3.336 \(\int \frac{\sqrt{c-c \sin (e+f x)}}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=36 \[ -\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^2 f} \]

[Out]

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

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Rubi [A]  time = 0.122419, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {2736, 2673} \[ -\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^2 f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c - c*Sin[e + f*x]]/(a + a*Sin[e + f*x])^3,x]

[Out]

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

Rule 2736

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Di
st[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&
EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0,
n, m] || LtQ[m, n, 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{\sqrt{c-c \sin (e+f x)}}{(a+a \sin (e+f x))^3} \, dx &=\frac{\int \sec ^6(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a^3 c^3}\\ &=-\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^3 c^2 f}\\ \end{align*}

Mathematica [B]  time = 0.1418, size = 73, normalized size = 2.03 \[ -\frac{2 \sqrt{c-c \sin (e+f x)}}{5 a^3 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.425, size = 49, normalized size = 1.4 \begin{align*}{\frac{2\,c \left ( -1+\sin \left ( fx+e \right ) \right ) }{5\,{a}^{3} \left ( 1+\sin \left ( fx+e \right ) \right ) ^{2}\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((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x)

[Out]

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

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Maxima [B]  time = 1.74812, size = 294, normalized size = 8.17 \begin{align*} \frac{2 \,{\left (\sqrt{c} + \frac{3 \, \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 )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{\sqrt{c} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{5 \,{\left (a^{3} + \frac{5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} f \sqrt{\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.07758, size = 153, normalized size = 4.25 \begin{align*} \frac{2 \, \sqrt{-c \sin \left (f x + e\right ) + c}}{5 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

2/5*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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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((c-c*sin(f*x+e))**(1/2)/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.70931, size = 875, normalized size = 24.31 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/20*((191*sqrt(2)*sqrt(c) - 270*sqrt(c))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(29*sqrt(2)*a^3 - 41*a^3) - 16*(5*(sq
rt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^9*c*sgn(tan(1/2*f*x + 1/2*e) - 1) + 15*(sqrt(
c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*c^(3/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 20*(sq
rt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*c^2*sgn(tan(1/2*f*x + 1/2*e) - 1) - 20*(sqr
t(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 34*(
sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) + 10*(s
qrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 20
*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) - 20*
(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) +
5*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) - c^(1
1/2)*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)^5*a^3))/f