3.3.46 \(\int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [246]

3.3.46.1 Optimal result

Integrand size = 26, antiderivative size = 1106 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {i a (e+f x)^2}{b \left (a^2-b^2\right ) d}+\frac {2 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 ) d^2}+\frac {i a^2 (e+f x)^2 \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}-\frac {i (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}+\frac {2 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 ) d^2}-\frac {i a^2 (e+f x)^2 \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}+\frac {i (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}-\frac {2 i 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 ) d^3}+\frac {2 a^2 f (e+f x) \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^2}-\frac {2 f (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^2}-\frac {2 i 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 ) d^3}-\frac {2 a^2 f (e+f x) \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^2}+\frac {2 f (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^2}+\frac {2 i a^2 f^2 \operatorname {PolyLog}\left (3,\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 {2 i f^2 \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^3}-\frac {2 i a^2 f^2 \operatorname {PolyLog}\left (3,\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 {2 i f^2 \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^3}-\frac {a (e+f x)^2 \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \]

output

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

3.3.46.2 Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(2279\) vs. \(2(1106)=2212\).

Time = 21.25 (sec) , antiderivative size = 2279, normalized size of antiderivative = 2.06 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Result too large to show} \]

input

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

output

((2*I)*E^(I*c)*(2*a*e*E^(I*c)*f*x + a*E^(I*c)*f^2*x^2 + (I*b^2*e^2*ArcTan[ 
(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*E^(I*c)) - (I 
*b^2*e^2*E^(I*c)*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]])/Sqrt[a 
^2 - b^2] + (2*a^2*e*f*ArcTan[(I*a + b*E^(I*(c + d*x)))/Sqrt[a^2 - b^2]])/ 
(Sqrt[a^2 - b^2]*d*E^(I*c)) - (a*e*E^(I*c)*f*ArcTan[(2*a*E^(I*(c + d*x)))/ 
(b*(-1 + E^((2*I)*(c + d*x))))])/d + ((2*I)*a^2*e*f*ArcTanh[(-a + I*b*E^(I 
*(c + d*x)))/Sqrt[a^2 - b^2]])/(Sqrt[a^2 - b^2]*d*E^(I*c)) - (I*a*e*f*Log[ 
b - (2*I)*a*E^(I*(c + d*x)) - b*E^((2*I)*(c + d*x))])/(d*E^(I*c)) + ((I/2) 
*a*e*E^(I*c)*f*Log[4*a^2*E^((2*I)*(c + d*x)) + b^2*(-1 + E^((2*I)*(c + d*x 
)))^2])/d + (I*b^2*e*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*b^2*e*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*a*f^2*x*Log[1 + (b* 
E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sqrt[(-a^2 + b^2)*E^((2*I)*c)])])/(d*E^( 
I*c)) + (I*a*E^(I*c)*f^2*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) - Sq 
rt[(-a^2 + b^2)*E^((2*I)*c)])])/d + ((I/2)*b^2*f^2*x^2*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)*b^2*E^((2*I)*c)*f^2*x^2*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*b^2*e*f*x*Log[1 + (b*E^(I*(2*c + d*x)))/(I*a*E^(I*c) +...
 

3.3.46.3 Rubi [A] (verified)

Time = 2.82 (sec) , antiderivative size = 1106, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {7293, 2009}

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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

input

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

output

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

3.3.46.3.1 Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 7293

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v] 
]
 

3.3.46.4 Maple [F]

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

input

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

output

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

3.3.46.5 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. 3122 vs. \(2 (958) = 1916\).

Time = 0.53 (sec) , antiderivative size = 3122, normalized size of antiderivative = 2.82 \[ \int \frac {(e+f x)^2 \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \]

input

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

output

-1/2*(2*(b^4*f^2*sin(d*x + c) + a*b^3*f^2)*sqrt(-(a^2 - b^2)/b^2)*polylog( 
3, -(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) - 2*(b^4*f^2*sin(d*x + c) + a*b^3*f^2)*sqrt( 
-(a^2 - b^2)/b^2)*polylog(3, -(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) + 2*(b^4*f^2*sin(d 
*x + c) + a*b^3*f^2)*sqrt(-(a^2 - b^2)/b^2)*polylog(3, -(-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) - 2*(b^4*f^2*sin(d*x + c) + a*b^3*f^2)*sqrt(-(a^2 - b^2)/b^2)*pol 
ylog(3, -(-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) + 2*((a^3*b - a*b^3)*d^2*f^2*x^2 + 2*( 
a^3*b - a*b^3)*d^2*e*f*x + (a^3*b - a*b^3)*d^2*e^2)*cos(d*x + c) + 2*(I*(a 
^3*b - a*b^3)*f^2*sin(d*x + c) + I*(a^4 - a^2*b^2)*f^2 + (I*a*b^3*d*f^2*x 
+ I*a*b^3*d*e*f + (I*b^4*d*f^2*x + I*b^4*d*e*f)*sin(d*x + c))*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) + 2*(I*(a^3*b - a*b^3) 
*f^2*sin(d*x + c) + I*(a^4 - a^2*b^2)*f^2 + (-I*a*b^3*d*f^2*x - I*a*b^3*d* 
e*f + (-I*b^4*d*f^2*x - I*b^4*d*e*f)*sin(d*x + c))*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) + 2*(-I*(a^3*b - a*b^3)*f^2*sin(d 
*x + c) - I*(a^4 - a^2*b^2)*f^2 + (-I*a*b^3*d*f^2*x - I*a*b^3*d*e*f + (...
 

3.3.46.6 Sympy [F(-1)]

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

input

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

output

Timed out
 

3.3.46.7 Maxima [F(-2)]

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

input

integrate((f*x+e)^2*sin(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
 

3.3.46.8 Giac [F]

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

input

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

output

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

3.3.46.9 Mupad [F(-1)]

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

input

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

output

\text{Hanged}