(d^3*e^2 - 2*c*d^2*e 
*f + c^2*d*f^2 + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cos(f*x + e) - (d^3*e 
^2 - 2*c*d^2*e*f + c^2*d*f^2)*sin(f*x + e))*log(cos(f*x + e) - I*sin(f*x + 
 e) + I) + 3*(d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f + (d^3*f 
^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*cos(f*x + e) - (d^3*f^2*x^ 
2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*sin(f*x + e))*log(I*cos(f*x + e 
) - sin(f*x + e) + 1) + 3*(d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2 
*e*f + (d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*cos(f*x + e) 
- (d^3*f^2*x^2 + 2*c*d^2*f^2*x - d^3*e^2 + 2*c*d^2*e*f)*sin(f*x + e))*log( 
-I*cos(f*x + e) - sin(f*x + e) + 1) + 3*(d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2 
 + (d^3*e^2 - 2*c*d^2*e*f + c^2*d*f^2)*cos(f*x + e) - (d^3*e^2 - 2*c*d^2*e 
*f + c^2*d*f^2)*sin(f*x + e))*log(-cos(f*x + e) - I*sin(f*x + e) + I) + 6* 
(d^3*cos(f*x + e) - d^3*sin(f*x + e) + d^3)*polylog(3, I*cos(f*x + e) + si 
n(f*x + e)) + 6*(d^3*cos(f*x + e) - d^3*sin(f*x + e) + d^3)*polylog(3, -I* 
cos(f*x + e) + sin(f*x + e)) + (d^3*f^3*x^3 + 3*c*d^2*f^3*x^2 + 3*c^2*d...
 

Sympy [F]

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

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

Output:

-(Integral(c**3/(sin(e + f*x) - 1), x) + Integral(d**3*x**3/(sin(e + f*x) 
- 1), x) + Integral(3*c*d**2*x**2/(sin(e + f*x) - 1), x) + Integral(3*c**2 
*d*x/(sin(e + f*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. 984 vs. \(2 (124) = 248\).

Time = 0.14 (sec) , antiderivative size = 984, normalized size of antiderivative = 6.69 \[ \int \frac {(c+d x)^3}{a-a \sin (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

-(6*(2*(f*x + e)*cos(f*x + e) + (cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f 
*x + e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1))*c* 
d^2*e/(a*f^2*cos(f*x + e)^2 + a*f^2*sin(f*x + e)^2 - 2*a*f^2*sin(f*x + e) 
+ a*f^2) - 6*c*d^2*e^2/(a*f^2 - a*f^2*sin(f*x + e)/(cos(f*x + e) + 1)) - 3 
*(2*(f*x + e)*cos(f*x + e) + (cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x 
+ e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1))*c^2*d 
/(a*f*cos(f*x + e)^2 + a*f*sin(f*x + e)^2 - 2*a*f*sin(f*x + e) + a*f) + 6* 
c^2*d*e/(a*f - a*f*sin(f*x + e)/(cos(f*x + e) + 1)) - 2*c^3/(a - a*sin(f*x 
 + e)/(cos(f*x + e) + 1)) - (2*I*d^3*e^3 + 6*(d^3*e^2*cos(f*x + e) + I*d^3 
*e^2*sin(f*x + e) - I*d^3*e^2)*arctan2(sin(f*x + e) - 1, cos(f*x + e)) - 6 
*(I*(f*x + e)^2*d^3 + 2*(-I*d^3*e + I*c*d^2*f)*(f*x + e) - ((f*x + e)^2*d^ 
3 - 2*(d^3*e - c*d^2*f)*(f*x + e))*cos(f*x + e) + (-I*(f*x + e)^2*d^3 + 2* 
(I*d^3*e - I*c*d^2*f)*(f*x + e))*sin(f*x + e))*arctan2(cos(f*x + e), -sin( 
f*x + e) + 1) - 2*((f*x + e)^3*d^3 + 3*(f*x + e)*d^3*e^2 - 3*(d^3*e - c*d^ 
2*f)*(f*x + e)^2)*cos(f*x + e) - 12*(-I*(f*x + e)*d^3 + I*d^3*e - I*c*d^2* 
f + ((f*x + e)*d^3 - d^3*e + c*d^2*f)*cos(f*x + e) + (I*(f*x + e)*d^3 - I* 
d^3*e + I*c*d^2*f)*sin(f*x + e))*dilog(-I*e^(I*f*x + I*e)) - 3*((f*x + e)^ 
2*d^3 + d^3*e^2 - 2*(d^3*e - c*d^2*f)*(f*x + e) + (I*(f*x + e)^2*d^3 + I*d 
^3*e^2 + 2*(-I*d^3*e + I*c*d^2*f)*(f*x + e))*cos(f*x + e) - ((f*x + e)^2*d 
^3 + d^3*e^2 - 2*(d^3*e - c*d^2*f)*(f*x + e))*sin(f*x + e))*log(cos(f*x...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

