3.246 \(\int \frac{(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^6} \, dx\)

Optimal. Leaf size=132 \[ \frac{a^2 c^2 \cos ^5(e+f x)}{11 f (c-c \sin (e+f x))^8}+\frac{2 a^2 \cos ^5(e+f x)}{1155 c f (c-c \sin (e+f x))^5}+\frac{2 a^2 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^6}+\frac{a^2 c \cos ^5(e+f x)}{33 f (c-c \sin (e+f x))^7} \]

[Out]

(a^2*c^2*Cos[e + f*x]^5)/(11*f*(c - c*Sin[e + f*x])^8) + (a^2*c*Cos[e + f*x]^5)/(33*f*(c - c*Sin[e + f*x])^7)
+ (2*a^2*Cos[e + f*x]^5)/(231*f*(c - c*Sin[e + f*x])^6) + (2*a^2*Cos[e + f*x]^5)/(1155*c*f*(c - c*Sin[e + f*x]
)^5)

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Rubi [A]  time = 0.232794, antiderivative size = 132, 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, 2671} \[ \frac{a^2 c^2 \cos ^5(e+f x)}{11 f (c-c \sin (e+f x))^8}+\frac{2 a^2 \cos ^5(e+f x)}{1155 c f (c-c \sin (e+f x))^5}+\frac{2 a^2 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^6}+\frac{a^2 c \cos ^5(e+f x)}{33 f (c-c \sin (e+f x))^7} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^6,x]

[Out]

(a^2*c^2*Cos[e + f*x]^5)/(11*f*(c - c*Sin[e + f*x])^8) + (a^2*c*Cos[e + f*x]^5)/(33*f*(c - c*Sin[e + f*x])^7)
+ (2*a^2*Cos[e + f*x]^5)/(231*f*(c - c*Sin[e + f*x])^6) + (2*a^2*Cos[e + f*x]^5)/(1155*c*f*(c - c*Sin[e + f*x]
)^5)

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 2671

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*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^2}{(c-c \sin (e+f x))^6} \, dx &=\left (a^2 c^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^8} \, dx\\ &=\frac{a^2 c^2 \cos ^5(e+f x)}{11 f (c-c \sin (e+f x))^8}+\frac{1}{11} \left (3 a^2 c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^7} \, dx\\ &=\frac{a^2 c^2 \cos ^5(e+f x)}{11 f (c-c \sin (e+f x))^8}+\frac{a^2 c \cos ^5(e+f x)}{33 f (c-c \sin (e+f x))^7}+\frac{1}{33} \left (2 a^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^6} \, dx\\ &=\frac{a^2 c^2 \cos ^5(e+f x)}{11 f (c-c \sin (e+f x))^8}+\frac{a^2 c \cos ^5(e+f x)}{33 f (c-c \sin (e+f x))^7}+\frac{2 a^2 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^6}+\frac{\left (2 a^2\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^5} \, dx}{231 c}\\ &=\frac{a^2 c^2 \cos ^5(e+f x)}{11 f (c-c \sin (e+f x))^8}+\frac{a^2 c \cos ^5(e+f x)}{33 f (c-c \sin (e+f x))^7}+\frac{2 a^2 \cos ^5(e+f x)}{231 f (c-c \sin (e+f x))^6}+\frac{2 a^2 \cos ^5(e+f x)}{1155 c f (c-c \sin (e+f x))^5}\\ \end{align*}

Mathematica [A]  time = 0.672689, size = 133, normalized size = 1.01 \[ \frac{a^2 \left (2541 \sin \left (\frac{1}{2} (e+f x)\right )+1155 \sin \left (\frac{3}{2} (e+f x)\right )-165 \sin \left (\frac{5}{2} (e+f x)\right )+11 \sin \left (\frac{9}{2} (e+f x)\right )+2079 \cos \left (\frac{1}{2} (e+f x)\right )-825 \cos \left (\frac{3}{2} (e+f x)\right )-55 \cos \left (\frac{7}{2} (e+f x)\right )+\cos \left (\frac{11}{2} (e+f x)\right )\right )}{9240 c^6 f \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^{11}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^2/(c - c*Sin[e + f*x])^6,x]

[Out]

(a^2*(2079*Cos[(e + f*x)/2] - 825*Cos[(3*(e + f*x))/2] - 55*Cos[(7*(e + f*x))/2] + Cos[(11*(e + f*x))/2] + 254
1*Sin[(e + f*x)/2] + 1155*Sin[(3*(e + f*x))/2] - 165*Sin[(5*(e + f*x))/2] + 11*Sin[(9*(e + f*x))/2]))/(9240*c^
6*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^11)

