\(\int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx\) [450]

Optimal result

Integrand size = 25, antiderivative size = 207 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx=\frac {a^2 (3 c-2 d) \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(c-d) (c+d)^3 \sqrt {c^2-d^2} f}+\frac {a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^3}-\frac {a^2 (c+6 d) \cos (e+f x)}{6 d (c+d)^2 f (c+d \sin (e+f x))^2}-\frac {a^2 \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{6 (c-d) d (c+d)^3 f (c+d \sin (e+f x))} \] Output:

a^2*(3*c-2*d)*arctan((d+c*tan(1/2*f*x+1/2*e))/(c^2-d^2)^(1/2))/(c-d)/(c+d) 
^3/(c^2-d^2)^(1/2)/f+1/3*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^3 
-1/6*a^2*(c+6*d)*cos(f*x+e)/d/(c+d)^2/f/(c+d*sin(f*x+e))^2-1/6*a^2*(c^2+6* 
c*d-10*d^2)*cos(f*x+e)/(c-d)/d/(c+d)^3/f/(c+d*sin(f*x+e))
 

Mathematica [A] (warning: unable to verify)

Time = 2.65 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.95 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx=\frac {a^2 \cos (e+f x) \left (-\frac {d (1+\sin (e+f x))^2}{(c+d \sin (e+f x))^3}-\frac {(3 c-2 d) \left (\frac {6 \text {arctanh}\left (\frac {\sqrt {c-d} \sqrt {1-\sin (e+f x)}}{\sqrt {-c-d} \sqrt {1+\sin (e+f x)}}\right )}{\sqrt {-c-d} \sqrt {c-d}}-\frac {\sqrt {\cos ^2(e+f x)} (4 c+d+(c+4 d) \sin (e+f x))}{(c+d \sin (e+f x))^2}\right )}{2 (c+d)^2 \sqrt {\cos ^2(e+f x)}}\right )}{3 (-c+d) (c+d) f} \] Input:

Integrate[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^4,x]
 

Output:

(a^2*Cos[e + f*x]*(-((d*(1 + Sin[e + f*x])^2)/(c + d*Sin[e + f*x])^3) - (( 
3*c - 2*d)*((6*ArcTanh[(Sqrt[c - d]*Sqrt[1 - Sin[e + f*x]])/(Sqrt[-c - d]* 
Sqrt[1 + Sin[e + f*x]])])/(Sqrt[-c - d]*Sqrt[c - d]) - (Sqrt[Cos[e + f*x]^ 
2]*(4*c + d + (c + 4*d)*Sin[e + f*x]))/(c + d*Sin[e + f*x])^2))/(2*(c + d) 
^2*Sqrt[Cos[e + f*x]^2])))/(3*(-c + d)*(c + d)*f)
 

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.05, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 3241, 25, 3042, 3233, 25, 3042, 3233, 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.

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c+d \sin (e+f x))^4} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a \sin (e+f x)+a)^2}{(c+d \sin (e+f x))^4}dx\)

\(\Big \downarrow \) 3241

\(\displaystyle \frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}-\frac {a \int -\frac {6 a d+a (c+5 d) \sin (e+f x)}{(c+d \sin (e+f x))^3}dx}{3 d (c+d)}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {a \int \frac {6 a d+a (c+5 d) \sin (e+f x)}{(c+d \sin (e+f x))^3}dx}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \int \frac {6 a d+a (c+5 d) \sin (e+f x)}{(c+d \sin (e+f x))^3}dx}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {a \left (-\frac {\int -\frac {10 a (c-d) d+a (c-d) (c+6 d) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {a \left (\frac {\int \frac {10 a (c-d) d+a (c-d) (c+6 d) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (\frac {\int \frac {10 a (c-d) d+a (c-d) (c+6 d) \sin (e+f x)}{(c+d \sin (e+f x))^2}dx}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 3233

\(\displaystyle \frac {a \left (\frac {-\frac {\int -\frac {3 a (3 c-2 d) (c-d) d}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {a \left (\frac {\frac {3 a d (3 c-2 d) (c-d) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \left (\frac {\frac {3 a d (3 c-2 d) (c-d) \int \frac {1}{c+d \sin (e+f x)}dx}{c^2-d^2}-\frac {a \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 3139

