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

Optimal. Leaf size=115 \[ -\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^2 c^2 f}-\frac{(A-7 B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}+\frac{4 c (A-7 B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 f} \]

[Out]

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

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Rubi [A]  time = 0.408315, antiderivative size = 115, 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} \[ -\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^2 c^2 f}-\frac{(A-7 B) \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}+\frac{4 c (A-7 B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 f} \]

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])^2,x]

[Out]

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

Mathematica [A]  time = 0.691932, size = 113, normalized size = 0.98 \[ \frac{c \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (6 (A-5 B) \sin (e+f x)+2 A+3 B \cos (2 (e+f x))-23 B)}{3 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(2*A - 23*B + 3*B*Cos[2*(e + f*x)] + 6*(A - 5*B)*Sin[e + f*x])*Sqrt[c
- c*Sin[e + f*x]])/(3*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 = 0.84, size = 81, 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 ( 3\,A-15\,B \right ) +3\,B \left ( \cos \left ( fx+e \right ) \right ) ^{2}+A-13\,B \right ) }{3\,{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))^(3/2)/(a+a*sin(f*x+e))^2,x)

[Out]

-2/3*c^2/a^2*(-1+sin(f*x+e))/(1+sin(f*x+e))*(sin(f*x+e)*(3*A-15*B)+3*B*cos(f*x+e)^2+A-13*B)/cos(f*x+e)/(c-c*si
n(f*x+e))^(1/2)/f

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Maxima [B]  time = 1.54192, size = 651, normalized size = 5.66 \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))^2,x, algorithm="maxima")

[Out]

-2/3*((c^(3/2) + 6*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1
2*c^(3/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 6*c^(3/2)*sin(
f*x + e)^5/(cos(f*x + e) + 1)^5 + c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*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)^(3/2)) - 2*(5*c^(3/2) + 15*c^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 21*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 + 21*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 + 5*c^(3/2)*sin(f*x + e)^6/(cos
(f*x + e) + 1)^6)*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)^(3/2)))/f

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Fricas [A]  time = 1.5179, size = 207, normalized size = 1.8 \begin{align*} \frac{2 \,{\left (3 \, B c \cos \left (f x + e\right )^{2} + 3 \,{\left (A - 5 \, B\right )} c \sin \left (f x + e\right ) +{\left (A - 13 \, B\right )} c\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{3 \,{\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))^(3/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

2/3*(3*B*c*cos(f*x + e)^2 + 3*(A - 5*B)*c*sin(f*x + e) + (A - 13*B)*c)*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))**(3/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.86378, size = 1110, normalized size = 9.65 \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))^2,x, algorithm="giac")

[Out]

-2/3*((2*sqrt(2)*A*c^(3/2) - 14*sqrt(2)*B*c^(3/2) - 3*A*c^(3/2) + 21*B*c^(3/2))*sgn(tan(1/2*f*x + 1/2*e) - 1)/
(5*sqrt(2)*a^2 - 7*a^2) - 3*(B*c^2*sgn(tan(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x + 1/2*e)/a^2 + B*c^2*sgn(tan(1/2*
f*x + 1/2*e) - 1)/a^2)/sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c) - 2*(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^2*sgn(tan(1/2*f*x + 1/2*e) - 1) + 3*(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^2*sgn(tan(1/2*f*x + 1/2*e) - 1) - 3*(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^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 21*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*t
an(1/2*f*x + 1/2*e)^2 + c))^4*B*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 2*(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^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 14*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqr
t(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*B*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) + 6*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqr
t(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*A*c^(7/2)*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*B*c^(7/2)*sgn(tan(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))*A*c^4*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))*B*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) + A*c^(9/2)*sgn(tan(1/2*f*x + 1/2
*e) - 1) - 7*B*c^(9/2)*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