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Maple [A]  time = 0.107, size = 178, normalized size = 1.4 \begin{align*} 2\,{\frac{{a}^{2}}{f{c}^{6}} \left ( -288\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-8}-{\frac{932}{5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-88\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}-{\frac{128}{11\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{11}}}-292\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-6}-30\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-3}-7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-64\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-10}-{\frac{2376}{7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-1}-{\frac{512}{3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{9}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^6,x)

[Out]

2/f*a^2/c^6*(-288/(tan(1/2*f*x+1/2*e)-1)^8-932/5/(tan(1/2*f*x+1/2*e)-1)^5-88/(tan(1/2*f*x+1/2*e)-1)^4-128/11/(
tan(1/2*f*x+1/2*e)-1)^11-292/(tan(1/2*f*x+1/2*e)-1)^6-30/(tan(1/2*f*x+1/2*e)-1)^3-7/(tan(1/2*f*x+1/2*e)-1)^2-6
4/(tan(1/2*f*x+1/2*e)-1)^10-2376/7/(tan(1/2*f*x+1/2*e)-1)^7-1/(tan(1/2*f*x+1/2*e)-1)-512/3/(tan(1/2*f*x+1/2*e)
-1)^9)

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Maxima [B]  time = 1.456, size = 1798, normalized size = 13.62 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^6,x, algorithm="maxima")

[Out]

-2/3465*(5*a^2*(913*sin(f*x + e)/(cos(f*x + e) + 1) - 4565*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 12540*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3 - 25080*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 33726*sin(f*x + e)^5/(cos(f*x + e)
+ 1)^5 - 33726*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 23100*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 11550*sin(f*x
+ e)^8/(cos(f*x + e) + 1)^8 + 3465*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 - 693*sin(f*x + e)^10/(cos(f*x + e) +
1)^10 - 146)/(c^6 - 11*c^6*sin(f*x + e)/(cos(f*x + e) + 1) + 55*c^6*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 165*
c^6*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 330*c^6*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 462*c^6*sin(f*x + e)^5
/(cos(f*x + e) + 1)^5 + 462*c^6*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 330*c^6*sin(f*x + e)^7/(cos(f*x + e) + 1
)^7 + 165*c^6*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 55*c^6*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 11*c^6*sin(f*
x + e)^10/(cos(f*x + e) + 1)^10 - c^6*sin(f*x + e)^11/(cos(f*x + e) + 1)^11) - 6*a^2*(671*sin(f*x + e)/(cos(f*
x + e) + 1) - 2200*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 6600*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 - 10890*sin(
f*x + e)^4/(cos(f*x + e) + 1)^4 + 15246*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 12936*sin(f*x + e)^6/(cos(f*x +
e) + 1)^6 + 9240*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 3465*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 + 1155*sin(f*x
+ e)^9/(cos(f*x + e) + 1)^9 - 61)/(c^6 - 11*c^6*sin(f*x + e)/(cos(f*x + e) + 1) + 55*c^6*sin(f*x + e)^2/(cos(
f*x + e) + 1)^2 - 165*c^6*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 330*c^6*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 -
462*c^6*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 462*c^6*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 330*c^6*sin(f*x +
e)^7/(cos(f*x + e) + 1)^7 + 165*c^6*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 55*c^6*sin(f*x + e)^9/(cos(f*x + e)
+ 1)^9 + 11*c^6*sin(f*x + e)^10/(cos(f*x + e) + 1)^10 - c^6*sin(f*x + e)^11/(cos(f*x + e) + 1)^11) + 4*a^2*(25
3*sin(f*x + e)/(cos(f*x + e) + 1) - 1265*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2640*sin(f*x + e)^3/(cos(f*x +
e) + 1)^3 - 5280*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5313*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 5313*sin(f*x
+ e)^6/(cos(f*x + e) + 1)^6 + 2310*sin(f*x + e)^7/(cos(f*x + e) + 1)^7 - 1155*sin(f*x + e)^8/(cos(f*x + e) +
1)^8 - 23)/(c^6 - 11*c^6*sin(f*x + e)/(cos(f*x + e) + 1) + 55*c^6*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 165*c^
6*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 330*c^6*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 462*c^6*sin(f*x + e)^5/(
cos(f*x + e) + 1)^5 + 462*c^6*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 330*c^6*sin(f*x + e)^7/(cos(f*x + e) + 1)^
7 + 165*c^6*sin(f*x + e)^8/(cos(f*x + e) + 1)^8 - 55*c^6*sin(f*x + e)^9/(cos(f*x + e) + 1)^9 + 11*c^6*sin(f*x
+ e)^10/(cos(f*x + e) + 1)^10 - c^6*sin(f*x + e)^11/(cos(f*x + e) + 1)^11))/f

