\(\int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx\) [710]

Optimal result

Integrand size = 25, antiderivative size = 117 \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} (b c-a d) f}-\frac {2 d \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{(b c-a d) \sqrt {c^2-d^2} f} \] Output:

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

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.89 \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\frac {2 d \arctan \left (\frac {d+c \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}}{b c f-a d f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.10, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3226, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3226

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

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

Defintions of rubi rules used

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 3226

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

Maple [A] (verified)

Time = 1.33 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a d -b c \right ) \sqrt {a^{2}-b^{2}}}+\frac {2 d \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (a d -b c \right ) \sqrt {c^{2}-d^{2}}}}{f}\)	\(114\)
default	\(\frac {-\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a d -b c \right ) \sqrt {a^{2}-b^{2}}}+\frac {2 d \arctan \left (\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 d}{2 \sqrt {c^{2}-d^{2}}}\right )}{\left (a d -b c \right ) \sqrt {c^{2}-d^{2}}}}{f}\)	\(114\)
risch	\(-\frac {b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right ) f}+\frac {b \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (a d -b c \right ) f}-\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c -c^{2}+d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (a d -b c \right ) f}+\frac {d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-c^{2}+d^{2}}\, c +c^{2}-d^{2}}{\sqrt {-c^{2}+d^{2}}\, d}\right )}{\sqrt {-c^{2}+d^{2}}\, \left (a d -b c \right ) f}\)	\(308\)

Input:

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

Output:

1/f*(-2*b/(a*d-b*c)/(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))+2/(a*d-b*c)*d/(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 [A] (verification not implemented)

Time = 0.95 (sec) , antiderivative size = 1057, normalized size of antiderivative = 9.03 \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*((a^2 - b^2)*sqrt(-c^2 + d^2)*d*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)) + (b*c^2 - b*d^2)*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*co 
s(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^2*b - b^3)*c^3 - (a^3 - a*b^2)*c^2*d - (a^2*b - b^3)*c*d^ 
2 + (a^3 - a*b^2)*d^3)*f), 1/2*(2*(a^2 - b^2)*sqrt(c^2 - d^2)*d*arctan(-(c 
*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) + (b*c^2 - b*d^2)*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^2*b - b^3)*c 
^3 - (a^3 - a*b^2)*c^2*d - (a^2*b - b^3)*c*d^2 + (a^3 - a*b^2)*d^3)*f), 1/ 
2*((a^2 - b^2)*sqrt(-c^2 + d^2)*d*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*(b*c^2 - b*d^2)*sqrt(a^2 - b^2)*arctan(-(a*sin(f*x + e) + b)/(sqrt( 
a^2 - b^2)*cos(f*x + e))))/(((a^2*b - b^3)*c^3 - (a^3 - a*b^2)*c^2*d - (a^ 
2*b - b^3)*c*d^2 + (a^3 - a*b^2)*d^3)*f), ((a^2 - b^2)*sqrt(c^2 - d^2)*d*a 
rctan(-(c*sin(f*x + e) + d)/(sqrt(c^2 - d^2)*cos(f*x + e))) - (b*c^2 - ...
                                                                                    
                                                                                    
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate(1/(a+b*sin(f*x+e))/(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 = 0.34 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.21 \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{\sqrt {a^{2} - b^{2}} {\left (b c - a d\right )}} - \frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (c\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + d}{\sqrt {c^{2} - d^{2}}}\right )\right )} d}{{\left (b c - a d\right )} \sqrt {c^{2} - d^{2}}}\right )}}{f} \] Input:

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

Output:

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)))*b/(sqrt(a^2 - b^2)*(b*c - a*d)) - (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/((b*c - a*d)*sqrt(c^2 - d^2)))/f
 

Mupad [B] (verification not implemented)

Time = 18.96 (sec) , antiderivative size = 3281, normalized size of antiderivative = 28.04 \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\text {Too large to display} \] Input:

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

Output:

