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

Optimal. Leaf size=57 \[ -\frac{2 c (A-B) \cos (e+f x)}{f (a \sin (e+f x)+a)}-\frac{c x (A-2 B)}{a}+\frac{B c \cos (e+f x)}{a f} \]

[Out]

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

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Rubi [A]  time = 0.155184, antiderivative size = 57, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.088, Rules used = {2967, 2857, 2638} \[ -\frac{2 c (A-B) \cos (e+f x)}{f (a \sin (e+f x)+a)}-\frac{c x (A-2 B)}{a}+\frac{B c \cos (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 0.561802, size = 127, normalized size = 2.23 \[ \frac{(c-c \sin (e+f x)) \left (\frac{4 (A-B) \sin \left (\frac{f x}{2}\right )}{f \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}+x (-(A-2 B))-\frac{B \sin (e) \sin (f x)}{f}+\frac{B \cos (e) \cos (f x)}{f}\right )}{a \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.096, size = 113, normalized size = 2. \begin{align*} 2\,{\frac{Bc}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{c\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) A}{af}}+4\,{\frac{c\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) B}{af}}-4\,{\frac{Ac}{af \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }}+4\,{\frac{Bc}{af \left ( \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))*(c-c*sin(f*x+e))/(a+a*sin(f*x+e)),x)

[Out]

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

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Maxima [B]  time = 1.45172, size = 346, normalized size = 6.07 \begin{align*} \frac{2 \,{\left (B c{\left (\frac{\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 2}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a}\right )} - A c{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + B c{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac{A c}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{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)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.42437, size = 289, normalized size = 5.07 \begin{align*} -\frac{{\left (A - 2 \, B\right )} c f x - B c \cos \left (f x + e\right )^{2} + 2 \,{\left (A - B\right )} c +{\left ({\left (A - 2 \, B\right )} c f x +{\left (2 \, A - 3 \, B\right )} c\right )} \cos \left (f x + e\right ) +{\left ({\left (A - 2 \, B\right )} c f x - B c \cos \left (f x + e\right ) - 2 \,{\left (A - B\right )} c\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a 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)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 7.1462, size = 830, normalized size = 14.56 \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)),x)

[Out]

Piecewise((-A*c*f*x*tan(e/2 + f*x/2)**3/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x
/2) + a*f) - A*c*f*x*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*
x/2) + a*f) - A*c*f*x*tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/
2) + a*f) - A*c*f*x/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - 4*A*c*t
an(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - 4*A*c/(a
*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*B*c*f*x*tan(e/2 + f*x/2)**3
/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*B*c*f*x*tan(e/2 + f*x/2)
**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*B*c*f*x*tan(e/2 + f*x
/2)/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*B*c*f*x/(a*f*tan(e/2
+ f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) - 2*B*c*tan(e/2 + f*x/2)**3/(a*f*tan(e/2 +
f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 2*B*c*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 +
f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f) + 4*B*c/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e
/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f), Ne(f, 0)), (x*(A + B*sin(e))*(-c*sin(e) + c)/(a*sin(e) + a), Tru
e))

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Giac [B]  time = 1.15299, size = 165, normalized size = 2.89 \begin{align*} -\frac{\frac{{\left (A c - 2 \, B c\right )}{\left (f x + e\right )}}{a} + \frac{2 \,{\left (2 \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 2 \, B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 2 \, A c - 3 \, B c\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} a}}{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)),x, algorithm="giac")

[Out]

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