\(\int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx\) [716]

Optimal result

Integrand size = 23, antiderivative size = 97 \[ \int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx=\frac {2 (a c-b d) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} f}+\frac {(b c-a d) \cos (e+f x)}{\left (a^2-b^2\right ) f (a+b \sin (e+f x))} \] Output:

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

Mathematica [A] (verified)

Time = 0.45 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.99 \[ \int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx=\frac {\frac {2 (a c-b d) \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {(b c-a d) \cos (e+f x)}{(a-b) (a+b) (a+b \sin (e+f x))}}{f} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3233

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3139

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

Input:

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

Output:

(2*(a*c - b*d)*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)*Cos[e + f*x])/((a^2 - b^2)*f*(a + b*Si 
n[e + f*x]))
 

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]
 

Maple [A] (verified)

Time = 1.02 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.48

method	result	size
derivativedivides	\(\frac {\frac {-\frac {2 b \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) a}-\frac {2 \left (a d -b c \right )}{a^{2}-b^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {2 \left (a c -b d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{f}\)	\(144\)
default	\(\frac {\frac {-\frac {2 b \left (a d -b c \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\left (a^{2}-b^{2}\right ) a}-\frac {2 \left (a d -b c \right )}{a^{2}-b^{2}}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} a +2 b \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+a}+\frac {2 \left (a c -b d \right ) \arctan \left (\frac {2 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (a^{2}-b^{2}\right )^{\frac {3}{2}}}}{f}\)	\(144\)
risch	\(\frac {2 i \left (a d -b c \right ) \left (i b +a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{b \left (a^{2}-b^{2}\right ) f \left (-i b \,{\mathrm e}^{2 i \left (f x +e \right )}+i b +2 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a c}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a -a^{2}+b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b d}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) a c}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i \sqrt {-a^{2}+b^{2}}\, a +a^{2}-b^{2}}{\sqrt {-a^{2}+b^{2}}\, b}\right ) b d}{\sqrt {-a^{2}+b^{2}}\, \left (a +b \right ) \left (a -b \right ) f}\)	\(396\)

Input:

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

Output:

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

Time = 0.10 (sec) , antiderivative size = 399, normalized size of antiderivative = 4.11 \[ \int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx=\left [-\frac {{\left (a^{2} c - a b d + {\left (a b c - b^{2} d\right )} \sin \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (f x + e\right ) \sin \left (f x + e\right ) + b \cos \left (f x + e\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left ({\left (a^{2} b - b^{3}\right )} c - {\left (a^{3} - a b^{2}\right )} d\right )} \cos \left (f x + e\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f\right )}}, -\frac {{\left (a^{2} c - a b d + {\left (a b c - b^{2} d\right )} \sin \left (f x + e\right )\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (f x + e\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (f x + e\right )}\right ) - {\left ({\left (a^{2} b - b^{3}\right )} c - {\left (a^{3} - a b^{2}\right )} d\right )} \cos \left (f x + e\right )}{{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} f \sin \left (f x + e\right ) + {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} f}\right ] \] Input:

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Time = 0.37 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.56 \[ \int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx=\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} {\left (a c - b d\right )}}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} + \frac {b^{2} c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a b d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a b c - a^{2} d}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + a\right )}}\right )}}{f} \] Input:

integrate((c+d*sin(f*x+e))/(a+b*sin(f*x+e))^2,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)))*(a*c - b*d)/(a^2 - b^2)^(3/2) + (b^2*c*tan(1/2*f 
*x + 1/2*e) - a*b*d*tan(1/2*f*x + 1/2*e) + a*b*c - a^2*d)/((a^3 - a*b^2)*( 
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 = 17.77 (sec) , antiderivative size = 215, normalized size of antiderivative = 2.22 \[ \int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx=\frac {2\,\mathrm {atan}\left (\frac {\left (\frac {2\,\left (a^2\,b-b^3\right )\,\left (a\,c-b\,d\right )}{{\left (a+b\right )}^{3/2}\,\left (a^2-b^2\right )\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,c-b\,d\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )\,\left (a^2-b^2\right )}{2\,\left (a\,c-b\,d\right )}\right )\,\left (a\,c-b\,d\right )}{f\,{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}-\frac {\frac {2\,\left (a\,d-b\,c\right )}{a^2-b^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (a\,d-b\,c\right )}{a\,\left (a^2-b^2\right )}}{f\,\left (a\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+a\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.98 \[ \int \frac {c+d \sin (e+f x)}{(a+b \sin (e+f x))^2} \, dx=\frac {2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (f x +e \right ) a b c -2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (f x +e \right ) b^{2} d +2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2} c -2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a b d -\cos \left (f x +e \right ) a^{3} d +\cos \left (f x +e \right ) a^{2} b c +\cos \left (f x +e \right ) a \,b^{2} d -\cos \left (f x +e \right ) b^{3} c}{f \left (\sin \left (f x +e \right ) a^{4} b -2 \sin \left (f x +e \right ) a^{2} b^{3}+\sin \left (f x +e \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}\right )} \] Input:

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

Output:

(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*c - 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a + b)/sqrt(a* 
*2 - b**2))*sin(e + f*x)*b**2*d + 2*sqrt(a**2 - b**2)*atan((tan((e + f*x)/ 
2)*a + b)/sqrt(a**2 - b**2))*a**2*c - 2*sqrt(a**2 - b**2)*atan((tan((e + f 
*x)/2)*a + b)/sqrt(a**2 - b**2))*a*b*d - cos(e + f*x)*a**3*d + cos(e + f*x 
)*a**2*b*c + cos(e + f*x)*a*b**2*d - cos(e + f*x)*b**3*c)/(f*(sin(e + f*x) 
*a**4*b - 2*sin(e + f*x)*a**2*b**3 + sin(e + f*x)*b**5 + a**5 - 2*a**3*b** 
2 + a*b**4))