3.254 \(\int \frac{(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx\)
Optimal. Leaf size=92 \[ -\frac{5 a^3 \cos (e+f x)}{c^2 f}+\frac{2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}+\frac{5 a^3 x}{c^2}-\frac{10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \]
[Out]
(5*a^3*x)/c^2 - (5*a^3*Cos[e + f*x])/(c^2*f) + (2*a^3*c^2*Cos[e + f*x]^5)/(3*f*(c - c*Sin[e + f*x])^4) - (10*a
^3*Cos[e + f*x]^3)/(3*f*(c - c*Sin[e + f*x])^2)
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Rubi [A] time = 0.18605, antiderivative size = 92, 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} \[ -\frac{5 a^3 \cos (e+f x)}{c^2 f}+\frac{2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}+\frac{5 a^3 x}{c^2}-\frac{10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^2,x]
[Out]
(5*a^3*x)/c^2 - (5*a^3*Cos[e + f*x])/(c^2*f) + (2*a^3*c^2*Cos[e + f*x]^5)/(3*f*(c - c*Sin[e + f*x])^4) - (10*a
^3*Cos[e + f*x]^3)/(3*f*(c - c*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{(a+a \sin (e+f x))^3}{(c-c \sin (e+f x))^2} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac{2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac{1}{3} \left (5 a^3 c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^3} \, dx\\ &=\frac{2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac{10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac{\left (5 a^3\right ) \int \frac{\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{c}\\ &=-\frac{5 a^3 \cos (e+f x)}{c^2 f}+\frac{2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac{10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}+\frac{\left (5 a^3\right ) \int 1 \, dx}{c^2}\\ &=\frac{5 a^3 x}{c^2}-\frac{5 a^3 \cos (e+f x)}{c^2 f}+\frac{2 a^3 c^2 \cos ^5(e+f x)}{3 f (c-c \sin (e+f x))^4}-\frac{10 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^2}\\ \end{align*}
Mathematica [A] time = 0.960648, size = 149, normalized size = 1.62 \[ \frac{a^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (6 (15 e+15 f x+23) \cos \left (\frac{1}{2} (e+f x)\right )-(30 e+30 f x+121) \cos \left (\frac{3}{2} (e+f x)\right )+3 \cos \left (\frac{5}{2} (e+f x)\right )-6 \sin \left (\frac{1}{2} (e+f x)\right ) (2 (5 e+5 f x-2) \cos (e+f x)-\cos (2 (e+f x))+20 e+20 f x+31)\right )}{12 c^2 f (\sin (e+f x)-1)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])^3/(c - c*Sin[e + f*x])^2,x]
[Out]
(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(6*(23 + 15*e + 15*f*x)*Cos[(e + f*x)/2] - (121 + 30*e + 30*f*x)*Co
s[(3*(e + f*x))/2] + 3*Cos[(5*(e + f*x))/2] - 6*(31 + 20*e + 20*f*x + 2*(-2 + 5*e + 5*f*x)*Cos[e + f*x] - Cos[
2*(e + f*x)])*Sin[(e + f*x)/2]))/(12*c^2*f*(-1 + Sin[e + f*x])^2)
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Maple [A] time = 0.086, size = 121, normalized size = 1.3 \begin{align*} -{\frac{32\,{a}^{3}}{3\,f{c}^{2}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-16\,{\frac{{a}^{3}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}+8\,{\frac{{a}^{3}}{f{c}^{2} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-2\,{\frac{{a}^{3}}{f{c}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}+10\,{\frac{{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x)
[Out]
-32/3/f*a^3/c^2/(tan(1/2*f*x+1/2*e)-1)^3-16/f*a^3/c^2/(tan(1/2*f*x+1/2*e)-1)^2+8/f*a^3/c^2/(tan(1/2*f*x+1/2*e)
-1)-2/f*a^3/c^2/(1+tan(1/2*f*x+1/2*e)^2)+10/f*a^3/c^2*arctan(tan(1/2*f*x+1/2*e))
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Maxima [B] time = 1.80274, size = 802, normalized size = 8.72 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="maxima")
[Out]
2/3*(2*a^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)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) +
4*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4*c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*c^2*sin(f*x + e)^4/
(cos(f*x + e) + 1)^4 - c^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^
2) + 3*a^3*((9*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 4)/(c^2 - 3*c^2*sin(f
*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c^2) - a^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 3*sin(f*x + e)
