3.279 \(\int \frac{(c-c \sin (e+f x))^5}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=161 \[ -\frac{63 c^5 \cos (e+f x)}{2 a^3 f}-\frac{2 a^4 c^5 \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}+\frac{6 a^2 c^5 \cos ^7(e+f x)}{5 f (a \sin (e+f x)+a)^5}-\frac{21 c^5 \cos ^3(e+f x)}{2 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{63 c^5 x}{2 a^3}-\frac{42 c^5 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^3} \]

[Out]

(-63*c^5*x)/(2*a^3) - (63*c^5*Cos[e + f*x])/(2*a^3*f) - (2*a^4*c^5*Cos[e + f*x]^9)/(5*f*(a + a*Sin[e + f*x])^7
) + (6*a^2*c^5*Cos[e + f*x]^7)/(5*f*(a + a*Sin[e + f*x])^5) - (42*c^5*Cos[e + f*x]^5)/(5*f*(a + a*Sin[e + f*x]
)^3) - (21*c^5*Cos[e + f*x]^3)/(2*f*(a^3 + a^3*Sin[e + f*x]))

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Rubi [A]  time = 0.27595, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2736, 2680, 2679, 2682, 8} \[ -\frac{63 c^5 \cos (e+f x)}{2 a^3 f}-\frac{2 a^4 c^5 \cos ^9(e+f x)}{5 f (a \sin (e+f x)+a)^7}+\frac{6 a^2 c^5 \cos ^7(e+f x)}{5 f (a \sin (e+f x)+a)^5}-\frac{21 c^5 \cos ^3(e+f x)}{2 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{63 c^5 x}{2 a^3}-\frac{42 c^5 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^5/(a + a*Sin[e + f*x])^3,x]

[Out]

(-63*c^5*x)/(2*a^3) - (63*c^5*Cos[e + f*x])/(2*a^3*f) - (2*a^4*c^5*Cos[e + f*x]^9)/(5*f*(a + a*Sin[e + f*x])^7
) + (6*a^2*c^5*Cos[e + f*x]^7)/(5*f*(a + a*Sin[e + f*x])^5) - (42*c^5*Cos[e + f*x]^5)/(5*f*(a + a*Sin[e + f*x]
)^3) - (21*c^5*Cos[e + f*x]^3)/(2*f*(a^3 + a^3*Sin[e + f*x]))

