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

Optimal. Leaf size=68 \[ \frac{8 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 c f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.195654, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ \frac{8 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*a^2*f) - (2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(5/2))/(a^2*c
*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 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{(c-c \sin (e+f x))^{3/2}}{(a+a \sin (e+f x))^2} \, dx &=\frac{\int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a^2 c^2}\\ &=-\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 c f}-\frac{4 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{a^2 c}\\ &=\frac{8 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 c f}\\ \end{align*}

Mathematica [A]  time = 0.351781, size = 92, normalized size = 1.35 \[ \frac{2 c (3 \sin (e+f x)+1) \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{3 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(1 + 3*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)

________________________________________________________________________________________

Maple [A]  time = 0.582, size = 61, normalized size = 0.9 \begin{align*} -{\frac{2\,{c}^{2} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 3\,\sin \left ( fx+e \right ) +1 \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((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))*(3*sin(f*x+e)+1)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

________________________________________________________________________________________

Maxima [B]  time = 1.87996, size = 323, normalized size = 4.75 \begin{align*} -\frac{2 \,{\left (c^{\frac{3}{2}} + \frac{6 \, c^{\frac{3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{12 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{3 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{6 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{c^{\frac{3}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\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{\left (\frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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 + 12
*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^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*(sin(f*
x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(3/2))

________________________________________________________________________________________

Fricas [A]  time = 1.03494, size = 147, normalized size = 2.16 \begin{align*} \frac{2 \,{\left (3 \, c \sin \left (f x + e\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((c-c*sin(f*x+e))^(3/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

2/3*(3*c*sin(f*x + e) + 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))

________________________________________________________________________________________

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))**(3/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.79874, size = 579, normalized size = 8.51 \begin{align*} -\frac{2 \,{\left (\frac{{\left (2 \, \sqrt{2} c^{\frac{3}{2}} - 3 \, c^{\frac{3}{2}}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{5 \, \sqrt{2} a^{2} - 7 \, a^{2}} - \frac{2 \,{\left (3 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{5} c^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) - 3 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{4} c^{\frac{5}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) + 2 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{3} c^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) + 6 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{2} c^{\frac{7}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) - 9 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )} c^{4} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) + c^{\frac{9}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )\right )}}{{\left ({\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )}^{2} + 2 \,{\left (\sqrt{c} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}\right )} \sqrt{c} - c\right )}^{3} a^{2}}\right )}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((2*sqrt(2)*c^(3/2) - 3*c^(3/2))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(5*sqrt(2)*a^2 - 7*a^2) - 2*(3*(sqrt(c)*ta
n(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*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*c^(5/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 2*(sqrt(c)*ta
n(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c^3*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*c^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 9*(sqrt(c)*ta
n(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*c^4*sgn(tan(1/2*f*x + 1/2*e) - 1) + 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