3.4.32 \(\int \frac {(c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^3} \, dx\) [332]

Optimal. Leaf size=174 \[ -\frac {4096 c^2 \sec ^5(e+f x) (c-c \sin (e+f x))^{5/2}}{15 a^3 f}+\frac {1024 c \sec ^5(e+f x) (c-c \sin (e+f x))^{7/2}}{3 a^3 f}-\frac {128 \sec ^5(e+f x) (c-c \sin (e+f x))^{9/2}}{a^3 f}+\frac {32 \sec ^5(e+f x) (c-c \sin (e+f x))^{11/2}}{3 a^3 c f}+\frac {2 \sec ^5(e+f x) (c-c \sin (e+f x))^{13/2}}{3 a^3 c^2 f} \]

[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

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Rubi [A]
time = 0.28, 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 = {2815, 2753, 2752} \begin {gather*} \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} \end {gather*}

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 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2753

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 2815

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*}

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Mathematica [A]
time = 1.90, size = 124, normalized size = 0.71 \begin {gather*} \frac {c^4 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) \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 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) (1+\sin (e+f x))^3} \end {gather*}

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 = 2.11, size = 91, normalized size = 0.52

method	result	size
default	\(-\frac {2 c^{5} \left (\sin \left (f x +e \right )-1\right ) \left (5 \left (\sin ^{4}\left (f x +e \right )\right )-100 \left (\sin ^{3}\left (f x +e \right )\right )-690 \left (\sin ^{2}\left (f x +e \right )\right )-900 \sin \left (f x +e \right )-363\right )}{15 a^{3} \left (1+\sin \left (f x +e \right )\right )^{2} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f}\)	\(91\)

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,method=_RETURNVERBOSE)

[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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 512 vs. \(2 (166) = 332\).
time = 0.52, size = 512, normalized size = 2.94 \begin {gather*} \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}}} \end {gather*}

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 = 0.36, size = 128, normalized size = 0.74 \begin {gather*} -\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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 7316 deep

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 450 vs. \(2 (166) = 332\).
time = 0.61, size = 450, normalized size = 2.59 \begin {gather*} -\frac {8 \, \sqrt {2} \sqrt {c} {\left (\frac {5 \, {\left (11 \, c^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) - \frac {24 \, c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {9 \, c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}}\right )}}{a^{3} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} - 1\right )}^{3}} - \frac {73 \, c^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right ) + \frac {320 \, c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + \frac {490 \, c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{2} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{2}} + \frac {240 \, c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{3} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{3}} + \frac {45 \, c^{4} {\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{4} \mathrm {sgn}\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}{{\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{4}}}{a^{3} {\left (\frac {\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1}{\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1} + 1\right )}^{5}}\right )}}{15 \, f} \end {gather*}

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]

-8/15*sqrt(2)*sqrt(c)*(5*(11*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e)) - 24*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e)
- 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) + 9*c^4*(cos(-1/4*pi + 1/2*f*x +
1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^2)/(a^3*((cos(-1/4*pi
+ 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1) - 1)^3) - (73*c^4*sgn(sin(-1/4*pi + 1/2*f*x + 1/2
*e)) + 320*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x
+ 1/2*e) + 1) + 490*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^2*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*e))/(cos(-1/4*
pi + 1/2*f*x + 1/2*e) + 1)^2 + 240*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^3*sgn(sin(-1/4*pi + 1/2*f*x + 1/2*
e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^3 + 45*c^4*(cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)^4*sgn(sin(-1/4*pi + 1
/2*f*x + 1/2*e))/(cos(-1/4*pi + 1/2*f*x + 1/2*e) + 1)^4)/(a^3*((cos(-1/4*pi + 1/2*f*x + 1/2*e) - 1)/(cos(-1/4*
pi + 1/2*f*x + 1/2*e) + 1) + 1)^5))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \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 \end {gather*}

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)

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