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 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{(c-c \sin (e+f x))^5}{(a+a \sin (e+f x))^3} \, dx &=\left (a^5 c^5\right ) \int \frac{\cos ^{10}(e+f x)}{(a+a \sin (e+f x))^8} \, dx\\ &=-\frac{2 a^4 c^5 \cos ^9(e+f x)}{5 f (a+a \sin (e+f x))^7}-\frac{1}{5} \left (9 a^3 c^5\right ) \int \frac{\cos ^8(e+f x)}{(a+a \sin (e+f x))^6} \, dx\\ &=-\frac{2 a^4 c^5 \cos ^9(e+f x)}{5 f (a+a \sin (e+f x))^7}+\frac{6 a^2 c^5 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^5}+\frac{1}{5} \left (21 a c^5\right ) \int \frac{\cos ^6(e+f x)}{(a+a \sin (e+f x))^4} \, dx\\ &=-\frac{2 a^4 c^5 \cos ^9(e+f x)}{5 f (a+a \sin (e+f x))^7}+\frac{6 a^2 c^5 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac{42 c^5 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{\left (21 c^5\right ) \int \frac{\cos ^4(e+f x)}{(a+a \sin (e+f x))^2} \, dx}{a}\\ &=-\frac{2 a^4 c^5 \cos ^9(e+f x)}{5 f (a+a \sin (e+f x))^7}+\frac{6 a^2 c^5 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac{42 c^5 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{21 c^5 \cos ^3(e+f x)}{2 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{\left (63 c^5\right ) \int \frac{\cos ^2(e+f x)}{a+a \sin (e+f x)} \, dx}{2 a^2}\\ &=-\frac{63 c^5 \cos (e+f x)}{2 a^3 f}-\frac{2 a^4 c^5 \cos ^9(e+f x)}{5 f (a+a \sin (e+f x))^7}+\frac{6 a^2 c^5 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac{42 c^5 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{21 c^5 \cos ^3(e+f x)}{2 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{\left (63 c^5\right ) \int 1 \, dx}{2 a^3}\\ &=-\frac{63 c^5 x}{2 a^3}-\frac{63 c^5 \cos (e+f x)}{2 a^3 f}-\frac{2 a^4 c^5 \cos ^9(e+f x)}{5 f (a+a \sin (e+f x))^7}+\frac{6 a^2 c^5 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^5}-\frac{42 c^5 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{21 c^5 \cos ^3(e+f x)}{2 f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.849494, size = 303, normalized size = 1.88 \[ \frac{(c-c \sin (e+f x))^5 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (256 \sin \left (\frac{1}{2} (e+f x)\right )-630 (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5-160 \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+5 \sin (2 (e+f x)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+2304 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+448 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-896 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-128 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{20 f (a \sin (e+f x)+a)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{10}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^5/(a + a*Sin[e + f*x])^3,x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^5*(256*Sin[(e + f*x)/2] - 128*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2]) - 896*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 448*(Cos[(e + f*x)/2] + Sin
[(e + f*x)/2])^3 + 2304*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 630*(e + f*x)*(Cos[(e + f*x
)/2] + Sin[(e + f*x)/2])^5 - 160*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 + 5*(Cos[(e + f*x)/2] +
Sin[(e + f*x)/2])^5*Sin[2*(e + f*x)]))/(20*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^10*(a + a*Sin[e + f*x])^3)

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Maple [A]  time = 0.108, size = 277, normalized size = 1.7 \begin{align*} -{\frac{{c}^{5}}{f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-16\,{\frac{{c}^{5} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}}{f{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{c}^{5}}{f{a}^{3}}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}-16\,{\frac{{c}^{5}}{f{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-63\,{\frac{{c}^{5}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}}}-{\frac{256\,{c}^{5}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}+128\,{\frac{{c}^{5}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-64\,{\frac{{c}^{5}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{3}}}-32\,{\frac{{c}^{5}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-64\,{\frac{{c}^{5}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sin(f*x+e))^5/(a+a*sin(f*x+e))^3,x)

[Out]

-1/f*c^5/a^3/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3-16/f*c^5/a^3/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f
*x+1/2*e)^2+1/f*c^5/a^3/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)-16/f*c^5/a^3/(1+tan(1/2*f*x+1/2*e)^2)^2-
63/f*c^5/a^3*arctan(tan(1/2*f*x+1/2*e))-256/5/f*c^5/a^3/(tan(1/2*f*x+1/2*e)+1)^5+128/f*c^5/a^3/(tan(1/2*f*x+1/
2*e)+1)^4-64/f*c^5/a^3/(tan(1/2*f*x+1/2*e)+1)^3-32/f*c^5/a^3/(tan(1/2*f*x+1/2*e)+1)^2-64/f*c^5/a^3/(tan(1/2*f*
x+1/2*e)+1)

