3.92 \(\int (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)} \, dx\)

Optimal. Leaf size=81 \[ \frac{2 a^2 c^3 (7 A+3 B) \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}-\frac{2 a^2 B c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}} \]

[Out]

(2*a^2*(7*A + 3*B)*c^3*Cos[e + f*x]^5)/(35*f*(c - c*Sin[e + f*x])^(5/2)) - (2*a^2*B*c^2*Cos[e + f*x]^5)/(7*f*(
c - c*Sin[e + f*x])^(3/2))

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Rubi [A]  time = 0.330345, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {2967, 2856, 2673} \[ \frac{2 a^2 c^3 (7 A+3 B) \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}-\frac{2 a^2 B c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(7*A + 3*B)*c^3*Cos[e + f*x]^5)/(35*f*(c - c*Sin[e + f*x])^(5/2)) - (2*a^2*B*c^2*Cos[e + f*x]^5)/(7*f*(
c - c*Sin[e + f*x])^(3/2))

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 2856

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[(a*d*m + b*c*(m + p + 1))/(b*(m + p + 1)), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^m, x], x] /; Fre
eQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[Simplify[(2*m + p + 1)/2], 0] && NeQ[m + p + 1
, 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 (a+a \sin (e+f x))^2 (A+B \sin (e+f x)) \sqrt{c-c \sin (e+f x)} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x) (A+B \sin (e+f x))}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=-\frac{2 a^2 B c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}+\frac{1}{7} \left (a^2 (7 A+3 B) c^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=\frac{2 a^2 (7 A+3 B) c^3 \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^{5/2}}-\frac{2 a^2 B c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.592306, size = 89, normalized size = 1.1 \[ \frac{2 a^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5 (7 A+5 B \sin (e+f x)-2 B)}{35 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(7*A - 2*B + 5*B*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]])/(35*f*
(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))

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Maple [A]  time = 0.865, size = 65, normalized size = 0.8 \begin{align*} -{\frac{ \left ( -2+2\,\sin \left ( fx+e \right ) \right ) c \left ( 1+\sin \left ( fx+e \right ) \right ) ^{3}{a}^{2} \left ( 5\,B\sin \left ( fx+e \right ) +7\,A-2\,B \right ) }{35\,f\cos \left ( fx+e \right ) }{\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+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x)

[Out]

-2/35*(-1+sin(f*x+e))*c*(1+sin(f*x+e))^3*a^2*(5*B*sin(f*x+e)+7*A-2*B)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sin \left (f x + e\right ) + A\right )}{\left (a \sin \left (f x + e\right ) + a\right )}^{2} \sqrt{-c \sin \left (f x + e\right ) + c}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)*(a*sin(f*x + e) + a)^2*sqrt(-c*sin(f*x + e) + c), x)

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Fricas [B]  time = 1.50575, size = 463, normalized size = 5.72 \begin{align*} \frac{2 \,{\left (5 \, B a^{2} \cos \left (f x + e\right )^{4} -{\left (7 \, A + 8 \, B\right )} a^{2} \cos \left (f x + e\right )^{3} -{\left (21 \, A + 19 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} + 2 \,{\left (7 \, A + 3 \, B\right )} a^{2} \cos \left (f x + e\right ) + 4 \,{\left (7 \, A + 3 \, B\right )} a^{2} -{\left (5 \, B a^{2} \cos \left (f x + e\right )^{3} +{\left (7 \, A + 13 \, B\right )} a^{2} \cos \left (f x + e\right )^{2} - 2 \,{\left (7 \, A + 3 \, B\right )} a^{2} \cos \left (f x + e\right ) - 4 \,{\left (7 \, A + 3 \, B\right )} a^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{35 \,{\left (f \cos \left (f x + e\right ) - f \sin \left (f x + e\right ) + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

2/35*(5*B*a^2*cos(f*x + e)^4 - (7*A + 8*B)*a^2*cos(f*x + e)^3 - (21*A + 19*B)*a^2*cos(f*x + e)^2 + 2*(7*A + 3*
B)*a^2*cos(f*x + e) + 4*(7*A + 3*B)*a^2 - (5*B*a^2*cos(f*x + e)^3 + (7*A + 13*B)*a^2*cos(f*x + e)^2 - 2*(7*A +
3*B)*a^2*cos(f*x + e) - 4*(7*A + 3*B)*a^2)*sin(f*x + e))*sqrt(-c*sin(f*x + e) + c)/(f*cos(f*x + e) - f*sin(f*
x + e) + f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int A \sqrt{- c \sin{\left (e + f x \right )} + c}\, dx + \int 2 A \sqrt{- c \sin{\left (e + f x \right )} + c} \sin{\left (e + f x \right )}\, dx + \int A \sqrt{- c \sin{\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\, dx + \int B \sqrt{- c \sin{\left (e + f x \right )} + c} \sin{\left (e + f x \right )}\, dx + \int 2 B \sqrt{- c \sin{\left (e + f x \right )} + c} \sin ^{2}{\left (e + f x \right )}\, dx + \int B \sqrt{- c \sin{\left (e + f x \right )} + c} \sin ^{3}{\left (e + f x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))**(1/2),x)

[Out]

a**2*(Integral(A*sqrt(-c*sin(e + f*x) + c), x) + Integral(2*A*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x) + Int
egral(A*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(B*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x), x)
+ Integral(2*B*sqrt(-c*sin(e + f*x) + c)*sin(e + f*x)**2, x) + Integral(B*sqrt(-c*sin(e + f*x) + c)*sin(e + f*
x)**3, x))

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Giac [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+a*sin(f*x+e))^2*(A+B*sin(f*x+e))*(c-c*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

Timed out