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

Optimal. Leaf size=72 \[ \frac{c (A-7 B) \cos (e+f x)}{3 a^2 f (\sin (e+f x)+1)}-\frac{B c x}{a^2}-\frac{2 c (A-B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

[Out]

-((B*c*x)/a^2) + ((A - 7*B)*c*Cos[e + f*x])/(3*a^2*f*(1 + Sin[e + f*x])) - (2*(A - B)*c*Cos[e + f*x])/(3*f*(a
+ a*Sin[e + f*x])^2)

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Rubi [A]  time = 0.207294, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {2967, 2857, 2735, 2648} \[ \frac{c (A-7 B) \cos (e+f x)}{3 a^2 f (\sin (e+f x)+1)}-\frac{B c x}{a^2}-\frac{2 c (A-B) \cos (e+f x)}{3 f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*c*x)/a^2) + ((A - 7*B)*c*Cos[e + f*x])/(3*a^2*f*(1 + Sin[e + f*x])) - (2*(A - B)*c*Cos[e + f*x])/(3*f*(a
+ a*Sin[e + f*x])^2)

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 2857

Int[cos[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_
)]), x_Symbol] :> Simp[(2*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(2*m + 3)), x] + Dist[
1/(b^3*(2*m + 3)), Int[(a + b*Sin[e + f*x])^(m + 2)*(b*c + 2*a*d*(m + 1) - b*d*(2*m + 3)*Sin[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -3/2]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

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)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^2} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx\\ &=-\frac{2 (A-B) c \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac{c \int \frac{a A-4 a B+3 a B \sin (e+f x)}{a+a \sin (e+f x)} \, dx}{3 a^2}\\ &=-\frac{B c x}{a^2}-\frac{2 (A-B) c \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}-\frac{((A-7 B) c) \int \frac{1}{a+a \sin (e+f x)} \, dx}{3 a}\\ &=-\frac{B c x}{a^2}-\frac{2 (A-B) c \cos (e+f x)}{3 f (a+a \sin (e+f x))^2}+\frac{(A-7 B) c \cos (e+f x)}{3 f \left (a^2+a^2 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.559679, size = 156, normalized size = 2.17 \[ \frac{c \left (-6 (A-3 B) \cos \left (e+\frac{f x}{2}\right )+2 A \cos \left (e+\frac{3 f x}{2}\right )-9 B f x \sin \left (e+\frac{f x}{2}\right )-3 B f x \sin \left (e+\frac{3 f x}{2}\right )-14 B \cos \left (e+\frac{3 f x}{2}\right )+3 B f x \cos \left (2 e+\frac{3 f x}{2}\right )+24 B \sin \left (\frac{f x}{2}\right )-9 B f x \cos \left (\frac{f x}{2}\right )\right )}{6 a^2 f \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(-9*B*f*x*Cos[(f*x)/2] - 6*(A - 3*B)*Cos[e + (f*x)/2] + 2*A*Cos[e + (3*f*x)/2] - 14*B*Cos[e + (3*f*x)/2] +
3*B*f*x*Cos[2*e + (3*f*x)/2] + 24*B*Sin[(f*x)/2] - 9*B*f*x*Sin[e + (f*x)/2] - 3*B*f*x*Sin[e + (3*f*x)/2]))/(6*
a^2*f*(Cos[e/2] + Sin[e/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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Maple [B]  time = 0.104, size = 160, normalized size = 2.2 \begin{align*} -2\,{\frac{Bc\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{{a}^{2}f}}+4\,{\frac{Ac}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-4\,{\frac{Bc}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-2\,{\frac{Ac}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-2\,{\frac{Bc}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}-{\frac{8\,Ac}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+{\frac{8\,Bc}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*c/a^2*B*arctan(tan(1/2*f*x+1/2*e))+4/f*c/a^2/(tan(1/2*f*x+1/2*e)+1)^2*A-4/f*c/a^2/(tan(1/2*f*x+1/2*e)+1)^
2*B-2/f*c/a^2/(tan(1/2*f*x+1/2*e)+1)*A-2/f*c/a^2/(tan(1/2*f*x+1/2*e)+1)*B-8/3/f*c/a^2/(tan(1/2*f*x+1/2*e)+1)^3
*A+8/3/f*c/a^2/(tan(1/2*f*x+1/2*e)+1)^3*B

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Maxima [B]  time = 1.48853, size = 610, normalized size = 8.47 \begin{align*} -\frac{2 \,{\left (B c{\left (\frac{\frac{9 \, \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}} + 4}{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{3 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} + \frac{A c{\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{A c{\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}}} + \frac{B c{\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))*(c-c*sin(f*x+e))/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/3*(B*c*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(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) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + A*c*(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) - A*c*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*s
in(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*c*(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 [B]  time = 1.6239, size = 401, normalized size = 5.57 \begin{align*} \frac{6 \, B c f x -{\left (3 \, B c f x +{\left (A - 7 \, B\right )} c\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (A - B\right )} c +{\left (3 \, B c f x +{\left (A + 5 \, B\right )} c\right )} \cos \left (f x + e\right ) +{\left (6 \, B c f x - 2 \,{\left (A - B\right )} c +{\left (3 \, B c f x -{\left (A - 7 \, B\right )} c\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{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))*(c-c*sin(f*x+e))/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

1/3*(6*B*c*f*x - (3*B*c*f*x + (A - 7*B)*c)*cos(f*x + e)^2 + 2*(A - B)*c + (3*B*c*f*x + (A + 5*B)*c)*cos(f*x +
e) + (6*B*c*f*x - 2*(A - B)*c + (3*B*c*f*x - (A - 7*B)*c)*cos(f*x + e))*sin(f*x + e))/(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 = 16.2594, size = 711, normalized size = 9.88 \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))*(c-c*sin(f*x+e))/(a+a*sin(f*x+e))**2,x)

[Out]

Piecewise((2*A*c*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*t
an(e/2 + f*x/2) + 3*a**2*f) + 6*A*c*tan(e/2 + f*x/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) - 3*B*c*f*x*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) - 9*B*c*f*x*tan(e/2 + f*x/2)**2/(3*a**2*f*t
an(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) - 9*B*c*f*x*tan(e/2
+ f*x/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)
- 3*B*c*f*x/(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*c*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) - 18*B*c*tan(e/2 + f*x/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) - 8*B*c/(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))*(-c*sin(e) + c)/(a*sin(e) + a)**2, True))

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Giac [A]  time = 1.19561, size = 124, normalized size = 1.72 \begin{align*} -\frac{\frac{3 \,{\left (f x + e\right )} B c}{a^{2}} + \frac{2 \,{\left (3 \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 12 \, B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + A c + 5 \, B c\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(f*x + e)*B*c/a^2 + 2*(3*A*c*tan(1/2*f*x + 1/2*e)^2 + 3*B*c*tan(1/2*f*x + 1/2*e)^2 + 12*B*c*tan(1/2*f*
x + 1/2*e) + A*c + 5*B*c)/(a^2*(tan(1/2*f*x + 1/2*e) + 1)^3))/f