3.476 \(\int \frac{1}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=83 \[ -\frac{2 \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{2 \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac{\cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.0469928, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2650, 2648} \[ -\frac{2 \cos (e+f x)}{15 f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{2 \cos (e+f x)}{15 a f (a \sin (e+f x)+a)^2}-\frac{\cos (e+f x)}{5 f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(-3),x]

[Out]

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

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^3} \, dx &=-\frac{\cos (e+f x)}{5 f (a+a \sin (e+f x))^3}+\frac{2 \int \frac{1}{(a+a \sin (e+f x))^2} \, dx}{5 a}\\ &=-\frac{\cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{2 \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}+\frac{2 \int \frac{1}{a+a \sin (e+f x)} \, dx}{15 a^2}\\ &=-\frac{\cos (e+f x)}{5 f (a+a \sin (e+f x))^3}-\frac{2 \cos (e+f x)}{15 a f (a+a \sin (e+f x))^2}-\frac{2 \cos (e+f x)}{15 f \left (a^3+a^3 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.122553, size = 76, normalized size = 0.92 \[ \frac{15 \sin (e+f x)-6 \sin (2 (e+f x))-\sin (3 (e+f x))-15 \cos (e+f x)-6 \cos (2 (e+f x))+\cos (3 (e+f x))+10}{30 a^3 f (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10 - 15*Cos[e + f*x] - 6*Cos[2*(e + f*x)] + Cos[3*(e + f*x)] + 15*Sin[e + f*x] - 6*Sin[2*(e + f*x)] - Sin[3*(
e + f*x)])/(30*a^3*f*(1 + Sin[e + f*x])^3)

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Maple [A]  time = 0.045, size = 85, normalized size = 1. \begin{align*} 2\,{\frac{1}{f{a}^{3}} \left ( -4/5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-5}+2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-4}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-1}+2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}-8/3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^3,x)

[Out]

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

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Maxima [B]  time = 1.7579, size = 274, normalized size = 3.3 \begin{align*} -\frac{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 )}}{15 \,{\left (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 )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(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)*f)

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Fricas [A]  time = 1.4969, size = 369, normalized size = 4.45 \begin{align*} -\frac{2 \, \cos \left (f x + e\right )^{3} - 4 \, \cos \left (f x + e\right )^{2} -{\left (2 \, \cos \left (f x + e\right )^{2} + 6 \, \cos \left (f x + e\right ) - 3\right )} \sin \left (f x + e\right ) - 9 \, \cos \left (f x + e\right ) - 3}{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(1/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(2*cos(f*x + e)^3 - 4*cos(f*x + e)^2 - (2*cos(f*x + e)^2 + 6*cos(f*x + e) - 3)*sin(f*x + e) - 9*cos(f*x
+ e) - 3)/(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 [A]  time = 4.77631, size = 558, normalized size = 6.72 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))**3,x)

[Out]

Piecewise((-30*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) - 60*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 + 1
50*a**3*f*tan(e/2 + f*x/2)**2 + 75*a**3*f*tan(e/2 + f*x/2) + 15*a**3*f) - 80*tan(e/2 + f*x/2)**2/(15*a**3*f*ta
n(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) - 40*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) - 14/(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/(a*sin(e
) + a)**3, True))

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Giac [A]  time = 1.2323, size = 105, normalized size = 1.27 \begin{align*} -\frac{2 \,{\left (15 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 30 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 40 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 20 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 7\right )}}{15 \, a^{3} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-2/15*(15*tan(1/2*f*x + 1/2*e)^4 + 30*tan(1/2*f*x + 1/2*e)^3 + 40*tan(1/2*f*x + 1/2*e)^2 + 20*tan(1/2*f*x + 1/
2*e) + 7)/(a^3*f*(tan(1/2*f*x + 1/2*e) + 1)^5)