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

Optimal. Leaf size=103 \[ -\frac{2 a^2 c^3 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5}-\frac{2 c^3 \cos (e+f x)}{f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{c^3 x}{a^3}+\frac{2 c^3 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.179353, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2680, 8} \[ -\frac{2 a^2 c^3 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5}-\frac{2 c^3 \cos (e+f x)}{f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{c^3 x}{a^3}+\frac{2 c^3 \cos ^3(e+f x)}{3 f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 0.427202, size = 239, normalized size = 2.32 \[ \frac{(c-c \sin (e+f x))^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (48 \sin \left (\frac{1}{2} (e+f x)\right )-15 (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+92 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+44 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-88 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-24 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{15 f (a \sin (e+f x)+a)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(48*Sin[(e + f*x)/2] - 24*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 88*Si
n[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 44*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 92*Sin[(
e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 15*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)*(c
- c*Sin[e + f*x])^3)/(15*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^6*(a + a*Sin[e + f*x])^3)

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Maple [A]  time = 0.092, size = 143, normalized size = 1.4 \begin{align*} -2\,{\frac{{c}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}}}-{\frac{64\,{c}^{3}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}+32\,{\frac{{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-{\frac{80\,{c}^{3}}{3\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+8\,{\frac{{c}^{3}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}-4\,{\frac{{c}^{3}}{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))^3/(a+a*sin(f*x+e))^3,x)

[Out]

-2/f*c^3/a^3*arctan(tan(1/2*f*x+1/2*e))-64/5/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^5+32/f*c^3/a^3/(tan(1/2*f*x+1/2*
e)+1)^4-80/3/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^3+8/f*c^3/a^3/(tan(1/2*f*x+1/2*e)+1)^2-4/f*c^3/a^3/(tan(1/2*f*x+
1/2*e)+1)

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Maxima [B]  time = 2.06637, size = 1054, normalized size = 10.23 \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))^3,x, algorithm="maxima")

[Out]

-2/15*(c^3*((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) + c^3*(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) + 6*c^3*(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*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) - 9*c^3*(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*sin(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 [B]  time = 1.3382, size = 551, normalized size = 5.35 \begin{align*} \frac{60 \, c^{3} f x -{\left (15 \, c^{3} f x + 46 \, c^{3}\right )} \cos \left (f x + e\right )^{3} + 24 \, c^{3} -{\left (45 \, c^{3} f x - 2 \, c^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (5 \, c^{3} f x + 12 \, c^{3}\right )} \cos \left (f x + e\right ) +{\left (60 \, c^{3} f x - 24 \, c^{3} -{\left (15 \, c^{3} f x - 46 \, c^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (5 \, c^{3} f x + 8 \, c^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \,{\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))^3/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(60*c^3*f*x - (15*c^3*f*x + 46*c^3)*cos(f*x + e)^3 + 24*c^3 - (45*c^3*f*x - 2*c^3)*cos(f*x + e)^2 + 6*(5*
c^3*f*x + 12*c^3)*cos(f*x + e) + (60*c^3*f*x - 24*c^3 - (15*c^3*f*x - 46*c^3)*cos(f*x + e)^2 + 6*(5*c^3*f*x +
8*c^3)*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))**3/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.89325, size = 150, normalized size = 1.46 \begin{align*} -\frac{\frac{15 \,{\left (f x + e\right )} c^{3}}{a^{3}} + \frac{4 \,{\left (15 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 30 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 100 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 50 \, c^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 13 \, c^{3}\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, 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))^3,x, algorithm="giac")

[Out]

-1/15*(15*(f*x + e)*c^3/a^3 + 4*(15*c^3*tan(1/2*f*x + 1/2*e)^4 + 30*c^3*tan(1/2*f*x + 1/2*e)^3 + 100*c^3*tan(1
/2*f*x + 1/2*e)^2 + 50*c^3*tan(1/2*f*x + 1/2*e) + 13*c^3)/(a^3*(tan(1/2*f*x + 1/2*e) + 1)^5))/f