3.332 \(\int \frac{(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx\)
Optimal. Leaf size=174 \[ \frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^3 c^2 f}-\frac{4096 c^2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 f}+\frac{32 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^3 c f}-\frac{128 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 f}+\frac{1024 c \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 f} \]
[Out]
(-4096*c^2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(15*a^3*f) + (1024*c*Sec[e + f*x]^5*(c - c*Sin[e + f*x])
^(7/2))/(3*a^3*f) - (128*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/(a^3*f) + (32*Sec[e + f*x]^5*(c - c*Sin[e
+ f*x])^(11/2))/(3*a^3*c*f) + (2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(13/2))/(3*a^3*c^2*f)
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Rubi [A] time = 0.403117, antiderivative size = 174, 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} \[ \frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^3 c^2 f}-\frac{4096 c^2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 f}+\frac{32 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^3 c f}-\frac{128 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 f}+\frac{1024 c \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 f} \]
Antiderivative was successfully verified.
[In]
Int[(c - c*Sin[e + f*x])^(9/2)/(a + a*Sin[e + f*x])^3,x]
[Out]
(-4096*c^2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(5/2))/(15*a^3*f) + (1024*c*Sec[e + f*x]^5*(c - c*Sin[e + f*x])
^(7/2))/(3*a^3*f) - (128*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(9/2))/(a^3*f) + (32*Sec[e + f*x]^5*(c - c*Sin[e
+ f*x])^(11/2))/(3*a^3*c*f) + (2*Sec[e + f*x]^5*(c - c*Sin[e + f*x])^(13/2))/(3*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))^{9/2}}{(a+a \sin (e+f x))^3} \, dx &=\frac{\int \sec ^6(e+f x) (c-c \sin (e+f x))^{15/2} \, dx}{a^3 c^3}\\ &=\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^3 c^2 f}+\frac{16 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{13/2} \, dx}{3 a^3 c^2}\\ &=\frac{32 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^3 c f}+\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^3 c^2 f}+\frac{64 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{11/2} \, dx}{a^3 c}\\ &=-\frac{128 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 f}+\frac{32 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^3 c f}+\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^3 c^2 f}-\frac{512 \int \sec ^6(e+f x) (c-c \sin (e+f x))^{9/2} \, dx}{a^3}\\ &=\frac{1024 c \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 f}-\frac{128 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 f}+\frac{32 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^3 c f}+\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^3 c^2 f}+\frac{(2048 c) \int \sec ^6(e+f x) (c-c \sin (e+f x))^{7/2} \, dx}{3 a^3}\\ &=-\frac{4096 c^2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 f}+\frac{1024 c \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 f}-\frac{128 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 f}+\frac{32 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^3 c f}+\frac{2 \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^3 c^2 f}\\ \end{align*}
Mathematica [A] time = 3.01368, size = 124, normalized size = 0.71 \[ \frac{c^4 \sqrt{c-c \sin (e+f x)} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (-7800 \sin (e+f x)+200 \sin (3 (e+f x))+2740 \cos (2 (e+f x))+5 \cos (4 (e+f x))-5649)}{60 a^3 f (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )} \]
Antiderivative was successfully verified.
