3.123 \(\int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=242 \[ -\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {2048 c^3 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {512 c^2 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {64 c (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f} \]

[Out]

-2048/15*(A-3*B)*c^3*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^3/f+512/5*(A-3*B)*c^2*sec(f*x+e)^3*(c-c*sin(f*x+e))
^(5/2)/a^3/f-64/5*(A-3*B)*c*sec(f*x+e)^3*(c-c*sin(f*x+e))^(7/2)/a^3/f-16/15*(A-3*B)*sec(f*x+e)^3*(c-c*sin(f*x+
e))^(9/2)/a^3/f-1/5*(A-3*B)*sec(f*x+e)^3*(c-c*sin(f*x+e))^(11/2)/a^3/c/f-1/5*(A-B)*sec(f*x+e)^5*(c-c*sin(f*x+e
))^(15/2)/a^3/c^3/f

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Rubi [A]  time = 0.65, antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2967, 2855, 2674, 2673} \[ -\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}+\frac {512 c^2 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {2048 c^3 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {64 c (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x])^3,x]

[Out]

(-2048*(A - 3*B)*c^3*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^3*f) + (512*(A - 3*B)*c^2*Sec[e + f*x]^3
*(c - c*Sin[e + f*x])^(5/2))/(5*a^3*f) - (64*(A - 3*B)*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^3*f)
- (16*(A - 3*B)*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(9/2))/(15*a^3*f) - ((A - 3*B)*Sec[e + f*x]^3*(c - c*Sin[e
+ f*x])^(11/2))/(5*a^3*c*f) - ((A - B)*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(15/2))/(5*a^3*c^3*f)

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]

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 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 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]))

Rubi steps

\begin {align*} \int \frac {(A+B \sin (e+f x)) (c-c \sin (e+f x))^{9/2}}{(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))^{15/2} \, dx}{a^3 c^3}\\ &=-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {(A-3 B) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{13/2} \, dx}{2 a^3 c^2}\\ &=-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {(8 (A-3 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{5 a^3 c}\\ &=-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {(32 (A-3 B)) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{5 a^3}\\ &=-\frac {64 (A-3 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}-\frac {(256 (A-3 B) c) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^3}\\ &=\frac {512 (A-3 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {64 (A-3 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}+\frac {\left (1024 (A-3 B) c^2\right ) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a^3}\\ &=-\frac {2048 (A-3 B) c^3 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^3 f}+\frac {512 (A-3 B) c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 f}-\frac {64 (A-3 B) c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^3 f}-\frac {16 (A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^3 f}-\frac {(A-3 B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^3 c f}-\frac {(A-B) \sec ^5(e+f x) (c-c \sin (e+f x))^{15/2}}{5 a^3 c^3 f}\\ \end {align*}

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Mathematica [A]  time = 4.19, size = 176, normalized size = 0.73 \[ -\frac {c^4 (\sin (e+f x)-1)^4 \sqrt {c-c \sin (e+f x)} (-40 (137 A-402 B) \cos (2 (e+f x))-10 (A-6 B) \cos (4 (e+f x))+15600 A \sin (e+f x)-400 A \sin (3 (e+f x))+11298 A-47430 B \sin (e+f x)+1335 B \sin (3 (e+f x))-3 B \sin (5 (e+f x))-33516 B)}{120 a^3 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9 \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x])^3,x]

[Out]

-1/120*(c^4*(-1 + Sin[e + f*x])^4*Sqrt[c - c*Sin[e + f*x]]*(11298*A - 33516*B - 40*(137*A - 402*B)*Cos[2*(e +
f*x)] - 10*(A - 6*B)*Cos[4*(e + f*x)] + 15600*A*Sin[e + f*x] - 47430*B*Sin[e + f*x] - 400*A*Sin[3*(e + f*x)] +
1335*B*Sin[3*(e + f*x)] - 3*B*Sin[5*(e + f*x)]))/(a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(Cos[(e + f*x
)/2] + Sin[(e + f*x)/2])^5)

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fricas [A]  time = 0.45, size = 166, normalized size = 0.69 \[ -\frac {2 \, {\left (5 \, {\left (A - 6 \, B\right )} c^{4} \cos \left (f x + e\right )^{4} + 20 \, {\left (34 \, A - 99 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} - 8 \, {\left (131 \, A - 387 \, B\right )} c^{4} + {\left (3 \, B c^{4} \cos \left (f x + e\right )^{4} + 4 \, {\left (25 \, A - 84 \, B\right )} c^{4} \cos \left (f x + e\right )^{2} - 8 \, {\left (125 \, A - 381 \, B\right )} c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/15*(5*(A - 6*B)*c^4*cos(f*x + e)^4 + 20*(34*A - 99*B)*c^4*cos(f*x + e)^2 - 8*(131*A - 387*B)*c^4 + (3*B*c^4
*cos(f*x + e)^4 + 4*(25*A - 84*B)*c^4*cos(f*x + e)^2 - 8*(125*A - 381*B)*c^4)*sin(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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 1.47, size = 143, normalized size = 0.59 \[ -\frac {2 c^{5} \left (\sin \left (f x +e \right )-1\right ) \left (3 B \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (100 A -336 B \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\left (-1000 A +3048 B \right ) \sin \left (f x +e \right )+\left (5 A -30 B \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\left (680 A -1980 B \right ) \left (\cos ^{2}\left (f x +e \right )\right )-1048 A +3096 B \right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x)

[Out]

-2/15*c^5/a^3*(sin(f*x+e)-1)/(1+sin(f*x+e))^2*(3*B*sin(f*x+e)*cos(f*x+e)^4+(100*A-336*B)*cos(f*x+e)^2*sin(f*x+
e)+(-1000*A+3048*B)*sin(f*x+e)+(5*A-30*B)*cos(f*x+e)^4+(680*A-1980*B)*cos(f*x+e)^2-1048*A+3096*B)/cos(f*x+e)/(
c-c*sin(f*x+e))^(1/2)/f

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maxima [B]  time = 0.49, size = 945, normalized size = 3.90 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/15*((363*c^(9/2) + 1800*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 5301*c^(9/2)*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 11600*c^(9/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 21343*c^(9/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
+ 30200*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 40065*c^(9/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 408
00*c^(9/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 40065*c^(9/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 30200*c^(
9/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 21343*c^(9/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 11600*c^(9/2)
*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 5301*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 1800*c^(9/2)*sin
(f*x + e)^13/(cos(f*x + e) + 1)^13 + 363*c^(9/2)*sin(f*x + e)^14/(cos(f*x + e) + 1)^14)*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)^(9/2)) - 6*(181*c^(9/2) + 905*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 2627*c^(9/2)*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5870*c^(9/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 10521*c^(9/2)*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + 15351*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 19695*c^(9/2)*sin(f*x + e)^
6/(cos(f*x + e) + 1)^6 + 20772*c^(9/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 19695*c^(9/2)*sin(f*x + e)^8/(cos
(f*x + e) + 1)^8 + 15351*c^(9/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 10521*c^(9/2)*sin(f*x + e)^10/(cos(f*x
+ e) + 1)^10 + 5870*c^(9/2)*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 2627*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e)
+ 1)^12 + 905*c^(9/2)*sin(f*x + e)^13/(cos(f*x + e) + 1)^13 + 181*c^(9/2)*sin(f*x + e)^14/(cos(f*x + e) + 1)^
14)*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)^(9/2)))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\left (A+B\,\sin \left (e+f\,x\right )\right )\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x))^3,x)

[Out]

int(((A + B*sin(e + f*x))*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x))^3, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(9/2)/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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