3.275 \(\int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=65 \[ -\frac{(A+2 B) \cos (e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )}-\frac{(A-B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

[Out]

-((A - B)*Cos[e + f*x])/(3*f*(a + a*Sin[e + f*x])^2) - ((A + 2*B)*Cos[e + f*x])/(3*f*(a^2 + a^2*Sin[e + f*x]))

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Rubi [A]  time = 0.0517211, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {2750, 2648} \[ -\frac{(A+2 B) \cos (e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )}-\frac{(A-B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x])^2,x]

[Out]

-((A - B)*Cos[e + f*x])/(3*f*(a + a*Sin[e + f*x])^2) - ((A + 2*B)*Cos[e + f*x])/(3*f*(a^2 + a^2*Sin[e + f*x]))

Rule 2750

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)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2648

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

Rubi steps

\begin{align*} \int \frac{A+B \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx &=-\frac{(A-B) \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}+\frac{(A+2 B) \int \frac{1}{a+a \sin (e+f x)} \, dx}{3 a}\\ &=-\frac{(A-B) \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac{(A+2 B) \cos (e+f x)}{3 f \left (a^2+a^2 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.0523491, size = 43, normalized size = 0.66 \[ -\frac{\cos (e+f x) ((A+2 B) \sin (e+f x)+2 A+B)}{3 a^2 f (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/(a + a*Sin[e + f*x])^2,x]

[Out]

-(Cos[e + f*x]*(2*A + B + (A + 2*B)*Sin[e + f*x]))/(3*a^2*f*(1 + Sin[e + f*x])^2)

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Maple [A]  time = 0.056, size = 70, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{{a}^{2}f} \left ( -1/2\,{\frac{-2\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-{\frac{A}{\tan \left ( 1/2\,fx+e/2 \right ) +1}}-1/3\,{\frac{2\,A-2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2,x)

[Out]

2/f/a^2*(-1/2*(-2*A+2*B)/(tan(1/2*f*x+1/2*e)+1)^2-A/(tan(1/2*f*x+1/2*e)+1)-1/3*(2*A-2*B)/(tan(1/2*f*x+1/2*e)+1
)^3)

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Maxima [B]  time = 0.977802, size = 289, normalized size = 4.45 \begin{align*} -\frac{2 \,{\left (\frac{A{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{B{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}}{a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/3*(A*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*sin(f*x +
e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)
+ B*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^
2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3))/f

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Fricas [A]  time = 1.8238, size = 288, normalized size = 4.43 \begin{align*} \frac{{\left (A + 2 \, B\right )} \cos \left (f x + e\right )^{2} +{\left (2 \, A + B\right )} \cos \left (f x + e\right ) +{\left ({\left (A + 2 \, B\right )} \cos \left (f x + e\right ) - A + B\right )} \sin \left (f x + e\right ) + A - B}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

1/3*((A + 2*B)*cos(f*x + e)^2 + (2*A + B)*cos(f*x + e) + ((A + 2*B)*cos(f*x + e) - A + B)*sin(f*x + e) + A - B
)/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))

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Sympy [A]  time = 4.62994, size = 309, normalized size = 4.75 \begin{align*} \begin{cases} \frac{2 A \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} - \frac{2 A}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} + \frac{2 B \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} + \frac{6 B \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{3 a^{2} f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 9 a^{2} f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + 3 a^{2} f} & \text{for}\: f \neq 0 \\\frac{x \left (A + B \sin{\left (e \right )}\right )}{\left (a \sin{\left (e \right )} + a\right )^{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**2,x)

[Out]

Piecewise((2*A*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan
(e/2 + f*x/2) + 3*a**2*f) - 2*A/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/
2 + f*x/2) + 3*a**2*f) + 2*B*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan(e/2 + f*x/2)**2
+ 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 6*B*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**3 + 9*a**2*f*tan
(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f), Ne(f, 0)), (x*(A + B*sin(e))/(a*sin(e) + a)**2, True
))

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Giac [A]  time = 1.23517, size = 92, normalized size = 1.42 \begin{align*} -\frac{2 \,{\left (3 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, A + B\right )}}{3 \, a^{2} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-2/3*(3*A*tan(1/2*f*x + 1/2*e)^2 + 3*A*tan(1/2*f*x + 1/2*e) + 3*B*tan(1/2*f*x + 1/2*e) + 2*A + B)/(a^2*f*(tan(
1/2*f*x + 1/2*e) + 1)^3)