Integrand size = 11, antiderivative size = 66 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \] Output:
-2*b*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/(a^2-b^2)^(3/2)-a*cos(x)/(a^ 2-b^2)/(a+b*sin(x))
Time = 0.16 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{(a-b) (a+b) (a+b \sin (x))} \] Input:
Integrate[Sin[x]/(a + b*Sin[x])^2,x]
Output:
(-2*b*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - (a*Cos [x])/((a - b)*(a + b)*(a + b*Sin[x]))
Time = 0.33 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.636, Rules used = {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.
\(\displaystyle \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (x)}{(a+b \sin (x))^2}dx\) |
\(\Big \downarrow \) 3233 |
\(\displaystyle -\frac {\int \frac {b}{a+b \sin (x)}dx}{a^2-b^2}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {b \int \frac {1}{a+b \sin (x)}dx}{a^2-b^2}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {b \int \frac {1}{a+b \sin (x)}dx}{a^2-b^2}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle -\frac {2 b \int \frac {1}{a \tan ^2\left (\frac {x}{2}\right )+2 b \tan \left (\frac {x}{2}\right )+a}d\tan \left (\frac {x}{2}\right )}{a^2-b^2}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {4 b \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle -\frac {2 b \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}\) |
Input:
Int[Sin[x]/(a + b*Sin[x])^2,x]
Output:
(-2*b*ArcTan[(2*b + 2*a*Tan[x/2])/(2*Sqrt[a^2 - b^2])])/(a^2 - b^2)^(3/2) - (a*Cos[x])/((a^2 - b^2)*(a + b*Sin[x]))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
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]
Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.50
method | result | size |
default | \(\frac {-8 b \tan \left (\frac {x}{2}\right )-8 a}{\left (4 a^{2}-4 b^{2}\right ) \left (a \tan \left (\frac {x}{2}\right )^{2}+2 b \tan \left (\frac {x}{2}\right )+a \right )}-\frac {8 b \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (4 a^{2}-4 b^{2}\right ) \sqrt {a^{2}-b^{2}}}\) | \(99\) |
risch | \(\frac {2 i a \left (-i a \,{\mathrm e}^{i x}+b \right )}{b \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )}-\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}+\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{b \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}\) | \(199\) |
Input:
int(sin(x)/(a+b*sin(x))^2,x,method=_RETURNVERBOSE)
Output:
4*(-2*b*tan(1/2*x)-2*a)/(4*a^2-4*b^2)/(a*tan(1/2*x)^2+2*b*tan(1/2*x)+a)-8* b/(4*a^2-4*b^2)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^ (1/2))
Time = 0.10 (sec) , antiderivative size = 266, normalized size of antiderivative = 4.03 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=\left [\frac {{\left (b^{2} \sin \left (x\right ) + a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )}}, \frac {{\left (b^{2} \sin \left (x\right ) + a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )}\right ] \] Input:
integrate(sin(x)/(a+b*sin(x))^2,x, algorithm="fricas")
Output:
[1/2*((b^2*sin(x) + a*b)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(x)^2 - 2* a*b*sin(x) - a^2 - b^2 + 2*(a*cos(x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/ (b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) - 2*(a^3 - a*b^2)*cos(x))/(a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x)), ((b^2*sin(x) + a* b)*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) - (a^3 - a*b^2)*cos(x))/(a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin (x))]
Timed out. \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=\text {Timed out} \] Input:
integrate(sin(x)/(a+b*sin(x))**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(sin(x)/(a+b*sin(x))^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
Time = 0.13 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.36 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, x\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )} {\left (a^{2} - b^{2}\right )}} \] Input:
integrate(sin(x)/(a+b*sin(x))^2,x, algorithm="giac")
Output:
-2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*b/(a^2 - b^2)^(3/2) - 2*(b*tan(1/2*x) + a)/((a*tan(1/2*x)^2 + 2*b* tan(1/2*x) + a)*(a^2 - b^2))
Time = 16.44 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.86 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {\frac {2\,a}{a^2-b^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2-b^2}}{a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}-\frac {2\,b\,\mathrm {atan}\left (\frac {\left (a^2-b^2\right )\,\left (\frac {2\,b^2}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )}{2\,b}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \] Input:
int(sin(x)/(a + b*sin(x))^2,x)
Output:
- ((2*a)/(a^2 - b^2) + (2*b*tan(x/2))/(a^2 - b^2))/(a + 2*b*tan(x/2) + a*t an(x/2)^2) - (2*b*atan(((a^2 - b^2)*((2*b^2)/((a + b)^(3/2)*(a - b)^(3/2)) + (2*a*b*tan(x/2))/((a + b)^(3/2)*(a - b)^(3/2))))/(2*b)))/((a + b)^(3/2) *(a - b)^(3/2))
Time = 0.18 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.02 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=\frac {-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) \sin \left (x \right ) b^{2}-2 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {x}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a b -\cos \left (x \right ) a^{3}+\cos \left (x \right ) a \,b^{2}}{\sin \left (x \right ) a^{4} b -2 \sin \left (x \right ) a^{2} b^{3}+\sin \left (x \right ) b^{5}+a^{5}-2 a^{3} b^{2}+a \,b^{4}} \] Input:
int(sin(x)/(a+b*sin(x))^2,x)
Output:
( - 2*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*sin(x)*b* *2 - 2*sqrt(a**2 - b**2)*atan((tan(x/2)*a + b)/sqrt(a**2 - b**2))*a*b - co s(x)*a**3 + cos(x)*a*b**2)/(sin(x)*a**4*b - 2*sin(x)*a**2*b**3 + sin(x)*b* *5 + a**5 - 2*a**3*b**2 + a*b**4)