3.116 \(\int \frac{(A+B \sin (e+f x)) (c-c \sin (e+f x))^{7/2}}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=201 \[ -\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac{32 c^2 (5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}+\frac{128 c^3 (5 A-11 B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{15 a^2 f}-\frac{(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac{4 c (5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f} \]

[Out]

(128*(5*A - 11*B)*c^3*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(15*a^2*f) - (32*(5*A - 11*B)*c^2*Sec[e + f*x]*(c
- c*Sin[e + f*x])^(3/2))/(15*a^2*f) - (4*(5*A - 11*B)*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*a^2*f) -
((5*A - 11*B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(15*a^2*f) - ((A - B)*Sec[e + f*x]^3*(c - c*Sin[e + f*
x])^(11/2))/(3*a^2*c^2*f)

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Rubi [A]  time = 0.558201, antiderivative size = 201, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {2967, 2855, 2674, 2673} \[ -\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac{32 c^2 (5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}+\frac{128 c^3 (5 A-11 B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{15 a^2 f}-\frac{(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac{4 c (5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^2,x]

[Out]

(128*(5*A - 11*B)*c^3*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(15*a^2*f) - (32*(5*A - 11*B)*c^2*Sec[e + f*x]*(c
- c*Sin[e + f*x])^(3/2))/(15*a^2*f) - (4*(5*A - 11*B)*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(15*a^2*f) -
((5*A - 11*B)*Sec[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(15*a^2*f) - ((A - B)*Sec[e + f*x]^3*(c - c*Sin[e + f*
x])^(11/2))/(3*a^2*c^2*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))^{7/2}}{(a+a \sin (e+f x))^2} \, dx &=\frac{\int \sec ^4(e+f x) (A+B \sin (e+f x)) (c-c \sin (e+f x))^{11/2} \, dx}{a^2 c^2}\\ &=-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac{(5 A-11 B) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{6 a^2 c}\\ &=-\frac{(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac{(2 (5 A-11 B)) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^2}\\ &=-\frac{4 (5 A-11 B) c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}-\frac{(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac{(16 (5 A-11 B) c) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{15 a^2}\\ &=-\frac{32 (5 A-11 B) c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}-\frac{4 (5 A-11 B) c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}-\frac{(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}-\frac{\left (64 (5 A-11 B) c^2\right ) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{15 a^2}\\ &=\frac{128 (5 A-11 B) c^3 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{15 a^2 f}-\frac{32 (5 A-11 B) c^2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}-\frac{4 (5 A-11 B) c \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^2 f}-\frac{(5 A-11 B) \sec (e+f x) (c-c \sin (e+f x))^{7/2}}{15 a^2 f}-\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^2 c^2 f}\\ \end{align*}

Mathematica [A]  time = 2.96615, size = 159, normalized size = 0.79 \[ -\frac{c^3 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (12 (25 A-62 B) \cos (2 (e+f x))-2730 A \sin (e+f x)-10 A \sin (3 (e+f x))-2100 A+5838 B \sin (e+f x)+46 B \sin (3 (e+f x))+3 B \cos (4 (e+f x))+4725 B)}{60 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*Sin[e + f*x])*(c - c*Sin[e + f*x])^(7/2))/(a + a*Sin[e + f*x])^2,x]

[Out]

-(c^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-2100*A + 4725*B + 12*(25*A - 62*B)*Cos[
2*(e + f*x)] + 3*B*Cos[4*(e + f*x)] - 2730*A*Sin[e + f*x] + 5838*B*Sin[e + f*x] - 10*A*Sin[3*(e + f*x)] + 46*B
*Sin[3*(e + f*x)]))/(60*a^2*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)

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Maple [A]  time = 0.915, size = 121, normalized size = 0.6 \begin{align*}{\frac{2\,{c}^{4} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \left ( -5\,A+23\,B \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}+ \left ( -340\,A+724\,B \right ) \sin \left ( fx+e \right ) +3\,B \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( 75\,A-189\,B \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}-300\,A+684\,B \right ) }{15\,{a}^{2} \left ( 1+\sin \left ( fx+e \right ) \right ) \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))^(7/2)/(a+a*sin(f*x+e))^2,x)

[Out]

