3.23 \(\int \frac{(a+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=104 \[ -\frac{a (A-4 B) \cos (e+f x)}{15 f \left (c^3-c^3 \sin (e+f x)\right )}-\frac{a c (A+11 B) \cos (e+f x)}{15 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3} \]
[Out]
(2*a*(A + B)*Cos[e + f*x])/(5*f*(c - c*Sin[e + f*x])^3) - (a*(A + 11*B)*c*Cos[e + f*x])/(15*f*(c^2 - c^2*Sin[e
+ f*x])^2) - (a*(A - 4*B)*Cos[e + f*x])/(15*f*(c^3 - c^3*Sin[e + f*x]))
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Rubi [A] time = 0.237488, antiderivative size = 104, 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} \[ -\frac{a (A-4 B) \cos (e+f x)}{15 f \left (c^3-c^3 \sin (e+f x)\right )}-\frac{a c (A+11 B) \cos (e+f x)}{15 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3} \]
Antiderivative was successfully verified.
[In]
Int[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^3,x]
[Out]
(2*a*(A + B)*Cos[e + f*x])/(5*f*(c - c*Sin[e + f*x])^3) - (a*(A + 11*B)*c*Cos[e + f*x])/(15*f*(c^2 - c^2*Sin[e
+ f*x])^2) - (a*(A - 4*B)*Cos[e + f*x])/(15*f*(c^3 - c^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+a \sin (e+f x)) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx &=(a c) \int \frac{\cos ^2(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac{2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}+\frac{a \int \frac{-A c-6 B c-5 B c \sin (e+f x)}{(c-c \sin (e+f x))^2} \, dx}{5 c^2}\\ &=\frac{2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}-\frac{a (A+11 B) \cos (e+f x)}{15 c f (c-c \sin (e+f x))^2}-\frac{(a (A-4 B)) \int \frac{1}{c-c \sin (e+f x)} \, dx}{15 c^2}\\ &=\frac{2 a (A+B) \cos (e+f x)}{5 f (c-c \sin (e+f x))^3}-\frac{a (A+11 B) \cos (e+f x)}{15 c f (c-c \sin (e+f x))^2}-\frac{a (A-4 B) \cos (e+f x)}{15 f \left (c^3-c^3 \sin (e+f x)\right )}\\ \end{align*}
Mathematica [A] time = 0.683176, size = 147, normalized size = 1.41 \[ \frac{a \left (15 (A-B) \cos \left (e+\frac{f x}{2}\right )-5 (A-B) \cos \left (e+\frac{3 f x}{2}\right )+A \sin \left (2 e+\frac{5 f x}{2}\right )+5 A \sin \left (\frac{f x}{2}\right )+15 B \sin \left (2 e+\frac{3 f x}{2}\right )-4 B \sin \left (2 e+\frac{5 f x}{2}\right )+25 B \sin \left (\frac{f x}{2}\right )\right )}{30 c^3 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + a*Sin[e + f*x])*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^3,x]
[Out]
(a*(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*S
in[2*e + (3*f*x)/2] + A*Sin[2*e + (5*f*x)/2] - 4*B*Sin[2*e + (5*f*x)/2]))/(30*c^3*f*(Cos[e/2] - Sin[e/2])*(Cos
[(e + f*x)/2] - Sin[(e + f*x)/2])^5)
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Maple [A] time = 0.108, size = 115, normalized size = 1.1 \begin{align*} 2\,{\frac{a}{f{c}^{3}} \left ( -1/5\,{\frac{8\,A+8\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-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}}}-{\frac{A}{\tan \left ( 1/2\,fx+e/2 \right ) -1}}-1/4\,{\frac{16\,A+16\,B}{ \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}} \right ) } \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))^3,x)
[Out]
2/f*a/c^3*(-1/5*(8*A+8*B)/(tan(1/2*f*x+1/2*e)-1)^5-1/3*(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-A/(tan(1/2*f*x+1/2*e)-1)-1/4*(16*A+16*B)/(tan(1/2*f*x+1/2*e)-1)^4)
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Maxima [B] time = 1.06084, size = 995, normalized size = 9.57 \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))^3,x, algorithm="maxima")
[Out]
-2/15*(A*a*(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)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1)
+ 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)
^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*A*a*(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)/(c^3 - 5*c^3*sin(f*x + e)
/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 3*B*a*(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)/(c^3
