3.278 \(\int \frac{1}{(a+a \sin (e+f x))^2 (c-c \sin (e+f x))^5} \, dx\)

Optimal. Leaf size=144 \[ \frac{8 \tan ^3(e+f x)}{63 a^2 c^5 f}+\frac{8 \tan (e+f x)}{21 a^2 c^5 f}+\frac{2 \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{2 \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{\sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3} \]

[Out]

Sec[e + f*x]^3/(9*a^2*c^2*f*(c - c*Sin[e + f*x])^3) + (2*Sec[e + f*x]^3)/(21*a^2*c^3*f*(c - c*Sin[e + f*x])^2)
+ (2*Sec[e + f*x]^3)/(21*a^2*f*(c^5 - c^5*Sin[e + f*x])) + (8*Tan[e + f*x])/(21*a^2*c^5*f) + (8*Tan[e + f*x]^
3)/(63*a^2*c^5*f)

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Rubi [A]  time = 0.215191, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2672, 3767} \[ \frac{8 \tan ^3(e+f x)}{63 a^2 c^5 f}+\frac{8 \tan (e+f x)}{21 a^2 c^5 f}+\frac{2 \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{2 \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{\sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^5),x]

[Out]

Sec[e + f*x]^3/(9*a^2*c^2*f*(c - c*Sin[e + f*x])^3) + (2*Sec[e + f*x]^3)/(21*a^2*c^3*f*(c - c*Sin[e + f*x])^2)
+ (2*Sec[e + f*x]^3)/(21*a^2*f*(c^5 - c^5*Sin[e + f*x])) + (8*Tan[e + f*x])/(21*a^2*c^5*f) + (8*Tan[e + f*x]^
3)/(63*a^2*c^5*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))^5} \, dx &=\frac{\int \frac{\sec ^4(e+f x)}{(c-c \sin (e+f x))^3} \, dx}{a^2 c^2}\\ &=\frac{\sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{2 \int \frac{\sec ^4(e+f x)}{(c-c \sin (e+f x))^2} \, dx}{3 a^2 c^3}\\ &=\frac{\sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{2 \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{10 \int \frac{\sec ^4(e+f x)}{c-c \sin (e+f x)} \, dx}{21 a^2 c^4}\\ &=\frac{\sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{2 \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{2 \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{8 \int \sec ^4(e+f x) \, dx}{21 a^2 c^5}\\ &=\frac{\sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{2 \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{2 \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}-\frac{8 \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (e+f x)\right )}{21 a^2 c^5 f}\\ &=\frac{\sec ^3(e+f x)}{9 a^2 c^2 f (c-c \sin (e+f x))^3}+\frac{2 \sec ^3(e+f x)}{21 a^2 c^3 f (c-c \sin (e+f x))^2}+\frac{2 \sec ^3(e+f x)}{21 a^2 f \left (c^5-c^5 \sin (e+f x)\right )}+\frac{8 \tan (e+f x)}{21 a^2 c^5 f}+\frac{8 \tan ^3(e+f x)}{63 a^2 c^5 f}\\ \end{align*}

Mathematica [A]  time = 1.13137, size = 193, normalized size = 1.34 \[ \frac{\left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) (18432 \sin (e+f x)+4185 \sin (2 (e+f x))+1024 \sin (3 (e+f x))+1860 \sin (4 (e+f x))-3072 \sin (5 (e+f x))-155 \sin (6 (e+f x))-5580 \cos (e+f x)+13824 \cos (2 (e+f x))-310 \cos (3 (e+f x))+6144 \cos (4 (e+f x))+930 \cos (5 (e+f x))-512 \cos (6 (e+f x)))}{64512 f (a \sin (e+f x)+a)^2 (c-c \sin (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^5),x]

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(-5580*Cos[e + f*x] + 13824*Cos[2
*(e + f*x)] - 310*Cos[3*(e + f*x)] + 6144*Cos[4*(e + f*x)] + 930*Cos[5*(e + f*x)] - 512*Cos[6*(e + f*x)] + 184
32*Sin[e + f*x] + 4185*Sin[2*(e + f*x)] + 1024*Sin[3*(e + f*x)] + 1860*Sin[4*(e + f*x)] - 3072*Sin[5*(e + f*x)
] - 155*Sin[6*(e + f*x)]))/(64512*f*(a + a*Sin[e + f*x])^2*(c - c*Sin[e + f*x])^5)

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Maple [A]  time = 0.067, size = 193, normalized size = 1.3 \begin{align*} 2\,{\frac{1}{{a}^{2}f{c}^{5}} \left ( -4/9\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-9}-2\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-8}-{\frac{34}{7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}-{\frac{23}{3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{6}}}-{\frac{35}{4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-{\frac{59}{8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-{\frac{19}{4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-9/4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-{\frac{57}{64\,\tan \left ( 1/2\,fx+e/2 \right ) -64}}-1/48\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-3}+1/32\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{-2}-{\frac{7}{64\,\tan \left ( 1/2\,fx+e/2 \right ) +64}} \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))^5,x)

[Out]

