\(\int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx\) [186]

Optimal result

Integrand size = 28, antiderivative size = 188 \[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {i (e+f x)^2}{a d}-\frac {(e+f x)^3}{3 a f}+\frac {2 f^2 \cos (c+d x)}{a d^3}-\frac {(e+f x)^2 \cos (c+d x)}{a d}-\frac {(e+f x)^2 \cot \left (\frac {c}{2}+\frac {\pi }{4}+\frac {d x}{2}\right )}{a d}+\frac {4 f (e+f x) \log \left (1-i e^{i (c+d x)}\right )}{a d^2}-\frac {4 i f^2 \operatorname {PolyLog}\left (2,i e^{i (c+d x)}\right )}{a d^3}+\frac {2 f (e+f x) \sin (c+d x)}{a d^2} \] Output:

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

Mathematica [A] (verified)

Time = 3.83 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.57 \[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {x \left (3 e^2+3 e f x+f^2 x^2\right )+\frac {3 \cos (d x) \left (\left (-2 f^2+d^2 (e+f x)^2\right ) \cos (c)-2 d f (e+f x) \sin (c)\right )}{d^3}+\frac {12 f (\cos (c)+i \sin (c)) \left (\frac {(e+f x)^2 (\cos (c)-i \sin (c))}{2 f}-\frac {(e+f x) \log (1+i \cos (c+d x)+\sin (c+d x)) (1+i \cos (c)+\sin (c))}{d}+\frac {f \operatorname {PolyLog}(2,-i \cos (c+d x)-\sin (c+d x)) (\cos (c)-i (1+\sin (c)))}{d^2}\right )}{d (\cos (c)+i (1+\sin (c)))}-\frac {3 \left (2 d f (e+f x) \cos (c)+\left (-2 f^2+d^2 (e+f x)^2\right ) \sin (c)\right ) \sin (d x)}{d^3}-\frac {6 (e+f x)^2 \sin \left (\frac {d x}{2}\right )}{d \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}}{3 a} \] Input:

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

Output:

-1/3*(x*(3*e^2 + 3*e*f*x + f^2*x^2) + (3*Cos[d*x]*((-2*f^2 + d^2*(e + f*x) 
^2)*Cos[c] - 2*d*f*(e + f*x)*Sin[c]))/d^3 + (12*f*(Cos[c] + I*Sin[c])*(((e 
 + f*x)^2*(Cos[c] - I*Sin[c]))/(2*f) - ((e + f*x)*Log[1 + I*Cos[c + d*x] + 
 Sin[c + d*x]]*(1 + I*Cos[c] + Sin[c]))/d + (f*PolyLog[2, (-I)*Cos[c + d*x 
] - Sin[c + d*x]]*(Cos[c] - I*(1 + Sin[c])))/d^2))/(d*(Cos[c] + I*(1 + Sin 
[c]))) - (3*(2*d*f*(e + f*x)*Cos[c] + (-2*f^2 + d^2*(e + f*x)^2)*Sin[c])*S 
in[d*x])/d^3 - (6*(e + f*x)^2*Sin[(d*x)/2])/(d*(Cos[c/2] + Sin[c/2])*(Cos[ 
(c + d*x)/2] + Sin[(c + d*x)/2])))/a
 

Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.08, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {5026, 3042, 3777, 3042, 3777, 25, 3042, 3118, 5026, 17, 3042, 3799, 3042, 4672, 3042, 25, 4202, 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.

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

\(\Big \downarrow \) 5026

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3777

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3118

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

\(\Big \downarrow \) 5026

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

\(\Big \downarrow \) 17

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3799

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4672

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 4202

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

\(\Big \downarrow \) 2620

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

\(\Big \downarrow \) 2715

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

\(\Big \downarrow \) 2838

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

Input:

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

Output:

-1/3*(e + f*x)^3/(a*f) + ((-2*(e + f*x)^2*Cot[c/2 + Pi/4 + (d*x)/2])/d - ( 
4*f*(((I/2)*(e + f*x)^2)/f - (2*I)*(((-I)*(e + f*x)*Log[1 + E^((I/2)*(2*c 
+ 3*Pi + 2*d*x))])/d - (f*PolyLog[2, -E^((I/2)*(2*c + 3*Pi + 2*d*x))])/d^2 
)))/d)/(2*a) + (-(((e + f*x)^2*Cos[c + d*x])/d) + (2*f*((f*Cos[c + d*x])/d 
^2 + ((e + f*x)*Sin[c + d*x])/d))/d)/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 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 3118

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Cos[c + d*x]/d, x] /; FreeQ 
[{c, d}, x]
 

rule 3777

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

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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 418 vs. \(2 (172 ) = 344\).

