3.324 \(\int \frac{(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=176 \[ -\frac{1024 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^2 f}+\frac{4096 c^3 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^2 c f}+\frac{32 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^2 f}+\frac{128 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^2 f} \]

[Out]

(4096*c^3*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^2*f) - (1024*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x]
)^(5/2))/(5*a^2*f) + (128*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^2*f) + (32*Sec[e + f*x]^3*(c - c*S
in[e + f*x])^(9/2))/(15*a^2*f) + (2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(11/2))/(5*a^2*c*f)

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Rubi [A]  time = 0.419948, antiderivative size = 176, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {2736, 2674, 2673} \[ -\frac{1024 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^2 f}+\frac{4096 c^3 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^2 c f}+\frac{32 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^2 f}+\frac{128 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^(9/2)/(a + a*Sin[e + f*x])^2,x]

[Out]

(4096*c^3*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(3/2))/(15*a^2*f) - (1024*c^2*Sec[e + f*x]^3*(c - c*Sin[e + f*x]
)^(5/2))/(5*a^2*f) + (128*c*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(7/2))/(5*a^2*f) + (32*Sec[e + f*x]^3*(c - c*S
in[e + f*x])^(9/2))/(15*a^2*f) + (2*Sec[e + f*x]^3*(c - c*Sin[e + f*x])^(11/2))/(5*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))^{9/2}}{(a+a \sin (e+f x))^2} \, dx &=\frac{\int \sec ^4(e+f x) (c-c \sin (e+f x))^{13/2} \, dx}{a^2 c^2}\\ &=\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^2 c f}+\frac{16 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{5 a^2 c}\\ &=\frac{32 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^2 c f}+\frac{64 \int \sec ^4(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{5 a^2}\\ &=\frac{128 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^2 f}+\frac{32 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^2 c f}+\frac{(512 c) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{5 a^2}\\ &=-\frac{1024 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^2 f}+\frac{128 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^2 f}+\frac{32 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^2 c f}-\frac{\left (2048 c^2\right ) \int \sec ^4(e+f x) (c-c \sin (e+f x))^{5/2} \, dx}{5 a^2}\\ &=\frac{4096 c^3 \sec ^3(e+f x) (c-c \sin (e+f x))^{3/2}}{15 a^2 f}-\frac{1024 c^2 \sec ^3(e+f x) (c-c \sin (e+f x))^{5/2}}{5 a^2 f}+\frac{128 c \sec ^3(e+f x) (c-c \sin (e+f x))^{7/2}}{5 a^2 f}+\frac{32 \sec ^3(e+f x) (c-c \sin (e+f x))^{9/2}}{15 a^2 f}+\frac{2 \sec ^3(e+f x) (c-c \sin (e+f x))^{11/2}}{5 a^2 c f}\\ \end{align*}

Mathematica [A]  time = 3.07125, size = 124, normalized size = 0.7 \[ \frac{c^4 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (8568 \sin (e+f x)+56 \sin (3 (e+f x))-1044 \cos (2 (e+f x))+3 \cos (4 (e+f x))+6825)}{60 a^2 f (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^(9/2)/(a + a*Sin[e + f*x])^2,x]

[Out]

(c^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(6825 - 1044*Cos[2*(e + f*x)] + 3*Cos[4*(e
+ f*x)] + 8568*Sin[e + f*x] + 56*Sin[3*(e + f*x)]))/(60*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 = 0.638, size = 91, normalized size = 0.5 \begin{align*} -{\frac{2\,{c}^{5} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 3\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}-28\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}+258\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}+1092\,\sin \left ( fx+e \right ) +723 \right ) }{15\,{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))^(9/2)/(a+a*sin(f*x+e))^2,x)

[Out]

-2/15*c^5/a^2*(-1+sin(f*x+e))/(1+sin(f*x+e))*(3*sin(f*x+e)^4-28*sin(f*x+e)^3+258*sin(f*x+e)^2+1092*sin(f*x+e)+
723)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f

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Maxima [B]  time = 1.73688, size = 513, normalized size = 2.91 \begin{align*} -\frac{2 \,{\left (723 \, c^{\frac{9}{2}} + \frac{2184 \, c^{\frac{9}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{5370 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{10696 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{15021 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{21168 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{20748 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{21168 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac{15021 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac{10696 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac{5370 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}} + \frac{2184 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{11}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{11}} + \frac{723 \, c^{\frac{9}{2}} \sin \left (f x + e\right )^{12}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{12}}\right )}}{15 \,{\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{9}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

-2/15*(723*c^(9/2) + 2184*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 5370*c^(9/2)*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + 10696*c^(9/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15021*c^(9/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
+ 21168*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 20748*c^(9/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 211
68*c^(9/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 15021*c^(9/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 10696*c^(
9/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 5370*c^(9/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 2184*c^(9/2)*s
in(f*x + e)^11/(cos(f*x + e) + 1)^11 + 723*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12)/((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)^(9/2))

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Fricas [A]  time = 1.06822, size = 261, normalized size = 1.48 \begin{align*} \frac{2 \,{\left (3 \, c^{4} \cos \left (f x + e\right )^{4} - 264 \, c^{4} \cos \left (f x + e\right )^{2} + 984 \, c^{4} + 28 \,{\left (c^{4} \cos \left (f x + e\right )^{2} + 38 \, c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt{-c \sin \left (f x + e\right ) + c}}{15 \,{\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))^(9/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

2/15*(3*c^4*cos(f*x + e)^4 - 264*c^4*cos(f*x + e)^2 + 984*c^4 + 28*(c^4*cos(f*x + e)^2 + 38*c^4)*sin(f*x + e))
*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))**(9/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

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Giac [B]  time = 2.09839, size = 887, normalized size = 5.04 \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))^(9/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

-1/60*(4*(511*sqrt(2)*a^12*c - 730*a^12*c - 920*sqrt(2)*c^10 + 1200*c^10)*sgn(tan(1/2*f*x + 1/2*e) - 1)/(5*sqr
t(2)*a^2*c^(11/2) - 7*a^2*c^(11/2)) + (323*a^10*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^2 + (255*a^10*sgn(tan(1/2*f*x
+ 1/2*e) - 1)/c^2 + (590*a^10*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^2 + (590*a^10*sgn(tan(1/2*f*x + 1/2*e) - 1)/c^2
+ 17*(19*a^10*sgn(tan(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x + 1/2*e)/c^2 + 15*a^10*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))*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)^(5/2) + 1280*(3*(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) + 15*(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) - 10*(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) - 30*(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) + 27*(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) - 5*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)^3*a^2))/f