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Fricas [B]  time = 1.39429, size = 832, normalized size = 6.3 \begin{align*} -\frac{2 \, a^{2} \cos \left (f x + e\right )^{6} + 12 \, a^{2} \cos \left (f x + e\right )^{5} - 25 \, a^{2} \cos \left (f x + e\right )^{4} - 70 \, a^{2} \cos \left (f x + e\right )^{3} - 245 \, a^{2} \cos \left (f x + e\right )^{2} + 210 \, a^{2} \cos \left (f x + e\right ) + 420 \, a^{2} -{\left (2 \, a^{2} \cos \left (f x + e\right )^{5} - 10 \, a^{2} \cos \left (f x + e\right )^{4} - 35 \, a^{2} \cos \left (f x + e\right )^{3} + 35 \, a^{2} \cos \left (f x + e\right )^{2} - 210 \, a^{2} \cos \left (f x + e\right ) - 420 \, a^{2}\right )} \sin \left (f x + e\right )}{1155 \,{\left (c^{6} f \cos \left (f x + e\right )^{6} - 5 \, c^{6} f \cos \left (f x + e\right )^{5} - 18 \, c^{6} f \cos \left (f x + e\right )^{4} + 20 \, c^{6} f \cos \left (f x + e\right )^{3} + 48 \, c^{6} f \cos \left (f x + e\right )^{2} - 16 \, c^{6} f \cos \left (f x + e\right ) - 32 \, c^{6} f +{\left (c^{6} f \cos \left (f x + e\right )^{5} + 6 \, c^{6} f \cos \left (f x + e\right )^{4} - 12 \, c^{6} f \cos \left (f x + e\right )^{3} - 32 \, c^{6} f \cos \left (f x + e\right )^{2} + 16 \, c^{6} f \cos \left (f x + e\right ) + 32 \, c^{6} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^6,x, algorithm="fricas")

[Out]

-1/1155*(2*a^2*cos(f*x + e)^6 + 12*a^2*cos(f*x + e)^5 - 25*a^2*cos(f*x + e)^4 - 70*a^2*cos(f*x + e)^3 - 245*a^
2*cos(f*x + e)^2 + 210*a^2*cos(f*x + e) + 420*a^2 - (2*a^2*cos(f*x + e)^5 - 10*a^2*cos(f*x + e)^4 - 35*a^2*cos
(f*x + e)^3 + 35*a^2*cos(f*x + e)^2 - 210*a^2*cos(f*x + e) - 420*a^2)*sin(f*x + e))/(c^6*f*cos(f*x + e)^6 - 5*
c^6*f*cos(f*x + e)^5 - 18*c^6*f*cos(f*x + e)^4 + 20*c^6*f*cos(f*x + e)^3 + 48*c^6*f*cos(f*x + e)^2 - 16*c^6*f*
cos(f*x + e) - 32*c^6*f + (c^6*f*cos(f*x + e)^5 + 6*c^6*f*cos(f*x + e)^4 - 12*c^6*f*cos(f*x + e)^3 - 32*c^6*f*
cos(f*x + e)^2 + 16*c^6*f*cos(f*x + e) + 32*c^6*f)*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((a+a*sin(f*x+e))**2/(c-c*sin(f*x+e))**6,x)

[Out]

Timed out

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Giac [A]  time = 2.1786, size = 265, normalized size = 2.01 \begin{align*} -\frac{2 \,{\left (1155 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{10} - 3465 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{9} + 13860 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{8} - 23100 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{7} + 37422 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 32802 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 27060 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 11220 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 4895 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 517 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 152 \, a^{2}\right )}}{1155 \, c^{6} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^2/(c-c*sin(f*x+e))^6,x, algorithm="giac")

[Out]

-2/1155*(1155*a^2*tan(1/2*f*x + 1/2*e)^10 - 3465*a^2*tan(1/2*f*x + 1/2*e)^9 + 13860*a^2*tan(1/2*f*x + 1/2*e)^8
- 23100*a^2*tan(1/2*f*x + 1/2*e)^7 + 37422*a^2*tan(1/2*f*x + 1/2*e)^6 - 32802*a^2*tan(1/2*f*x + 1/2*e)^5 + 27
060*a^2*tan(1/2*f*x + 1/2*e)^4 - 11220*a^2*tan(1/2*f*x + 1/2*e)^3 + 4895*a^2*tan(1/2*f*x + 1/2*e)^2 - 517*a^2*
tan(1/2*f*x + 1/2*e) + 152*a^2)/(c^6*f*(tan(1/2*f*x + 1/2*e) - 1)^11)