\(\displaystyle \frac {a \left (\frac {\frac {6 a d (3 c-2 d) (c-d) \int \frac {1}{c \tan ^2\left (\frac {1}{2} (e+f x)\right )+2 d \tan \left (\frac {1}{2} (e+f x)\right )+c}d\tan \left (\frac {1}{2} (e+f x)\right )}{f \left (c^2-d^2\right )}-\frac {a \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {a \left (\frac {-\frac {12 a d (3 c-2 d) (c-d) \int \frac {1}{-\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )^2-4 \left (c^2-d^2\right )}d\left (2 d+2 c \tan \left (\frac {1}{2} (e+f x)\right )\right )}{f \left (c^2-d^2\right )}-\frac {a \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}+\frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^3}+\frac {a \left (\frac {\frac {6 a d (3 c-2 d) (c-d) \arctan \left (\frac {2 c \tan \left (\frac {1}{2} (e+f x)\right )+2 d}{2 \sqrt {c^2-d^2}}\right )}{f \left (c^2-d^2\right )^{3/2}}-\frac {a \left (c^2+6 c d-10 d^2\right ) \cos (e+f x)}{f (c+d) (c+d \sin (e+f x))}}{2 \left (c^2-d^2\right )}-\frac {a (c+6 d) \cos (e+f x)}{2 f (c+d) (c+d \sin (e+f x))^2}\right )}{3 d (c+d)}\)

Input:

Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^4,x]
 

Output:

(a^2*(c - d)*Cos[e + f*x])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^3) + (a*(-1 
/2*(a*(c + 6*d)*Cos[e + f*x])/((c + d)*f*(c + d*Sin[e + f*x])^2) + ((6*a*( 
3*c - 2*d)*(c - d)*d*ArcTan[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2 
])])/((c^2 - d^2)^(3/2)*f) - (a*(c^2 + 6*c*d - 10*d^2)*Cos[e + f*x])/((c + 
 d)*f*(c + d*Sin[e + f*x])))/(2*(c^2 - d^2))))/(3*d*(c + d))
 

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 3241

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b 
*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c + a* 
d))), x] + Simp[b^2/(d*(n + 1)*(b*c + a*d))   Int[(a + b*Sin[e + f*x])^(m - 
 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b* 
c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, 
d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] 
 && GtQ[m, 1] && LtQ[n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || 
 (IntegerQ[m] && EqQ[c, 0]))
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(522\) vs. \(2(196)=392\).

Time = 1.76 (sec) , antiderivative size = 523, normalized size of antiderivative = 2.53

