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

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

[Out]

-2/3*sec(f*x+e)^3*(c-c*sin(f*x+e))^(3/2)/a^2/c/f

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Rubi [A]  time = 0.14, antiderivative size = 36, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]))

Rubi steps

\begin {align*} \int \frac {\sqrt {c-c \sin (e+f x)}}{(a+a \sin (e+f x))^2} \, dx &=\frac {\int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{a^2 c^2}\\ &=-\frac {2 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 c f}\\ \end {align*}

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Mathematica [B]  time = 0.13, size = 73, normalized size = 2.03 \[ -\frac {2 \sqrt {c-c \sin (e+f x)}}{3 a^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*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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fricas [A]  time = 0.42, size = 46, normalized size = 1.28 \[ -\frac {2 \, \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, {\left (a^{2} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) + a^{2} f \cos \left (f x + e\right )\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

-2/3*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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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))^2,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)sqrt(2*c)/3*(-3*((-cos(1/4*(2*f*x+2*exp(1)-pi))+1)/(cos(1/4*(2
*f*x+2*exp(1)-pi))+1))^2*sign(sin(1/2*(f*x+exp(1))-1/4*pi))-sign(sin(1/2*(f*x+exp(1))-1/4*pi)))/a^2/((-cos(1/4
*(2*f*x+2*exp(1)-pi))+1)/(cos(1/4*(2*f*x+2*exp(1)-pi))+1)-1)^3/f

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maple [A]  time = 0.75, size = 49, normalized size = 1.36 \[ \frac {2 c \left (\sin \left (f x +e \right )-1\right )}{3 a^{2} \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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))^2,x)

[Out]

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

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maxima [B]  time = 0.98, size = 149, normalized size = 4.14 \[ \frac {2 \, {\left (\sqrt {c} + \frac {2 \, \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {\sqrt {c} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}\right )}}{3 \, {\left (a^{2} + \frac {3 \, a^{2} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {3 \, a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )} f \sqrt {\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1}} \]

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))^2,x, algorithm="maxima")

[Out]

2/3*(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)*f*sqrt(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1))

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mupad [B]  time = 9.35, size = 227, normalized size = 6.31 \[ -\frac {4\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (\sin \left (2\,e+2\,f\,x\right )-4\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-{\sin \left (e+f\,x\right )}^2\,2{}\mathrm {i}+2+2{}\mathrm {i}\right )}{3\,a^2\,f\,\left (-4\,{\sin \left (e+f\,x\right )}^2+\sin \left (e+f\,x\right )+\sin \left (3\,e+3\,f\,x\right )+4\right )}+\frac {4\,\sqrt {-c\,\left (\sin \left (e+f\,x\right )-1\right )}\,\left (-{\sin \left (e+f\,x\right )}^2\,4{}\mathrm {i}+\sin \left (e+f\,x\right )\,1{}\mathrm {i}-2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+2\,{\sin \left (\frac {3\,e}{2}+\frac {3\,f\,x}{2}\right )}^2+2\,\sin \left (2\,e+2\,f\,x\right )+\sin \left (3\,e+3\,f\,x\right )\,1{}\mathrm {i}+4{}\mathrm {i}\right )}{3\,a^2\,f\,\left (-8\,{\sin \left (e+f\,x\right )}^2+4\,\sin \left (e+f\,x\right )+2\,{\sin \left (2\,e+2\,f\,x\right )}^2+4\,\sin \left (3\,e+3\,f\,x\right )+8\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*sin(e + f*x))^(1/2)/(a + a*sin(e + f*x))^2,x)

[Out]

(4*(-c*(sin(e + f*x) - 1))^(1/2)*(sin(e + f*x)*1i + 2*sin(2*e + 2*f*x) + sin(3*e + 3*f*x)*1i - 2*sin(e/2 + (f*
x)/2)^2 + 2*sin((3*e)/2 + (3*f*x)/2)^2 - sin(e + f*x)^2*4i + 4i))/(3*a^2*f*(4*sin(e + f*x) + 4*sin(3*e + 3*f*x
) + 2*sin(2*e + 2*f*x)^2 - 8*sin(e + f*x)^2 + 8)) - (4*(-c*(sin(e + f*x) - 1))^(1/2)*(sin(2*e + 2*f*x) - 4*sin
(e/2 + (f*x)/2)^2 - sin(e + f*x)^2*2i + (2 + 2i)))/(3*a^2*f*(sin(e + f*x) + sin(3*e + 3*f*x) - 4*sin(e + f*x)^
2 + 4))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {\sqrt {- c \sin {\left (e + f x \right )} + c}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

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))**2,x)

[Out]

Integral(sqrt(-c*sin(e + f*x) + c)/(sin(e + f*x)**2 + 2*sin(e + f*x) + 1), x)/a**2

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