\(\int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx\) [718]

Optimal result

Integrand size = 25, antiderivative size = 181 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {2 b \left (a b c-2 a^2 d+b^2 d\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} (b c-a d)^2 f}+\frac {2 d^2 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(b c-a d)^2 \sqrt {c^2-d^2} f}+\frac {b^2 \cos (e+f x)}{\left (a^2-b^2\right ) (b c-a d) f (a+b \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.94 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {\frac {2 b \left (a b c-2 a^2 d+b^2 d\right ) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} (b c-a d)^2}+\frac {2 d^2 \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(b c-a d)^2 \sqrt {c^2-d^2}}-\frac {b^2 \cos (e+f x)}{(a-b) (a+b) (-b c+a d) (a+b \sin (e+f x))}}{f} \] Input:

Integrate[1/((a + b*Sin[e + f*x])^2*(c + d*Sin[e + f*x])),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.81 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.24, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3042, 3281, 25, 3042, 3480, 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.

\(\displaystyle \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))}dx\)

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3480

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

Int[1/((a + b*Sin[e + f*x])^2*(c + d*Sin[e + f*x])),x]
 

Output:

((2*b*(a*b*c - 2*a^2*d + b^2*d)*ArcTan[(2*b + 2*a*Tan[(e + f*x)/2])/(2*Sqr 
t[a^2 - b^2])])/(Sqrt[a^2 - b^2]*(b*c - a*d)*f) + (2*(a^2 - b^2)*d^2*ArcTa 
n[(2*d + 2*c*Tan[(e + f*x)/2])/(2*Sqrt[c^2 - d^2])])/((b*c - a*d)*Sqrt[c^2 
 - d^2]*f))/((a^2 - b^2)*(b*c - a*d)) + (b^2*Cos[e + f*x])/((a^2 - b^2)*(b 
*c - a*d)*f*(a + b*Sin[e + f*x]))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))
 

rule 3480

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

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.30

method	result	size
derivativedivides	\(\frac {\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {c^{2}-d^{2}}}-\frac {2 b \left (\frac {\frac {b^{2} \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {b \left (a d -b c \right )}{a^{2}-b^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (2 a^{2} d -a b c -b^{2} d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a d -b c \right )^{2}}}{f}\)	\(236\)
default	\(\frac {\frac {2 d^{2} \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right ) \sqrt {c^{2}-d^{2}}}-\frac {2 b \left (\frac {\frac {b^{2} \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a \left (a^{2}-b^{2}\right )}+\frac {b \left (a d -b c \right )}{a^{2}-b^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {\left (2 a^{2} d -a b c -b^{2} d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}\right )}{\left (a d -b c \right )^{2}}}{f}\)	\(236\)
risch	\(-\frac {2 b \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{\left (a^{2}-b^{2}\right ) \left (a d -b c \right ) f \left (b \,{\mathrm e}^{2 i \left (f x +e \right )}-b +2 i a \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (a d -b c \right )^{2} f}+\frac {d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (a d -b c \right )^{2} f}-\frac {2 b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{2} d}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}+\frac {b^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a c}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}+\frac {b^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) d}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}+\frac {2 b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a^{2} d}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}-\frac {b^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a c}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}-\frac {b^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) d}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right )^{2} \left (a +b \right ) \left (a -b \right ) f}\)	\(786\)

Input:

int(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e)),x,method=_RETURNVERBOSE)
 

Output:

1/f*(2*d^2/(a^2*d^2-2*a*b*c*d+b^2*c^2)/(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))-2*b/(a*d-b*c)^2*((b^2*(a*d-b*c)/a/(a 
^2-b^2)*tan(1/2*f*x+1/2*e)+b*(a*d-b*c)/(a^2-b^2))/(tan(1/2*f*x+1/2*e)^2*a+ 
2*b*tan(1/2*f*x+1/2*e)+a)+(2*a^2*d-a*b*c-b^2*d)/(a^2-b^2)^(3/2)*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. 634 vs. \(2 (171) = 342\).

Time = 71.47 (sec) , antiderivative size = 2871, normalized size of antiderivative = 15.86 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

integrate(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Timed out} \] Input:

integrate(1/(a+b*sin(f*x+e))**2/(c+d*sin(f*x+e)),x)
 

Output:

Timed out
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e)),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 [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.64 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} d^{2}}{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \sqrt {c^{2} - d^{2}}} + \frac {{\left (a b^{2} c - 2 \, a^{2} b d + b^{3} d\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} c^{2} - b^{4} c^{2} - 2 \, a^{3} b c d + 2 \, a b^{3} c d + a^{4} d^{2} - a^{2} b^{2} d^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {b^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a b^{2}}{{\left (a^{3} b c - a b^{3} c - a^{4} d + a^{2} b^{2} d\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}}\right )}}{f} \] Input:

integrate(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e)),x, algorithm="giac")
 

Output:

2*((pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(c) + arctan((c*tan(1/2*f*x + 1/2* 
e) + d)/sqrt(c^2 - d^2)))*d^2/((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sqrt(c^2 - 
d^2)) + (a*b^2*c - 2*a^2*b*d + b^3*d)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sg 
n(a) + arctan((a*tan(1/2*f*x + 1/2*e) + b)/sqrt(a^2 - b^2)))/((a^2*b^2*c^2 
 - b^4*c^2 - 2*a^3*b*c*d + 2*a*b^3*c*d + a^4*d^2 - a^2*b^2*d^2)*sqrt(a^2 - 
 b^2)) + (b^3*tan(1/2*f*x + 1/2*e) + a*b^2)/((a^3*b*c - a*b^3*c - a^4*d + 
a^2*b^2*d)*(a*tan(1/2*f*x + 1/2*e)^2 + 2*b*tan(1/2*f*x + 1/2*e) + a)))/f
 