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (\frac {\frac {\left (c^{4}-6 c^{3} d +4 c \,d^{3}+2 d^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2 c \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {\left (4 c^{5}-3 c^{4} d +18 c^{3} d^{2}-8 c^{2} d^{3}-12 c \,d^{4}-4 d^{5}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2 c^{2} \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {d \left (36 c^{5}-21 c^{4} d +6 c^{3} d^{2}-20 c^{2} d^{3}-12 c \,d^{4}-4 d^{5}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 c^{3} \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {\left (4 c^{5}-2 c^{4} d +12 c^{3} d^{2}-11 c^{2} d^{3}-6 c \,d^{4}-2 d^{5}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c^{2} \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {\left (c^{4}+18 c^{3} d -14 c^{2} d^{2}-8 c \,d^{3}-2 d^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {12 c^{3}-7 c^{2} d -6 c \,d^{2}-2 d^{3}}{6 \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )^{3}}+\frac {\left (3 c -2 d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(523\)
default	\(\frac {2 a^{2} \left (\frac {\frac {\left (c^{4}-6 c^{3} d +4 c \,d^{3}+2 d^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{2 c \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {\left (4 c^{5}-3 c^{4} d +18 c^{3} d^{2}-8 c^{2} d^{3}-12 c \,d^{4}-4 d^{5}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{2 c^{2} \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {d \left (36 c^{5}-21 c^{4} d +6 c^{3} d^{2}-20 c^{2} d^{3}-12 c \,d^{4}-4 d^{5}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 c^{3} \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {\left (4 c^{5}-2 c^{4} d +12 c^{3} d^{2}-11 c^{2} d^{3}-6 c \,d^{4}-2 d^{5}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{c^{2} \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {\left (c^{4}+18 c^{3} d -14 c^{2} d^{2}-8 c \,d^{3}-2 d^{4}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 c \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}-\frac {12 c^{3}-7 c^{2} d -6 c \,d^{2}-2 d^{3}}{6 \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right )}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +2 d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+c \right )^{3}}+\frac {\left (3 c -2 d \right ) \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{2 \left (c^{4}+2 c^{3} d -2 c \,d^{3}-d^{4}\right ) \sqrt {c^{2}-d^{2}}}\right )}{f}\)	\(523\)
risch	\(\frac {i a^{2} \left (-12 i c^{3} d^{2} {\mathrm e}^{4 i \left (f x +e \right )}+6 i d^{5} {\mathrm e}^{4 i \left (f x +e \right )}-72 i c^{2} d^{3} {\mathrm e}^{2 i \left (f x +e \right )}-24 i d^{5} {\mathrm e}^{2 i \left (f x +e \right )}-6 i c \,d^{4}+9 c \,d^{4} {\mathrm e}^{5 i \left (f x +e \right )}-6 d^{5} {\mathrm e}^{5 i \left (f x +e \right )}-18 i c \,d^{4} {\mathrm e}^{4 i \left (f x +e \right )}+6 i c^{4} d \,{\mathrm e}^{2 i \left (f x +e \right )}+10 i d^{5}+60 i c^{3} d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+4 c^{5} {\mathrm e}^{3 i \left (f x +e \right )}+24 c^{4} d \,{\mathrm e}^{3 i \left (f x +e \right )}-34 c^{3} d^{2} {\mathrm e}^{3 i \left (f x +e \right )}+36 c^{2} d^{3} {\mathrm e}^{3 i \left (f x +e \right )}-60 c \,d^{4} {\mathrm e}^{3 i \left (f x +e \right )}-i c^{2} d^{3}-6 i c^{4} d \,{\mathrm e}^{4 i \left (f x +e \right )}+45 i c^{2} d^{3} {\mathrm e}^{4 i \left (f x +e \right )}-6 c^{3} d^{2} {\mathrm e}^{i \left (f x +e \right )}-36 c^{2} d^{3} {\mathrm e}^{i \left (f x +e \right )}+51 c \,d^{4} {\mathrm e}^{i \left (f x +e \right )}+6 d^{5} {\mathrm e}^{i \left (f x +e \right )}\right )}{3 \left (-i d \,{\mathrm e}^{2 i \left (f x +e \right )}+i d +2 c \,{\mathrm e}^{i \left (f x +e \right )}\right )^{3} \left (c +d \right )^{3} \left (c -d \right ) f \,d^{2}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{3} \left (c -d \right ) f}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}-c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{3} \left (c -d \right ) f}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) c}{2 \sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{3} \left (c -d \right ) f}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c \sqrt {-c^{2}+d^{2}}+c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right ) d}{\sqrt {-c^{2}+d^{2}}\, \left (c +d \right )^{3} \left (c -d \right ) f}\)	\(728\)

Input:

int((a+sin(f*x+e)*a)^2/(c+d*sin(f*x+e))^4,x,method=_RETURNVERBOSE)
 

Output:

2/f*a^2*((1/2*(c^4-6*c^3*d+4*c*d^3+2*d^4)/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan( 
1/2*f*x+1/2*e)^5-1/2*(4*c^5-3*c^4*d+18*c^3*d^2-8*c^2*d^3-12*c*d^4-4*d^5)/c 
^2/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^4-1/3/c^3*d*(36*c^5-21*c^4 
*d+6*c^3*d^2-20*c^2*d^3-12*c*d^4-4*d^5)/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2* 
f*x+1/2*e)^3-(4*c^5-2*c^4*d+12*c^3*d^2-11*c^2*d^3-6*c*d^4-2*d^5)/c^2/(c^4+ 
2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)^2-1/2*(c^4+18*c^3*d-14*c^2*d^2-8*c 
*d^3-2*d^4)/c/(c^4+2*c^3*d-2*c*d^3-d^4)*tan(1/2*f*x+1/2*e)-1/6*(12*c^3-7*c 
^2*d-6*c*d^2-2*d^3)/(c^4+2*c^3*d-2*c*d^3-d^4))/(tan(1/2*f*x+1/2*e)^2*c+2*d 
*tan(1/2*f*x+1/2*e)+c)^3+1/2*(3*c-2*d)/(c^4+2*c^3*d-2*c*d^3-d^4)/(c^2-d^2) 
^(1/2)*arctan(1/2*(2*c*tan(1/2*f*x+1/2*e)+2*d)/(c^2-d^2)^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 641 vs. \(2 (196) = 392\).

