3.126 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=121 \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 a^3 c^3 f}-\frac{(A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c f}+\frac{4 (A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f} \]
[Out]
(4*(A + 9*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) - ((A + 9*B)*Sec[e + f*x]^3*(c - c*Sin[e +
f*x])^(5/2))/(5*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/(5*a^3*c^3*f)
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Rubi [A] time = 0.412688, antiderivative size = 121, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used =
{2967, 2855, 2674, 2673} \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 a^3 c^3 f}-\frac{(A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c f}+\frac{4 (A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))/(a + a*Sin[e + f*x])^3,x]
[Out]
(4*(A + 9*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) - ((A + 9*B)*Sec[e + f*x]^3*(c - c*Sin[e +
f*x])^(5/2))/(5*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/(5*a^3*c^3*f)
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 2855
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]
Rule 2674
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[Simplify[(2*m + p - 1)/2], 0] && NeQ[m + p, 0]
Rule 2673
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]
Rubi steps
\begin{align*} \int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^3} \, dx &=\frac{\int \sec ^6(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2} \, dx}{a^3 c^3}\\ &=-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 a^3 c^3 f}+\frac{(A+9 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{10 a^3 c^2}\\ &=-\frac{(A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 a^3 c^3 f}-\frac{(2 (A+9 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a^3 c}\\ &=\frac{4 (A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}-\frac{(A+9 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{5 a^3 c^3 f}\\ \end{align*}
Mathematica [A] time = 0.713139, size = 113, normalized size = 0.93 \[ \frac{c \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (10 (A+3 B) \sin (e+f x)-2 A-15 B \cos (2 (e+f x))+27 B)}{15 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(3/2))/(a + a*Sin[e + f*x])^3,x]
[Out]
(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-2*A + 27*B - 15*B*Cos[2*(e + f*x)] + 10*(A + 3*B)*Sin[e + f*x])*Sqr
t[c - c*Sin[e + f*x]])/(15*a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3)
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Maple [A] time = 0.993, size = 83, normalized size = 0.7 \begin{align*} -{\frac{2\,{c}^{2} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \sin \left ( fx+e \right ) \left ( 5\,A+15\,B \right ) -15\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}-A+21\,B \right ) }{15\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\cos \left ( fx+e \right ) f}{\frac{1}{\sqrt{c-c\sin \left ( fx+e \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))^(3/2)/(a+a*sin(f*x+e))^3,x)
[Out]
-2/15*c^2/a^3*(-1+sin(f*x+e))/(1+sin(f*x+e))^2*(sin(f*x+e)*(5*A+15*B)-15*B*cos(f*x+e)^2-A+21*B)/cos(f*x+e)/(c-
c*sin(f*x+e))^(1/2)/f
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Maxima [B] time = 1.58665, size = 895, normalized size = 7.4 \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))^(3/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
2/15*((c^(3/2) - 10*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 4*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 -
30*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 6*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 30*c^(3/2)*si
n(f*x + e)^5/(cos(f*x + e) + 1)^5 + 4*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 10*c^(3/2)*sin(f*x + e)^7/
(cos(f*x + e) + 1)^7 + c^(3/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*A/((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)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 1)^(3/2)) - 6*(c^(3/2) + 5*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 14*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 15*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 26*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 15
*c^(3/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 14*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 5*c^(3/2)*sin(
f*x + e)^7/(cos(f*x + e) + 1)^7 + c^(3/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*B/((a^3 + 5*a^3*sin(f*x + e)/(c
os(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)*(sin(f*x + e)^2/(cos(f*x +
e) + 1)^2 + 1)^(3/2)))/f
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Fricas [A] time = 1.77413, size = 246, normalized size = 2.03 \begin{align*} \frac{2 \,{\left (15 \, B c \cos \left (f x + e\right )^{2} - 5 \,{\left (A + 3 \, B\right )} c \sin \left (f x + e\right ) +{\left (A - 21 \, B\right )} c\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \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))^(3/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
2/15*(15*B*c*cos(f*x + e)^2 - 5*(A + 3*B)*c*sin(f*x + e) + (A - 21*B)*c)*sqrt(-c*sin(f*x + e) + c)/(a^3*f*cos(
f*x + e)^3 - 2*a^3*f*cos(f*x + e)*sin(f*x + e) - 2*a^3*f*cos(f*x + e))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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))**(3/2)/(a+a*sin(f*x+e))**3,x)
[Out]
Timed out
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Giac [B] time = 1.99072, size = 1391, normalized size = 11.5 \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))^(3/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")
[Out]
-1/15*((99*sqrt(2)*A*c^(3/2) + 21*sqrt(2)*B*c^(3/2) - 140*A*c^(3/2) - 30*B*c^(3/2))*sgn(tan(1/2*f*x + 1/2*e) -
1)/(29*sqrt(2)*a^3 - 41*a^3) - 4*(15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^9*A*
c^2*sgn(tan(1/2*f*x + 1/2*e) - 1) + 15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*A
*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 30*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c)
)^8*B*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 40*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2
+ c))^7*A*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 60*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^
2 + c))^7*B*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 20*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)
^2 + c))^6*A*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 34*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1
/2*e)^2 + c))^5*A*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) + 204*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x +
1/2*e)^2 + c))^5*B*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) + 10*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x
+ 1/2*e)^2 + c))^4*A*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 180*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/
2*f*x + 1/2*e)^2 + c))^4*B*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 180*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*
tan(1/2*f*x + 1/2*e)^2 + c))^3*B*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - 20*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c
*tan(1/2*f*x + 1/2*e)^2 + c))^2*A*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 240*(sqrt(c)*tan(1/2*f*x + 1/2*e) -
sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*B*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 5*(sqrt(c)*tan(1/2*f*x + 1/2
*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1) - 60*(sqrt(c)*tan(1/2*f*x + 1/2*
e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*B*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1) - A*c^(13/2)*sgn(tan(1/2*f*x + 1/
2*e) - 1) + 6*B*c^(13/2)*sgn(tan(1/2*f*x + 1/2*e) - 1))/(((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x +
1/2*e)^2 + c))^2 + 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*sqrt(c) - c)^5*a^3))
/f