\(\int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx\) [245]

Optimal result

Integrand size = 24, antiderivative size = 574 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}-\frac {i (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}-\frac {i a^2 (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d}+\frac {i (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d}+\frac {a f \log (a+b \sin (c+d x))}{b \left (a^2-b^2\right ) d^2}+\frac {a^2 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}-\frac {f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {a^2 f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \left (a^2-b^2\right )^{3/2} d^2}+\frac {f \operatorname {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b \sqrt {a^2-b^2} d^2}-\frac {a (e+f x) \cos (c+d x)}{\left (a^2-b^2\right ) d (a+b \sin (c+d x))} \] Output:

I*a^2*(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-I*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(1 
/2)/d-I*a^2*(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+I*(f*x+e)*ln(1-I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b 
^2)^(1/2)/d+a*f*ln(a+b*sin(d*x+c))/b/(a^2-b^2)/d^2+a^2*f*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-f*polylog(2,I*b*exp 
(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(1/2)/d^2-a^2*f*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+f*polylog(2,I*b 
*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/(a^2-b^2)^(1/2)/d^2-a*(f*x+e)*cos(d 
*x+c)/(a^2-b^2)/d/(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. \(2286\) vs. \(2(574)=1148\).

Time = 17.04 (sec) , antiderivative size = 2286, normalized size of antiderivative = 3.98 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Result too large to show} \] Input:

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

Output:

(-(a*d*e*Cos[c + d*x]) + a*c*f*Cos[c + d*x] - a*f*(c + d*x)*Cos[c + d*x])/ 
((a - b)*(a + b)*d^2*(a + b*Sin[c + d*x])) + (((-2*b*d*e*ArcTan[(b + a*Tan 
[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + (2*b*c*f*ArcTan[(b + a* 
Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] - (a*f*Log[Sec[(c + d* 
x)/2]^2])/b + (a*f*Log[Sec[(c + d*x)/2]^2*(a + b*Sin[c + d*x])])/b + (I*b* 
f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/ 
2])/(I*a + b - Sqrt[-a^2 + b^2])])/Sqrt[-a^2 + b^2] - (I*b*f*Log[1 - I*Tan 
[(c + d*x)/2]]*Log[-((b - Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a - b 
+ Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] + (I*b*f*Log[1 - I*Tan[(c + d*x)/2 
]]*Log[(b + Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/((-I)*a + b + Sqrt[-a^2 
 + b^2])])/Sqrt[-a^2 + b^2] - (I*b*f*Log[1 + I*Tan[(c + d*x)/2]]*Log[(b + 
Sqrt[-a^2 + b^2] + a*Tan[(c + d*x)/2])/(I*a + b + Sqrt[-a^2 + b^2])])/Sqrt 
[-a^2 + b^2] + (I*b*f*PolyLog[2, (a*(1 - I*Tan[(c + d*x)/2]))/(a + I*(b + 
Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] - (I*b*f*PolyLog[2, (a*(1 + I*Tan[(c 
 + d*x)/2]))/(a - I*(b + Sqrt[-a^2 + b^2]))])/Sqrt[-a^2 + b^2] - (I*b*f*Po 
lyLog[2, (a*(I + Tan[(c + d*x)/2]))/(I*a - b + Sqrt[-a^2 + b^2])])/Sqrt[-a 
^2 + b^2] + (I*b*f*PolyLog[2, (a + I*a*Tan[(c + d*x)/2])/(a + I*(-b + Sqrt 
[-a^2 + b^2]))])/Sqrt[-a^2 + b^2])*(-((b*e)/((a^2 - b^2)*(a + b*Sin[c + d* 
x]))) + (b*c*f)/((a^2 - b^2)*d*(a + b*Sin[c + d*x])) - (b*f*(c + d*x))/((a 
^2 - b^2)*d*(a + b*Sin[c + d*x])) + (a*f*Cos[c + d*x])/((a^2 - b^2)*d*(...
 

Rubi [A] (verified)

Time = 1.91 (sec) , antiderivative size = 574, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, 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.

