3.57 \(\int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=285 \[ -\frac {80 c^5 \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {40 c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {10 c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {10 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a \sin (e+f x)+a)^{3/2}} \]

[Out]

-cos(f*x+e)*(c-c*sin(f*x+e))^(9/2)/a/f/(a+a*sin(f*x+e))^(3/2)-10*c^3*cos(f*x+e)*(c-c*sin(f*x+e))^(3/2)/a^2/f/(
a+a*sin(f*x+e))^(1/2)-10/3*c^2*cos(f*x+e)*(c-c*sin(f*x+e))^(5/2)/a^2/f/(a+a*sin(f*x+e))^(1/2)-5/4*c*cos(f*x+e)
*(c-c*sin(f*x+e))^(7/2)/a^2/f/(a+a*sin(f*x+e))^(1/2)-80*c^5*cos(f*x+e)*ln(1+sin(f*x+e))/a^2/f/(a+a*sin(f*x+e))
^(1/2)/(c-c*sin(f*x+e))^(1/2)-40*c^4*cos(f*x+e)*(c-c*sin(f*x+e))^(1/2)/a^2/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.86, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2841, 2739, 2740, 2737, 2667, 31} \[ -\frac {40 c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {10 c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {10 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {80 c^5 \cos (e+f x) \log (\sin (e+f x)+1)}{a^2 f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}}-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a \sin (e+f x)+a}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a \sin (e+f x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[e + f*x]^2*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(-80*c^5*Cos[e + f*x]*Log[1 + Sin[e + f*x]])/(a^2*f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]]) - (40*c
^4*Cos[e + f*x]*Sqrt[c - c*Sin[e + f*x]])/(a^2*f*Sqrt[a + a*Sin[e + f*x]]) - (10*c^3*Cos[e + f*x]*(c - c*Sin[e
+ f*x])^(3/2))/(a^2*f*Sqrt[a + a*Sin[e + f*x]]) - (10*c^2*Cos[e + f*x]*(c - c*Sin[e + f*x])^(5/2))/(3*a^2*f*S
qrt[a + a*Sin[e + f*x]]) - (5*c*Cos[e + f*x]*(c - c*Sin[e + f*x])^(7/2))/(4*a^2*f*Sqrt[a + a*Sin[e + f*x]]) -
(Cos[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(a*f*(a + a*Sin[e + f*x])^(3/2))

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2737

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2739

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(-2*b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(2*n + 1)), x] - Dist[(b*(2*m - 1)
)/(d*(2*n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e
, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] && LtQ[n, -1] &&  !(ILtQ[m + n, 0] && G
tQ[2*m + n + 1, 0])

Rule 2740

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n)/(f*(m + n)), x] + Dist[(a*(2*m - 1))/(m
+ n), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E
qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[m - 1/2, 0] &&  !LtQ[n, -1] &&  !(IGtQ[n - 1/2, 0] && LtQ[n, m])
&&  !(ILtQ[m + n, 0] && GtQ[2*m + n + 1, 0])

Rule 2841

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*
(x_)])^(n_.), x_Symbol] :> Dist[1/(a^(p/2)*c^(p/2)), Int[(a + b*Sin[e + f*x])^(m + p/2)*(c + d*Sin[e + f*x])^(
n + p/2), x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[p
/2]

Rubi steps

\begin {align*} \int \frac {\cos ^2(e+f x) (c-c \sin (e+f x))^{9/2}}{(a+a \sin (e+f x))^{5/2}} \, dx &=\frac {\int \frac {(c-c \sin (e+f x))^{11/2}}{(a+a \sin (e+f x))^{3/2}} \, dx}{a c}\\ &=-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {5 \int \frac {(c-c \sin (e+f x))^{9/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {(10 c) \int \frac {(c-c \sin (e+f x))^{7/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {10 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\left (20 c^2\right ) \int \frac {(c-c \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {10 c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\left (40 c^3\right ) \int \frac {(c-c \sin (e+f x))^{3/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {40 c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\left (80 c^4\right ) \int \frac {\sqrt {c-c \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)}} \, dx}{a^2}\\ &=-\frac {40 c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\left (80 c^5 \cos (e+f x)\right ) \int \frac {\cos (e+f x)}{a+a \sin (e+f x)} \, dx}{a \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {40 c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a+a \sin (e+f x))^{3/2}}-\frac {\left (80 c^5 \cos (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,a \sin (e+f x)\right )}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}\\ &=-\frac {80 c^5 \cos (e+f x) \log (1+\sin (e+f x))}{a^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c-c \sin (e+f x)}}-\frac {40 c^4 \cos (e+f x) \sqrt {c-c \sin (e+f x)}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^3 \cos (e+f x) (c-c \sin (e+f x))^{3/2}}{a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {10 c^2 \cos (e+f x) (c-c \sin (e+f x))^{5/2}}{3 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {5 c \cos (e+f x) (c-c \sin (e+f x))^{7/2}}{4 a^2 f \sqrt {a+a \sin (e+f x)}}-\frac {\cos (e+f x) (c-c \sin (e+f x))^{9/2}}{a f (a+a \sin (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 6.64, size = 553, normalized size = 1.94 \[ \frac {203 \sin (e+f x) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{4 f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}+\frac {47 \cos (2 (e+f x)) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{8 f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}-\frac {\cos (4 (e+f x)) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{32 f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}-\frac {7 \sin (3 (e+f x)) (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5}{12 f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}-\frac {32 (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3}{f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9}-\frac {160 (c-c \sin (e+f x))^{9/2} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^5 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{f (a (\sin (e+f x)+1))^{5/2} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^9} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[e + f*x]^2*(c - c*Sin[e + f*x])^(9/2))/(a + a*Sin[e + f*x])^(5/2),x]