Mupad [B] (verification not implemented)

Time = 27.40 (sec) , antiderivative size = 24123, normalized size of antiderivative = 133.28 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

int(1/((a + b*sin(e + f*x))^2*(c + d*sin(e + f*x))),x)
                                                                                    
                                                                                    
 

Output:

(d^2*atan(((d^2*(d^2 - c^2)^(1/2)*((32*tan(e/2 + (f*x)/2)*(a^3*b^5*c^6 - a 
^8*c*d^5 - 4*a*b^7*c^2*d^4 + a*b^7*c^4*d^2 + 4*a^2*b^6*c*d^5 + 2*a^2*b^6*c 
^5*d - 13*a^4*b^4*c*d^5 - 5*a^4*b^4*c^5*d + 12*a^6*b^2*c*d^5 + a^7*b*c^2*d 
^4 - 5*a^2*b^6*c^3*d^3 + 17*a^3*b^5*c^2*d^4 - 8*a^3*b^5*c^4*d^2 + 14*a^4*b 
^4*c^3*d^3 - 20*a^5*b^3*c^2*d^4 + 8*a^5*b^3*c^4*d^2 - 4*a^6*b^2*c^3*d^3))/ 
(a^7*d^3 - b^7*c^3 + 2*a^2*b^5*c^3 - a^4*b^3*c^3 + a^3*b^4*d^3 - 2*a^5*b^2 
*d^3 - 3*a^2*b^5*c*d^2 - 6*a^3*b^4*c^2*d + 6*a^4*b^3*c*d^2 + 3*a^5*b^2*c^2 
*d + 3*a*b^6*c^2*d - 3*a^6*b*c*d^2) - (32*(a*b^7*c^3*d^3 - a^3*b^5*c*d^5 + 
 a^3*b^5*c^5*d + 2*a^5*b^3*c*d^5 + 2*a^2*b^6*c^4*d^2 - 6*a^3*b^5*c^3*d^3 + 
 2*a^4*b^4*c^2*d^4 - 5*a^4*b^4*c^4*d^2 + 8*a^5*b^3*c^3*d^3 - 3*a^6*b^2*c^2 
*d^4 - a^7*b*c*d^5))/(a^7*d^3 - b^7*c^3 + 2*a^2*b^5*c^3 - a^4*b^3*c^3 + a^ 
3*b^4*d^3 - 2*a^5*b^2*d^3 - 3*a^2*b^5*c*d^2 - 6*a^3*b^4*c^2*d + 6*a^4*b^3* 
c*d^2 + 3*a^5*b^2*c^2*d + 3*a*b^6*c^2*d - 3*a^6*b*c*d^2) + (d^2*(d^2 - c^2 
)^(1/2)*((32*(a^3*b^7*c^7 - a^5*b^5*c^7 + a^10*c^2*d^5 + a*b^9*c^5*d^2 + a 
^2*b^8*c^6*d - 6*a^4*b^6*c^6*d + a^5*b^5*c*d^6 + 5*a^6*b^4*c^6*d - 3*a^7*b 
^3*c*d^6 - 5*a^9*b*c^3*d^4 - 4*a^2*b^8*c^4*d^3 + 6*a^3*b^7*c^3*d^4 - 7*a^3 
*b^7*c^5*d^2 - 4*a^4*b^6*c^2*d^5 + 18*a^4*b^6*c^4*d^3 - 22*a^5*b^5*c^3*d^4 
 + 16*a^5*b^5*c^5*d^2 + 13*a^6*b^4*c^2*d^5 - 24*a^6*b^4*c^4*d^3 + 21*a^7*b 
^3*c^3*d^4 - 10*a^7*b^3*c^5*d^2 - 10*a^8*b^2*c^2*d^5 + 10*a^8*b^2*c^4*d^3 
+ 2*a^9*b*c*d^6))/(a^7*d^3 - b^7*c^3 + 2*a^2*b^5*c^3 - a^4*b^3*c^3 + a^...
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1480, normalized size of antiderivative = 8.18 \[ \int \frac {1}{(a+b \sin (e+f x))^2 (c+d \sin (e+f x))} \, dx =\text {Too large to display} \] Input:

int(1/(a+b*sin(f*x+e))^2/(c+d*sin(f*x+e)),x)
 

Output:

( - 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*s 
in(e + f*x)*a**2*b**2*c**2*d + 4*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**2*d**3 + 2*sqrt(a**2 - b**2 
)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*a*b**3*c** 
3 - 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*s 
in(e + f*x)*a*b**3*c*d**2 + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + 
 b)/sqrt(a**2 - b**2))*sin(e + f*x)*b**4*c**2*d - 2*sqrt(a**2 - b**2)*atan 
((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*sin(e + f*x)*b**4*d**3 - 4*sq 
rt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a**3*b*c* 
*2*d + 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2) 
)*a**3*b*d**3 + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a** 
2 - b**2))*a**2*b**2*c**3 - 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + 
 b)/sqrt(a**2 - b**2))*a**2*b**2*c*d**2 + 2*sqrt(a**2 - b**2)*atan((tan((e 
 + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a*b**3*c**2*d - 2*sqrt(a**2 - b**2)*a 
tan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))*a*b**3*d**3 + 2*sqrt(c**2 
- d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*a**4 
*b*d**2 - 4*sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d* 
*2))*sin(e + f*x)*a**2*b**3*d**2 + 2*sqrt(c**2 - d**2)*atan((tan((e + f*x) 
/2)*c + d)/sqrt(c**2 - d**2))*sin(e + f*x)*b**5*d**2 + 2*sqrt(c**2 - d**2) 
*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d**2))*a**5*d**2 - 4*sqrt(c*...