3.276 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=76 \[ \frac{4 \tan ^3(e+f x)}{15 a^2 c^3 f}+\frac{4 \tan (e+f x)}{5 a^2 c^3 f}+\frac{\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )} \]

[Out]

Sec[e + f*x]^3/(5*a^2*f*(c^3 - c^3*Sin[e + f*x])) + (4*Tan[e + f*x])/(5*a^2*c^3*f) + (4*Tan[e + f*x]^3)/(15*a^
2*c^3*f)

________________________________________________________________________________________

Rubi [A]  time = 0.113286, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2672, 3767} \[ \frac{4 \tan ^3(e+f x)}{15 a^2 c^3 f}+\frac{4 \tan (e+f x)}{5 a^2 c^3 f}+\frac{\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^3),x]

[Out]

Sec[e + f*x]^3/(5*a^2*f*(c^3 - c^3*Sin[e + f*x])) + (4*Tan[e + f*x])/(5*a^2*c^3*f) + (4*Tan[e + f*x]^3)/(15*a^
2*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]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^3} \, dx &=\frac{\int \frac{\sec ^4(e+f x)}{c-c \sin (e+f x)} \, dx}{a^2 c^2}\\ &=\frac{\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}+\frac{4 \int \sec ^4(e+f x) \, dx}{5 a^2 c^3}\\ &=\frac{\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}-\frac{4 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{5 a^2 c^3 f}\\ &=\frac{\sec ^3(e+f x)}{5 a^2 f \left (c^3-c^3 \sin (e+f x)\right )}+\frac{4 \tan (e+f x)}{5 a^2 c^3 f}+\frac{4 \tan ^3(e+f x)}{15 a^2 c^3 f}\\ \end{align*}

Mathematica [A]  time = 0.82248, size = 131, normalized size = 1.72 \[ -\frac{18 \sin (e+f x)+512 \sin (2 (e+f x))+27 \sin (3 (e+f x))+128 \sin (4 (e+f x))+9 \sin (5 (e+f x))+128 \cos (e+f x)-72 \cos (2 (e+f x))+192 \cos (3 (e+f x))-18 \cos (4 (e+f x))+64 \cos (5 (e+f x))-54}{1920 a^2 c^3 f (\sin (e+f x)-1)^3 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(-54 + 128*Cos[e + f*x] - 72*Cos[2*(e + f*x)] + 192*Cos[3*(e + f*x)] - 18*Cos[4*(e + f*x)] + 64*Cos[5*(e + f*
x)] + 18*Sin[e + f*x] + 512*Sin[2*(e + f*x)] + 27*Sin[3*(e + f*x)] + 128*Sin[4*(e + f*x)] + 9*Sin[5*(e + f*x)]
)/(1920*a^2*c^3*f*(-1 + Sin[e + f*x])^3*(1 + Sin[e + f*x])^2)

________________________________________________________________________________________

Maple [A]  time = 0.059, size = 133, normalized size = 1.8 \begin{align*} 2\,{\frac{1}{f{c}^{3}{a}^{2}} \left ( -1/5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-5}-1/2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}-5/6\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-3}-3/4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-{\frac{11}{16\,\tan \left ( 1/2\,fx+e/2 \right ) -16}}-1/12\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3}+1/8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}-{\frac{5}{16\,\tan \left ( 1/2\,fx+e/2 \right ) +16}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x)

[Out]

2/f/c^3/a^2*(-1/5/(tan(1/2*f*x+1/2*e)-1)^5-1/2/(tan(1/2*f*x+1/2*e)-1)^4-5/6/(tan(1/2*f*x+1/2*e)-1)^3-3/4/(tan(
1/2*f*x+1/2*e)-1)^2-11/16/(tan(1/2*f*x+1/2*e)-1)-1/12/(tan(1/2*f*x+1/2*e)+1)^3+1/8/(tan(1/2*f*x+1/2*e)+1)^2-5/
16/(tan(1/2*f*x+1/2*e)+1))

________________________________________________________________________________________

Maxima [B]  time = 1.37926, size = 452, normalized size = 5.95 \begin{align*} \frac{2 \,{\left (\frac{9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{13 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{25 \, \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}} - \frac{15 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{15 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + 3\right )}}{15 \,{\left (a^{2} c^{3} - \frac{2 \, a^{2} c^{3} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{2 \, a^{2} c^{3} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{6 \, a^{2} c^{3} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{6 \, a^{2} c^{3} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{2 \, a^{2} c^{3} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{2 \, a^{2} c^{3} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac{a^{2} c^{3} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

2/15*(9*sin(f*x + e)/(cos(f*x + e) + 1) - 21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 13*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 + 25*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 15*sin(f*x + e)
^6/(cos(f*x + e) + 1)^6 + 15*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 3)/((a^2*c^3 - 2*a^2*c^3*sin(f*x + e)/(cos(
f*x + e) + 1) - 2*a^2*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 6*a^2*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
- 6*a^2*c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 2*a^2*c^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 2*a^2*c^3*si
n(f*x + e)^7/(cos(f*x + e) + 1)^7 - a^2*c^3*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*f)

________________________________________________________________________________________

Fricas [A]  time = 1.29684, size = 211, normalized size = 2.78 \begin{align*} -\frac{8 \, \cos \left (f x + e\right )^{4} - 4 \, \cos \left (f x + e\right )^{2} + 4 \,{\left (2 \, \cos \left (f x + e\right )^{2} + 1\right )} \sin \left (f x + e\right ) - 1}{15 \,{\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - a^{2} c^{3} f \cos \left (f x + e\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(8*cos(f*x + e)^4 - 4*cos(f*x + e)^2 + 4*(2*cos(f*x + e)^2 + 1)*sin(f*x + e) - 1)/(a^2*c^3*f*cos(f*x + e
)^3*sin(f*x + e) - a^2*c^3*f*cos(f*x + e)^3)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**3,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 2.02454, size = 180, normalized size = 2.37 \begin{align*} -\frac{\frac{5 \,{\left (15 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 24 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 13\right )}}{a^{2} c^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}} + \frac{165 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 480 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 650 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 400 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 113}{a^{2} c^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{5}}}{120 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/120*(5*(15*tan(1/2*f*x + 1/2*e)^2 + 24*tan(1/2*f*x + 1/2*e) + 13)/(a^2*c^3*(tan(1/2*f*x + 1/2*e) + 1)^3) +
(165*tan(1/2*f*x + 1/2*e)^4 - 480*tan(1/2*f*x + 1/2*e)^3 + 650*tan(1/2*f*x + 1/2*e)^2 - 400*tan(1/2*f*x + 1/2*
e) + 113)/(a^2*c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f