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

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

[Out]

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

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Rubi [A]  time = 0.147634, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2672, 3767, 8} \[ \frac{2 \tan (e+f x)}{5 a^3 c f}-\frac{\sec (e+f x)}{5 c f \left (a^3 \sin (e+f x)+a^3\right )}-\frac{\sec (e+f x)}{5 a c f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])),x]

[Out]

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

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 2672

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*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m
, p}, x] && EqQ[a^2 - b^2, 0] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.58773, size = 111, normalized size = 1.34 \[ \frac{-12 \sin (e+f x)-32 \sin (2 (e+f x))-12 \sin (3 (e+f x))+8 \sin (4 (e+f x))+32 \cos (e+f x)-12 \cos (2 (e+f x))+32 \cos (3 (e+f x))+3 \cos (4 (e+f x))-15}{160 a^3 c f (\sin (e+f x)-1) (\sin (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^3*(c - c*Sin[e + f*x])),x]

[Out]

(-15 + 32*Cos[e + f*x] - 12*Cos[2*(e + f*x)] + 32*Cos[3*(e + f*x)] + 3*Cos[4*(e + f*x)] - 12*Sin[e + f*x] - 32
*Sin[2*(e + f*x)] - 12*Sin[3*(e + f*x)] + 8*Sin[4*(e + f*x)])/(160*a^3*c*f*(-1 + Sin[e + f*x])*(1 + Sin[e + f*
x])^3)

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Maple [A]  time = 0.06, size = 101, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{{a}^{3}cf} \left ( -1/8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-1}-2/5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-5}+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-4}-3/2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3}+5/4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}-{\frac{7}{8\,\tan \left ( 1/2\,fx+e/2 \right ) +8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/f/a^3/c*(-1/8/(tan(1/2*f*x+1/2*e)-1)-2/5/(tan(1/2*f*x+1/2*e)+1)^5+1/(tan(1/2*f*x+1/2*e)+1)^4-3/2/(tan(1/2*f*
x+1/2*e)+1)^3+5/4/(tan(1/2*f*x+1/2*e)+1)^2-7/8/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.13727, size = 285, normalized size = 3.43 \begin{align*} -\frac{2 \,{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{10 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{5 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + 2\right )}}{5 \,{\left (a^{3} c + \frac{4 \, a^{3} c \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{5 \, a^{3} c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{5 \, a^{3} c \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{4 \, a^{3} c \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac{a^{3} c \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(3*sin(f*x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 10*sin(f*x + e)^4/(cos(f*x
+ e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2)/((a^3*c + 4*a^3*c*sin(f*x + e)/(cos(f*x + e) + 1) + 5
*a^3*c*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*a^3*c*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 4*a^3*c*sin(f*x + e
)^5/(cos(f*x + e) + 1)^5 - a^3*c*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*f)

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Fricas [A]  time = 1.37743, size = 208, normalized size = 2.51 \begin{align*} \frac{4 \, \cos \left (f x + e\right )^{2} +{\left (2 \, \cos \left (f x + e\right )^{2} - 3\right )} \sin \left (f x + e\right ) - 2}{5 \,{\left (a^{3} c f \cos \left (f x + e\right )^{3} - 2 \, a^{3} c f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a^{3} c f \cos \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/(c-c*sin(f*x+e)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 17.9375, size = 738, normalized size = 8.89 \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/(c-c*sin(f*x+e)),x)

[Out]

Piecewise((3*tan(e/2 + f*x/2)**6/(10*a**3*c*f*tan(e/2 + f*x/2)**6 + 40*a**3*c*f*tan(e/2 + f*x/2)**5 + 50*a**3*
c*f*tan(e/2 + f*x/2)**4 - 50*a**3*c*f*tan(e/2 + f*x/2)**2 - 40*a**3*c*f*tan(e/2 + f*x/2) - 10*a**3*c*f) - 8*ta
n(e/2 + f*x/2)**5/(10*a**3*c*f*tan(e/2 + f*x/2)**6 + 40*a**3*c*f*tan(e/2 + f*x/2)**5 + 50*a**3*c*f*tan(e/2 + f
*x/2)**4 - 50*a**3*c*f*tan(e/2 + f*x/2)**2 - 40*a**3*c*f*tan(e/2 + f*x/2) - 10*a**3*c*f) - 25*tan(e/2 + f*x/2)
**4/(10*a**3*c*f*tan(e/2 + f*x/2)**6 + 40*a**3*c*f*tan(e/2 + f*x/2)**5 + 50*a**3*c*f*tan(e/2 + f*x/2)**4 - 50*
a**3*c*f*tan(e/2 + f*x/2)**2 - 40*a**3*c*f*tan(e/2 + f*x/2) - 10*a**3*c*f) - 40*tan(e/2 + f*x/2)**3/(10*a**3*c
*f*tan(e/2 + f*x/2)**6 + 40*a**3*c*f*tan(e/2 + f*x/2)**5 + 50*a**3*c*f*tan(e/2 + f*x/2)**4 - 50*a**3*c*f*tan(e
/2 + f*x/2)**2 - 40*a**3*c*f*tan(e/2 + f*x/2) - 10*a**3*c*f) - 15*tan(e/2 + f*x/2)**2/(10*a**3*c*f*tan(e/2 + f
*x/2)**6 + 40*a**3*c*f*tan(e/2 + f*x/2)**5 + 50*a**3*c*f*tan(e/2 + f*x/2)**4 - 50*a**3*c*f*tan(e/2 + f*x/2)**2
- 40*a**3*c*f*tan(e/2 + f*x/2) - 10*a**3*c*f) + 5/(10*a**3*c*f*tan(e/2 + f*x/2)**6 + 40*a**3*c*f*tan(e/2 + f*
x/2)**5 + 50*a**3*c*f*tan(e/2 + f*x/2)**4 - 50*a**3*c*f*tan(e/2 + f*x/2)**2 - 40*a**3*c*f*tan(e/2 + f*x/2) - 1
0*a**3*c*f), Ne(f, 0)), (x/((a*sin(e) + a)**3*(-c*sin(e) + c)), True))

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Giac [A]  time = 2.1182, size = 142, normalized size = 1.71 \begin{align*} -\frac{\frac{5}{a^{3} c{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}} + \frac{35 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 90 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 120 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 70 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 21}{a^{3} c{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}}}{20 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(5/(a^3*c*(tan(1/2*f*x + 1/2*e) - 1)) + (35*tan(1/2*f*x + 1/2*e)^4 + 90*tan(1/2*f*x + 1/2*e)^3 + 120*tan
(1/2*f*x + 1/2*e)^2 + 70*tan(1/2*f*x + 1/2*e) + 21)/(a^3*c*(tan(1/2*f*x + 1/2*e) + 1)^5))/f