- 5*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(
cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 2
*B*a*(5*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(c^3 - 5*c^3*sin(f*x + e
)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f
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Fricas [A] time = 1.35311, size = 466, normalized size = 4.48 \begin{align*} -\frac{{\left (A - 4 \, B\right )} a \cos \left (f x + e\right )^{3} -{\left (2 \, A + 7 \, B\right )} a \cos \left (f x + e\right )^{2} + 3 \,{\left (A + B\right )} a \cos \left (f x + e\right ) + 6 \,{\left (A + B\right )} a +{\left ({\left (A - 4 \, B\right )} a \cos \left (f x + e\right )^{2} + 3 \,{\left (A + B\right )} a \cos \left (f x + e\right ) + 6 \,{\left (A + B\right )} a\right )} \sin \left (f x + e\right )}{15 \,{\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f -{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
-1/15*((A - 4*B)*a*cos(f*x + e)^3 - (2*A + 7*B)*a*cos(f*x + e)^2 + 3*(A + B)*a*cos(f*x + e) + 6*(A + B)*a + ((
A - 4*B)*a*cos(f*x + e)^2 + 3*(A + B)*a*cos(f*x + e) + 6*(A + B)*a)*sin(f*x + e))/(c^3*f*cos(f*x + e)^3 + 3*c^
3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^3*f)*
sin(f*x + e))
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Sympy [A] time = 29.8805, size = 1035, normalized size = 9.95 \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))**3,x)
[Out]
Piecewise((-30*A*a*tan(e/2 + f*x/2)**4/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c*
*3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 30*A*a*t
an(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2
)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 50*A*a*tan(e/2 + f*x/2)**2/(
15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*ta
n(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 10*A*a*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2 + f*x
/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*
c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 8*A*a/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 +
150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 30
*B*a*tan(e/2 + f*x/2)**3/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 +
f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 10*B*a*tan(e/2 + f*x/2
)**2/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**
3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) - 10*B*a*tan(e/2 + f*x/2)/(15*c**3*f*tan(e/2
+ f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2
+ 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f) + 2*B*a/(15*c**3*f*tan(e/2 + f*x/2)**5 - 75*c**3*f*tan(e/2 + f*x/2)
**4 + 150*c**3*f*tan(e/2 + f*x/2)**3 - 150*c**3*f*tan(e/2 + f*x/2)**2 + 75*c**3*f*tan(e/2 + f*x/2) - 15*c**3*f
), Ne(f, 0)), (x*(A + B*sin(e))*(a*sin(e) + a)/(-c*sin(e) + c)**3, True))
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Giac [A] time = 1.16847, size = 188, normalized size = 1.81 \begin{align*} -\frac{2 \,{\left (15 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 15 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 15 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 25 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 5 \, A a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 5 \, B a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, A a - B a\right )}}{15 \, c^{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+a*sin(f*x+e))*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="giac")
[Out]
-2/15*(15*A*a*tan(1/2*f*x + 1/2*e)^4 - 15*A*a*tan(1/2*f*x + 1/2*e)^3 + 15*B*a*tan(1/2*f*x + 1/2*e)^3 + 25*A*a*
tan(1/2*f*x + 1/2*e)^2 + 5*B*a*tan(1/2*f*x + 1/2*e)^2 - 5*A*a*tan(1/2*f*x + 1/2*e) + 5*B*a*tan(1/2*f*x + 1/2*e
) + 4*A*a - B*a)/(c^3*f*(tan(1/2*f*x + 1/2*e) - 1)^5)