\(\int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx\) [724]

Optimal result

Integrand size = 25, antiderivative size = 196 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx=-\frac {\left (6 a b c d-a^2 \left (2 c^2+d^2\right )-b^2 \left (c^2+2 d^2\right )\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2} f}+\frac {(b c-a d)^2 \cos (e+f x)}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac {(b c-a d) \left (3 a b c+a^2 d-4 b^2 d\right ) \cos (e+f x)}{2 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))} \] Output:

-(6*a*b*c*d-a^2*(2*c^2+d^2)-b^2*(c^2+2*d^2))*arctan((b+a*tan(1/2*f*x+1/2*e 
))/(a^2-b^2)^(1/2))/(a^2-b^2)^(5/2)/f+1/2*(-a*d+b*c)^2*cos(f*x+e)/b/(a^2-b 
^2)/f/(a+b*sin(f*x+e))^2+1/2*(-a*d+b*c)*(a^2*d+3*a*b*c-4*b^2*d)*cos(f*x+e) 
/b/(a^2-b^2)^2/f/(a+b*sin(f*x+e))
 

Mathematica [A] (verified)

Time = 1.34 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.04 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx=\frac {\frac {2 \left (-6 a b c d+a^2 \left (2 c^2+d^2\right )+b^2 \left (c^2+2 d^2\right )\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac {(b c-a d)^2 \cos (e+f x)}{(a-b) b (a+b) (a+b \sin (e+f x))^2}-\frac {\left (2 a^2 b c d+4 b^3 c d+a^3 d^2-a b^2 \left (3 c^2+4 d^2\right )\right ) \cos (e+f x)}{(a-b)^2 b (a+b)^2 (a+b \sin (e+f x))}}{2 f} \] Input:

Integrate[(c + d*Sin[e + f*x])^2/(a + b*Sin[e + f*x])^3,x]
 

Output:

((2*(-6*a*b*c*d + a^2*(2*c^2 + d^2) + b^2*(c^2 + 2*d^2))*ArcTan[(b + a*Tan 
[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + ((b*c - a*d)^2*Cos[e 
+ f*x])/((a - b)*b*(a + b)*(a + b*Sin[e + f*x])^2) - ((2*a^2*b*c*d + 4*b^3 
*c*d + a^3*d^2 - a*b^2*(3*c^2 + 4*d^2))*Cos[e + f*x])/((a - b)^2*b*(a + b) 
^2*(a + b*Sin[e + f*x])))/(2*f)
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.11, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.440, Rules used = {3042, 3269, 25, 3042, 3233, 25, 27, 3042, 3139, 1083, 217}

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.

\(\displaystyle \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3269

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

\(\displaystyle \frac {(b c-a d)^2 \cos (e+f x)}{2 b f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2}-\frac {\frac {2 b \left (-\left (a^2 \left (2 c^2+d^2\right )\right )+6 a b c d-b^2 \left (c^2+2 d^2\right )\right ) \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (e+f x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{f \left (a^2-b^2\right )^{3/2}}-\frac {(b c-a d) \left (a^2 d+3 a b c-4 b^2 d\right ) \cos (e+f x)}{f \left (a^2-b^2\right ) (a+b \sin (e+f x))}}{2 b \left (a^2-b^2\right )}\)

Input:

Int[(c + d*Sin[e + f*x])^2/(a + b*Sin[e + f*x])^3,x]
 

Output:

((b*c - a*d)^2*Cos[e + f*x])/(2*b*(a^2 - b^2)*f*(a + b*Sin[e + f*x])^2) - 
((2*b*(6*a*b*c*d - a^2*(2*c^2 + d^2) - b^2*(c^2 + 2*d^2))*ArcTan[(2*b + 2* 
a*Tan[(e + f*x)/2])/(2*Sqrt[a^2 - b^2])])/((a^2 - b^2)^(3/2)*f) - ((b*c - 
a*d)*(3*a*b*c + a^2*d - 4*b^2*d)*Cos[e + f*x])/((a^2 - b^2)*f*(a + b*Sin[e 
 + f*x])))/(2*b*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 1083