Time = 2.41 (sec) , antiderivative size = 419, normalized size of antiderivative = 2.23

method	result	size
risch	\(-\frac {f^{2} x^{3}}{3 a}-\frac {f e \,x^{2}}{a}-\frac {e^{2} x}{a}-\frac {e^{3}}{3 a f}-\frac {\left (d^{2} x^{2} f^{2}+2 e f x \,d^{2}+2 i d \,f^{2} x +d^{2} e^{2}+2 i d e f -2 f^{2}\right ) {\mathrm e}^{i \left (d x +c \right )}}{2 a \,d^{3}}-\frac {\left (d^{2} x^{2} f^{2}+2 e f x \,d^{2}-2 i d \,f^{2} x +d^{2} e^{2}-2 i d e f -2 f^{2}\right ) {\mathrm e}^{-i \left (d x +c \right )}}{2 a \,d^{3}}-\frac {2 \left (x^{2} f^{2}+2 e f x +e^{2}\right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}+\frac {4 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) e f}{d^{2} a}-\frac {4 f e \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{2}}-\frac {2 i f^{2} x^{2}}{d a}-\frac {4 i f^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}-\frac {2 i f^{2} c^{2}}{d^{3} a}+\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) x}{a \,d^{2}}+\frac {4 f^{2} \ln \left (1-i {\mathrm e}^{i \left (d x +c \right )}\right ) c}{a \,d^{3}}-\frac {4 i f^{2} c x}{a \,d^{2}}-\frac {4 c \,f^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d^{3} a}+\frac {4 f^{2} c \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{a \,d^{3}}\)	\(419\)

Input:

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

Output:

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

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. 716 vs. \(2 (167) = 334\).

Time = 0.11 (sec) , antiderivative size = 716, normalized size of antiderivative = 3.81 \[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/3*(d^3*f^2*x^3 + 3*d^2*e^2 - 6*d*e*f + 3*(d^3*e*f + d^2*f^2)*x^2 + 3*(d 
^2*f^2*x^2 + d^2*e^2 + 2*d*e*f - 2*f^2 + 2*(d^2*e*f + d*f^2)*x)*cos(d*x + 
c)^2 + 3*(d^3*e^2 + 2*d^2*e*f - 2*d*f^2)*x + (d^3*f^2*x^3 + 6*d^2*e^2 + 3* 
(d^3*e*f + 2*d^2*f^2)*x^2 - 6*f^2 + 3*(d^3*e^2 + 4*d^2*e*f)*x)*cos(d*x + c 
) + 6*(I*f^2*cos(d*x + c) + I*f^2*sin(d*x + c) + I*f^2)*dilog(I*cos(d*x + 
c) - sin(d*x + c)) + 6*(-I*f^2*cos(d*x + c) - I*f^2*sin(d*x + c) - I*f^2)* 
dilog(-I*cos(d*x + c) - sin(d*x + c)) - 6*(d*e*f - c*f^2 + (d*e*f - c*f^2) 
*cos(d*x + c) + (d*e*f - c*f^2)*sin(d*x + c))*log(cos(d*x + c) + I*sin(d*x 
 + c) + I) - 6*(d*f^2*x + c*f^2 + (d*f^2*x + c*f^2)*cos(d*x + c) + (d*f^2* 
x + c*f^2)*sin(d*x + c))*log(I*cos(d*x + c) + sin(d*x + c) + 1) - 6*(d*f^2 
*x + c*f^2 + (d*f^2*x + c*f^2)*cos(d*x + c) + (d*f^2*x + c*f^2)*sin(d*x + 
c))*log(-I*cos(d*x + c) + sin(d*x + c) + 1) - 6*(d*e*f - c*f^2 + (d*e*f - 
c*f^2)*cos(d*x + c) + (d*e*f - c*f^2)*sin(d*x + c))*log(-cos(d*x + c) + I* 
sin(d*x + c) + I) + (d^3*f^2*x^3 - 3*d^2*e^2 - 6*d*e*f + 3*(d^3*e*f - d^2* 
f^2)*x^2 + 3*(d^3*e^2 - 2*d^2*e*f - 2*d*f^2)*x + 3*(d^2*f^2*x^2 + d^2*e^2 
- 2*d*e*f - 2*f^2 + 2*(d^2*e*f - d*f^2)*x)*cos(d*x + c))*sin(d*x + c))/(a* 
d^3*cos(d*x + c) + a*d^3*sin(d*x + c) + a*d^3)
 

Sympy [F]

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

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

Output:

(Integral(e**2*sin(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(f**2*x**2 
*sin(c + d*x)**2/(sin(c + d*x) + 1), x) + Integral(2*e*f*x*sin(c + d*x)**2 
/(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. 602 vs. \(2 (167) = 334\).