( - 6*int(x**2/(tan((e + f*x)/2)**2 - 2*tan((e + f*x)/2) + 1),x)*tan((e + 
f*x)/2)*d**3*f**3 + 6*int(x**2/(tan((e + f*x)/2)**2 - 2*tan((e + f*x)/2) + 
 1),x)*d**3*f**3 - 12*int(x/(tan((e + f*x)/2)**2 - 2*tan((e + f*x)/2) + 1) 
,x)*tan((e + f*x)/2)*c*d**2*f**3 - 12*int(x/(tan((e + f*x)/2)**2 - 2*tan(( 
e + f*x)/2) + 1),x)*tan((e + f*x)/2)*d**3*f**2 + 12*int(x/(tan((e + f*x)/2 
)**2 - 2*tan((e + f*x)/2) + 1),x)*c*d**2*f**3 + 12*int(x/(tan((e + f*x)/2) 
**2 - 2*tan((e + f*x)/2) + 1),x)*d**3*f**2 - 3*log(tan((e + f*x)/2)**2 + 1 
)*tan((e + f*x)/2)*c**2*d*f**2 - 6*log(tan((e + f*x)/2)**2 + 1)*tan((e + f 
*x)/2)*c*d**2*f - 6*log(tan((e + f*x)/2)**2 + 1)*tan((e + f*x)/2)*d**3 + 3 
*log(tan((e + f*x)/2)**2 + 1)*c**2*d*f**2 + 6*log(tan((e + f*x)/2)**2 + 1) 
*c*d**2*f + 6*log(tan((e + f*x)/2)**2 + 1)*d**3 + 6*log(tan((e + f*x)/2) - 
 1)*tan((e + f*x)/2)*c**2*d*f**2 + 12*log(tan((e + f*x)/2) - 1)*tan((e + f 
*x)/2)*c*d**2*f + 12*log(tan((e + f*x)/2) - 1)*tan((e + f*x)/2)*d**3 - 6*l 
og(tan((e + f*x)/2) - 1)*c**2*d*f**2 - 12*log(tan((e + f*x)/2) - 1)*c*d**2 
*f - 12*log(tan((e + f*x)/2) - 1)*d**3 - 2*tan((e + f*x)/2)*c**3*f**3 - 3* 
tan((e + f*x)/2)*c**2*d*f**3*x - 3*tan((e + f*x)/2)*c*d**2*f**3*x**2 - 6*t 
an((e + f*x)/2)*c*d**2*f**2*x - tan((e + f*x)/2)*d**3*f**3*x**3 - 3*tan((e 
 + f*x)/2)*d**3*f**2*x**2 - 6*tan((e + f*x)/2)*d**3*f*x - 3*c**2*d*f**3*x 
- 3*c*d**2*f**3*x**2 - 6*c*d**2*f**2*x - d**3*f**3*x**3 - 3*d**3*f**2*x**2 
 - 6*d**3*f*x)/(a*f**4*(tan((e + f*x)/2) - 1))