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

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

[Out]

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

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Rubi [A]  time = 0.224578, 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} \[ -\frac{a (A+7 B) \cos (e+f x)}{3 c^2 f (1-\sin (e+f x))}+\frac{2 a (A+B) \cos (e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac{a B x}{c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 0.607064, size = 160, normalized size = 2.22 \[ -\frac{a \left (-6 (A+3 B) \cos \left (e+\frac{f x}{2}\right )+2 A \cos \left (e+\frac{3 f x}{2}\right )+9 B f x \sin \left (e+\frac{f x}{2}\right )+3 B f x \sin \left (e+\frac{3 f x}{2}\right )+14 B \cos \left (e+\frac{3 f x}{2}\right )+3 B f x \cos \left (2 e+\frac{3 f x}{2}\right )+24 B \sin \left (\frac{f x}{2}\right )-9 B f x \cos \left (\frac{f x}{2}\right )\right )}{6 c^2 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*(-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
*c^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 = 0.102, size = 160, normalized size = 2.2 \begin{align*} -2\,{\frac{aA}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}+2\,{\frac{Ba}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-{\frac{8\,aA}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-{\frac{8\,Ba}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-4\,{\frac{aA}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-4\,{\frac{Ba}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}+2\,{\frac{Ba\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.48382, size = 616, normalized size = 8.56 \begin{align*} \frac{2 \,{\left (B a{\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}{c^{2} - \frac{3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{c^{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 )}{c^{2}}\right )} - \frac{A 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 )}}{c^{2} - \frac{3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{A a{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac{3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{c^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{B a{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}}{c^{2} - \frac{3 \, c^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, c^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{c^{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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

2/3*(B*a*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3*c^2*sin(f*x
+ e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) - A*a*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2
/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + A*a*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^2*si
n(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) +
1)^3) + B*a*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(
f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3))/f

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Fricas [B]  time = 1.42143, size = 402, normalized size = 5.58 \begin{align*} -\frac{6 \, B a f x -{\left (3 \, B a f x +{\left (A + 7 \, B\right )} a\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (A + B\right )} a +{\left (3 \, B a f x +{\left (A - 5 \, B\right )} a\right )} \cos \left (f x + e\right ) -{\left (6 \, B a f x - 2 \,{\left (A + B\right )} a +{\left (3 \, B a f x -{\left (A + 7 \, B\right )} a\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f +{\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-1/3*(6*B*a*f*x - (3*B*a*f*x + (A + 7*B)*a)*cos(f*x + e)^2 + 2*(A + B)*a + (3*B*a*f*x + (A - 5*B)*a)*cos(f*x +
e) - (6*B*a*f*x - 2*(A + B)*a + (3*B*a*f*x - (A + 7*B)*a)*cos(f*x + e))*sin(f*x + e))/(c^2*f*cos(f*x + e)^2 -
c^2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))

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Sympy [A]  time = 19.9417, 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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**2,x)

[Out]

Piecewise((-2*A*a*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*
tan(e/2 + f*x/2) - 3*c**2*f) - 6*A*a*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2
)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 3*B*a*f*x*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9
*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 9*B*a*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*
tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 9*B*a*f*x*tan(e/2
+ f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f)
- 3*B*a*f*x/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2
*f) + 2*B*a*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/
2 + f*x/2) - 3*c**2*f) - 18*B*a*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2
+ 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 8*B*a/(3*c**2*f*tan(e/2 + f*x/2)**3 - 9*c**2*f*tan(e/2 + f*x/2)**2 +
9*c**2*f*tan(e/2 + f*x/2) - 3*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 = 1.14274, size = 124, normalized size = 1.72 \begin{align*} \frac{\frac{3 \,{\left (f x + e\right )} B a}{c^{2}} - \frac{2 \,{\left (3 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 12 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + A a - 5 \, B a\right )}}{c^{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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^2,x, algorithm="giac")

[Out]

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