3.277 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx\)

Optimal. Leaf size=111 \[ \frac{4 \tan ^3(e+f x)}{21 a^2 c^4 f}+\frac{4 \tan (e+f x)}{7 a^2 c^4 f}+\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2} \]

[Out]

Sec[e + f*x]^3/(7*a^2*f*(c^2 - c^2*Sin[e + f*x])^2) + Sec[e + f*x]^3/(7*a^2*f*(c^4 - c^4*Sin[e + f*x])) + (4*T
an[e + f*x])/(7*a^2*c^4*f) + (4*Tan[e + f*x]^3)/(21*a^2*c^4*f)

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Rubi [A]  time = 0.159483, antiderivative size = 111, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2672, 3767} \[ \frac{4 \tan ^3(e+f x)}{21 a^2 c^4 f}+\frac{4 \tan (e+f x)}{7 a^2 c^4 f}+\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^4),x]

[Out]

Sec[e + f*x]^3/(7*a^2*f*(c^2 - c^2*Sin[e + f*x])^2) + Sec[e + f*x]^3/(7*a^2*f*(c^4 - c^4*Sin[e + f*x])) + (4*T
an[e + f*x])/(7*a^2*c^4*f) + (4*Tan[e + f*x]^3)/(21*a^2*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]

Rubi steps

\begin{align*} \int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^4} \, dx &=\frac{\int \frac{\sec ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{a^2 c^2}\\ &=\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{5 \int \frac{\sec ^4(e+f x)}{c-c \sin (e+f x)} \, dx}{7 a^2 c^3}\\ &=\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{4 \int \sec ^4(e+f x) \, dx}{7 a^2 c^4}\\ &=\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}-\frac{4 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{7 a^2 c^4 f}\\ &=\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^2-c^2 \sin (e+f x)\right )^2}+\frac{\sec ^3(e+f x)}{7 a^2 f \left (c^4-c^4 \sin (e+f x)\right )}+\frac{4 \tan (e+f x)}{7 a^2 c^4 f}+\frac{4 \tan ^3(e+f x)}{21 a^2 c^4 f}\\ \end{align*}

Mathematica [A]  time = 0.934362, size = 151, normalized size = 1.36 \[ \frac{120 \sin (e+f x)+1088 \sin (2 (e+f x))+180 \sin (3 (e+f x))+128 \sin (4 (e+f x))+60 \sin (5 (e+f x))-64 \sin (6 (e+f x))+512 \cos (e+f x)-255 \cos (2 (e+f x))+768 \cos (3 (e+f x))-30 \cos (4 (e+f x))+256 \cos (5 (e+f x))+15 \cos (6 (e+f x))-210}{5376 a^2 c^4 f (\sin (e+f x)-1)^4 (\sin (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-210 + 512*Cos[e + f*x] - 255*Cos[2*(e + f*x)] + 768*Cos[3*(e + f*x)] - 30*Cos[4*(e + f*x)] + 256*Cos[5*(e +
f*x)] + 15*Cos[6*(e + f*x)] + 120*Sin[e + f*x] + 1088*Sin[2*(e + f*x)] + 180*Sin[3*(e + f*x)] + 128*Sin[4*(e +
f*x)] + 60*Sin[5*(e + f*x)] - 64*Sin[6*(e + f*x)])/(5376*a^2*c^4*f*(-1 + Sin[e + f*x])^4*(1 + Sin[e + f*x])^2
)

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Maple [A]  time = 0.066, size = 163, normalized size = 1.5 \begin{align*} 2\,{\frac{1}{{a}^{2}f{c}^{4}} \left ( -2/7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-7}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-6}-2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-5}-5/2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}-{\frac{55}{24\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-{\frac{23}{16\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-{\frac{13}{16\,\tan \left ( 1/2\,fx+e/2 \right ) -16}}-1/24\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3}+1/16\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}-3/16\, \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))^2/(c-c*sin(f*x+e))^4,x)

[Out]

