3.271 \(\int \frac{(c-c \sin (e+f x))^3}{(a+a \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=90 \[ \frac{5 c^3 \cos (e+f x)}{a^2 f}-\frac{2 a^2 c^3 \cos ^5(e+f x)}{3 f (a \sin (e+f x)+a)^4}+\frac{5 c^3 x}{a^2}+\frac{10 c^3 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^2} \]
[Out]
(5*c^3*x)/a^2 + (5*c^3*Cos[e + f*x])/(a^2*f) - (2*a^2*c^3*Cos[e + f*x]^5)/(3*f*(a + a*Sin[e + f*x])^4) + (10*c
^3*Cos[e + f*x]^3)/(3*f*(a + a*Sin[e + f*x])^2)
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Rubi [A] time = 0.175377, antiderivative size = 90, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used =
{2736, 2680, 2682, 8} \[ \frac{5 c^3 \cos (e+f x)}{a^2 f}-\frac{2 a^2 c^3 \cos ^5(e+f x)}{3 f (a \sin (e+f x)+a)^4}+\frac{5 c^3 x}{a^2}+\frac{10 c^3 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^2} \]
Antiderivative was successfully verified.
[In]
Int[(c - c*Sin[e + f*x])^3/(a + a*Sin[e + f*x])^2,x]
[Out]
(5*c^3*x)/a^2 + (5*c^3*Cos[e + f*x])/(a^2*f) - (2*a^2*c^3*Cos[e + f*x]^5)/(3*f*(a + a*Sin[e + f*x])^4) + (10*c
^3*Cos[e + f*x]^3)/(3*f*(a + a*Sin[e + f*x])^2)
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 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))^3}{(a+a \sin (e+f x))^2} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(a+a \sin (e+f x))^5} \, dx\\ &=-\frac{2 a^2 c^3 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^4}-\frac{1}{3} \left (5 a c^3\right ) \int \frac{\cos ^4(e+f x)}{(a+a \sin (e+f x))^3} \, dx\\ &=-\frac{2 a^2 c^3 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^4}+\frac{10 c^3 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^2}+\frac{\left (5 c^3\right ) \int \frac{\cos ^2(e+f x)}{a+a \sin (e+f x)} \, dx}{a}\\ &=\frac{5 c^3 \cos (e+f x)}{a^2 f}-\frac{2 a^2 c^3 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^4}+\frac{10 c^3 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^2}+\frac{\left (5 c^3\right ) \int 1 \, dx}{a^2}\\ &=\frac{5 c^3 x}{a^2}+\frac{5 c^3 \cos (e+f x)}{a^2 f}-\frac{2 a^2 c^3 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^4}+\frac{10 c^3 \cos ^3(e+f x)}{3 f (a+a \sin (e+f x))^2}\\ \end{align*}
Mathematica [B] time = 0.360541, size = 210, normalized size = 2.33 \[ \frac{(c-c \sin (e+f x))^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (16 \sin \left (\frac{1}{2} (e+f x)\right )+15 (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+3 \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-56 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-8 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{3 f (a \sin (e+f x)+a)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6} \]
Antiderivative was successfully verified.
