3.318 \(\int \frac{(c-c \sin (e+f x))^{5/2}}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=98 \[ -\frac{64 c^2 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}+\frac{16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f} \]

[Out]

(-64*c^2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(3*a*f) + (16*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(3*a*
f) + (2*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*a*f)

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Rubi [A]  time = 0.269496, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ -\frac{64 c^2 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}+\frac{16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*c^2*Sec[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(3*a*f) + (16*c*Sec[e + f*x]*(c - c*Sin[e + f*x])^(3/2))/(3*a*
f) + (2*Sec[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*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))^{5/2}}{a+a \sin (e+f x)} \, dx &=\frac{\int \sec ^2(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a c}\\ &=\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}+\frac{8 \int \sec ^2(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{3 a}\\ &=\frac{16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}+\frac{(32 c) \int \sec ^2(e+f x) (c-c \sin (e+f x))^{3/2} \, dx}{3 a}\\ &=-\frac{64 c^2 \sec (e+f x) \sqrt{c-c \sin (e+f x)}}{3 a f}+\frac{16 c \sec (e+f x) (c-c \sin (e+f x))^{3/2}}{3 a f}+\frac{2 \sec (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a f}\\ \end{align*}

Mathematica [A]  time = 0.704548, size = 102, normalized size = 1.04 \[ -\frac{c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (20 \sin (e+f x)+\cos (2 (e+f x))+45)}{3 a f (\sin (e+f x)+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(45 + Cos[2*(e + f*x)] + 20*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]
)/(3*a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x]))

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Maple [A]  time = 0.403, size = 59, normalized size = 0.6 \begin{align*} -{\frac{2\,{c}^{3} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( \left ( \sin \left ( fx+e \right ) \right ) ^{2}-10\,\sin \left ( fx+e \right ) -23 \right ) }{3\,af\cos \left ( fx+e \right ) }{\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))^(5/2)/(a+a*sin(f*x+e)),x)

[Out]

-2/3*c^3/a*(-1+sin(f*x+e))*(sin(f*x+e)^2-10*sin(f*x+e)-23)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [B]  time = 1.79514, size = 259, normalized size = 2.64 \begin{align*} \frac{2 \,{\left (23 \, c^{\frac{5}{2}} + \frac{20 \, c^{\frac{5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{65 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{40 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{65 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{20 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{23 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )}}{3 \,{\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{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

2/3*(23*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 65*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 40*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 65*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 20*c^(5/2)
*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 23*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)/((a + a*sin(f*x + e)/(c
os(f*x + e) + 1))*f*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)^(5/2))

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Fricas [A]  time = 1.07028, size = 139, normalized size = 1.42 \begin{align*} -\frac{2 \,{\left (c^{2} \cos \left (f x + e\right )^{2} + 10 \, c^{2} \sin \left (f x + e\right ) + 22 \, c^{2}\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{3 \, 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))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

-2/3*(c^2*cos(f*x + e)^2 + 10*c^2*sin(f*x + e) + 22*c^2)*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))**(5/2)/(a+a*sin(f*x+e)),x)

[Out]

Timed out

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Giac [B]  time = 1.6757, size = 486, normalized size = 4.96 \begin{align*} -\frac{\frac{2 \,{\left (6 \, \sqrt{2} c^{7} - 5 \, \sqrt{2} a^{4} c + 10 \, a^{4} c\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{\sqrt{2} a c^{\frac{9}{2}} - a c^{\frac{9}{2}}} - \frac{\frac{11 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{2}} +{\left (\frac{9 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{2}} +{\left (\frac{11 \, a^{3} \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 )}{c^{2}} + \frac{9 \, a^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}{c^{2}}\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{{\left (c \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c\right )}^{\frac{3}{2}}} - \frac{48 \,{\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^{3} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right ) - c^{\frac{7}{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}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

-1/3*(2*(6*sqrt(2)*c^7 - 5*sqrt(2)*a^4*c + 10*a^4*c)*sgn(tan(1/2*f*x + 1/2*e) - 1)/(sqrt(2)*a*c^(9/2) - a*c^(9
/2)) - (11*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^2 + (9*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^2 + (11*a^3*sgn(tan(
1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x + 1/2*e)/c^2 + 9*a^3*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^2)*tan(1/2*f*x + 1/2*e)
)*tan(1/2*f*x + 1/2*e))/(c*tan(1/2*f*x + 1/2*e)^2 + c)^(3/2) - 48*((sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(
1/2*f*x + 1/2*e)^2 + c))*c^3*sgn(tan(1/2*f*x + 1/2*e) - 1) - 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)*a))/f