3.125 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=160 \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}-\frac{(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}+\frac{8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac{32 c (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f} \]
[Out]
(-32*(A - 11*B)*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) + (8*(A - 11*B)*Sec[e + f*x]^3*(c - c*
Sin[e + f*x])^(5/2))/(5*a^3*f) - ((A - 11*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^3*c*f) - ((A - B)
*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(11/2))/(5*a^3*c^3*f)
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Rubi [A] time = 0.479844, antiderivative size = 160, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used =
{2967, 2855, 2674, 2673} \[ -\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}-\frac{(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}+\frac{8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac{32 c (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f} \]
Antiderivative was successfully verified.
[In]
Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x])^3,x]
[Out]
(-32*(A - 11*B)*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) + (8*(A - 11*B)*Sec[e + f*x]^3*(c - c*
Sin[e + f*x])^(5/2))/(5*a^3*f) - ((A - 11*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^3*c*f) - ((A - B)
*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(11/2))/(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))^{5/2}}{(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))^{11/2} \, dx}{a^3 c^3}\\ &=-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}-\frac{(A-11 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{10 a^3 c^2}\\ &=-\frac{(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}-\frac{(4 (A-11 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^3 c}\\ &=\frac{8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac{(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}+\frac{(16 (A-11 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a^3}\\ &=-\frac{32 (A-11 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac{8 (A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac{(A-11 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 c f}-\frac{(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c^3 f}\\ \end{align*}
Mathematica [A] time = 1.26005, size = 132, normalized size = 0.82 \[ -\frac{c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (5 (8 A-133 B) \sin (e+f x)-30 (A-8 B) \cos (2 (e+f x))+58 A+15 B \sin (3 (e+f x))-488 B)}{30 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(5/2))/(a + a*Sin[e + f*x])^3,x]
[Out]
-(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(58*A - 488*B - 30*(A - 8*B)*Cos[2*(e + f
*x)] + 5*(8*A - 133*B)*Sin[e + f*x] + 15*B*Sin[3*(e + f*x)]))/(30*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 = 1.176, size = 105, normalized size = 0.7 \begin{align*}{\frac{2\,{c}^{3} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 15\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) + \left ( 10\,A-170\,B \right ) \sin \left ( fx+e \right ) + \left ( -15\,A+120\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+22\,A-182\,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))^(5/2)/(a+a*sin(f*x+e))^3,x)
[Out]
2/15*c^3/a^3*(-1+sin(f*x+e))/(1+sin(f*x+e))^2*(15*B*cos(f*x+e)^2*sin(f*x+e)+(10*A-170*B)*sin(f*x+e)+(-15*A+120
*B)*cos(f*x+e)^2+22*A-182*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
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Maxima [B] time = 1.60501, size = 1027, normalized size = 6.42 \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))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")
[Out]
2/15*((7*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 95*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 80*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 250*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 120*c^(5
/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 250*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 80*c^(5/2)*sin(f*x
+ e)^7/(cos(f*x + e) + 1)^7 + 95*c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 20*c^(5/2)*sin(f*x + e)^9/(cos
(f*x + e) + 1)^9 + 7*c^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*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)^(5/2)) - 2*(31*c^(5/2) + 155*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 395*c^(5/2)*sin(f*x + e)^2/(cos(f*
x + e) + 1)^2 + 680*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1030*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + 1050*c^(5/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 1030*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 6
80*c^(5/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 395*c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 155*c^(5/2)
*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 31*c^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*B/((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)^(5/2)))/f
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Fricas [A] time = 1.72745, size = 313, normalized size = 1.96 \begin{align*} -\frac{2 \,{\left (15 \,{\left (A - 8 \, B\right )} c^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (11 \, A - 91 \, B\right )} c^{2} - 5 \,{\left (3 \, B c^{2} \cos \left (f x + e\right )^{2} + 2 \,{\left (A - 17 \, B\right )} c^{2}\right )} \sin \left (f x + e\right )\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))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
-2/15*(15*(A - 8*B)*c^2*cos(f*x + e)^2 - 2*(11*A - 91*B)*c^2 - 5*(3*B*c^2*cos(f*x + e)^2 + 2*(A - 17*B)*c^2)*s
in(f*x + e))*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))**(5/2)/(a+a*sin(f*x+e))**3,x)
[Out]
Timed out
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Giac [B] time = 2.53042, size = 1716, normalized size = 10.72 \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))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")
[Out]
-2/15*((39*sqrt(2)*A*c^(5/2) + 441*sqrt(2)*B*c^(5/2) - 55*A*c^(5/2) - 625*B*c^(5/2))*sgn(tan(1/2*f*x + 1/2*e)
- 1)/(29*sqrt(2)*a^3 - 41*a^3) + 15*(B*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x + 1/2*e)/a^3 + B*c^3*sgn(
tan(1/2*f*x + 1/2*e) - 1)/a^3)/sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c) - 2*(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^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqr
t(c*tan(1/2*f*x + 1/2*e)^2 + c))^9*B*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 15*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sq
rt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*A*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 105*(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^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 100*(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^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 500*(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^4*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^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 340*(sqrt(c)*tan(
1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*B*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 238*(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^5*sgn(tan(1/2*f*x + 1/2*e) - 1) + 1598*(sqr
t(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*B*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) + 190*(s
qrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*A*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) -
1070*(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^(11/2)*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*A*c^6*sgn(tan(1/2*f*x + 1/2
*e) - 1) - 1380*(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^6*sgn(tan(1/2*f*x +
1/2*e) - 1) - 260*(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^(13/2)*sgn(tan(1/2
*f*x + 1/2*e) - 1) + 1540*(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^(13/2)*sgn
(tan(1/2*f*x + 1/2*e) - 1) + 55*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*A*c^7*sgn(
tan(1/2*f*x + 1/2*e) - 1) - 455*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*B*c^7*sgn(
tan(1/2*f*x + 1/2*e) - 1) - 7*A*c^(15/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 47*B*c^(15/2)*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