2/f/a^2/c^4*(-2/7/(tan(1/2*f*x+1/2*e)-1)^7-1/(tan(1/2*f*x+1/2*e)-1)^6-2/(tan(1/2*f*x+1/2*e)-1)^5-5/2/(tan(1/2*
f*x+1/2*e)-1)^4-55/24/(tan(1/2*f*x+1/2*e)-1)^3-23/16/(tan(1/2*f*x+1/2*e)-1)^2-13/16/(tan(1/2*f*x+1/2*e)-1)-1/2
4/(tan(1/2*f*x+1/2*e)+1)^3+1/16/(tan(1/2*f*x+1/2*e)+1)^2-3/16/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.61514, size = 576, normalized size = 5.19 \begin{align*} -\frac{2 \,{\left (\frac{3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{24 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{76 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac{28 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{42 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}} - \frac{56 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac{28 \, \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} + \frac{42 \, \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - \frac{21 \, \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} - 6\right )}}{21 \,{\left (a^{2} c^{4} - \frac{4 \, a^{2} c^{4} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac{3 \, a^{2} c^{4} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{8 \, a^{2} c^{4} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} - \frac{14 \, a^{2} c^{4} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac{14 \, a^{2} c^{4} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac{8 \, a^{2} c^{4} \sin \left (f x + e\right )^{7}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{7}} - \frac{3 \, a^{2} c^{4} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac{4 \, a^{2} c^{4} \sin \left (f x + e\right )^{9}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{9}} - \frac{a^{2} c^{4} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}\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))^4,x, algorithm="maxima")

[Out]

-2/21*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 24*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 76*sin(f*x + e)^3/(cos(f*x
+ e) + 1)^3 + 28*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 42*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 56*sin(f*x +
e)^6/(cos(f*x + e) + 1)^6 - 28*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 42*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 -
21*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 6)/((a^2*c^4 - 4*a^2*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*c^4*
sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 8*a^2*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 14*a^2*c^4*sin(f*x + e)^
4/(cos(f*x + e) + 1)^4 + 14*a^2*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 8*a^2*c^4*sin(f*x + e)^7/(cos(f*x +
e) + 1)^7 - 3*a^2*c^4*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 4*a^2*c^4*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - a^
2*c^4*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)*f)

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Fricas [A]  time = 1.38188, size = 278, normalized size = 2.5 \begin{align*} -\frac{16 \, \cos \left (f x + e\right )^{4} - 8 \, \cos \left (f x + e\right )^{2} -{\left (8 \, \cos \left (f x + e\right )^{4} - 12 \, \cos \left (f x + e\right )^{2} - 5\right )} \sin \left (f x + e\right ) - 2}{21 \,{\left (a^{2} c^{4} f \cos \left (f x + e\right )^{5} + 2 \, a^{2} c^{4} f \cos \left (f x + e\right )^{3} \sin \left (f x + e\right ) - 2 \, a^{2} c^{4} 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))^4,x, algorithm="fricas")

[Out]

-1/21*(16*cos(f*x + e)^4 - 8*cos(f*x + e)^2 - (8*cos(f*x + e)^4 - 12*cos(f*x + e)^2 - 5)*sin(f*x + e) - 2)/(a^
2*c^4*f*cos(f*x + e)^5 + 2*a^2*c^4*f*cos(f*x + e)^3*sin(f*x + e) - 2*a^2*c^4*f*cos(f*x + e)^3)

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

[Out]

Timed out

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Giac [A]  time = 2.18122, size = 217, normalized size = 1.95 \begin{align*} -\frac{\frac{7 \,{\left (9 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 15 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 8\right )}}{a^{2} c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}} + \frac{273 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 1155 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 2450 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 2870 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2037 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 791 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 152}{a^{2} c^{4}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}}}{168 \, 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))^4,x, algorithm="giac")

[Out]

-1/168*(7*(9*tan(1/2*f*x + 1/2*e)^2 + 15*tan(1/2*f*x + 1/2*e) + 8)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) + 1)^3) + (2
73*tan(1/2*f*x + 1/2*e)^6 - 1155*tan(1/2*f*x + 1/2*e)^5 + 2450*tan(1/2*f*x + 1/2*e)^4 - 2870*tan(1/2*f*x + 1/2
*e)^3 + 2037*tan(1/2*f*x + 1/2*e)^2 - 791*tan(1/2*f*x + 1/2*e) + 152)/(a^2*c^4*(tan(1/2*f*x + 1/2*e) - 1)^7))/
f