3.263 \(\int \frac{(c-c \sin (e+f x))^2}{a+a \sin (e+f x)} \, dx\)

Optimal. Leaf size=56 \[ -\frac{3 c^2 \cos (e+f x)}{a f}-\frac{2 a c^2 \cos ^3(e+f x)}{f (a \sin (e+f x)+a)^2}-\frac{3 c^2 x}{a} \]

[Out]

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

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Rubi [A]  time = 0.135655, antiderivative size = 56, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2680, 2682, 8} \[ -\frac{3 c^2 \cos (e+f x)}{a f}-\frac{2 a c^2 \cos ^3(e+f x)}{f (a \sin (e+f x)+a)^2}-\frac{3 c^2 x}{a} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^2/(a + a*Sin[e + f*x]),x]

[Out]

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

Mathematica [B]  time = 0.366895, size = 129, normalized size = 2.3 \[ -\frac{c^2 (\sin (e+f x)-1)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right ) (3 (e+f x)+\cos (e+f x))+\sin \left (\frac{1}{2} (e+f x)\right ) (\cos (e+f x)+3 e+3 f x-8)\right )}{a f (\sin (e+f x)+1) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^2/(a + a*Sin[e + f*x]),x]

[Out]

-((c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(Cos[(e + f*x)/2]*(3*(e + f*x) + Cos[e + f*x]) + (-8 + 3*e + 3*f*
x + Cos[e + f*x])*Sin[(e + f*x)/2])*(-1 + Sin[e + f*x])^2)/(a*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*(1 + S
in[e + f*x])))

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Maple [A]  time = 0.071, size = 73, normalized size = 1.3 \begin{align*} -2\,{\frac{{c}^{2}}{af \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}-6\,{\frac{{c}^{2}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{af}}-8\,{\frac{{c}^{2}}{af \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))^2/(a+a*sin(f*x+e)),x)

[Out]

-2/f*c^2/a/(1+tan(1/2*f*x+1/2*e)^2)-6/f*c^2/a*arctan(tan(1/2*f*x+1/2*e))-8/f*c^2/a/(tan(1/2*f*x+1/2*e)+1)

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Maxima [B]  time = 2.14845, size = 284, normalized size = 5.07 \begin{align*} -\frac{2 \,{\left (c^{2}{\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}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{a \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{a \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 )}{a}\right )} + 2 \, c^{2}{\left (\frac{\arctan \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a} + \frac{1}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )} + \frac{c^{2}}{a + \frac{a \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.36464, size = 238, normalized size = 4.25 \begin{align*} -\frac{3 \, c^{2} f x + c^{2} \cos \left (f x + e\right )^{2} + 4 \, c^{2} +{\left (3 \, c^{2} f x + 5 \, c^{2}\right )} \cos \left (f x + e\right ) +{\left (3 \, c^{2} f x + c^{2} \cos \left (f x + e\right ) - 4 \, c^{2}\right )} \sin \left (f x + e\right )}{a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 7.78863, size = 456, normalized size = 8.14 \begin{align*} \begin{cases} - \frac{3 c^{2} f x \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{a f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} - \frac{3 c^{2} f x \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{a f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} - \frac{3 c^{2} f x \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{a f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} - \frac{3 c^{2} f x}{a f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} + \frac{2 c^{2} \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{a f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} - \frac{6 c^{2} \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )}}{a f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} - \frac{8 c^{2}}{a f \tan ^{3}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan ^{2}{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f \tan{\left (\frac{e}{2} + \frac{f x}{2} \right )} + a f} & \text{for}\: f \neq 0 \\\frac{x \left (- c \sin{\left (e \right )} + c\right )^{2}}{a \sin{\left (e \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))**2/(a+a*sin(f*x+e)),x)

[Out]

Piecewise((-3*c**2*f*x*tan(e/2 + f*x/2)**3/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 +
f*x/2) + a*f) - 3*c**2*f*x*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/
2 + f*x/2) + a*f) - 3*c**2*f*x*tan(e/2 + f*x/2)/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e
/2 + f*x/2) + a*f) - 3*c**2*f*x/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*
f) + 2*c**2*tan(e/2 + f*x/2)**3/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*
f) - 6*c**2*tan(e/2 + f*x/2)**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*
f) - 8*c**2/(a*f*tan(e/2 + f*x/2)**3 + a*f*tan(e/2 + f*x/2)**2 + a*f*tan(e/2 + f*x/2) + a*f), Ne(f, 0)), (x*(-
c*sin(e) + c)**2/(a*sin(e) + a), True))

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Giac [A]  time = 1.93332, size = 135, normalized size = 2.41 \begin{align*} -\frac{\frac{3 \,{\left (f x + e\right )} c^{2}}{a} + \frac{2 \,{\left (4 \, c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 5 \, c^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} a}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^2/(a+a*sin(f*x+e)),x, algorithm="giac")

[Out]

-(3*(f*x + e)*c^2/a + 2*(4*c^2*tan(1/2*f*x + 1/2*e)^2 + c^2*tan(1/2*f*x + 1/2*e) + 5*c^2)/((tan(1/2*f*x + 1/2*
e)^3 + tan(1/2*f*x + 1/2*e)^2 + tan(1/2*f*x + 1/2*e) + 1)*a))/f