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3139

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre 
eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + 2*b*e*x + a 
*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ 
[a^2 - b^2, 0]
 

rule 3233

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

rule 3269

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^2, x_Symbol] :> Simp[(-(b^2*c^2 - 2*a*b*c*d + a^2*d^2))*Cos[e 
+ f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[ 
1/(b*(m + 1)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1) 
*(2*b*c*d - a*(c^2 + d^2)) + (a^2*d^2 - 2*a*b*c*d*(m + 2) + b^2*(d^2*(m + 1 
) + c^2*(m + 2)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] 
&& NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(464\) vs. \(2(187)=374\).

Time = 1.75 (sec) , antiderivative size = 465, normalized size of antiderivative = 2.37

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (a^{4} d^{2}-6 a^{3} b c d +5 a^{2} b^{2} c^{2}+2 a^{2} b^{2} d^{2}-2 b^{4} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}-\frac {\left (4 a^{5} c d -4 a^{4} b \,c^{2}-3 a^{4} b \,d^{2}+10 a^{3} b^{2} c d -7 a^{2} b^{3} c^{2}-6 a^{2} b^{3} d^{2}+4 a \,b^{4} c d +2 b^{5} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}-\frac {\left (a^{4} d^{2}+10 a^{3} b c d -11 a^{2} b^{2} c^{2}-10 a^{2} b^{2} d^{2}+8 a \,b^{3} c d +2 b^{4} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {4 a^{3} c d -4 a^{2} b \,c^{2}-3 a^{2} b \,d^{2}+2 a \,b^{2} c d +b^{3} c^{2}}{a^{4}-2 a^{2} b^{2}+b^{4}}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2} c^{2}+a^{2} d^{2}-6 a b c d +b^{2} c^{2}+2 d^{2} b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}}{f}\)	\(465\)
default	\(\frac {\frac {\frac {\left (a^{4} d^{2}-6 a^{3} b c d +5 a^{2} b^{2} c^{2}+2 a^{2} b^{2} d^{2}-2 b^{4} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a}-\frac {\left (4 a^{5} c d -4 a^{4} b \,c^{2}-3 a^{4} b \,d^{2}+10 a^{3} b^{2} c d -7 a^{2} b^{3} c^{2}-6 a^{2} b^{3} d^{2}+4 a \,b^{4} c d +2 b^{5} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) a^{2}}-\frac {\left (a^{4} d^{2}+10 a^{3} b c d -11 a^{2} b^{2} c^{2}-10 a^{2} b^{2} d^{2}+8 a \,b^{3} c d +2 b^{4} c^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {4 a^{3} c d -4 a^{2} b \,c^{2}-3 a^{2} b \,d^{2}+2 a \,b^{2} c d +b^{3} c^{2}}{a^{4}-2 a^{2} b^{2}+b^{4}}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a \right )^{2}}+\frac {\left (2 a^{2} c^{2}+a^{2} d^{2}-6 a b c d +b^{2} c^{2}+2 d^{2} b^{2}\right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {a^{2}-b^{2}}}}{f}\)	\(465\)
risch	\(\text {Expression too large to display}\)	\(1294\)

Input:

int((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*(2*(1/2*(a^4*d^2-6*a^3*b*c*d+5*a^2*b^2*c^2+2*a^2*b^2*d^2-2*b^4*c^2)/(a 
^4-2*a^2*b^2+b^4)/a*tan(1/2*f*x+1/2*e)^3-1/2*(4*a^5*c*d-4*a^4*b*c^2-3*a^4* 
b*d^2+10*a^3*b^2*c*d-7*a^2*b^3*c^2-6*a^2*b^3*d^2+4*a*b^4*c*d+2*b^5*c^2)/(a 
^4-2*a^2*b^2+b^4)/a^2*tan(1/2*f*x+1/2*e)^2-1/2*(a^4*d^2+10*a^3*b*c*d-11*a^ 
2*b^2*c^2-10*a^2*b^2*d^2+8*a*b^3*c*d+2*b^4*c^2)/a/(a^4-2*a^2*b^2+b^4)*tan( 
1/2*f*x+1/2*e)-1/2*(4*a^3*c*d-4*a^2*b*c^2-3*a^2*b*d^2+2*a*b^2*c*d+b^3*c^2) 
/(a^4-2*a^2*b^2+b^4))/(tan(1/2*f*x+1/2*e)^2*a+2*b*tan(1/2*f*x+1/2*e)+a)^2+ 
(2*a^2*c^2+a^2*d^2-6*a*b*c*d+b^2*c^2+2*b^2*d^2)/(a^4-2*a^2*b^2+b^4)/(a^2-b 
^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*f*x+1/2*e)+2*b)/(a^2-b^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 470 vs. \(2 (187) = 374\).

