3.268 \(\int \frac{1}{(a+a \sin (e+f x)) (c-c \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=118 \[ \frac{8 \tan (e+f x)}{35 a c^4 f}+\frac{4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3} \]

[Out]

Sec[e + f*x]/(7*a*c*f*(c - c*Sin[e + f*x])^3) + (4*Sec[e + f*x])/(35*a*f*(c^2 - c^2*Sin[e + f*x])^2) + (4*Sec[
e + f*x])/(35*a*f*(c^4 - c^4*Sin[e + f*x])) + (8*Tan[e + f*x])/(35*a*c^4*f)

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Rubi [A]  time = 0.206891, antiderivative size = 118, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2672, 3767, 8} \[ \frac{8 \tan (e+f x)}{35 a c^4 f}+\frac{4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^4),x]

[Out]

Sec[e + f*x]/(7*a*c*f*(c - c*Sin[e + f*x])^3) + (4*Sec[e + f*x])/(35*a*f*(c^2 - c^2*Sin[e + f*x])^2) + (4*Sec[
e + f*x])/(35*a*f*(c^4 - c^4*Sin[e + f*x])) + (8*Tan[e + f*x])/(35*a*c^4*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))^4} \, dx &=\frac{\int \frac{\sec ^2(e+f x)}{(c-c \sin (e+f x))^3} \, dx}{a c}\\ &=\frac{\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac{4 \int \frac{\sec ^2(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{7 a c^2}\\ &=\frac{\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac{4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{12 \int \frac{\sec ^2(e+f x)}{c-c \sin (e+f x)} \, dx}{35 a c^3}\\ &=\frac{\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac{4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{8 \int \sec ^2(e+f x) \, dx}{35 a c^4}\\ &=\frac{\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac{4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{8 \operatorname{Subst}(\int 1 \, dx,x,-\tan (e+f x))}{35 a c^4 f}\\ &=\frac{\sec (e+f x)}{7 a c f (c-c \sin (e+f x))^3}+\frac{4 \sec (e+f x)}{35 a f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{4 \sec (e+f x)}{35 a f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{8 \tan (e+f x)}{35 a c^4 f}\\ \end{align*}

Mathematica [A]  time = 0.740573, size = 131, normalized size = 1.11 \[ \frac{406 \sin (e+f x)+512 \sin (2 (e+f x))+377 \sin (3 (e+f x))-384 \sin (4 (e+f x))-29 \sin (5 (e+f x))+896 \cos (e+f x)-232 \cos (2 (e+f x))+832 \cos (3 (e+f x))+174 \cos (4 (e+f x))-64 \cos (5 (e+f x))-406}{4480 a c^4 f (\sin (e+f x)-1)^4 (\sin (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])*(c - c*Sin[e + f*x])^4),x]

[Out]

(-406 + 896*Cos[e + f*x] - 232*Cos[2*(e + f*x)] + 832*Cos[3*(e + f*x)] + 174*Cos[4*(e + f*x)] - 64*Cos[5*(e +
f*x)] + 406*Sin[e + f*x] + 512*Sin[2*(e + f*x)] + 377*Sin[3*(e + f*x)] - 384*Sin[4*(e + f*x)] - 29*Sin[5*(e +
f*x)])/(4480*a*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x]))

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Maple [A]  time = 0.059, size = 133, normalized size = 1.1 \begin{align*} 2\,{\frac{1}{af{c}^{4}} \left ( -4/7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-7}-2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-6}-{\frac{19}{5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-9/2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}-{\frac{15}{4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-{\frac{17}{8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{15}{16\,\tan \left ( 1/2\,fx+e/2 \right ) -16}}-1/16\, \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))^4,x)

[Out]

2/f/a/c^4*(-4/7/(tan(1/2*f*x+1/2*e)-1)^7-2/(tan(1/2*f*x+1/2*e)-1)^6-19/5/(tan(1/2*f*x+1/2*e)-1)^5-9/2/(tan(1/2
*f*x+1/2*e)-1)^4-15/4/(tan(1/2*f*x+1/2*e)-1)^3-17/8/(tan(1/2*f*x+1/2*e)-1)^2-15/16/(tan(1/2*f*x+1/2*e)-1)-1/16
/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.56919, size = 431, normalized size = 3.65 \begin{align*} -\frac{2 \,{\left (\frac{43 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac{77 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{7 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{105 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac{175 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} + \frac{105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac{35 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - 13\right )}}{35 \,{\left (a c^{4} - \frac{6 \, a c^{4} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{14 \, a c^{4} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{14 \, a c^{4} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{14 \, a c^{4} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac{14 \, a c^{4} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac{6 \, a c^{4} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac{a c^{4} \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))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")

[Out]

-2/35*(43*sin(f*x + e)/(cos(f*x + e) + 1) - 77*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 7*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 + 105*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 175*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 105*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 - 35*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 13)/((a*c^4 - 6*a*c^4*sin(f*x + e)/(co
s(f*x + e) + 1) + 14*a*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 14*a*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3
+ 14*a*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 14*a*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 6*a*c^4*sin(f*
x + e)^7/(cos(f*x + e) + 1)^7 - a*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8)*f)

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Fricas [A]  time = 1.3308, size = 279, normalized size = 2.36 \begin{align*} \frac{8 \, \cos \left (f x + e\right )^{4} - 36 \, \cos \left (f x + e\right )^{2} + 4 \,{\left (6 \, \cos \left (f x + e\right )^{2} - 5\right )} \sin \left (f x + e\right ) + 15}{35 \,{\left (3 \, a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right ) -{\left (a c^{4} f \cos \left (f x + e\right )^{3} - 4 \, a c^{4} f \cos \left (f x + e\right )\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))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")

[Out]

1/35*(8*cos(f*x + e)^4 - 36*cos(f*x + e)^2 + 4*(6*cos(f*x + e)^2 - 5)*sin(f*x + e) + 15)/(3*a*c^4*f*cos(f*x +
e)^3 - 4*a*c^4*f*cos(f*x + e) - (a*c^4*f*cos(f*x + e)^3 - 4*a*c^4*f*cos(f*x + e))*sin(f*x + e))

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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))/(c-c*sin(f*x+e))**4,x)

[Out]

Timed out

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Giac [A]  time = 2.17976, size = 180, normalized size = 1.53 \begin{align*} -\frac{\frac{35}{a c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}} + \frac{525 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 1960 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 4025 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 4480 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3143 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 1176 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 243}{a c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}}}{280 \, 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))^4,x, algorithm="giac")

[Out]

-1/280*(35/(a*c^4*(tan(1/2*f*x + 1/2*e) + 1)) + (525*tan(1/2*f*x + 1/2*e)^6 - 1960*tan(1/2*f*x + 1/2*e)^5 + 40
25*tan(1/2*f*x + 1/2*e)^4 - 4480*tan(1/2*f*x + 1/2*e)^3 + 3143*tan(1/2*f*x + 1/2*e)^2 - 1176*tan(1/2*f*x + 1/2
*e) + 243)/(a*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/f