[In]
Integrate[(c - c*Sin[e + f*x])^(9/2)/(a + a*Sin[e + f*x])^3,x]
[Out]
(c^4*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*Sqrt[c - c*Sin[e + f*x]]*(-5649 + 2740*Cos[2*(e + f*x)] + 5*Cos[4*(
e + f*x)] - 7800*Sin[e + f*x] + 200*Sin[3*(e + f*x)]))/(60*a^3*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Si
n[e + f*x])^3)
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Maple [A] time = 0.761, size = 91, normalized size = 0.5 \begin{align*} -{\frac{2\,{c}^{5} \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 5\, \left ( \sin \left ( fx+e \right ) \right ) ^{4}-100\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}-690\, \left ( \sin \left ( fx+e \right ) \right ) ^{2}-900\,\sin \left ( fx+e \right ) -363 \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))^(9/2)/(a+a*sin(f*x+e))^3,x)
[Out]
-2/15*c^5/a^3*(-1+sin(f*x+e))/(1+sin(f*x+e))^2*(5*sin(f*x+e)^4-100*sin(f*x+e)^3-690*sin(f*x+e)^2-900*sin(f*x+e
)-363)/cos(f*x+e)/(c-c*sin(f*x+e))^(1/2)/f
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Maxima [B] time = 1.82271, size = 637, normalized size = 3.66 \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))^3,x, algorithm="maxima")
[Out]
2/15*(363*c^(9/2) + 1800*c^(9/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 5301*c^(9/2)*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 11600*c^(9/2)*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 21343*c^(9/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4
+ 30200*c^(9/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 40065*c^(9/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 4080
0*c^(9/2)*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 40065*c^(9/2)*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 30200*c^(9
/2)*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 21343*c^(9/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 + 11600*c^(9/2)*
sin(f*x + e)^11/(cos(f*x + e) + 1)^11 + 5301*c^(9/2)*sin(f*x + e)^12/(cos(f*x + e) + 1)^12 + 1800*c^(9/2)*sin(
f*x + e)^13/(cos(f*x + e) + 1)^13 + 363*c^(9/2)*sin(f*x + e)^14/(cos(f*x + e) + 1)^14)/((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/(c
os(f*x + e) + 1)^2 + 1)^(9/2))
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Fricas [A] time = 1.14366, size = 301, normalized size = 1.73 \begin{align*} -\frac{2 \,{\left (5 \, c^{4} \cos \left (f x + e\right )^{4} + 680 \, c^{4} \cos \left (f x + e\right )^{2} - 1048 \, c^{4} + 100 \,{\left (c^{4} \cos \left (f x + e\right )^{2} - 10 \, c^{4}\right )} \sin \left (f x + e\right )\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))^(9/2)/(a+a*sin(f*x+e))^3,x, algorithm="fricas")
[Out]
-2/15*(5*c^4*cos(f*x + e)^4 + 680*c^4*cos(f*x + e)^2 - 1048*c^4 + 100*(c^4*cos(f*x + e)^2 - 10*c^4)*sin(f*x +
e))*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))**(9/2)/(a+a*sin(f*x+e))**3,x)
[Out]
Timed out
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Giac [B] time = 2.34077, size = 1084, normalized size = 6.23 \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))^3,x, algorithm="giac")
[Out]
1/15*(2*(2255*sqrt(2)*a^12*sqrt(c) - 3190*a^12*sqrt(c) - 1206*sqrt(2)*c^(13/2) + 1700*c^(13/2))*sgn(tan(1/2*f*
x + 1/2*e) - 1)/(29*sqrt(2)*a^3*c^2 - 41*a^3*c^2) + 5*(23*a^9*sgn(tan(1/2*f*x + 1/2*e) - 1) + (21*a^9*sgn(tan(
1/2*f*x + 1/2*e) - 1) + (23*a^9*sgn(tan(1/2*f*x + 1/2*e) - 1)*tan(1/2*f*x + 1/2*e) + 21*a^9*sgn(tan(1/2*f*x +
1/2*e) - 1))*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) + 32*(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^5*sgn(tan(1/2*f*x + 1/2*e) - 1) + 105*(sqrt(c
)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^8*c^(11/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) + 340*(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^6*sgn(tan(1/2*f*x + 1/2*e) - 1) - 260*(s
qrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^6*c^(13/2)*sgn(tan(1/2*f*x + 1/2*e) - 1) - 1
054*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^5*c^7*sgn(tan(1/2*f*x + 1/2*e) - 1) +
670*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^4*c^(15/2)*sgn(tan(1/2*f*x + 1/2*e) -
1) + 900*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^3*c^8*sgn(tan(1/2*f*x + 1/2*e) -
1) - 980*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))^2*c^(17/2)*sgn(tan(1/2*f*x + 1/2*
e) - 1) + 295*(sqrt(c)*tan(1/2*f*x + 1/2*e) - sqrt(c*tan(1/2*f*x + 1/2*e)^2 + c))*c^9*sgn(tan(1/2*f*x + 1/2*e)
- 1) - 31*c^(19/2)*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