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

Optimal. Leaf size=60 \[ \frac{2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a f}-\frac{8 c \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a f} \]

[Out]

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

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Rubi [A]  time = 0.203895, antiderivative size = 60, 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} \[ \frac{2 \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{a f}-\frac{8 c \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.29723, size = 88, normalized size = 1.47 \[ -\frac{2 c (\sin (e+f x)+3) \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{a f (\sin (e+f x)+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.454, size = 49, normalized size = 0.8 \begin{align*} 2\,{\frac{{c}^{2} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 3+\sin \left ( fx+e \right ) \right ) }{\cos \left ( fx+e \right ) a\sqrt{c-c\sin \left ( fx+e \right ) }f}} \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)),x)

[Out]

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

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Maxima [B]  time = 2.43061, size = 197, normalized size = 3.28 \begin{align*} \frac{2 \,{\left (3 \, c^{\frac{3}{2}} + \frac{2 \, c^{\frac{3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{6 \, c^{\frac{3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{2 \, 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}}\right )}}{{\left (a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\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)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.00194, size = 97, normalized size = 1.62 \begin{align*} -\frac{2 \,{\left (c \sin \left (f x + e\right ) + 3 \, c\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{a f \cos \left (f x + e\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)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.57878, size = 358, normalized size = 5.97 \begin{align*} -\frac{2 \,{\left (\frac{2 \, c^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{\sqrt{2} a - a} - \frac{\frac{c^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a} + \frac{c^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{a}}{\sqrt{c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c}} - \frac{4 \,{\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 )} c^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) - c^{\frac{5}{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 )} a}\right )}}{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)),x, algorithm="giac")

[Out]

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