[Out]

(-32*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3*(c - c*Sin[e + f*x])^(9/2))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2])^9*(a*(1 + Sin[e + f*x]))^(5/2)) + (47*Cos[2*(e + f*x)]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(c - c*Sin[
e + f*x])^(9/2))/(8*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(5/2)) - (Cos[4*(e + f*x)
]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(c - c*Sin[e + f*x])^(9/2))/(32*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/
2])^9*(a*(1 + Sin[e + f*x]))^(5/2)) - (160*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]*(Cos[(e + f*x)/2] + Sin[(e
+ f*x)/2])^5*(c - c*Sin[e + f*x])^(9/2))/(f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(5
/2)) + (203*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*Sin[e + f*x]*(c - c*Sin[e + f*x])^(9/2))/(4*f*(Cos[(e + f*
x)/2] - Sin[(e + f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(5/2)) - (7*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5*(c - c*
Sin[e + f*x])^(9/2)*Sin[3*(e + f*x)])/(12*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^9*(a*(1 + Sin[e + f*x]))^(5/
2))

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fricas [F]  time = 2.26, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (c^{4} \cos \left (f x + e\right )^{6} - 8 \, c^{4} \cos \left (f x + e\right )^{4} + 8 \, c^{4} \cos \left (f x + e\right )^{2} + 4 \, {\left (c^{4} \cos \left (f x + e\right )^{4} - 2 \, c^{4} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} + {\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(-(c^4*cos(f*x + e)^6 - 8*c^4*cos(f*x + e)^4 + 8*c^4*cos(f*x + e)^2 + 4*(c^4*cos(f*x + e)^4 - 2*c^4*co
s(f*x + e)^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(3*a^3*cos(f*x + e)^2 - 4*a^3 +
(a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e)), x)

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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(cos(f*x+e)^2*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.36, size = 349, normalized size = 1.22 \[ \frac {\left (3 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )-25 \left (\cos ^{4}\left (f x +e \right )\right )-116 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+500 \left (\cos ^{2}\left (f x +e \right )\right )-960 \sin \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+1920 \sin \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-859 \sin \left (f x +e \right )-960 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+1920 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-475\right ) \left (-c \left (\sin \left (f x +e \right )-1\right )\right )^{\frac {9}{2}} \left (\cos ^{2}\left (f x +e \right )+\sin \left (f x +e \right ) \cos \left (f x +e \right )+\cos \left (f x +e \right )-2 \sin \left (f x +e \right )-2\right )}{12 f \left (\sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+\cos ^{5}\left (f x +e \right )+4 \sin \left (f x +e \right ) \left (\cos ^{3}\left (f x +e \right )\right )-5 \left (\cos ^{4}\left (f x +e \right )\right )-12 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-8 \left (\cos ^{3}\left (f x +e \right )\right )-8 \sin \left (f x +e \right ) \cos \left (f x +e \right )+20 \left (\cos ^{2}\left (f x +e \right )\right )+16 \sin \left (f x +e \right )+8 \cos \left (f x +e \right )-16\right ) \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)^2*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x)

[Out]

1/12/f*(3*sin(f*x+e)*cos(f*x+e)^4-25*cos(f*x+e)^4-116*cos(f*x+e)^2*sin(f*x+e)+500*cos(f*x+e)^2-960*sin(f*x+e)*
ln(2/(cos(f*x+e)+1))+1920*sin(f*x+e)*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-859*sin(f*x+e)-960*ln(2/(cos(f
*x+e)+1))+1920*ln(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))-475)*(-c*(sin(f*x+e)-1))^(9/2)*(cos(f*x+e)^2+sin(f*x
+e)*cos(f*x+e)+cos(f*x+e)-2*sin(f*x+e)-2)/(sin(f*x+e)*cos(f*x+e)^4+cos(f*x+e)^5+4*sin(f*x+e)*cos(f*x+e)^3-5*co
s(f*x+e)^4-12*cos(f*x+e)^2*sin(f*x+e)-8*cos(f*x+e)^3-8*sin(f*x+e)*cos(f*x+e)+20*cos(f*x+e)^2+16*sin(f*x+e)+8*c
os(f*x+e)-16)/(a*(1+sin(f*x+e)))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (-c \sin \left (f x + e\right ) + c\right )}^{\frac {9}{2}} \cos \left (f x + e\right )^{2}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)^2*(c-c*sin(f*x+e))^(9/2)/(a+a*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((-c*sin(f*x + e) + c)^(9/2)*cos(f*x + e)^2/(a*sin(f*x + e) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (e+f\,x\right )}^2\,{\left (c-c\,\sin \left (e+f\,x\right )\right )}^{9/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(e + f*x)^2*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x))^(5/2),x)

[Out]

int((cos(e + f*x)^2*(c - c*sin(e + f*x))^(9/2))/(a + a*sin(e + f*x))^(5/2), x)

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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(cos(f*x+e)**2*(c-c*sin(f*x+e))**(9/2)/(a+a*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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