3.189 \(\int \frac{1}{(a+b \sin (x))^2} \, dx\)

Optimal. Leaf size=65 \[ \frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]

[Out]

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

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Rubi [A]  time = 0.0508134, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 8, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.625, Rules used = {2664, 12, 2660, 618, 204} \[ \frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b \cos (x)}{\left (a^2-b^2\right ) (a+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x])^(-2),x]

[Out]

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

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

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

Rule 2660

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 618

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.0959064, size = 66, normalized size = 1.02 \[ \frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b \cos (x)}{(a-b) (a+b) (a+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x])^(-2),x]

[Out]

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

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Maple [A]  time = 0.038, size = 98, normalized size = 1.5 \begin{align*} 2\,{\frac{1}{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a} \left ({\frac{{b}^{2}\tan \left ( x/2 \right ) }{a \left ({a}^{2}-{b}^{2} \right ) }}+{\frac{b}{{a}^{2}-{b}^{2}}} \right ) }+2\,{\frac{a}{ \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(x))^2,x)

[Out]

2*(b^2/a/(a^2-b^2)*tan(1/2*x)+1/(a^2-b^2)*b)/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)+2*a/(a^2-b^2)^(3/2)*arctan(1/2*
(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.81296, size = 614, normalized size = 9.45 \begin{align*} \left [\frac{{\left (a b \sin \left (x\right ) + a^{2}\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^{2} b - b^{3}\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 (a b \sin \left (x\right ) + a^{2}\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^{2} b - b^{3}\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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a*b*sin(x) + a^2)*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^2*b - b^3)*cos(x))/(a
^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x)), -((a*b*sin(x) + a^2)*sqrt(a^2 - b^2)*arctan(-(a*si
n(x) + b)/(sqrt(a^2 - b^2)*cos(x))) - (a^2*b - b^3)*cos(x))/(a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^
5)*sin(x))]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(x))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.66169, size = 128, normalized size = 1.97 \begin{align*} \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 )} a}{{\left (a^{2} - b^{2}\right )}^{\frac{3}{2}}} + \frac{2 \,{\left (b^{2} \tan \left (\frac{1}{2} \, x\right ) + a b\right )}}{{\left (a^{3} - a b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(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)))*a/(a^2 - b^2)^(3/2) + 2*(b^2*
tan(1/2*x) + a*b)/((a^3 - a*b^2)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a))