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Maxima [B]  time = 3.68809, size = 2020, normalized size = 12.55 \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))^5/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/15*(c^5*((1325*sin(f*x + e)/(cos(f*x + e) + 1) + 2673*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 3805*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3 + 4329*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 3575*sin(f*x + e)^5/(cos(f*x + e) + 1)^
5 + 2275*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 975*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 195*sin(f*x + e)^8/(c
os(f*x + e) + 1)^8 + 304)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 12*a^3*sin(f*x + e)^2/(cos(f*x + e) +
1)^2 + 20*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 26*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 26*a^3*sin(f
*x + e)^5/(cos(f*x + e) + 1)^5 + 20*a^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 12*a^3*sin(f*x + e)^7/(cos(f*x +
e) + 1)^7 + 5*a^3*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + a^3*sin(f*x + e)^9/(cos(f*x + e) + 1)^9) + 195*arctan
(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + 30*c^5*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 189*sin(f*x + e)^2/(co
s(f*x + e) + 1)^2 + 200*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 160*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 75*sin
(f*x + e)^5/(cos(f*x + e) + 1)^5 + 15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 24)/(a^3 + 5*a^3*sin(f*x + e)/(cos
(f*x + e) + 1) + 11*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 15*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*
a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 11*a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 5*a^3*sin(f*x + e)^6/(c
os(f*x + e) + 1)^6 + a^3*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3
) + 20*c^5*((95*sin(f*x + e)/(cos(f*x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3/
(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 22)/(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) + 15*arctan(sin(f*x + e)/(cos(f*x + e) +
1))/a^3) + 2*c^5*(20*sin(f*x + e)/(cos(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x +
e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 7)/(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) + 40*c^5*(5*sin(f*x + e)/(cos(f*x +
e) + 1) + 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*si
n(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) - 30*c^5*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*
x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(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*si
n(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [A]  time = 1.38678, size = 707, normalized size = 4.39 \begin{align*} -\frac{5 \, c^{5} \cos \left (f x + e\right )^{5} + 70 \, c^{5} \cos \left (f x + e\right )^{4} - 1260 \, c^{5} f x - 64 \, c^{5} + 7 \,{\left (45 \, c^{5} f x + 113 \, c^{5}\right )} \cos \left (f x + e\right )^{3} +{\left (945 \, c^{5} f x - 502 \, c^{5}\right )} \cos \left (f x + e\right )^{2} - 2 \,{\left (315 \, c^{5} f x + 646 \, c^{5}\right )} \cos \left (f x + e\right ) -{\left (5 \, c^{5} \cos \left (f x + e\right )^{4} - 65 \, c^{5} \cos \left (f x + e\right )^{3} + 1260 \, c^{5} f x - 64 \, c^{5} - 3 \,{\left (105 \, c^{5} f x - 242 \, c^{5}\right )} \cos \left (f x + e\right )^{2} + 2 \,{\left (315 \, c^{5} f x + 614 \, c^{5}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{10 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \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))^5/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/10*(5*c^5*cos(f*x + e)^5 + 70*c^5*cos(f*x + e)^4 - 1260*c^5*f*x - 64*c^5 + 7*(45*c^5*f*x + 113*c^5)*cos(f*x
+ e)^3 + (945*c^5*f*x - 502*c^5)*cos(f*x + e)^2 - 2*(315*c^5*f*x + 646*c^5)*cos(f*x + e) - (5*c^5*cos(f*x + e
)^4 - 65*c^5*cos(f*x + e)^3 + 1260*c^5*f*x - 64*c^5 - 3*(105*c^5*f*x - 242*c^5)*cos(f*x + e)^2 + 2*(315*c^5*f*
x + 614*c^5)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e)
- 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f)*sin(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))**5/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 2.13942, size = 251, normalized size = 1.56 \begin{align*} -\frac{\frac{315 \,{\left (f x + e\right )} c^{5}}{a^{3}} + \frac{10 \,{\left (c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 16 \, c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 16 \, c^{5}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} a^{3}} + \frac{64 \,{\left (10 \, c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 45 \, c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 85 \, c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 55 \, c^{5} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 13 \, c^{5}\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}}}{10 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^5/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/10*(315*(f*x + e)*c^5/a^3 + 10*(c^5*tan(1/2*f*x + 1/2*e)^3 + 16*c^5*tan(1/2*f*x + 1/2*e)^2 - c^5*tan(1/2*f*
x + 1/2*e) + 16*c^5)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*a^3) + 64*(10*c^5*tan(1/2*f*x + 1/2*e)^4 + 45*c^5*tan(1/2
*f*x + 1/2*e)^3 + 85*c^5*tan(1/2*f*x + 1/2*e)^2 + 55*c^5*tan(1/2*f*x + 1/2*e) + 13*c^5)/(a^3*(tan(1/2*f*x + 1/
2*e) + 1)^5))/f