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

Optimal. Leaf size=94 \[ \frac{2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{15 a^3 \cos (e+f x)}{2 c f}+\frac{5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac{15 a^3 x}{2 c} \]

[Out]

(-15*a^3*x)/(2*c) + (15*a^3*Cos[e + f*x])/(2*c*f) + (2*a^3*c^2*Cos[e + f*x]^5)/(f*(c - c*Sin[e + f*x])^3) + (5
*a^3*Cos[e + f*x]^3)/(2*f*(c - c*Sin[e + f*x]))

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Rubi [A]  time = 0.182553, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 5, 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} \[ \frac{2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{15 a^3 \cos (e+f x)}{2 c f}+\frac{5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac{15 a^3 x}{2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-15*a^3*x)/(2*c) + (15*a^3*Cos[e + f*x])/(2*c*f) + (2*a^3*c^2*Cos[e + f*x]^5)/(f*(c - c*Sin[e + f*x])^3) + (5
*a^3*Cos[e + f*x]^3)/(2*f*(c - c*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{(a+a \sin (e+f x))^3}{c-c \sin (e+f x)} \, dx &=\left (a^3 c^3\right ) \int \frac{\cos ^6(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac{2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}-\left (5 a^3 c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx\\ &=\frac{2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac{1}{2} \left (15 a^3\right ) \int \frac{\cos ^2(e+f x)}{c-c \sin (e+f x)} \, dx\\ &=\frac{15 a^3 \cos (e+f x)}{2 c f}+\frac{2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}-\frac{\left (15 a^3\right ) \int 1 \, dx}{2 c}\\ &=-\frac{15 a^3 x}{2 c}+\frac{15 a^3 \cos (e+f x)}{2 c f}+\frac{2 a^3 c^2 \cos ^5(e+f x)}{f (c-c \sin (e+f x))^3}+\frac{5 a^3 \cos ^3(e+f x)}{2 f (c-c \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 0.50658, size = 153, normalized size = 1.63 \[ \frac{a^3 (\sin (e+f x)+1)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right ) (30 (e+f x)-\sin (2 (e+f x))-16 \cos (e+f x))+\sin \left (\frac{1}{2} (e+f x)\right ) (\sin (2 (e+f x))+16 \cos (e+f x)-30 e-30 f x-64)\right )}{4 c f (\sin (e+f x)-1) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^3*(Cos[(e + f*x)/2]*(30*(e + f*x) - 16*Cos[e + f
*x] - Sin[2*(e + f*x)]) + Sin[(e + f*x)/2]*(-64 - 30*e - 30*f*x + 16*Cos[e + f*x] + Sin[2*(e + f*x)])))/(4*c*f
*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(-1 + Sin[e + f*x]))

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Maple [B]  time = 0.08, size = 181, normalized size = 1.9 \begin{align*} -16\,{\frac{{a}^{3}}{cf \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-{\frac{{a}^{3}}{cf} \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}}+8\,{\frac{{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}}{cf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{a}^{3}}{cf}\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}}+8\,{\frac{{a}^{3}}{cf \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-15\,{\frac{{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{cf}} \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)),x)

[Out]

-16/f*a^3/c/(tan(1/2*f*x+1/2*e)-1)-1/f*a^3/c/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3+8/f*a^3/c/(1+tan(
1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^2+1/f*a^3/c/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)+8/f*a^3/c/(1+
tan(1/2*f*x+1/2*e)^2)^2-15/f*a^3/c*arctan(tan(1/2*f*x+1/2*e))

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Maxima [B]  time = 2.1269, size = 585, normalized size = 6.22 \begin{align*} -\frac{6 \, a^{3}{\left (\frac{\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - 2}{c - \frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}} + \frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + a^{3}{\left (\frac{\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{5 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{3 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{3 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 4}{c - \frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{2 \, c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{2 \, c \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac{3 \, \arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c}\right )} + 6 \, a^{3}{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c} - \frac{1}{c - \frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} - \frac{2 \, a^{3}}{c - \frac{c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}{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)),x, algorithm="maxima")

[Out]

