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

Optimal. Leaf size=63 \[ \frac{(2 A-B) \tan (e+f x)}{3 a c^2 f}+\frac{(A+B) \sec (e+f x)}{3 a f \left (c^2-c^2 \sin (e+f x)\right )} \]

[Out]

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

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Rubi [A]  time = 0.202196, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2967, 2859, 3767, 8} \[ \frac{(2 A-B) \tan (e+f x)}{3 a c^2 f}+\frac{(A+B) \sec (e+f x)}{3 a f \left (c^2-c^2 \sin (e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 2859

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(2*m +
p + 1)), x] + Dist[(a*d*m + b*c*(m + p + 1))/(a*b*(2*m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^
(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && (LtQ[m, -1] || ILtQ[Simplify[
m + p], 0]) && NeQ[2*m + p + 1, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.569136, size = 108, normalized size = 1.71 \[ \frac{\cos (e+f x) (-2 (A+B) \cos (e+f x)+(4 A-2 B) \cos (2 (e+f x))+8 A \sin (e+f x)+A \sin (2 (e+f x))-4 B \sin (e+f x)+B \sin (2 (e+f x))+6 B)}{12 a c^2 f (\sin (e+f x)-1)^2 (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(6*B - 2*(A + B)*Cos[e + f*x] + (4*A - 2*B)*Cos[2*(e + f*x)] + 8*A*Sin[e + f*x] - 4*B*Sin[e + f*
x] + A*Sin[2*(e + f*x)] + B*Sin[2*(e + f*x)]))/(12*a*c^2*f*(-1 + Sin[e + f*x])^2*(1 + Sin[e + f*x]))

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Maple [A]  time = 0.076, size = 93, normalized size = 1.5 \begin{align*} 2\,{\frac{1}{af{c}^{2}} \left ( -1/3\,{\frac{A+B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-1/2\,{\frac{A+B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{3/4\,A+B/4}{\tan \left ( 1/2\,fx+e/2 \right ) -1}}-{\frac{A/4-B/4}{\tan \left ( 1/2\,fx+e/2 \right ) +1}} \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))/(c-c*sin(f*x+e))^2,x)

[Out]

2/f/a/c^2*(-1/3*(A+B)/(tan(1/2*f*x+1/2*e)-1)^3-1/2*(A+B)/(tan(1/2*f*x+1/2*e)-1)^2-(3/4*A+1/4*B)/(tan(1/2*f*x+1
/2*e)-1)-(1/4*A-1/4*B)/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 0.99171, size = 359, normalized size = 5.7 \begin{align*} -\frac{2 \,{\left (\frac{B{\left (\frac{2 \, \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}} - 1\right )}}{a c^{2} - \frac{2 \, a c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \, a c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{a c^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}} - \frac{A{\left (\frac{\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}} + \frac{3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a c^{2} - \frac{2 \, a c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \, a c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{a c^{2} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}\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))/(c-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/3*(B*(2*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(a*c^2 - 2*a*c^2*sin(f
*x + e)/(cos(f*x + e) + 1) + 2*a*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - a*c^2*sin(f*x + e)^4/(cos(f*x + e)
+ 1)^4) - A*(sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 + 1)/(a*c^2 - 2*a*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 2*a*c^2*sin(f*x + e)^3/(cos(f*x + e) +
1)^3 - a*c^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4))/f

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Fricas [A]  time = 1.40067, size = 171, normalized size = 2.71 \begin{align*} -\frac{{\left (2 \, A - B\right )} \cos \left (f x + e\right )^{2} +{\left (2 \, A - B\right )} \sin \left (f x + e\right ) - A + 2 \, B}{3 \,{\left (a c^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - a c^{2} f \cos \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))/(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 16.2408, size = 578, normalized size = 9.17 \begin{align*} \text{result too large to display} \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))/(c-c*sin(f*x+e))**2,x)

[Out]

Piecewise((-6*A*tan(e/2 + f*x/2)**3/(3*a*c**2*f*tan(e/2 + f*x/2)**4 - 6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**
2*f*tan(e/2 + f*x/2) - 3*a*c**2*f) + 6*A*tan(e/2 + f*x/2)**2/(3*a*c**2*f*tan(e/2 + f*x/2)**4 - 6*a*c**2*f*tan(
e/2 + f*x/2)**3 + 6*a*c**2*f*tan(e/2 + f*x/2) - 3*a*c**2*f) - 2*A*tan(e/2 + f*x/2)/(3*a*c**2*f*tan(e/2 + f*x/2
)**4 - 6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**2*f*tan(e/2 + f*x/2) - 3*a*c**2*f) - 2*A/(3*a*c**2*f*tan(e/2 +
f*x/2)**4 - 6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**2*f*tan(e/2 + f*x/2) - 3*a*c**2*f) - 6*B*tan(e/2 + f*x/2)*
*2/(3*a*c**2*f*tan(e/2 + f*x/2)**4 - 6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**2*f*tan(e/2 + f*x/2) - 3*a*c**2*f
) + 4*B*tan(e/2 + f*x/2)/(3*a*c**2*f*tan(e/2 + f*x/2)**4 - 6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**2*f*tan(e/2
+ f*x/2) - 3*a*c**2*f) - 2*B/(3*a*c**2*f*tan(e/2 + f*x/2)**4 - 6*a*c**2*f*tan(e/2 + f*x/2)**3 + 6*a*c**2*f*ta
n(e/2 + f*x/2) - 3*a*c**2*f), Ne(f, 0)), (x*(A + B*sin(e))/((a*sin(e) + a)*(-c*sin(e) + c)**2), True))

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Giac [A]  time = 1.19162, size = 138, normalized size = 2.19 \begin{align*} -\frac{\frac{3 \,{\left (A - B\right )}}{a c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}} + \frac{9 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, A \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7 \, A + B}{a c^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{3}}}{6 \, 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))/(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/6*(3*(A - B)/(a*c^2*(tan(1/2*f*x + 1/2*e) + 1)) + (9*A*tan(1/2*f*x + 1/2*e)^2 + 3*B*tan(1/2*f*x + 1/2*e)^2
- 12*A*tan(1/2*f*x + 1/2*e) + 7*A + B)/(a*c^2*(tan(1/2*f*x + 1/2*e) - 1)^3))/f