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

Optimal. Leaf size=85 \[ \frac{2 \tan (e+f x)}{5 a c^3 f}+\frac{\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac{\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2} \]

[Out]

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

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Rubi [A]  time = 0.15787, antiderivative size = 85, 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} \[ \frac{2 \tan (e+f x)}{5 a c^3 f}+\frac{\sec (e+f x)}{5 a f \left (c^3-c^3 \sin (e+f x)\right )}+\frac{\sec (e+f x)}{5 a c f (c-c \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.669206, size = 111, normalized size = 1.31 \[ -\frac{12 \sin (e+f x)+32 \sin (2 (e+f x))+12 \sin (3 (e+f x))-8 \sin (4 (e+f x))+32 \cos (e+f x)-12 \cos (2 (e+f x))+32 \cos (3 (e+f x))+3 \cos (4 (e+f x))-15}{160 a c^3 f (\sin (e+f x)-1)^3 (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

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Maple [A]  time = 0.063, size = 103, normalized size = 1.2 \begin{align*} 2\,{\frac{1}{af{c}^{3}} \left ( -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}}-1/8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.46496, size = 285, normalized size = 3.35 \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 c^{3} - \frac{4 \, a c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{5 \, a c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{5 \, a c^{3} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{4 \, a c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac{a c^{3} \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))/(c-c*sin(f*x+e))^3,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*c^3 - 4*a*c^3*sin(f*x + e)/(cos(f*x + e) + 1) + 5
*a*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 5*a*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 4*a*c^3*sin(f*x + e
)^5/(cos(f*x + e) + 1)^5 - a*c^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6)*f)

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Fricas [A]  time = 1.30729, size = 209, normalized size = 2.46 \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 c^{3} f \cos \left (f x + e\right )^{3} + 2 \, a c^{3} f \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a c^{3} 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))/(c-c*sin(f*x+e))^3,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*c^3*f*cos(f*x + e)^3 + 2*a*c^3*f*cos(f*x
+ e)*sin(f*x + e) - 2*a*c^3*f*cos(f*x + e))

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Sympy [A]  time = 17.9411, size = 738, normalized size = 8.68 \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))/(c-c*sin(f*x+e))**3,x)

[Out]

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

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Giac [A]  time = 2.01864, size = 142, normalized size = 1.67 \begin{align*} -\frac{\frac{5}{a c^{3}{\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 c^{3}{\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))/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/20*(5/(a*c^3*(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*c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f