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

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

[Out]

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

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Rubi [A]  time = 0.18, antiderivative size = 103, normalized size of antiderivative = 1.00, 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 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

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 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))^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*}

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Mathematica [B]  time = 0.44, 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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fricas [B]  time = 0.43, size = 233, normalized size = 2.26 \[ \frac {60 \, c^{3} f x - {\left (15 \, c^{3} f x + 46 \, c^{3}\right )} \cos \left (f x + e\right )^{3} + 24 \, c^{3} - {\left (45 \, c^{3} f x - 2 \, c^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, c^{3} f x + 12 \, c^{3}\right )} \cos \left (f x + e\right ) + {\left (60 \, c^{3} f x - 24 \, c^{3} - {\left (15 \, c^{3} f x - 46 \, c^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \, {\left (5 \, c^{3} f x + 8 \, c^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f + {\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \]

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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giac [A]  time = 0.23, size = 111, normalized size = 1.08 \[ -\frac {\frac {15 \, {\left (f x + e\right )} c^{3}}{a^{3}} + \frac {4 \, {\left (15 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 100 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 50 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 13 \, c^{3}\right )}}{a^{3} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \]

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

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maple [A]  time = 0.30, size = 143, normalized size = 1.39 \[ -\frac {2 c^{3} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{a^{3} f}-\frac {64 c^{3}}{5 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{5}}+\frac {32 c^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{4}}-\frac {80 c^{3}}{3 a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{3}}+\frac {8 c^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}-\frac {4 c^{3}}{a^{3} f \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \]

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

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maxima [B]  time = 1.46, size = 781, normalized size = 7.58 \[ \text {result too large to display} \]

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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mupad [B]  time = 8.94, size = 200, normalized size = 1.94 \[ \frac {c^3\,\left (e+f\,x\right )+\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (5\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (75\,e+75\,f\,x+200\right )}{15}\right )-\frac {c^3\,\left (15\,e+15\,f\,x+52\right )}{15}+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\,\left (5\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (75\,e+75\,f\,x+60\right )}{15}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (10\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (150\,e+150\,f\,x+120\right )}{15}\right )+{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (10\,c^3\,\left (e+f\,x\right )-\frac {c^3\,\left (150\,e+150\,f\,x+400\right )}{15}\right )}{a^3\,f\,{\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )+1\right )}^5}-\frac {c^3\,x}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c^3*(e + f*x) + tan(e/2 + (f*x)/2)*(5*c^3*(e + f*x) - (c^3*(75*e + 75*f*x + 200))/15) - (c^3*(15*e + 15*f*x +
52))/15 + tan(e/2 + (f*x)/2)^4*(5*c^3*(e + f*x) - (c^3*(75*e + 75*f*x + 60))/15) + tan(e/2 + (f*x)/2)^3*(10*c
^3*(e + f*x) - (c^3*(150*e + 150*f*x + 120))/15) + tan(e/2 + (f*x)/2)^2*(10*c^3*(e + f*x) - (c^3*(150*e + 150*
f*x + 400))/15))/(a^3*f*(tan(e/2 + (f*x)/2) + 1)^5) - (c^3*x)/a^3

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sympy [A]  time = 26.24, size = 1284, normalized size = 12.47 \[ \text {result too large to display} \]

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]

Piecewise((-15*c**3*f*x*tan(e/2 + f*x/2)**5/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 1
50*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 75*
c**3*f*x*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e
/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 150*c**3*f*x*tan(e
/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3
+ 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 150*c**3*f*x*tan(e/2 + f*x/2)**2
/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*
tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 75*c**3*f*x*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/
2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**
2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 15*c**3*f*x/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 +
f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15
*a**3*f) - 60*c**3*tan(e/2 + f*x/2)**4/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a*
*3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 120*c**3
*tan(e/2 + f*x/2)**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x
/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 400*c**3*tan(e/2 + f*x/2)*
*2/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*
f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 200*c**3*tan(e/2 + f*x/2)/(15*a**3*f*tan(e/2
+ f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2
+ 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 52*c**3/(15*a**3*f*tan(e/2 + f*x/2)**5 + 75*a**3*f*tan(e/2 + f*x/
2)**4 + 150*a**3*f*tan(e/2 + f*x/2)**3 + 150*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3
*f), Ne(f, 0)), (x*(-c*sin(e) + c)**3/(a*sin(e) + a)**3, True))

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