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]

-4096/15*c^2*sec(f*x+e)^5*(c-c*sin(f*x+e))^(5/2)/a^3/f+1024/3*c*sec(f*x+e)^5*(c-c*sin(f*x+e))^(7/2)/a^3/f-128*
sec(f*x+e)^5*(c-c*sin(f*x+e))^(9/2)/a^3/f+32/3*sec(f*x+e)^5*(c-c*sin(f*x+e))^(11/2)/a^3/c/f+2/3*sec(f*x+e)^5*(
c-c*sin(f*x+e))^(13/2)/a^3/c^2/f

________________________________________________________________________________________

Rubi [A]  time = 0.40, antiderivative size = 174, normalized size of antiderivative = 1.00, 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 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]

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 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]))

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 = 3.15, 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)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 119, normalized size = 0.68 \[ -\frac {2 \, {\left (5 \, c^{4} \cos \left (f x + e\right )^{4} + 680 \, c^{4} \cos \left (f x + e\right )^{2} - 1048 \, c^{4} + 100 \, {\left (c^{4} \cos \left (f x + e\right )^{2} - 10 \, c^{4}\right )} \sin \left (f x + e\right )\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} - 2 \, a^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} f \cos \left (f x + e\right )\right )}} \]

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)
)

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

________________________________________________________________________________________

maple [A]  time = 0.79, size = 91, normalized size = 0.52 \[ -\frac {2 c^{5} \left (\sin \left (f x +e \right )-1\right ) \left (5 \left (\sin ^{4}\left (f x +e \right )\right )-100 \left (\sin ^{3}\left (f x +e \right )\right )-690 \left (\sin ^{2}\left (f x +e \right )\right )-900 \sin \left (f x +e \right )-363\right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

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*(sin(f*x+e)-1)/(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

________________________________________________________________________________________

maxima [B]  time = 0.84, size = 472, normalized size = 2.71 \[ \frac {2 \, {\left (363 \, c^{\frac {9}{2}} + \frac {1800 \, c^{\frac {9}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {5301 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {11600 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {21343 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {30200 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac {40065 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {40800 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac {40065 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac {30200 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} + \frac {21343 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}} + \frac {11600 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{11}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{11}} + \frac {5301 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{12}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{12}} + \frac {1800 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{13}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{13}} + \frac {363 \, c^{\frac {9}{2}} \sin \left (f x + e\right )^{14}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{14}}\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 {9}{2}}} \]

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))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*sin(e + f*x))^(9/2)/(a + a*sin(e + f*x))^3,x)

[Out]

int((c - c*sin(e + f*x))^(9/2)/(a + a*sin(e + f*x))^3, x)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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

________________________________________________________________________________________