3.334 \(\int \frac{(c-c \sin (e+f x))^{5/2}}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=104 \[ -\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c^2 f}+\frac{16 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c f}-\frac{64 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 f} \]

[Out]

(-64*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(15*a^3*f) + (16*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(7/2))/(3
*a^3*c*f) - (2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/(a^3*c^2*f)

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Rubi [A]  time = 0.269207, antiderivative size = 104, 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} \[ -\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c^2 f}+\frac{16 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c f}-\frac{64 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-64*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(15*a^3*f) + (16*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(7/2))/(3
*a^3*c*f) - (2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/(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 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))^3} \, dx &=\frac{\int \sec ^6(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{a^3 c^3}\\ &=-\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c^2 f}-\frac{8 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a^3 c^2}\\ &=\frac{16 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c f}-\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c^2 f}+\frac{32 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{3 a^3 c}\\ &=-\frac{64 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 f}+\frac{16 \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 c f}-\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 c^2 f}\\ \end{align*}

Mathematica [A]  time = 0.841201, size = 104, normalized size = 1. \[ \frac{c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (-20 \sin (e+f x)+15 \cos (2 (e+f x))-29)}{15 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-29 + 15*Cos[2*(e + f*x)] - 20*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*
x]])/(15*a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3)

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Maple [A]  time = 0.504, size = 71, normalized size = 0.7 \begin{align*}{\frac{2\,{c}^{3} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 15\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+10\,\sin \left ( fx+e \right ) +7 \right ) }{15\,{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))^(5/2)/(a+a*sin(f*x+e))^3,x)

[Out]

2/15*c^3/a^3*(-1+sin(f*x+e))/(1+sin(f*x+e))^2*(15*sin(f*x+e)^2+10*sin(f*x+e)+7)/cos(f*x+e)/(c-c*sin(f*x+e))^(1
/2)/f

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Maxima [B]  time = 1.82795, size = 513, normalized size = 4.93 \begin{align*} \frac{2 \,{\left (7 \, c^{\frac{5}{2}} + \frac{20 \, c^{\frac{5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{95 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{80 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{250 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{120 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{250 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{80 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac{95 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac{20 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac{7 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}\right )}}{15 \,{\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{\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))^3,x, algorithm="maxima")

[Out]

2/15*(7*c^(5/2) + 20*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 95*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 80*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 250*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 120*c^(5/
2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 250*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 80*c^(5/2)*sin(f*x
+ e)^7/(cos(f*x + e) + 1)^7 + 95*c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 20*c^(5/2)*sin(f*x + e)^9/(cos(
f*x + e) + 1)^9 + 7*c^(5/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)/((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*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 +
1)^(5/2))

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Fricas [A]  time = 1.12084, size = 230, normalized size = 2.21 \begin{align*} -\frac{2 \,{\left (15 \, 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}}{15 \,{\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))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/15*(15*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^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))**(5/2)/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [B]  time = 1.94364, size = 878, normalized size = 8.44 \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))^(5/2)/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-2/15*((39*sqrt(2)*c^(5/2) - 55*c^(5/2))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(29*sqrt(2)*a^3 - 41*a^3) - 2*(15*(sqrt
(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^9*c^3*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^(7/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 100*(s
qrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^7*c^4*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))^6*c^(9/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 238
*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*c^5*sgn(tan(1/2*f*x + 1/2*e) - 1) + 190
*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1)
+ 180*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c^6*sgn(tan(1/2*f*x + 1/2*e) - 1)
- 260*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*c^(13/2)*sgn(tan(1/2*f*x + 1/2*e)
- 1) + 55*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*c^7*sgn(tan(1/2*f*x + 1/2*e) - 1
) - 7*c^(15/2)*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