2/f/a^2/c^5*(-4/9/(tan(1/2*f*x+1/2*e)-1)^9-2/(tan(1/2*f*x+1/2*e)-1)^8-34/7/(tan(1/2*f*x+1/2*e)-1)^7-23/3/(tan(
1/2*f*x+1/2*e)-1)^6-35/4/(tan(1/2*f*x+1/2*e)-1)^5-59/8/(tan(1/2*f*x+1/2*e)-1)^4-19/4/(tan(1/2*f*x+1/2*e)-1)^3-
9/4/(tan(1/2*f*x+1/2*e)-1)^2-57/64/(tan(1/2*f*x+1/2*e)-1)-1/48/(tan(1/2*f*x+1/2*e)+1)^3+1/32/(tan(1/2*f*x+1/2*
e)+1)^2-7/64/(tan(1/2*f*x+1/2*e)+1))

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Maxima [B]  time = 1.73964, size = 701, normalized size = 4.87 \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))^2/(c-c*sin(f*x+e))^5,x, algorithm="maxima")

[Out]

-2/63*(51*sin(f*x + e)/(cos(f*x + e) + 1) - 39*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 235*sin(f*x + e)^3/(cos(f
*x + e) + 1)^3 + 450*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 306*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 294*sin(f
*x + e)^6/(cos(f*x + e) + 1)^6 + 378*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 63*sin(f*x + e)^8/(cos(f*x + e) + 1
)^8 - 273*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 189*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - 63*sin(f*x + e)^11
/(cos(f*x + e) + 1)^11 - 19)/((a^2*c^5 - 6*a^2*c^5*sin(f*x + e)/(cos(f*x + e) + 1) + 12*a^2*c^5*sin(f*x + e)^2
/(cos(f*x + e) + 1)^2 - 2*a^2*c^5*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 27*a^2*c^5*sin(f*x + e)^4/(cos(f*x + e
) + 1)^4 + 36*a^2*c^5*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 36*a^2*c^5*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 + 2
7*a^2*c^5*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 2*a^2*c^5*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 12*a^2*c^5*sin
(f*x + e)^10/(cos(f*x + e) + 1)^10 + 6*a^2*c^5*sin(f*x + e)^11/(cos(f*x + e) + 1)^11 - a^2*c^5*sin(f*x + e)^12
/(cos(f*x + e) + 1)^12)*f)

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Fricas [A]  time = 1.38318, size = 351, normalized size = 2.44 \begin{align*} \frac{16 \, \cos \left (f x + e\right )^{6} - 72 \, \cos \left (f x + e\right )^{4} + 30 \, \cos \left (f x + e\right )^{2} + 2 \,{\left (24 \, \cos \left (f x + e\right )^{4} - 20 \, \cos \left (f x + e\right )^{2} - 7\right )} \sin \left (f x + e\right ) + 7}{63 \,{\left (3 \, a^{2} c^{5} f \cos \left (f x + e\right )^{5} - 4 \, a^{2} c^{5} f \cos \left (f x + e\right )^{3} -{\left (a^{2} c^{5} f \cos \left (f x + e\right )^{5} - 4 \, a^{2} c^{5} f \cos \left (f x + e\right )^{3}\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))^2/(c-c*sin(f*x+e))^5,x, algorithm="fricas")

[Out]

1/63*(16*cos(f*x + e)^6 - 72*cos(f*x + e)^4 + 30*cos(f*x + e)^2 + 2*(24*cos(f*x + e)^4 - 20*cos(f*x + e)^2 - 7
)*sin(f*x + e) + 7)/(3*a^2*c^5*f*cos(f*x + e)^5 - 4*a^2*c^5*f*cos(f*x + e)^3 - (a^2*c^5*f*cos(f*x + e)^5 - 4*a
^2*c^5*f*cos(f*x + e)^3)*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))**2/(c-c*sin(f*x+e))**5,x)

[Out]

Timed out

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Giac [A]  time = 2.04002, size = 255, normalized size = 1.77 \begin{align*} -\frac{\frac{21 \,{\left (21 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 36 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 19\right )}}{a^{2} c^{5}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}} + \frac{3591 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 19656 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 56196 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 95760 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 107730 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 79464 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 38484 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 10944 \, \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1615}{a^{2} c^{5}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{9}}}{2016 \, 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))^5,x, algorithm="giac")

[Out]

-1/2016*(21*(21*tan(1/2*f*x + 1/2*e)^2 + 36*tan(1/2*f*x + 1/2*e) + 19)/(a^2*c^5*(tan(1/2*f*x + 1/2*e) + 1)^3)
+ (3591*tan(1/2*f*x + 1/2*e)^8 - 19656*tan(1/2*f*x + 1/2*e)^7 + 56196*tan(1/2*f*x + 1/2*e)^6 - 95760*tan(1/2*f
*x + 1/2*e)^5 + 107730*tan(1/2*f*x + 1/2*e)^4 - 79464*tan(1/2*f*x + 1/2*e)^3 + 38484*tan(1/2*f*x + 1/2*e)^2 -
10944*tan(1/2*f*x + 1/2*e) + 1615)/(a^2*c^5*(tan(1/2*f*x + 1/2*e) - 1)^9))/f