(b*d^2*atan((b^4*c^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(3/2)*3i - a^6*d^2*tan 
(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i - a^3*b^3*d^2*(b^2 - a^2)^(1/2)*2i - 
b^6*c^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i - b^4*d^2*tan(e/2 + (f*x)/ 
2)*(b^2 - a^2)^(3/2)*4i + b^6*d^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*4i 
- b^4*c*d*(b^2 - a^2)^(3/2)*1i + b^6*c*d*(b^2 - a^2)^(1/2)*1i + a*b^3*c^2* 
(b^2 - a^2)^(3/2)*1i - a*b^3*d^2*(b^2 - a^2)^(3/2)*1i + a*b^5*d^2*(b^2 - a 
^2)^(1/2)*1i + a^5*b*d^2*(b^2 - a^2)^(1/2)*1i - a^2*b^2*c^2*tan(e/2 + (f*x 
)/2)*(b^2 - a^2)^(3/2)*2i + a^2*b^4*c^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/ 
2)*2i - a^4*b^2*c^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i + a^2*b^2*d^2* 
tan(e/2 + (f*x)/2)*(b^2 - a^2)^(3/2)*3i - a^2*b^4*d^2*tan(e/2 + (f*x)/2)*( 
b^2 - a^2)^(1/2)*9i + a^4*b^2*d^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*6i 
+ a^2*b^2*c*d*(b^2 - a^2)^(3/2)*1i - a^2*b^4*c*d*(b^2 - a^2)^(1/2)*2i + a^ 
4*b^2*c*d*(b^2 - a^2)^(1/2)*1i)/(a^7*d^2 - a*b^6*c^2 + 2*a^3*b^4*c^2 - a^5 
*b^2*c^2 + a^3*b^4*d^2 - 2*a^5*b^2*d^2 - 2*b^7*c^2*tan(e/2 + (f*x)/2) + 2* 
a^6*b*d^2*tan(e/2 + (f*x)/2) + 4*a^2*b^5*c^2*tan(e/2 + (f*x)/2) - 2*a^4*b^ 
3*c^2*tan(e/2 + (f*x)/2) + 2*a^2*b^5*d^2*tan(e/2 + (f*x)/2) - 4*a^4*b^3*d^ 
2*tan(e/2 + (f*x)/2)))*(b^2 - a^2)^(1/2)*2i)/(f*(a^3*d^3 - b^3*c^3 + a^2*b 
*c^3 - a*b^2*d^3 - a^3*c^2*d + b^3*c*d^2 + a*b^2*c^2*d - a^2*b*c*d^2)) - ( 
b*c^2*atan((b^4*c^2*tan(e/2 + (f*x)/2)*(b^2 - a^2)^(3/2)*3i - a^6*d^2*tan( 
e/2 + (f*x)/2)*(b^2 - a^2)^(1/2)*1i - a^3*b^3*d^2*(b^2 - a^2)^(1/2)*2i ...
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.11 \[ \int \frac {1}{(a+b \sin (e+f x)) (c+d \sin (e+f x))} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) b \,c^{2}+2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) b \,d^{2}+2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) a^{2} d -2 \sqrt {c^{2}-d^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c +d}{\sqrt {c^{2}-d^{2}}}\right ) b^{2} d}{f \left (a^{3} c^{2} d -a^{3} d^{3}-a^{2} b \,c^{3}+a^{2} b c \,d^{2}-a \,b^{2} c^{2} d +a \,b^{2} d^{3}+b^{3} c^{3}-b^{3} c \,d^{2}\right )} \] Input:

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

Output:

(2*( - sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2))* 
b*c**2 + sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a**2 - b**2) 
)*b*d**2 + sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d** 
2))*a**2*d - sqrt(c**2 - d**2)*atan((tan((e + f*x)/2)*c + d)/sqrt(c**2 - d 
**2))*b**2*d))/(f*(a**3*c**2*d - a**3*d**3 - a**2*b*c**3 + a**2*b*c*d**2 - 
 a*b**2*c**2*d + a*b**2*d**3 + b**3*c**3 - b**3*c*d**2))