Time = 0.20 (sec) , antiderivative size = 1025, normalized size of antiderivative = 5.23 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx=\text {Too large to display} \] Input:

integrate((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,x, algorithm="fricas")
 

Output:

[-1/4*(2*(3*(a^3*b^2 - a*b^4)*c^2 - 2*(a^4*b + a^2*b^3 - 2*b^5)*c*d - (a^5 
 - 5*a^3*b^2 + 4*a*b^4)*d^2)*cos(f*x + e)*sin(f*x + e) - ((2*a^4 + 3*a^2*b 
^2 + b^4)*c^2 - 6*(a^3*b + a*b^3)*c*d + (a^4 + 3*a^2*b^2 + 2*b^4)*d^2 + (6 
*a*b^3*c*d - (2*a^2*b^2 + b^4)*c^2 - (a^2*b^2 + 2*b^4)*d^2)*cos(f*x + e)^2 
 - 2*(6*a^2*b^2*c*d - (2*a^3*b + a*b^3)*c^2 - (a^3*b + 2*a*b^3)*d^2)*sin(f 
*x + e))*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(f*x + e)^2 - 2*a*b*sin(f* 
x + e) - a^2 - b^2 + 2*(a*cos(f*x + e)*sin(f*x + e) + b*cos(f*x + e))*sqrt 
(-a^2 + b^2))/(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)) + 2*( 
(4*a^4*b - 5*a^2*b^3 + b^5)*c^2 - 2*(2*a^5 - a^3*b^2 - a*b^4)*c*d + 3*(a^4 
*b - a^2*b^3)*d^2)*cos(f*x + e))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)* 
f*cos(f*x + e)^2 - 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*f*sin(f*x + e 
) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*f), -1/2*((3*(a^3*b^2 - a*b^4)*c^2 
 - 2*(a^4*b + a^2*b^3 - 2*b^5)*c*d - (a^5 - 5*a^3*b^2 + 4*a*b^4)*d^2)*cos( 
f*x + e)*sin(f*x + e) - ((2*a^4 + 3*a^2*b^2 + b^4)*c^2 - 6*(a^3*b + a*b^3) 
*c*d + (a^4 + 3*a^2*b^2 + 2*b^4)*d^2 + (6*a*b^3*c*d - (2*a^2*b^2 + b^4)*c^ 
2 - (a^2*b^2 + 2*b^4)*d^2)*cos(f*x + e)^2 - 2*(6*a^2*b^2*c*d - (2*a^3*b + 
a*b^3)*c^2 - (a^3*b + 2*a*b^3)*d^2)*sin(f*x + e))*sqrt(a^2 - b^2)*arctan(- 
(a*sin(f*x + e) + b)/(sqrt(a^2 - b^2)*cos(f*x + e))) + ((4*a^4*b - 5*a^2*b 
^3 + b^5)*c^2 - 2*(2*a^5 - a^3*b^2 - a*b^4)*c*d + 3*(a^4*b - a^2*b^3)*d^2) 
*cos(f*x + e))/((a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*f*cos(f*x + e)^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx=\text {Timed out} \] Input:

integrate((c+d*sin(f*x+e))**2/(a+b*sin(f*x+e))**3,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 586 vs. \(2 (187) = 374\).

