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

Optimal. Leaf size=77 \[ -\frac{2 c \tan ^{-1}\left (\frac{b \tan \left (\frac{a x}{2}\right )+c}{\sqrt{b^2-c^2}}\right )}{a \left (b^2-c^2\right )^{3/2}}-\frac{b \cos (a x)}{a \left (b^2-c^2\right ) (c \sin (a x)+b)} \]

[Out]

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

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Rubi [A]  time = 0.0977043, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2754, 12, 2660, 618, 204} \[ -\frac{2 c \tan ^{-1}\left (\frac{b \tan \left (\frac{a x}{2}\right )+c}{\sqrt{b^2-c^2}}\right )}{a \left (b^2-c^2\right )^{3/2}}-\frac{b \cos (a x)}{a \left (b^2-c^2\right ) (c \sin (a x)+b)} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a*x]/(b + c*Sin[a*x])^2,x]

[Out]

(-2*c*ArcTan[(c + b*Tan[(a*x)/2])/Sqrt[b^2 - c^2]])/(a*(b^2 - c^2)^(3/2)) - (b*Cos[a*x])/(a*(b^2 - c^2)*(b + c
*Sin[a*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 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{\sin (a x)}{(b+c \sin (a x))^2} \, dx &=-\frac{b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}+\frac{\int \frac{c}{b+c \sin (a x)} \, dx}{-b^2+c^2}\\ &=-\frac{b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}-\frac{c \int \frac{1}{b+c \sin (a x)} \, dx}{b^2-c^2}\\ &=-\frac{b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}-\frac{(2 c) \operatorname{Subst}\left (\int \frac{1}{b+2 c x+b x^2} \, dx,x,\tan \left (\frac{a x}{2}\right )\right )}{a \left (b^2-c^2\right )}\\ &=-\frac{b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}+\frac{(4 c) \operatorname{Subst}\left (\int \frac{1}{-4 \left (b^2-c^2\right )-x^2} \, dx,x,2 c+2 b \tan \left (\frac{a x}{2}\right )\right )}{a \left (b^2-c^2\right )}\\ &=-\frac{2 c \tan ^{-1}\left (\frac{c+b \tan \left (\frac{a x}{2}\right )}{\sqrt{b^2-c^2}}\right )}{a \left (b^2-c^2\right )^{3/2}}-\frac{b \cos (a x)}{a \left (b^2-c^2\right ) (b+c \sin (a x))}\\ \end{align*}

Mathematica [A]  time = 0.246516, size = 76, normalized size = 0.99 \[ -\frac{\frac{2 c \tan ^{-1}\left (\frac{b \tan \left (\frac{a x}{2}\right )+c}{\sqrt{b^2-c^2}}\right )}{\left (b^2-c^2\right )^{3/2}}+\frac{b \cos (a x)}{(b-c) (b+c) (c \sin (a x)+b)}}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 143, normalized size = 1.9 \begin{align*} -8\,{\frac{c\tan \left ( 1/2\,ax \right ) }{a \left ( 4\,{b}^{2}-4\,{c}^{2} \right ) \left ( b \left ( \tan \left ( 1/2\,ax \right ) \right ) ^{2}+2\,c\tan \left ( 1/2\,ax \right ) +b \right ) }}-8\,{\frac{b}{a \left ( 4\,{b}^{2}-4\,{c}^{2} \right ) \left ( b \left ( \tan \left ( 1/2\,ax \right ) \right ) ^{2}+2\,c\tan \left ( 1/2\,ax \right ) +b \right ) }}-8\,{\frac{c}{a \left ( 4\,{b}^{2}-4\,{c}^{2} \right ) \sqrt{{b}^{2}-{c}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( 1/2\,ax \right ) +2\,c}{\sqrt{{b}^{2}-{c}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/a/(4*b^2-4*c^2)/(b*tan(1/2*a*x)^2+2*c*tan(1/2*a*x)+b)*c*tan(1/2*a*x)-8/a/(4*b^2-4*c^2)/(b*tan(1/2*a*x)^2+2*
c*tan(1/2*a*x)+b)*b-8/a*c/(4*b^2-4*c^2)/(b^2-c^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*a*x)+2*c)/(b^2-c^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(sin(a*x)/(b+c*sin(a*x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.92501, size = 684, normalized size = 8.88 \begin{align*} \left [\frac{{\left (c^{2} \sin \left (a x\right ) + b c\right )} \sqrt{-b^{2} + c^{2}} \log \left (\frac{{\left (2 \, b^{2} - c^{2}\right )} \cos \left (a x\right )^{2} - 2 \, b c \sin \left (a x\right ) - b^{2} - c^{2} + 2 \,{\left (b \cos \left (a x\right ) \sin \left (a x\right ) + c \cos \left (a x\right )\right )} \sqrt{-b^{2} + c^{2}}}{c^{2} \cos \left (a x\right )^{2} - 2 \, b c \sin \left (a x\right ) - b^{2} - c^{2}}\right ) - 2 \,{\left (b^{3} - b c^{2}\right )} \cos \left (a x\right )}{2 \,{\left (a b^{5} - 2 \, a b^{3} c^{2} + a b c^{4} +{\left (a b^{4} c - 2 \, a b^{2} c^{3} + a c^{5}\right )} \sin \left (a x\right )\right )}}, \frac{{\left (c^{2} \sin \left (a x\right ) + b c\right )} \sqrt{b^{2} - c^{2}} \arctan \left (-\frac{b \sin \left (a x\right ) + c}{\sqrt{b^{2} - c^{2}} \cos \left (a x\right )}\right ) -{\left (b^{3} - b c^{2}\right )} \cos \left (a x\right )}{a b^{5} - 2 \, a b^{3} c^{2} + a b c^{4} +{\left (a b^{4} c - 2 \, a b^{2} c^{3} + a c^{5}\right )} \sin \left (a x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.10437, size = 132, normalized size = 1.71 \begin{align*} -\frac{2 \,{\left (\frac{{\left (\pi \left \lfloor \frac{a x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, a x\right ) + c}{\sqrt{b^{2} - c^{2}}}\right )\right )} c}{{\left (b^{2} - c^{2}\right )}^{\frac{3}{2}}} + \frac{c \tan \left (\frac{1}{2} \, a x\right ) + b}{{\left (b \tan \left (\frac{1}{2} \, a x\right )^{2} + 2 \, c \tan \left (\frac{1}{2} \, a x\right ) + b\right )}{\left (b^{2} - c^{2}\right )}}\right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*((pi*floor(1/2*a*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*a*x) + c)/sqrt(b^2 - c^2)))*c/(b^2 - c^2)^(3/2) + (
c*tan(1/2*a*x) + b)/((b*tan(1/2*a*x)^2 + 2*c*tan(1/2*a*x) + b)*(b^2 - c^2)))/a