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

Optimal. Leaf size=78 \[ -\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 c^2 f}-\frac{(A+5 B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 f} \]

[Out]

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

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Rubi [A]  time = 0.312557, antiderivative size = 78, 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, 2855, 2673} \[ -\frac{(A-B) \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 c^2 f}-\frac{(A+5 B) \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.280934, size = 87, normalized size = 1.12 \[ -\frac{2 \sqrt{c-c \sin (e+f x)} (A+3 B \sin (e+f x)+2 B)}{3 a^2 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(A + 2*B + 3*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])*(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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Maple [A]  time = 1.041, size = 63, normalized size = 0.8 \begin{align*}{\frac{2\,c \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 3\,B\sin \left ( fx+e \right ) +A+2\,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))^(1/2)/(a+a*sin(f*x+e))^2,x)

[Out]

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

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Maxima [B]  time = 1.52507, size = 463, normalized size = 5.94 \begin{align*} \frac{2 \,{\left (\frac{2 \, B{\left (\sqrt{c} + \frac{3 \, \sqrt{c} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \, \sqrt{c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{3 \, \sqrt{c} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{\sqrt{c} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} + \frac{A{\left (\sqrt{c} + \frac{2 \, \sqrt{c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{\sqrt{c} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{{\left (a^{2} + \frac{3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}}\right )}}{3 \, f} \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))^(1/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

2/3*(2*B*(sqrt(c) + 3*sqrt(c)*sin(f*x + e)/(cos(f*x + e) + 1) + 2*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 3*sqrt(c)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + sqrt(c)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)/((a^2 + 3*a^2*s
in(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)*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)) + A*(sqrt(c) + 2*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + sqrt(c)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4)/((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)*sqrt(sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 1)))/f

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Fricas [A]  time = 1.59285, size = 157, normalized size = 2.01 \begin{align*} -\frac{2 \,{\left (3 \, B \sin \left (f x + e\right ) + A + 2 \, B\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))^(1/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-2/3*(3*B*sin(f*x + e) + A + 2*B)*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))**(1/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 1.70291, size = 934, normalized size = 11.97 \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))^(1/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/6*((13*sqrt(2)*A*sqrt(c) + 5*sqrt(2)*B*sqrt(c) - 18*A*sqrt(c) - 6*B*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e) - 1)/
(5*sqrt(2)*a^2 - 7*a^2) - 8*(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*sgn(t
an(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^(3/2)*s
gn(tan(1/2*f*x + 1/2*e) - 1) + 6*(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^(3/
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
^2*sgn(tan(1/2*f*x + 1/2*e) - 1) - 4*(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
^2*sgn(tan(1/2*f*x + 1/2*e) - 1) - 6*(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
^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 12*(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^(5/2)*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))*A*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) + 12*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c
))*B*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - A*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 2*B*c^(7/2)*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