\(\int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx\) [188]

Optimal result

Integrand size = 11, antiderivative size = 66 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]

[Out]

-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))

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {2833, 12, 2739, 632, 210} \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 b \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]

[In]

Int[Sin[x]/(a + b*Sin[x])^2,x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 210

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 632

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[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 2833

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] + Dist[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]

Rubi steps \begin{align*} \text {integral}& = -\frac {a \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))}+\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))}-\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))}-\frac {(2 b) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\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))}+\frac {(4 b) \text {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {x}{2}\right )\right )}{a^2-b^2} \\ & = -\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))} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 b \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac {a \cos (x)}{(a-b) (a+b) (a+b \sin (x))} \]

[In]

Integrate[Sin[x]/(a + b*Sin[x])^2,x]

[Out]

(-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])
)

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.50

method	result	size
default	\(\frac {-8 b \tan \left (\frac {x}{2}\right )-8 a}{\left (4 a^{2}-4 b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+2 b \tan \left (\frac {x}{2}\right )+a \right )}-\frac {8 b \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{\left (4 a^{2}-4 b^{2}\right ) \sqrt {a^{2}-b^{2}}}\)	\(99\)
risch	\(\frac {2 i a \left (-i a \,{\mathrm e}^{i x}+b \right )}{b \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i x}-b +2 i a \,{\mathrm e}^{i x}\right )}-\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a +a^{2}-b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}+\frac {i b \ln \left ({\mathrm e}^{i x}+\frac {i \left (\sqrt {a^{2}-b^{2}}\, a -a^{2}+b^{2}\right )}{\sqrt {a^{2}-b^{2}}\, b}\right )}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right )}\)	\(199\)

[In]

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

[Out]

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)*arct
an(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 266, normalized size of antiderivative = 4.03 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=\left [\frac {{\left (b^{2} \sin \left (x\right ) + a b\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{2 \, {\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )}}, \frac {{\left (b^{2} \sin \left (x\right ) + a b\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (x\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (x\right )}\right ) - {\left (a^{3} - a b^{2}\right )} \cos \left (x\right )}{a^{5} - 2 \, a^{3} b^{2} + a b^{4} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )}\right ] \]

[In]

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

[Out]

[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))]

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=\text {Timed out} \]

[In]

integrate(sin(x)/(a+b*sin(x))**2,x)

[Out]

Timed out

Maxima [F(-2)]

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

[In]

integrate(sin(x)/(a+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.36 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b}{{\left (a^{2} - b^{2}\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, x\right ) + a\right )}}{{\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )} {\left (a^{2} - b^{2}\right )}} \]

[In]

integrate(sin(x)/(a+b*sin(x))^2,x, algorithm="giac")

[Out]

-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*t
an(1/2*x) + a)/((a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)*(a^2 - b^2))

Mupad [B] (verification not implemented)

Time = 6.44 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.86 \[ \int \frac {\sin (x)}{(a+b \sin (x))^2} \, dx=-\frac {\frac {2\,a}{a^2-b^2}+\frac {2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^2-b^2}}{a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )+a}-\frac {2\,b\,\mathrm {atan}\left (\frac {\left (a^2-b^2\right )\,\left (\frac {2\,b^2}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}+\frac {2\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}}\right )}{2\,b}\right )}{{\left (a+b\right )}^{3/2}\,{\left (a-b\right )}^{3/2}} \]

[In]

int(sin(x)/(a + b*sin(x))^2,x)

[Out]

- ((2*a)/(a^2 - b^2) + (2*b*tan(x/2))/(a^2 - b^2))/(a + 2*b*tan(x/2) + a*tan(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))