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

\(\Big \downarrow \) 7293

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(I*a^2*(e + f*x)*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)*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqr 
t[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d) - (I*a^2*(e + f*x)*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 
)*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2] 
*d) + (a*f*Log[a + b*Sin[c + d*x]])/(b*(a^2 - b^2)*d^2) + (a^2*f*PolyLog[2 
, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*(a^2 - b^2)^(3/2)*d^2) 
- (f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*Sqrt[a^2 
- b^2]*d^2) - (a^2*f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2] 
)])/(b*(a^2 - b^2)^(3/2)*d^2) + (f*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + S 
qrt[a^2 - b^2])])/(b*Sqrt[a^2 - b^2]*d^2) - (a*(e + f*x)*Cos[c + d*x])/((a 
^2 - b^2)*d*(a + b*Sin[c + d*x]))
 

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] 
]
 

Maple [A] (verified)

Time = 3.82 (sec) , antiderivative size = 750, normalized size of antiderivative = 1.31

method	result	size
risch	\(\frac {2 i a \left (f x +e \right ) \left (b -i a \,{\mathrm e}^{i \left (d x +c \right )}\right )}{b \left (-a^{2}+b^{2}\right ) d \left (2 i a \,{\mathrm e}^{i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}-b \right )}+\frac {a f \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 \,d^{2} \left (a^{2}-b^{2}\right )}-\frac {2 a f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{b \,d^{2} \left (a^{2}-b^{2}\right )}+\frac {2 i b c f \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}-\frac {b f \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}{d \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {b f \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}{d \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}-\frac {b f \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}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {b f \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}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}+\frac {i b f \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 )}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}-\frac {i b f \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 )}{d^{2} \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}-\frac {2 i b e \arctan \left (\frac {2 i b \,{\mathrm e}^{i \left (d x +c \right )}-2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d \left (a^{2}-b^{2}\right ) \sqrt {-a^{2}+b^{2}}}\)	\(750\)

Input:

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

Output:

2*I*a*(f*x+e)*(b-I*a*exp(I*(d*x+c)))/b/(-a^2+b^2)/d/(2*I*a*exp(I*(d*x+c))+ 
b*exp(2*I*(d*x+c))-b)+1/b/d^2/(a^2-b^2)*a*f*ln(I*b*exp(2*I*(d*x+c))-2*a*ex 
p(I*(d*x+c))-I*b)-2/b/d^2/(a^2-b^2)*a*f*ln(exp(I*(d*x+c)))+2*I*b/d^2/(a^2- 
b^2)*c*f/(-a^2+b^2)^(1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*a)/(-a^2+b^2) 
^(1/2))-b/d/(a^2-b^2)*f/(-a^2+b^2)^(1/2)*ln((I*a+exp(I*(d*x+c))*b-(-a^2+b^ 
2)^(1/2))/(I*a-(-a^2+b^2)^(1/2)))*x+b/d/(a^2-b^2)*f/(-a^2+b^2)^(1/2)*ln((I 
*a+exp(I*(d*x+c))*b+(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))*x-b/d^2/(a^2 
-b^2)*f/(-a^2+b^2)^(1/2)*ln((I*a+exp(I*(d*x+c))*b-(-a^2+b^2)^(1/2))/(I*a-( 
-a^2+b^2)^(1/2)))*c+b/d^2/(a^2-b^2)*f/(-a^2+b^2)^(1/2)*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/d^2/(a^2-b^2)*f/(-a^ 
2+b^2)^(1/2)*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/d^2/(a^2-b^2)*f/(-a^2+b^2)^(1/2)*dilog((I*a+exp(I*(d*x+c))*b+ 
(-a^2+b^2)^(1/2))/(I*a+(-a^2+b^2)^(1/2)))-2*I*b/d/(a^2-b^2)*e/(-a^2+b^2)^( 
1/2)*arctan(1/2*(2*I*b*exp(I*(d*x+c))-2*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. 1506 vs. \(2 (503) = 1006\).

