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

Optimal. Leaf size=12 \[ -\frac{\cos (x)}{a \sin (x)+b} \]

[Out]

-(Cos[x]/(b + a*Sin[x]))

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Rubi [A]  time = 0.0288894, antiderivative size = 12, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2754, 8} \[ -\frac{\cos (x)}{a \sin (x)+b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x])/(b + a*Sin[x])^2,x]

[Out]

-(Cos[x]/(b + a*Sin[x]))

Rule 2754

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]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{a+b \sin (x)}{(b+a \sin (x))^2} \, dx &=-\frac{\cos (x)}{b+a \sin (x)}+\frac{\int 0 \, dx}{a^2-b^2}\\ &=-\frac{\cos (x)}{b+a \sin (x)}\\ \end{align*}

Mathematica [A]  time = 0.0300018, size = 12, normalized size = 1. \[ -\frac{\cos (x)}{a \sin (x)+b} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x])/(b + a*Sin[x])^2,x]

[Out]

-(Cos[x]/(b + a*Sin[x]))

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Maple [B]  time = 0.04, size = 34, normalized size = 2.8 \begin{align*} 2\,{\frac{1}{ \left ( \tan \left ( x/2 \right ) \right ) ^{2}b+2\,a\tan \left ( x/2 \right ) +b} \left ( -{\frac{a\tan \left ( x/2 \right ) }{b}}-1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.36419, size = 32, normalized size = 2.67 \begin{align*} -\frac{\cos \left (x\right )}{a \sin \left (x\right ) + b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-cos(x)/(a*sin(x) + b)

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

[Out]

Timed out

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Giac [B]  time = 1.25038, size = 43, normalized size = 3.58 \begin{align*} -\frac{2 \,{\left (a \tan \left (\frac{1}{2} \, x\right ) + b\right )}}{{\left (b \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, a \tan \left (\frac{1}{2} \, x\right ) + b\right )} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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