2/15*c^4/a^2*(-1+sin(f*x+e))/(1+sin(f*x+e))*((-5*A+23*B)*sin(f*x+e)*cos(f*x+e)^2+(-340*A+724*B)*sin(f*x+e)+3*B
*cos(f*x+e)^4+(75*A-189*B)*cos(f*x+e)^2-300*A+684*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [B]  time = 1.56883, size = 905, normalized size = 4.5 \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))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/15*(5*(45*c^(7/2) + 138*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 285*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 544*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 630*c^(7/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 8
12*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 630*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 544*c^(7/2)
*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 285*c^(7/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 138*c^(7/2)*sin(f*x +
e)^9/(cos(f*x + e) + 1)^9 + 45*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*A/((a^2 + 3*a^2*sin(f*x + e)/(c
os(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*(sin(f
*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(7/2)) - 2*(249*c^(7/2) + 747*c^(7/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 16
11*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2896*c^(7/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3612*c^(7/
2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4298*c^(7/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3612*c^(7/2)*sin(f
*x + e)^6/(cos(f*x + e) + 1)^6 + 2896*c^(7/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 1611*c^(7/2)*sin(f*x + e)^
8/(cos(f*x + e) + 1)^8 + 747*c^(7/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 249*c^(7/2)*sin(f*x + e)^10/(cos(f*
x + e) + 1)^10)*B/((a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 +
a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(7/2)))/f

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Fricas [A]  time = 1.51195, size = 331, normalized size = 1.65 \begin{align*} -\frac{2 \,{\left (3 \, B c^{3} \cos \left (f x + e\right )^{4} + 3 \,{\left (25 \, A - 63 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} - 12 \,{\left (25 \, A - 57 \, B\right )} c^{3} -{\left ({\left (5 \, A - 23 \, B\right )} c^{3} \cos \left (f x + e\right )^{2} + 4 \,{\left (85 \, A - 181 \, B\right )} c^{3}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{15 \,{\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} 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))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-2/15*(3*B*c^3*cos(f*x + e)^4 + 3*(25*A - 63*B)*c^3*cos(f*x + e)^2 - 12*(25*A - 57*B)*c^3 - ((5*A - 23*B)*c^3*
cos(f*x + e)^2 + 4*(85*A - 181*B)*c^3)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(a^2*f*cos(f*x + e)*sin(f*x + e
) + a^2*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))**(7/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 2.36949, size = 1589, normalized size = 7.91 \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))^(7/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/60*(4*(140*sqrt(2)*A*a^12*sqrt(c) - 371*sqrt(2)*B*a^12*sqrt(c) - 200*A*a^12*sqrt(c) + 530*B*a^12*sqrt(c) -
280*sqrt(2)*A*c^(19/2) + 640*sqrt(2)*B*c^(19/2) + 360*A*c^(19/2) - 840*B*c^(19/2))*sgn(tan(1/2*f*x + 1/2*e) -
1)/(5*sqrt(2)*a^2*c^6 - 7*a^2*c^6) + (((((17*(5*A*a^10*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1) - 14*B*a^10*c^6*sgn(t
an(1/2*f*x + 1/2*e) - 1))*tan(1/2*f*x + 1/2*e)/c^9 + 15*(5*A*a^10*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1) - 12*B*a^1
0*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^9)*tan(1/2*f*x + 1/2*e) + 10*(16*A*a^10*c^6*sgn(tan(1/2*f*x + 1/2*e) -
1) - 43*B*a^10*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^9)*tan(1/2*f*x + 1/2*e) + 10*(16*A*a^10*c^6*sgn(tan(1/2*f*
x + 1/2*e) - 1) - 43*B*a^10*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^9)*tan(1/2*f*x + 1/2*e) + 15*(5*A*a^10*c^6*sg
n(tan(1/2*f*x + 1/2*e) - 1) - 12*B*a^10*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^9)*tan(1/2*f*x + 1/2*e) + 17*(5*A
*a^10*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1) - 14*B*a^10*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1))/c^9)/(c*tan(1/2*f*x + 1
/2*e)^2 + c)^(5/2) + 320*(3*(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(ta
n(1/2*f*x + 1/2*e) - 1) - 9*(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(ta
n(1/2*f*x + 1/2*e) - 1) + 21*(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)*s
gn(tan(1/2*f*x + 1/2*e) - 1) - 39*(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) - 14*(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^5*sgn(tan(1/2*f*x + 1/2*e) - 1) + 26*(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) - 42*(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) + 78*(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) + 39*(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) - 69*(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) - 7*A*c^(13/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 13*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)^3*a^2))/f