Time = 0.37 (sec) , antiderivative size = 1506, normalized size of antiderivative = 2.62 \[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*((-I*b^4*f*sin(d*x + c) - I*a*b^3*f)*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*b^4*f*sin(d*x + c) + I*a*b^3*f)*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*b^4*f*sin(d*x 
 + c) + I*a*b^3*f)*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*b^4*f*sin(d*x + c) - I*a*b^3*f)*sqrt(-(a^2 - b^2)/b^2)*dilo 
g((-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*b^3*d*f*x + a*b^3*c*f + (b^4*d*f 
*x + b^4*c*f)*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*b^3*d*f*x + a*b^3*c*f + (b^4*d*f*x + b^4*c*f)*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*b^3*d*f* 
x + a*b^3*c*f + (b^4*d*f*x + b^4*c*f)*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*b^3*d*f*x + a*b^3*c*f + (b^4*d*f 
*x + b^4*c*f)*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^...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((f*x+e)*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
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int \frac {(e+f x) \sin (c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (d x +c \right ) b^{2} e -2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a b e -\cos \left (d x +c \right ) a^{3} e +\cos \left (d x +c \right ) a \,b^{2} e +\left (\int \frac {\sin \left (d x +c \right ) x}{\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +a^{2}}d x \right ) \sin \left (d x +c \right ) a^{4} b d f -2 \left (\int \frac {\sin \left (d x +c \right ) x}{\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +a^{2}}d x \right ) \sin \left (d x +c \right ) a^{2} b^{3} d f +\left (\int \frac {\sin \left (d x +c \right ) x}{\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +a^{2}}d x \right ) \sin \left (d x +c \right ) b^{5} d f +\left (\int \frac {\sin \left (d x +c \right ) x}{\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +a^{2}}d x \right ) a^{5} d f -2 \left (\int \frac {\sin \left (d x +c \right ) x}{\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +a^{2}}d x \right ) a^{3} b^{2} d f +\left (\int \frac {\sin \left (d x +c \right ) x}{\sin \left (d x +c \right )^{2} b^{2}+2 \sin \left (d x +c \right ) a b +a^{2}}d x \right ) a \,b^{4} d f}{d \left (\sin \left (d x +c \right ) a^{4} b -2 \sin \left (d x +c \right ) a^{2} b^{3}+\sin \left (d x +c \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

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

Output:

( - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*s 
in(c + d*x)*b**2*e - 2*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqr 
t(a**2 - b**2))*a*b*e - cos(c + d*x)*a**3*e + cos(c + d*x)*a*b**2*e + int( 
(sin(c + d*x)*x)/(sin(c + d*x)**2*b**2 + 2*sin(c + d*x)*a*b + a**2),x)*sin 
(c + d*x)*a**4*b*d*f - 2*int((sin(c + d*x)*x)/(sin(c + d*x)**2*b**2 + 2*si 
n(c + d*x)*a*b + a**2),x)*sin(c + d*x)*a**2*b**3*d*f + int((sin(c + d*x)*x 
)/(sin(c + d*x)**2*b**2 + 2*sin(c + d*x)*a*b + a**2),x)*sin(c + d*x)*b**5* 
d*f + int((sin(c + d*x)*x)/(sin(c + d*x)**2*b**2 + 2*sin(c + d*x)*a*b + a* 
*2),x)*a**5*d*f - 2*int((sin(c + d*x)*x)/(sin(c + d*x)**2*b**2 + 2*sin(c + 
 d*x)*a*b + a**2),x)*a**3*b**2*d*f + int((sin(c + d*x)*x)/(sin(c + d*x)**2 
*b**2 + 2*sin(c + d*x)*a*b + a**2),x)*a*b**4*d*f)/(d*(sin(c + d*x)*a**4*b 
- 2*sin(c + d*x)*a**2*b**3 + sin(c + d*x)*b**5 + a**5 - 2*a**3*b**2 + a*b* 
*4))