^2/(cos(f*x + e) + 1)^2 - 2)/(c^2 - 3*c^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e)
+ 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*a^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 1)/(c^2 - 3*c^
2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*c^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c^2*sin(f*x + e)^3/(cos(f*x +
e) + 1)^3))/f
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Fricas [B] time = 1.30488, size = 435, normalized size = 4.73 \begin{align*} -\frac{3 \, a^{3} \cos \left (f x + e\right )^{3} + 30 \, a^{3} f x + 8 \, a^{3} -{\left (15 \, a^{3} f x + 31 \, a^{3}\right )} \cos \left (f x + e\right )^{2} +{\left (15 \, a^{3} f x - 26 \, a^{3}\right )} \cos \left (f x + e\right ) -{\left (30 \, a^{3} f x - 3 \, a^{3} \cos \left (f x + e\right )^{2} - 8 \, a^{3} +{\left (15 \, a^{3} f x - 34 \, a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{3 \,{\left (c^{2} f \cos \left (f x + e\right )^{2} - c^{2} f \cos \left (f x + e\right ) - 2 \, c^{2} f +{\left (c^{2} f \cos \left (f x + e\right ) + 2 \, c^{2} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="fricas")
[Out]
-1/3*(3*a^3*cos(f*x + e)^3 + 30*a^3*f*x + 8*a^3 - (15*a^3*f*x + 31*a^3)*cos(f*x + e)^2 + (15*a^3*f*x - 26*a^3)
*cos(f*x + e) - (30*a^3*f*x - 3*a^3*cos(f*x + e)^2 - 8*a^3 + (15*a^3*f*x - 34*a^3)*cos(f*x + e))*sin(f*x + e))
/(c^2*f*cos(f*x + e)^2 - c^2*f*cos(f*x + e) - 2*c^2*f + (c^2*f*cos(f*x + e) + 2*c^2*f)*sin(f*x + e))
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Sympy [A] time = 49.1151, size = 1282, normalized size = 13.93 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**2,x)
[Out]
Piecewise((15*a**3*f*x*tan(e/2 + f*x/2)**5/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c
**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 45*a**3*f*
x*tan(e/2 + f*x/2)**4/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2
)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 60*a**3*f*x*tan(e/2 + f*x/2)**3
/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(
e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 60*a**3*f*x*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 +
f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c
**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 45*a**3*f*x*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan
(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) -
3*c**2*f) - 15*a**3*f*x/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*
x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 8*a**3*tan(e/2 + f*x/2)**5/(
3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/
2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) - 70*a**3*tan(e/2 + f*x/2)**3/(3*c**2*f*tan(e/2 + f*x/2)
**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*
tan(e/2 + f*x/2) - 3*c**2*f) + 50*a**3*tan(e/2 + f*x/2)**2/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 +
f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2
*f) - 90*a**3*tan(e/2 + f*x/2)/(3*c**2*f*tan(e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/
2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2 + 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f) + 38*a**3/(3*c**2*f*tan(
e/2 + f*x/2)**5 - 9*c**2*f*tan(e/2 + f*x/2)**4 + 12*c**2*f*tan(e/2 + f*x/2)**3 - 12*c**2*f*tan(e/2 + f*x/2)**2
+ 9*c**2*f*tan(e/2 + f*x/2) - 3*c**2*f), Ne(f, 0)), (x*(a*sin(e) + a)**3/(-c*sin(e) + c)**2, True))
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Giac [A] time = 1.75864, size = 136, normalized size = 1.48 \begin{align*} \frac{\frac{15 \,{\left (f x + e\right )} a^{3}}{c^{2}} - \frac{6 \, a^{3}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} c^{2}} + \frac{8 \,{\left (3 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 12 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 5 \, a^{3}\right )}}{c^{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((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^2,x, algorithm="giac")
[Out]
1/3*(15*(f*x + e)*a^3/c^2 - 6*a^3/((tan(1/2*f*x + 1/2*e)^2 + 1)*c^2) + 8*(3*a^3*tan(1/2*f*x + 1/2*e)^2 - 12*a^
3*tan(1/2*f*x + 1/2*e) + 5*a^3)/(c^2*(tan(1/2*f*x + 1/2*e) - 1)^3))/f