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

Optimal. Leaf size=100 \[ \frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}-\frac{16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac{64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f} \]

[Out]

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

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Rubi [A]  time = 0.26357, antiderivative size = 100, 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} \[ \frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}-\frac{16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac{64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(64*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(3*a^2*f) - (16*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(5/2))/(a
^2*f) + (2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(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))^{5/2}}{(a+a \sin (e+f x))^2} \, dx &=\frac{\int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a^2 c^2}\\ &=\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}+\frac{8 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{a^2 c}\\ &=-\frac{16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}-\frac{32 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{a^2}\\ &=\frac{64 c \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{3 a^2 f}-\frac{16 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{a^2 c f}\\ \end{align*}

Mathematica [A]  time = 0.780483, size = 104, normalized size = 1.04 \[ \frac{c^2 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (36 \sin (e+f x)-3 \cos (2 (e+f x))+25)}{3 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(25 - 3*Cos[2*(e + f*x)] + 36*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)

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Maple [A]  time = 0.601, size = 71, normalized size = 0.7 \begin{align*} -{\frac{2\,{c}^{3} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 3\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+18\,\sin \left ( fx+e \right ) +11 \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))^(5/2)/(a+a*sin(f*x+e))^2,x)

[Out]

-2/3*c^3/a^2*(-1+sin(f*x+e))/(1+sin(f*x+e))*(3*sin(f*x+e)^2+18*sin(f*x+e)+11)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2
)/f

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Maxima [B]  time = 1.77455, size = 389, normalized size = 3.89 \begin{align*} -\frac{2 \,{\left (11 \, c^{\frac{5}{2}} + \frac{36 \, c^{\frac{5}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{56 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{108 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{90 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{108 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{56 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{36 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac{11 \, c^{\frac{5}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\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{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))^2,x, algorithm="maxima")

[Out]

-2/3*(11*c^(5/2) + 36*c^(5/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 56*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 108*c^(5/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 90*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 108*c^(5
/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 56*c^(5/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 36*c^(5/2)*sin(f*x
+ e)^7/(cos(f*x + e) + 1)^7 + 11*c^(5/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)/((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)^(5/2))

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Fricas [A]  time = 1.07377, size = 190, normalized size = 1.9 \begin{align*} -\frac{2 \,{\left (3 \, c^{2} \cos \left (f x + e\right )^{2} - 18 \, c^{2} \sin \left (f x + e\right ) - 14 \, c^{2}\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))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

-2/3*(3*c^2*cos(f*x + e)^2 - 18*c^2*sin(f*x + e) - 14*c^2)*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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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))**2,x)

[Out]

Timed out

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Giac [B]  time = 2.49808, size = 603, normalized size = 6.03 \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))^2,x, algorithm="giac")

[Out]

-2/3*(8*(2*sqrt(2)*c^(5/2) - 3*c^(5/2))*sgn(tan(1/2*f*x + 1/2*e) - 1)/(5*sqrt(2)*a^2 - 7*a^2) + 3*(c^3*sgn(tan
(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x + 1/2*e)/a^2 + c^3*sgn(tan(1/2*f*x + 1/2*e) - 1)/a^2)/sqrt(c*tan(1/2*f*x +
1/2*e)^2 + c) + 16*(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^(7/2)*sgn(tan(1/
2*f*x + 1/2*e) - 1) - 2*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c^4*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^(9/2)*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))*c^5*sgn(tan(1/2*f*
x + 1/2*e) - 1) - c^(11/2)*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