[In]
Integrate[(c - c*Sin[e + f*x])^3/(a + a*Sin[e + f*x])^2,x]
[Out]
((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(16*Sin[(e + f*x)/2] - 8*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 56*Sin
[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 15*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 +
3*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)*(c - c*Sin[e + f*x])^3)/(3*f*(Cos[(e + f*x)/2] - Sin[
(e + f*x)/2])^6*(a + a*Sin[e + f*x])^2)
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Maple [A] time = 0.089, size = 121, normalized size = 1.3 \begin{align*} 2\,{\frac{{c}^{3}}{{a}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}+10\,{\frac{{c}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{{a}^{2}f}}-{\frac{32\,{c}^{3}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+16\,{\frac{{c}^{3}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+8\,{\frac{{c}^{3}}{{a}^{2}f \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))^3/(a+a*sin(f*x+e))^2,x)
[Out]
2/f*c^3/a^2/(1+tan(1/2*f*x+1/2*e)^2)+10/f*c^3/a^2*arctan(tan(1/2*f*x+1/2*e))-32/3/f*c^3/a^2/(tan(1/2*f*x+1/2*e
)+1)^3+16/f*c^3/a^2/(tan(1/2*f*x+1/2*e)+1)^2+8/f*c^3/a^2/(tan(1/2*f*x+1/2*e)+1)
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Maxima [B] time = 3.91568, size = 797, normalized size = 8.86 \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))^3/(a+a*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
2/3*(2*c^3*((12*sin(f*x + e)/(cos(f*x + e) + 1) + 11*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) +
4*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/
(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^
2) + 3*c^3*((9*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f
*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) - c^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)
^2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*c^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^
2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x +
e) + 1)^3))/f
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Fricas [B] time = 1.37893, size = 433, normalized size = 4.81 \begin{align*} \frac{3 \, c^{3} \cos \left (f x + e\right )^{3} - 30 \, c^{3} f x + 8 \, c^{3} +{\left (15 \, c^{3} f x - 31 \, c^{3}\right )} \cos \left (f x + e\right )^{2} -{\left (15 \, c^{3} f x + 26 \, c^{3}\right )} \cos \left (f x + e\right ) -{\left (30 \, c^{3} f x + 3 \, c^{3} \cos \left (f x + e\right )^{2} + 8 \, c^{3} +{\left (15 \, c^{3} f x + 34 \, c^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} 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))^3/(a+a*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
1/3*(3*c^3*cos(f*x + e)^3 - 30*c^3*f*x + 8*c^3 + (15*c^3*f*x - 31*c^3)*cos(f*x + e)^2 - (15*c^3*f*x + 26*c^3)*
cos(f*x + e) - (30*c^3*f*x + 3*c^3*cos(f*x + e)^2 + 8*c^3 + (15*c^3*f*x + 34*c^3)*cos(f*x + e))*sin(f*x + e))/
(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*f - (a^2*f*cos(f*x + e) + 2*a^2*f)*sin(f*x + e))
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Sympy [A] time = 51.7506, size = 1282, normalized size = 14.24 \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))**3/(a+a*sin(f*x+e))**2,x)
[Out]
Piecewise((15*c**3*f*x*tan(e/2 + f*x/2)**5/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a
**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 45*c**3*f*
x*tan(e/2 + f*x/2)**4/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2
)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 60*c**3*f*x*tan(e/2 + f*x/2)**3
/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(
e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 60*c**3*f*x*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 +
f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a
**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 45*c**3*f*x*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan
(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) +
3*a**2*f) + 15*c**3*f*x/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*
x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) - 8*c**3*tan(e/2 + f*x/2)**5/(
3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/
2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 70*c**3*tan(e/2 + f*x/2)**3/(3*a**2*f*tan(e/2 + f*x/2)
**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*
tan(e/2 + f*x/2) + 3*a**2*f) + 50*c**3*tan(e/2 + f*x/2)**2/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 +
f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2
*f) + 90*c**3*tan(e/2 + f*x/2)/(3*a**2*f*tan(e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/
2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2 + 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f) + 38*c**3/(3*a**2*f*tan(
e/2 + f*x/2)**5 + 9*a**2*f*tan(e/2 + f*x/2)**4 + 12*a**2*f*tan(e/2 + f*x/2)**3 + 12*a**2*f*tan(e/2 + f*x/2)**2
+ 9*a**2*f*tan(e/2 + f*x/2) + 3*a**2*f), Ne(f, 0)), (x*(-c*sin(e) + c)**3/(a*sin(e) + a)**2, True))
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Giac [A] time = 2.16426, size = 136, normalized size = 1.51 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )} c^{3}}{a^{2}} + \frac{6 \, c^{3}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} a^{2}} + \frac{8 \,{\left (3 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 12 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 5 \, c^{3}\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}}{3 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c-c*sin(f*x+e))^3/(a+a*sin(f*x+e))^2,x, algorithm="giac")
[Out]
1/3*(15*(f*x + e)*c^3/a^2 + 6*c^3/((tan(1/2*f*x + 1/2*e)^2 + 1)*a^2) + 8*(3*c^3*tan(1/2*f*x + 1/2*e)^2 + 12*c^
3*tan(1/2*f*x + 1/2*e) + 5*c^3)/(a^2*(tan(1/2*f*x + 1/2*e) + 1)^3))/f