Time = 0.38 (sec) , antiderivative size = 586, normalized size of antiderivative = 2.99 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

integrate((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,x, algorithm="giac")
 

Output:

((2*a^2*c^2 + b^2*c^2 - 6*a*b*c*d + a^2*d^2 + 2*b^2*d^2)*(pi*floor(1/2*(f* 
x + e)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b 
^2)))/((a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)) + (5*a^3*b^2*c^2*tan(1/2*f 
*x + 1/2*e)^3 - 2*a*b^4*c^2*tan(1/2*f*x + 1/2*e)^3 - 6*a^4*b*c*d*tan(1/2*f 
*x + 1/2*e)^3 + a^5*d^2*tan(1/2*f*x + 1/2*e)^3 + 2*a^3*b^2*d^2*tan(1/2*f*x 
 + 1/2*e)^3 + 4*a^4*b*c^2*tan(1/2*f*x + 1/2*e)^2 + 7*a^2*b^3*c^2*tan(1/2*f 
*x + 1/2*e)^2 - 2*b^5*c^2*tan(1/2*f*x + 1/2*e)^2 - 4*a^5*c*d*tan(1/2*f*x + 
 1/2*e)^2 - 10*a^3*b^2*c*d*tan(1/2*f*x + 1/2*e)^2 - 4*a*b^4*c*d*tan(1/2*f* 
x + 1/2*e)^2 + 3*a^4*b*d^2*tan(1/2*f*x + 1/2*e)^2 + 6*a^2*b^3*d^2*tan(1/2* 
f*x + 1/2*e)^2 + 11*a^3*b^2*c^2*tan(1/2*f*x + 1/2*e) - 2*a*b^4*c^2*tan(1/2 
*f*x + 1/2*e) - 10*a^4*b*c*d*tan(1/2*f*x + 1/2*e) - 8*a^2*b^3*c*d*tan(1/2* 
f*x + 1/2*e) - a^5*d^2*tan(1/2*f*x + 1/2*e) + 10*a^3*b^2*d^2*tan(1/2*f*x + 
 1/2*e) + 4*a^4*b*c^2 - a^2*b^3*c^2 - 4*a^5*c*d - 2*a^3*b^2*c*d + 3*a^4*b* 
d^2)/((a^6 - 2*a^4*b^2 + a^2*b^4)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2* 
f*x + 1/2*e) + a)^2))/f
 

Mupad [B] (verification not implemented)

Time = 19.31 (sec) , antiderivative size = 641, normalized size of antiderivative = 3.27 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx=\frac {\mathrm {atan}\left (\frac {\left (\frac {\left (2\,a^4\,b-4\,a^2\,b^3+2\,b^5\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{2\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2}\right )\,\left (2\,a^2\,c^2+a^2\,d^2-6\,a\,b\,c\,d+b^2\,c^2+2\,b^2\,d^2\right )}{f\,{\left (a+b\right )}^{5/2}\,{\left (a-b\right )}^{5/2}}-\frac {\frac {4\,a^3\,c\,d-4\,a^2\,b\,c^2-3\,a^2\,b\,d^2+2\,a\,b^2\,c\,d+b^3\,c^2}{a^4-2\,a^2\,b^2+b^4}+\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a^4\,d^2+10\,a^3\,b\,c\,d-11\,a^2\,b^2\,c^2-10\,a^2\,b^2\,d^2+8\,a\,b^3\,c\,d+2\,b^4\,c^2\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (a^4\,d^2-6\,a^3\,b\,c\,d+5\,a^2\,b^2\,c^2+2\,a^2\,b^2\,d^2-2\,b^4\,c^2\right )}{a\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (a^2+2\,b^2\right )\,\left (4\,a^3\,c\,d-4\,a^2\,b\,c^2-3\,a^2\,b\,d^2+2\,a\,b^2\,c\,d+b^3\,c^2\right )}{a^2\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{f\,\left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (2\,a^2+4\,b^2\right )+a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+a^2+4\,a\,b\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3+4\,a\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )} \] Input:

int((c + d*sin(e + f*x))^2/(a + b*sin(e + f*x))^3,x)
 

Output:

(atan(((((2*a^4*b + 2*b^5 - 4*a^2*b^3)*(2*a^2*c^2 + a^2*d^2 + b^2*c^2 + 2* 
b^2*d^2 - 6*a*b*c*d))/(2*(a + b)^(5/2)*(a - b)^(5/2)*(a^4 + b^4 - 2*a^2*b^ 
2)) + (a*tan(e/2 + (f*x)/2)*(2*a^2*c^2 + a^2*d^2 + b^2*c^2 + 2*b^2*d^2 - 6 
*a*b*c*d))/((a + b)^(5/2)*(a - b)^(5/2)))*(a^4 + b^4 - 2*a^2*b^2))/(2*a^2* 
c^2 + a^2*d^2 + b^2*c^2 + 2*b^2*d^2 - 6*a*b*c*d))*(2*a^2*c^2 + a^2*d^2 + b 
^2*c^2 + 2*b^2*d^2 - 6*a*b*c*d))/(f*(a + b)^(5/2)*(a - b)^(5/2)) - ((b^3*c 
^2 - 4*a^2*b*c^2 - 3*a^2*b*d^2 + 4*a^3*c*d + 2*a*b^2*c*d)/(a^4 + b^4 - 2*a 
^2*b^2) + (tan(e/2 + (f*x)/2)*(a^4*d^2 + 2*b^4*c^2 - 11*a^2*b^2*c^2 - 10*a 
^2*b^2*d^2 + 8*a*b^3*c*d + 10*a^3*b*c*d))/(a*(a^4 + b^4 - 2*a^2*b^2)) - (t 
an(e/2 + (f*x)/2)^3*(a^4*d^2 - 2*b^4*c^2 + 5*a^2*b^2*c^2 + 2*a^2*b^2*d^2 - 
 6*a^3*b*c*d))/(a*(a^4 + b^4 - 2*a^2*b^2)) + (tan(e/2 + (f*x)/2)^2*(a^2 + 
2*b^2)*(b^3*c^2 - 4*a^2*b*c^2 - 3*a^2*b*d^2 + 4*a^3*c*d + 2*a*b^2*c*d))/(a 
^2*(a^4 + b^4 - 2*a^2*b^2)))/(f*(tan(e/2 + (f*x)/2)^2*(2*a^2 + 4*b^2) + a^ 
2*tan(e/2 + (f*x)/2)^4 + a^2 + 4*a*b*tan(e/2 + (f*x)/2)^3 + 4*a*b*tan(e/2 
+ (f*x)/2)))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1590, normalized size of antiderivative = 8.11 \[ \int \frac {(c+d \sin (e+f x))^2}{(a+b \sin (e+f x))^3} \, dx =\text {Too large to display} \] Input:

int((c+d*sin(f*x+e))^2/(a+b*sin(f*x+e))^3,x)
 

Output:

(8*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin( 
e + f*x)**2*a**3*b**3*c**2 + 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a 
+ b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*a**3*b**3*d**2 - 24*sqrt(a**2 - b* 
*2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*a**2* 
b**4*c*d + 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b 
**2))*sin(e + f*x)**2*a*b**5*c**2 + 8*sqrt(a**2 - b**2)*atan((tan((e + f*x 
)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)**2*a*b**5*d**2 + 16*sqrt(a**2 
- b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a**4 
*b**2*c**2 + 8*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - 
 b**2))*sin(e + f*x)*a**4*b**2*d**2 - 48*sqrt(a**2 - b**2)*atan((tan((e + 
f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a**3*b**3*c*d + 8*sqrt(a**2 
 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a** 
2*b**4*c**2 + 16*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 
 - b**2))*sin(e + f*x)*a**2*b**4*d**2 + 8*sqrt(a**2 - b**2)*atan((tan((e + 
 f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**5*b*c**2 + 4*sqrt(a**2 - b**2)*atan( 
(tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**5*b*d**2 - 24*sqrt(a**2 - b 
**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**4*b**2*c*d + 4*sq 
rt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**3*b**3 
*c**2 + 8*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2 
))*a**3*b**3*d**2 - 2*cos(e + f*x)*sin(e + f*x)*a**6*b*d**2 - 4*cos(e +...