Time = 0.31 (sec) , antiderivative size = 602, normalized size of antiderivative = 3.20 \[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=-\frac {2 \, d^{3} f^{2} x^{3} - 15 i \, d^{2} e^{2} - 6 \, d e f + 3 \, {\left (2 \, d^{3} e f - i \, d^{2} f^{2}\right )} x^{2} + 6 i \, f^{2} + 6 \, {\left (d^{3} e^{2} - i \, d^{2} e f - d f^{2}\right )} x - 24 \, {\left (d e f \cos \left (d x + c\right ) + i \, d e f \sin \left (d x + c\right ) + i \, d e f\right )} \arctan \left (\sin \left (d x + c\right ) + 1, \cos \left (d x + c\right )\right ) + 24 \, {\left (d f^{2} x \cos \left (d x + c\right ) + i \, d f^{2} x \sin \left (d x + c\right ) + i \, d f^{2} x\right )} \arctan \left (\cos \left (d x + c\right ), \sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (-i \, d^{2} f^{2} x^{2} - i \, d^{2} e^{2} + 2 \, d e f + 2 i \, f^{2} + 2 \, {\left (-i \, d^{2} e f + d f^{2}\right )} x\right )} \cos \left (2 \, d x + 2 \, c\right ) - {\left (2 i \, d^{3} f^{2} x^{3} - 3 \, d^{2} e^{2} - 6 i \, d e f - 3 \, {\left (-2 i \, d^{3} e f + 5 \, d^{2} f^{2}\right )} x^{2} + 6 \, f^{2} - 6 \, {\left (-i \, d^{3} e^{2} + 5 \, d^{2} e f + i \, d f^{2}\right )} x\right )} \cos \left (d x + c\right ) + 24 \, {\left (f^{2} \cos \left (d x + c\right ) + i \, f^{2} \sin \left (d x + c\right ) + i \, f^{2}\right )} {\rm Li}_2\left (i \, e^{\left (i \, d x + i \, c\right )}\right ) - 12 \, {\left (d f^{2} x + d e f - {\left (i \, d f^{2} x + i \, d e f\right )} \cos \left (d x + c\right ) + {\left (d f^{2} x + d e f\right )} \sin \left (d x + c\right )\right )} \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (d^{2} f^{2} x^{2} + d^{2} e^{2} + 2 i \, d e f - 2 \, f^{2} + 2 \, {\left (d^{2} e f + i \, d f^{2}\right )} x\right )} \sin \left (2 \, d x + 2 \, c\right ) + {\left (2 \, d^{3} f^{2} x^{3} + 3 i \, d^{2} e^{2} - 6 \, d e f + 3 \, {\left (2 \, d^{3} e f + 5 i \, d^{2} f^{2}\right )} x^{2} - 6 i \, f^{2} + 6 \, {\left (d^{3} e^{2} + 5 i \, d^{2} e f - d f^{2}\right )} x\right )} \sin \left (d x + c\right )}{-6 i \, a d^{3} \cos \left (d x + c\right ) + 6 \, a d^{3} \sin \left (d x + c\right ) + 6 \, a d^{3}} \] Input:

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

Output:

-(2*d^3*f^2*x^3 - 15*I*d^2*e^2 - 6*d*e*f + 3*(2*d^3*e*f - I*d^2*f^2)*x^2 + 
 6*I*f^2 + 6*(d^3*e^2 - I*d^2*e*f - d*f^2)*x - 24*(d*e*f*cos(d*x + c) + I* 
d*e*f*sin(d*x + c) + I*d*e*f)*arctan2(sin(d*x + c) + 1, cos(d*x + c)) + 24 
*(d*f^2*x*cos(d*x + c) + I*d*f^2*x*sin(d*x + c) + I*d*f^2*x)*arctan2(cos(d 
*x + c), sin(d*x + c) + 1) + 3*(-I*d^2*f^2*x^2 - I*d^2*e^2 + 2*d*e*f + 2*I 
*f^2 + 2*(-I*d^2*e*f + d*f^2)*x)*cos(2*d*x + 2*c) - (2*I*d^3*f^2*x^3 - 3*d 
^2*e^2 - 6*I*d*e*f - 3*(-2*I*d^3*e*f + 5*d^2*f^2)*x^2 + 6*f^2 - 6*(-I*d^3* 
e^2 + 5*d^2*e*f + I*d*f^2)*x)*cos(d*x + c) + 24*(f^2*cos(d*x + c) + I*f^2* 
sin(d*x + c) + I*f^2)*dilog(I*e^(I*d*x + I*c)) - 12*(d*f^2*x + d*e*f - (I* 
d*f^2*x + I*d*e*f)*cos(d*x + c) + (d*f^2*x + d*e*f)*sin(d*x + c))*log(cos( 
d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) + 3*(d^2*f^2*x^2 + d^2*e 
^2 + 2*I*d*e*f - 2*f^2 + 2*(d^2*e*f + I*d*f^2)*x)*sin(2*d*x + 2*c) + (2*d^ 
3*f^2*x^3 + 3*I*d^2*e^2 - 6*d*e*f + 3*(2*d^3*e*f + 5*I*d^2*f^2)*x^2 - 6*I* 
f^2 + 6*(d^3*e^2 + 5*I*d^2*e*f - d*f^2)*x)*sin(d*x + c))/(-6*I*a*d^3*cos(d 
*x + c) + 6*a*d^3*sin(d*x + c) + 6*a*d^3)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(e+f x)^2 \sin ^2(c+d x)}{a+a \sin (c+d x)} \, dx=\frac {-12 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) x}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}d x \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} f^{2}-12 \left (\int \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) x}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}d x \right ) d^{2} f^{2}+6 \cos \left (d x +c \right ) f^{2}-6 \cos \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} e f x -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{3} f^{2} x^{3}-d^{3} f^{2} x^{3}-3 \cos \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} e^{2}-3 \cos \left (d x +c \right ) d^{2} f^{2} x^{2}-6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) d e f +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) d e f +6 \sin \left (d x +c \right ) d e f +6 \sin \left (d x +c \right ) d \,f^{2} x -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{3} e^{2} x +6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} f^{2} x^{2}-3 d^{3} e f \,x^{2}-6 d^{2} e f x +6 \cos \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) f^{2}-3 \cos \left (d x +c \right ) d^{2} e^{2}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} e^{2}-3 d^{3} e^{2} x -3 \cos \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} f^{2} x^{2}-6 \cos \left (d x +c \right ) d^{2} e f x -6 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d e f +12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d e f +6 \sin \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d e f +6 \sin \left (d x +c \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d \,f^{2} x -3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{3} e f \,x^{2}+6 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) d^{2} e f x}{3 a \,d^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )} \] Input:

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

Output:

( - 3*cos(c + d*x)*tan((c + d*x)/2)*d**2*e**2 - 6*cos(c + d*x)*tan((c + d* 
x)/2)*d**2*e*f*x - 3*cos(c + d*x)*tan((c + d*x)/2)*d**2*f**2*x**2 + 6*cos( 
c + d*x)*tan((c + d*x)/2)*f**2 - 3*cos(c + d*x)*d**2*e**2 - 6*cos(c + d*x) 
*d**2*e*f*x - 3*cos(c + d*x)*d**2*f**2*x**2 + 6*cos(c + d*x)*f**2 - 12*int 
((tan((c + d*x)/2)*x)/(tan((c + d*x)/2) + 1),x)*tan((c + d*x)/2)*d**2*f**2 
 - 12*int((tan((c + d*x)/2)*x)/(tan((c + d*x)/2) + 1),x)*d**2*f**2 - 6*log 
(tan((c + d*x)/2)**2 + 1)*tan((c + d*x)/2)*d*e*f - 6*log(tan((c + d*x)/2)* 
*2 + 1)*d*e*f + 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)*d*e*f + 12*l 
og(tan((c + d*x)/2) + 1)*d*e*f + 6*sin(c + d*x)*tan((c + d*x)/2)*d*e*f + 6 
*sin(c + d*x)*tan((c + d*x)/2)*d*f**2*x + 6*sin(c + d*x)*d*e*f + 6*sin(c + 
 d*x)*d*f**2*x - 3*tan((c + d*x)/2)*d**3*e**2*x - 3*tan((c + d*x)/2)*d**3* 
e*f*x**2 - tan((c + d*x)/2)*d**3*f**2*x**3 + 6*tan((c + d*x)/2)*d**2*e**2 
+ 6*tan((c + d*x)/2)*d**2*e*f*x + 6*tan((c + d*x)/2)*d**2*f**2*x**2 - 3*d* 
*3*e**2*x - 3*d**3*e*f*x**2 - d**3*f**2*x**3 - 6*d**2*e*f*x)/(3*a*d**3*(ta 
n((c + d*x)/2) + 1))