3.282 \(\int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=33 \[ -\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 33, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2736, 2671} \[ -\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (a \sin (e+f x)+a)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2671

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b*(g*
Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

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]))

Rubi steps

\begin {align*} \int \frac {(c-c \sin (e+f x))^2}{(a+a \sin (e+f x))^3} \, dx &=\left (a^2 c^2\right ) \int \frac {\cos ^4(e+f x)}{(a+a \sin (e+f x))^5} \, dx\\ &=-\frac {a^2 c^2 \cos ^5(e+f x)}{5 f (a+a \sin (e+f x))^5}\\ \end {align*}

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Mathematica [B]  time = 0.41, size = 81, normalized size = 2.45 \[ \frac {c^2 \left (10 \sin \left (\frac {1}{2} (e+f x)\right )+5 \sin \left (\frac {3}{2} (e+f x)\right )-\sin \left (\frac {5}{2} (e+f x)\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )}{10 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(10*Sin[(e + f*x)/2] + 5*Sin[(3*(e + f*x))/2] - Sin[(5*(e + f*x))/2
]))/(10*a^3*f*(1 + Sin[e + f*x])^3)

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fricas [B]  time = 0.42, size = 168, normalized size = 5.09 \[ -\frac {c^{2} \cos \left (f x + e\right )^{3} + 3 \, c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) - 4 \, c^{2} - {\left (c^{2} \cos \left (f x + e\right )^{2} - 2 \, c^{2} \cos \left (f x + e\right ) - 4 \, c^{2}\right )} \sin \left (f x + e\right )}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

-1/5*(c^2*cos(f*x + e)^3 + 3*c^2*cos(f*x + e)^2 - 2*c^2*cos(f*x + e) - 4*c^2 - (c^2*cos(f*x + e)^2 - 2*c^2*cos
(f*x + e) - 4*c^2)*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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giac [A]  time = 0.26, size = 60, normalized size = 1.82 \[ -\frac {2 \, {\left (5 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 10 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c^{2}\right )}}{5 \, a^{3} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}} \]

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))^3,x, algorithm="giac")

[Out]

-2/5*(5*c^2*tan(1/2*f*x + 1/2*e)^4 + 10*c^2*tan(1/2*f*x + 1/2*e)^2 + c^2)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)

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maple [B]  time = 0.27, size = 88, normalized size = 2.67 \[ \frac {2 c^{2} \left (\frac {4}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1}-\frac {8}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}-\frac {16}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}\right )}{f \,a^{3}} \]

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))^3,x)

[Out]

2/f*c^2/a^3*(4/(tan(1/2*f*x+1/2*e)+1)^2+8/(tan(1/2*f*x+1/2*e)+1)^4-1/(tan(1/2*f*x+1/2*e)+1)-8/(tan(1/2*f*x+1/2
*e)+1)^3-16/5/(tan(1/2*f*x+1/2*e)+1)^5)

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maxima [B]  time = 0.71, size = 554, normalized size = 16.79 \[ -\frac {2 \, {\left (\frac {c^{2} {\left (\frac {20 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {40 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {30 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + 7\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} + \frac {2 \, c^{2} {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}} - \frac {6 \, c^{2} {\left (\frac {5 \, \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 {5 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + 1\right )}}{a^{3} + \frac {5 \, a^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {10 \, a^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {5 \, a^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {a^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}\right )}}{15 \, f} \]

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))^3,x, algorithm="maxima")

[Out]

-2/15*(c^2*(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/(c
os(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) + 2*c^2*(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) - 6*c^2*(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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mupad [B]  time = 7.21, size = 90, normalized size = 2.73 \[ -\frac {2\,c^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left ({\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4+10\,{\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+5\,{\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{5\,a^3\,f\,{\left (\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )+\sin \left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c*sin(e + f*x))^2/(a + a*sin(e + f*x))^3,x)

[Out]

-(2*c^2*cos(e/2 + (f*x)/2)*(cos(e/2 + (f*x)/2)^4 + 5*sin(e/2 + (f*x)/2)^4 + 10*cos(e/2 + (f*x)/2)^2*sin(e/2 +
(f*x)/2)^2))/(5*a^3*f*(cos(e/2 + (f*x)/2) + sin(e/2 + (f*x)/2))^5)

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sympy [A]  time = 14.47, size = 354, normalized size = 10.73 \[ \begin {cases} - \frac {10 c^{2} \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} - \frac {20 c^{2} \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} - \frac {2 c^{2}}{5 a^{3} f \tan ^{5}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan ^{4}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{3}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 50 a^{3} f \tan ^{2}{\left (\frac {e}{2} + \frac {f x}{2} \right )} + 25 a^{3} f \tan {\left (\frac {e}{2} + \frac {f x}{2} \right )} + 5 a^{3} f} & \text {for}\: f \neq 0 \\\frac {x \left (- c \sin {\relax (e )} + c\right )^{2}}{\left (a \sin {\relax (e )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

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))**3,x)

[Out]

Piecewise((-10*c**2*tan(e/2 + f*x/2)**4/(5*a**3*f*tan(e/2 + f*x/2)**5 + 25*a**3*f*tan(e/2 + f*x/2)**4 + 50*a**
3*f*tan(e/2 + f*x/2)**3 + 50*a**3*f*tan(e/2 + f*x/2)**2 + 25*a**3*f*tan(e/2 + f*x/2) + 5*a**3*f) - 20*c**2*tan
(e/2 + f*x/2)**2/(5*a**3*f*tan(e/2 + f*x/2)**5 + 25*a**3*f*tan(e/2 + f*x/2)**4 + 50*a**3*f*tan(e/2 + f*x/2)**3
+ 50*a**3*f*tan(e/2 + f*x/2)**2 + 25*a**3*f*tan(e/2 + f*x/2) + 5*a**3*f) - 2*c**2/(5*a**3*f*tan(e/2 + f*x/2)*
*5 + 25*a**3*f*tan(e/2 + f*x/2)**4 + 50*a**3*f*tan(e/2 + f*x/2)**3 + 50*a**3*f*tan(e/2 + f*x/2)**2 + 25*a**3*f
*tan(e/2 + f*x/2) + 5*a**3*f), Ne(f, 0)), (x*(-c*sin(e) + c)**2/(a*sin(e) + a)**3, True))

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