-(6*a^3*((sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2)/(c - c*sin(f*x + e)/(cos(
f*x + e) + 1) + c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + arctan(sin(f*
x + e)/(cos(f*x + e) + 1))/c) + a^3*((sin(f*x + e)/(cos(f*x + e) + 1) - 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2
+ 3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 4)/(c - c*sin(f*x + e)/(cos(
f*x + e) + 1) + 2*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*c*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + c*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - c*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1
))/c) + 6*a^3*(arctan(sin(f*x + e)/(cos(f*x + e) + 1))/c - 1/(c - c*sin(f*x + e)/(cos(f*x + e) + 1))) - 2*a^3/
(c - c*sin(f*x + e)/(cos(f*x + e) + 1)))/f

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Fricas [A]  time = 1.409, size = 312, normalized size = 3.32 \begin{align*} \frac{a^{3} \cos \left (f x + e\right )^{3} - 15 \, a^{3} f x + 8 \, a^{3} \cos \left (f x + e\right )^{2} + 16 \, a^{3} -{\left (15 \, a^{3} f x - 23 \, a^{3}\right )} \cos \left (f x + e\right ) +{\left (15 \, a^{3} f x + a^{3} \cos \left (f x + e\right )^{2} - 7 \, a^{3} \cos \left (f x + e\right ) + 16 \, a^{3}\right )} \sin \left (f x + e\right )}{2 \,{\left (c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f\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)),x, algorithm="fricas")

[Out]

1/2*(a^3*cos(f*x + e)^3 - 15*a^3*f*x + 8*a^3*cos(f*x + e)^2 + 16*a^3 - (15*a^3*f*x - 23*a^3)*cos(f*x + e) + (1
5*a^3*f*x + a^3*cos(f*x + e)^2 - 7*a^3*cos(f*x + e) + 16*a^3)*sin(f*x + e))/(c*f*cos(f*x + e) - c*f*sin(f*x +
e) + c*f)

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Sympy [A]  time = 17.9005, size = 1170, normalized size = 12.45 \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)),x)

[Out]

Piecewise((-15*a**3*f*x*tan(e/2 + f*x/2)**5/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan
(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 15*a**3*f*x*tan(e/2 + f*x/2)*
*4/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)
**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 30*a**3*f*x*tan(e/2 + f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan
(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) + 3
0*a**3*f*x*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)
**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 15*a**3*f*x*tan(e/2 + f*x/2)/(2*c*f*tan(e/
2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(
e/2 + f*x/2) - 2*c*f) + 15*a**3*f*x/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f
*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 14*a**3*tan(e/2 + f*x/2)**5/(2*c*f*ta
n(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*
tan(e/2 + f*x/2) - 2*c*f) - 20*a**3*tan(e/2 + f*x/2)**4/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4
+ 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 10*a**3*tan(e/2 +
f*x/2)**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2
+ f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f) - 50*a**3*tan(e/2 + f*x/2)**2/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f
*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2 + f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f)
- 34*a**3/(2*c*f*tan(e/2 + f*x/2)**5 - 2*c*f*tan(e/2 + f*x/2)**4 + 4*c*f*tan(e/2 + f*x/2)**3 - 4*c*f*tan(e/2
+ f*x/2)**2 + 2*c*f*tan(e/2 + f*x/2) - 2*c*f), Ne(f, 0)), (x*(a*sin(e) + a)**3/(-c*sin(e) + c), True))

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Giac [A]  time = 2.29381, size = 158, normalized size = 1.68 \begin{align*} -\frac{\frac{15 \,{\left (f x + e\right )} a^{3}}{c} + \frac{32 \, a^{3}}{c{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}} + \frac{2 \,{\left (a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 8 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 8 \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} c}}{2 \, 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)),x, algorithm="giac")

[Out]

-1/2*(15*(f*x + e)*a^3/c + 32*a^3/(c*(tan(1/2*f*x + 1/2*e) - 1)) + 2*(a^3*tan(1/2*f*x + 1/2*e)^3 - 8*a^3*tan(1
/2*f*x + 1/2*e)^2 - a^3*tan(1/2*f*x + 1/2*e) - 8*a^3)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*c))/f