Time = 0.15 (sec) , antiderivative size = 1366, normalized size of antiderivative = 6.60 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx=\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x, algorithm="fricas")
 

Output:

[-1/12*(2*(a^2*c^4*d + 6*a^2*c^3*d^2 - 11*a^2*c^2*d^3 - 6*a^2*c*d^4 + 10*a 
^2*d^5)*cos(f*x + e)^3 - 6*(a^2*c^5 + 6*a^2*c^4*d - 8*a^2*c^3*d^2 - 8*a^2* 
c^2*d^3 + 7*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x + e)*sin(f*x + e) - 3*(3*a^2*c^ 
4 - 2*a^2*c^3*d + 9*a^2*c^2*d^2 - 6*a^2*c*d^3 - 3*(3*a^2*c^2*d^2 - 2*a^2*c 
*d^3)*cos(f*x + e)^2 + (9*a^2*c^3*d - 6*a^2*c^2*d^2 + 3*a^2*c*d^3 - 2*a^2* 
d^4 - (3*a^2*c*d^3 - 2*a^2*d^4)*cos(f*x + e)^2)*sin(f*x + e))*sqrt(-c^2 + 
d^2)*log(((2*c^2 - d^2)*cos(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2 + 
2*(c*cos(f*x + e)*sin(f*x + e) + d*cos(f*x + e))*sqrt(-c^2 + d^2))/(d^2*co 
s(f*x + e)^2 - 2*c*d*sin(f*x + e) - c^2 - d^2)) - 12*(2*a^2*c^5 - a^2*c^4* 
d - 2*a^2*c^3*d^2 - a^2*c^2*d^3 + 2*a^2*d^5)*cos(f*x + e))/(3*(c^7*d^2 + 2 
*c^6*d^3 - c^5*d^4 - 4*c^4*d^5 - c^3*d^6 + 2*c^2*d^7 + c*d^8)*f*cos(f*x + 
e)^2 - (c^9 + 2*c^8*d + 2*c^7*d^2 + 2*c^6*d^3 - 4*c^5*d^4 - 10*c^4*d^5 - 2 
*c^3*d^6 + 6*c^2*d^7 + 3*c*d^8)*f + ((c^6*d^3 + 2*c^5*d^4 - c^4*d^5 - 4*c^ 
3*d^6 - c^2*d^7 + 2*c*d^8 + d^9)*f*cos(f*x + e)^2 - (3*c^8*d + 6*c^7*d^2 - 
 2*c^6*d^3 - 10*c^5*d^4 - 4*c^4*d^5 + 2*c^3*d^6 + 2*c^2*d^7 + 2*c*d^8 + d^ 
9)*f)*sin(f*x + e)), -1/6*((a^2*c^4*d + 6*a^2*c^3*d^2 - 11*a^2*c^2*d^3 - 6 
*a^2*c*d^4 + 10*a^2*d^5)*cos(f*x + e)^3 - 3*(a^2*c^5 + 6*a^2*c^4*d - 8*a^2 
*c^3*d^2 - 8*a^2*c^2*d^3 + 7*a^2*c*d^4 + 2*a^2*d^5)*cos(f*x + e)*sin(f*x + 
 e) - 3*(3*a^2*c^4 - 2*a^2*c^3*d + 9*a^2*c^2*d^2 - 6*a^2*c*d^3 - 3*(3*a^2* 
c^2*d^2 - 2*a^2*c*d^3)*cos(f*x + e)^2 + (9*a^2*c^3*d - 6*a^2*c^2*d^2 + ...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx=\text {Timed out} \] Input:

integrate((a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**4,x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx=\text {Exception raised: ValueError} \] Input:

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,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*d^2-4*c^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. 750 vs. \(2 (196) = 392\).

