3.74 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=103 \[ \frac{c (A+4 B) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{a c (A-11 B) \cos (e+f x)}{15 f \left (a^2 \sin (e+f x)+a^2\right )^2}-\frac{2 c (A-B) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]
[Out]
(-2*(A - B)*c*Cos[e + f*x])/(5*f*(a + a*Sin[e + f*x])^3) + (a*(A - 11*B)*c*Cos[e + f*x])/(15*f*(a^2 + a^2*Sin[
e + f*x])^2) + ((A + 4*B)*c*Cos[e + f*x])/(15*f*(a^3 + a^3*Sin[e + f*x]))
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Rubi [A] time = 0.225012, antiderivative size = 103, 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, 2750, 2648} \[ \frac{c (A+4 B) \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}+\frac{a c (A-11 B) \cos (e+f x)}{15 f \left (a^2 \sin (e+f x)+a^2\right )^2}-\frac{2 c (A-B) \cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x]))/(a + a*Sin[e + f*x])^3,x]
[Out]
(-2*(A - B)*c*Cos[e + f*x])/(5*f*(a + a*Sin[e + f*x])^3) + (a*(A - 11*B)*c*Cos[e + f*x])/(15*f*(a^2 + a^2*Sin[
e + f*x])^2) + ((A + 4*B)*c*Cos[e + f*x])/(15*f*(a^3 + a^3*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 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)) (c-c \sin (e+f x))}{(a+a \sin (e+f x))^3} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(a+a \sin (e+f x))^4} \, dx\\ &=-\frac{2 (A-B) c \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{c \int \frac{a A-6 a B+5 a B \sin (e+f x)}{(a+a \sin (e+f x))^2} \, dx}{5 a^2}\\ &=-\frac{2 (A-B) c \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}+\frac{(A-11 B) c \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{((A+4 B) c) \int \frac{1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=-\frac{2 (A-B) c \cos (e+f x)}{5 f (a+a \sin (e+f x))^3}+\frac{(A-11 B) c \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}+\frac{(A+4 B) c \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 0.772458, size = 139, normalized size = 1.35 \[ \frac{c \left (-15 (A+B) \cos \left (e+\frac{f x}{2}\right )+5 (A+B) \cos \left (e+\frac{3 f x}{2}\right )+A \sin \left (2 e+\frac{5 f x}{2}\right )+5 A \sin \left (\frac{f x}{2}\right )-15 B \sin \left (2 e+\frac{3 f x}{2}\right )+4 B \sin \left (2 e+\frac{5 f x}{2}\right )-25 B \sin \left (\frac{f x}{2}\right )\right )}{30 a^3 f \left (\sin \left (\frac{e}{2}\right )+\cos \left (\frac{e}{2}\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x]))/(a + a*Sin[e + f*x])^3,x]
[Out]
(c*(-15*(A + B)*Cos[e + (f*x)/2] + 5*(A + B)*Cos[e + (3*f*x)/2] + 5*A*Sin[(f*x)/2] - 25*B*Sin[(f*x)/2] - 15*B*
Sin[2*e + (3*f*x)/2] + A*Sin[2*e + (5*f*x)/2] + 4*B*Sin[2*e + (5*f*x)/2]))/(30*a^3*f*(Cos[e/2] + Sin[e/2])*(Co
s[(e + f*x)/2] + Sin[(e + f*x)/2])^5)
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Maple [A] time = 0.11, size = 115, normalized size = 1.1 \begin{align*} 2\,{\frac{c}{f{a}^{3}} \left ( -1/4\,{\frac{-16\,A+16\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-1/5\,{\frac{8\,A-8\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{5}}}-{\frac{A}{\tan \left ( 1/2\,fx+e/2 \right ) +1}}-1/3\,{\frac{14\,A-10\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}-1/2\,{\frac{-6\,A+2\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}} \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))^3,x)
[Out]
2/f*c/a^3*(-1/4*(-16*A+16*B)/(tan(1/2*f*x+1/2*e)+1)^4-1/5*(8*A-8*B)/(tan(1/2*f*x+1/2*e)+1)^5-A/(tan(1/2*f*x+1/
2*e)+1)-1/3*(14*A-10*B)/(tan(1/2*f*x+1/2*e)+1)^3-1/2*(-6*A+2*B)/(tan(1/2*f*x+1/2*e)+1)^2)
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Maxima [B] time = 1.04638, size = 990, normalized size = 9.61 \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))^3,x, algorithm="maxima")
[Out]
-2/15*(A*c*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 2*B*c*(5*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*A*c*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/
(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e
)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*B*c*(5*sin(f*x + e)/(cos(f*x + e) + 1)
+ 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e
)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f
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Fricas [A] time = 1.61019, size = 464, normalized size = 4.5 \begin{align*} \frac{{\left (A + 4 \, B\right )} c \cos \left (f x + e\right )^{3} -{\left (2 \, A - 7 \, B\right )} c \cos \left (f x + e\right )^{2} + 3 \,{\left (A - B\right )} c \cos \left (f x + e\right ) + 6 \,{\left (A - B\right )} c -{\left ({\left (A + 4 \, B\right )} c \cos \left (f x + e\right )^{2} + 3 \,{\left (A - B\right )} c \cos \left (f x + e\right ) + 6 \,{\left (A - B\right )} c\right )} \sin \left (f x + e\right )}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} 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))^3,x, algorithm="fricas")
[Out]
1/15*((A + 4*B)*c*cos(f*x + e)^3 - (2*A - 7*B)*c*cos(f*x + e)^2 + 3*(A - B)*c*cos(f*x + e) + 6*(A - B)*c - ((A
+ 4*B)*c*cos(f*x + e)^2 + 3*(A - B)*c*cos(f*x + e) + 6*(A - B)*c)*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3
*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*s
in(f*x + e))
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Sympy [A] time = 34.1097, size = 1035, normalized size = 10.05 \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))**3,x)
[Out]
Piecewise((-30*A*c*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a*
*3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30*A*c*t
an(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2
)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 50*A*c*tan(e/2 + f*x/2)**2/(
15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*ta
n(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 10*A*c*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2 + f*x
/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*
a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 8*A*c/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 +
150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 30
*B*c*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 +
f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) + 10*B*c*tan(e/2 + f*x/2
)**2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**
3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 10*B*c*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2
+ f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2
+ 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 2*B*c/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)
**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f
), Ne(f, 0)), (x*(A + B*sin(e))*(-c*sin(e) + c)/(a*sin(e) + a)**3, True))
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Giac [A] time = 1.20507, size = 186, normalized size = 1.81 \begin{align*} -\frac{2 \,{\left (15 \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 15 \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 25 \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5 \, B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, A c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 5 \, B c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, A c + B c\right )}}{15 \, a^{3} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}} \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))^3,x, algorithm="giac")
[Out]
-2/15*(15*A*c*tan(1/2*f*x + 1/2*e)^4 + 15*A*c*tan(1/2*f*x + 1/2*e)^3 + 15*B*c*tan(1/2*f*x + 1/2*e)^3 + 25*A*c*
tan(1/2*f*x + 1/2*e)^2 - 5*B*c*tan(1/2*f*x + 1/2*e)^2 + 5*A*c*tan(1/2*f*x + 1/2*e) + 5*B*c*tan(1/2*f*x + 1/2*e
) + 4*A*c + B*c)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)