Time = 0.19 (sec) , antiderivative size = 750, normalized size of antiderivative = 3.62 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx =\text {Too large to display} \] Input:

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x, algorithm="giac")
 

Output:

1/3*(3*(3*a^2*c - 2*a^2*d)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arct 
an((c*tan(1/2*f*x + 1/2*e) + d)/sqrt(c^2 - d^2)))/((c^4 + 2*c^3*d - 2*c*d^ 
3 - d^4)*sqrt(c^2 - d^2)) + (3*a^2*c^6*tan(1/2*f*x + 1/2*e)^5 - 18*a^2*c^5 
*d*tan(1/2*f*x + 1/2*e)^5 + 12*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^5 + 6*a^2* 
c^2*d^4*tan(1/2*f*x + 1/2*e)^5 - 12*a^2*c^6*tan(1/2*f*x + 1/2*e)^4 + 9*a^2 
*c^5*d*tan(1/2*f*x + 1/2*e)^4 - 54*a^2*c^4*d^2*tan(1/2*f*x + 1/2*e)^4 + 24 
*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^4 + 36*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^ 
4 + 12*a^2*c*d^5*tan(1/2*f*x + 1/2*e)^4 - 72*a^2*c^5*d*tan(1/2*f*x + 1/2*e 
)^3 + 42*a^2*c^4*d^2*tan(1/2*f*x + 1/2*e)^3 - 12*a^2*c^3*d^3*tan(1/2*f*x + 
 1/2*e)^3 + 40*a^2*c^2*d^4*tan(1/2*f*x + 1/2*e)^3 + 24*a^2*c*d^5*tan(1/2*f 
*x + 1/2*e)^3 + 8*a^2*d^6*tan(1/2*f*x + 1/2*e)^3 - 24*a^2*c^6*tan(1/2*f*x 
+ 1/2*e)^2 + 12*a^2*c^5*d*tan(1/2*f*x + 1/2*e)^2 - 72*a^2*c^4*d^2*tan(1/2* 
f*x + 1/2*e)^2 + 66*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e)^2 + 36*a^2*c^2*d^4*ta 
n(1/2*f*x + 1/2*e)^2 + 12*a^2*c*d^5*tan(1/2*f*x + 1/2*e)^2 - 3*a^2*c^6*tan 
(1/2*f*x + 1/2*e) - 54*a^2*c^5*d*tan(1/2*f*x + 1/2*e) + 42*a^2*c^4*d^2*tan 
(1/2*f*x + 1/2*e) + 24*a^2*c^3*d^3*tan(1/2*f*x + 1/2*e) + 6*a^2*c^2*d^4*ta 
n(1/2*f*x + 1/2*e) - 12*a^2*c^6 + 7*a^2*c^5*d + 6*a^2*c^4*d^2 + 2*a^2*c^3* 
d^3)/((c^7 + 2*c^6*d - 2*c^4*d^3 - c^3*d^4)*(c*tan(1/2*f*x + 1/2*e)^2 + 2* 
d*tan(1/2*f*x + 1/2*e) + c)^3))/f
 

Mupad [B] (verification not implemented)

Time = 18.99 (sec) , antiderivative size = 735, normalized size of antiderivative = 3.55 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx =\text {Too large to display} \] Input:

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^4,x)
 

Output:

- ((2*a^2*d^3 - 12*a^2*c^3 + 6*a^2*c*d^2 + 7*a^2*c^2*d)/(3*(2*c*d^3 - 2*c^ 
3*d - c^4 + d^4)) + (a^2*tan(e/2 + (f*x)/2)^5*(4*c*d^3 - 6*c^3*d + c^4 + 2 
*d^4))/(c*(2*c*d^3 - 2*c^3*d - c^4 + d^4)) + (2*a^2*tan(e/2 + (f*x)/2)^2*( 
6*c*d^4 + 2*c^4*d - 4*c^5 + 2*d^5 + 11*c^2*d^3 - 12*c^3*d^2))/(c^2*(2*c*d^ 
3 - 2*c^3*d - c^4 + d^4)) + (a^2*tan(e/2 + (f*x)/2)^4*(12*c*d^4 + 3*c^4*d 
- 4*c^5 + 4*d^5 + 8*c^2*d^3 - 18*c^3*d^2))/(c^2*(2*c*d^3 - 2*c^3*d - c^4 + 
 d^4)) + (a^2*tan(e/2 + (f*x)/2)*(8*c*d^3 - 18*c^3*d - c^4 + 2*d^4 + 14*c^ 
2*d^2))/(c*(2*c*d^3 - 2*c^3*d - c^4 + d^4)) + (2*a^2*d*tan(e/2 + (f*x)/2)^ 
3*(3*c^2 + 2*d^2)*(6*c*d^2 + 7*c^2*d - 12*c^3 + 2*d^3))/(3*c^3*(2*c*d^3 - 
2*c^3*d - c^4 + d^4)))/(f*(c^3*tan(e/2 + (f*x)/2)^6 + tan(e/2 + (f*x)/2)^2 
*(12*c*d^2 + 3*c^3) + tan(e/2 + (f*x)/2)^4*(12*c*d^2 + 3*c^3) + tan(e/2 + 
(f*x)/2)^3*(12*c^2*d + 8*d^3) + c^3 + 6*c^2*d*tan(e/2 + (f*x)/2) + 6*c^2*d 
*tan(e/2 + (f*x)/2)^5)) - (a^2*atan((((a^2*(3*c - 2*d)*(4*c*d^4 - 2*c^4*d 
+ 2*d^5 - 4*c^3*d^2))/(2*(c + d)^(7/2)*(c - d)^(3/2)*(2*c*d^3 - 2*c^3*d - 
c^4 + d^4)) + (a^2*c*tan(e/2 + (f*x)/2)*(3*c - 2*d))/((c + d)^(7/2)*(c - d 
)^(3/2)))*(2*c*d^3 - 2*c^3*d - c^4 + d^4))/(3*a^2*c - 2*a^2*d))*(3*c - 2*d 
))/(f*(c + d)^(7/2)*(c - d)^(3/2))
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1399, normalized size of antiderivative = 6.76 \[ \int \frac {(a+a \sin (e+f x))^2}{(c+d \sin (e+f x))^4} \, dx =\text {Too large to display} \] Input:

int((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^4,x)
 

Output:

(a**2*(18*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2 
))*sin(e + f*x)**3*c**3*d**4 - 12*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2) 
*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)**3*c**2*d**5 + 54*sqrt(c**2 - d**2 
)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)**2*c**4*d* 
*3 - 36*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2)) 
*sin(e + f*x)**2*c**3*d**4 + 54*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c 
 + d)/sqrt(c**2 - d**2))*sin(e + f*x)*c**5*d**2 - 36*sqrt(c**2 - d**2)*ata 
n((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*c**4*d**3 + 18* 
sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*c**6*d 
- 12*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*c* 
*5*d**2 - cos(e + f*x)*sin(e + f*x)**2*c**6*d**2 - 6*cos(e + f*x)*sin(e + 
f*x)**2*c**5*d**3 + 11*cos(e + f*x)*sin(e + f*x)**2*c**4*d**4 + 6*cos(e + 
f*x)*sin(e + f*x)**2*c**3*d**5 - 10*cos(e + f*x)*sin(e + f*x)**2*c**2*d**6 
 - 3*cos(e + f*x)*sin(e + f*x)*c**7*d - 18*cos(e + f*x)*sin(e + f*x)*c**6* 
d**2 + 24*cos(e + f*x)*sin(e + f*x)*c**5*d**3 + 24*cos(e + f*x)*sin(e + f* 
x)*c**4*d**4 - 21*cos(e + f*x)*sin(e + f*x)*c**3*d**5 - 6*cos(e + f*x)*sin 
(e + f*x)*c**2*d**6 - 12*cos(e + f*x)*c**7*d + 7*cos(e + f*x)*c**6*d**2 + 
18*cos(e + f*x)*c**5*d**3 - 5*cos(e + f*x)*c**4*d**4 - 6*cos(e + f*x)*c**3 
*d**5 - 2*cos(e + f*x)*c**2*d**6 - sin(e + f*x)**3*c**5*d**3 - 6*sin(e + f 
*x)**3*c**4*d**4 + 8*sin(e + f*x)**3*c**3*d**5 + 8*sin(e + f*x)**3*c**2...