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

Optimal. Leaf size=67 \[ \frac{a^2 c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac{a^2 c \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^5} \]

[Out]

(a^2*c^2*Cos[e + f*x]^5)/(7*f*(c - c*Sin[e + f*x])^6) + (a^2*c*Cos[e + f*x]^5)/(35*f*(c - c*Sin[e + f*x])^5)

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Rubi [A]  time = 0.135929, antiderivative size = 67, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2672, 2671} \[ \frac{a^2 c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac{a^2 c \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*c^2*Cos[e + f*x]^5)/(7*f*(c - c*Sin[e + f*x])^6) + (a^2*c*Cos[e + f*x]^5)/(35*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))^4} \, dx &=\left (a^2 c^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)}{7 f (c-c \sin (e+f x))^6}+\frac{1}{7} \left (a^2 c\right ) \int \frac{\cos ^4(e+f x)}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac{a^2 c^2 \cos ^5(e+f x)}{7 f (c-c \sin (e+f x))^6}+\frac{a^2 c \cos ^5(e+f x)}{35 f (c-c \sin (e+f x))^5}\\ \end{align*}

Mathematica [A]  time = 0.600629, size = 117, normalized size = 1.75 \[ -\frac{a^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (-70 \sin \left (\frac{1}{2} (e+f x)\right )-35 \sin \left (\frac{3}{2} (e+f x)\right )+7 \sin \left (\frac{5}{2} (e+f x)\right )-35 \cos \left (\frac{1}{2} (e+f x)\right )+14 \cos \left (\frac{3}{2} (e+f x)\right )+\cos \left (\frac{7}{2} (e+f x)\right )\right )}{140 c^4 f (\sin (e+f x)-1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(-35*Cos[(e + f*x)/2] + 14*Cos[(3*(e + f*x))/2] + Cos[(7*(e + f*x)
)/2] - 70*Sin[(e + f*x)/2] - 35*Sin[(3*(e + f*x))/2] + 7*Sin[(5*(e + f*x))/2]))/(140*c^4*f*(-1 + Sin[e + f*x])
^4)

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Maple [A]  time = 0.092, size = 118, normalized size = 1.8 \begin{align*} 2\,{\frac{{a}^{2}}{f{c}^{4}} \left ( -{\frac{128}{5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}-16\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-6}-5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-14\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-3}-24\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-1}-{\frac{32}{7\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{7}}} \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))^4,x)

[Out]

2/f*a^2/c^4*(-128/5/(tan(1/2*f*x+1/2*e)-1)^5-16/(tan(1/2*f*x+1/2*e)-1)^6-5/(tan(1/2*f*x+1/2*e)-1)^2-14/(tan(1/
2*f*x+1/2*e)-1)^3-24/(tan(1/2*f*x+1/2*e)-1)^4-1/(tan(1/2*f*x+1/2*e)-1)-32/7/(tan(1/2*f*x+1/2*e)-1)^7)

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Maxima [B]  time = 1.27222, size = 1102, normalized size = 16.45 \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))^4,x, algorithm="maxima")

[Out]

2/105*(2*a^2*(91*sin(f*x + e)/(cos(f*x + e) + 1) - 168*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 280*sin(f*x + e)^
3/(cos(f*x + e) + 1)^3 - 175*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 1
3)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x
+ e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e)
+ 1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7) - 3*a^2*(49*sin
(f*x + e)/(cos(f*x + e) + 1) - 147*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 210*sin(f*x + e)^3/(cos(f*x + e) + 1)
^3 - 210*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 105*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 - 35*sin(f*x + e)^6/(co
s(f*x + e) + 1)^6 - 12)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) + 1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1
)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x
+ e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) +
1)^7) - 4*a^2*(14*sin(f*x + e)/(cos(f*x + e) + 1) - 42*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^3
/(cos(f*x + e) + 1)^3 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2)/(c^4 - 7*c^4*sin(f*x + e)/(cos(f*x + e) +
1) + 21*c^4*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 35*c^4*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 35*c^4*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - 21*c^4*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 7*c^4*sin(f*x + e)^6/(cos(f*x + e)
+ 1)^6 - c^4*sin(f*x + e)^7/(cos(f*x + e) + 1)^7))/f

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Fricas [B]  time = 1.33499, size = 537, normalized size = 8.01 \begin{align*} -\frac{a^{2} \cos \left (f x + e\right )^{4} + 4 \, a^{2} \cos \left (f x + e\right )^{3} + 13 \, a^{2} \cos \left (f x + e\right )^{2} - 10 \, a^{2} \cos \left (f x + e\right ) - 20 \, a^{2} -{\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 10 \, a^{2} \cos \left (f x + e\right ) + 20 \, a^{2}\right )} \sin \left (f x + e\right )}{35 \,{\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f +{\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} 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))^4,x, algorithm="fricas")

[Out]

-1/35*(a^2*cos(f*x + e)^4 + 4*a^2*cos(f*x + e)^3 + 13*a^2*cos(f*x + e)^2 - 10*a^2*cos(f*x + e) - 20*a^2 - (a^2
*cos(f*x + e)^3 - 3*a^2*cos(f*x + e)^2 + 10*a^2*cos(f*x + e) + 20*a^2)*sin(f*x + e))/(c^4*f*cos(f*x + e)^4 - 3
*c^4*f*cos(f*x + e)^3 - 8*c^4*f*cos(f*x + e)^2 + 4*c^4*f*cos(f*x + e) + 8*c^4*f + (c^4*f*cos(f*x + e)^3 + 4*c^
4*f*cos(f*x + e)^2 - 4*c^4*f*cos(f*x + e) - 8*c^4*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))**4,x)

[Out]

Timed out

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Giac [A]  time = 2.12167, size = 173, normalized size = 2.58 \begin{align*} -\frac{2 \,{\left (35 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 35 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 140 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 70 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 91 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 7 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 6 \, a^{2}\right )}}{35 \, c^{4} f{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1\right )}^{7}} \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))^4,x, algorithm="giac")

[Out]

-2/35*(35*a^2*tan(1/2*f*x + 1/2*e)^6 - 35*a^2*tan(1/2*f*x + 1/2*e)^5 + 140*a^2*tan(1/2*f*x + 1/2*e)^4 - 70*a^2
*tan(1/2*f*x + 1/2*e)^3 + 91*a^2*tan(1/2*f*x + 1/2*e)^2 - 7*a^2*tan(1/2*f*x + 1/2*e) + 6*a^2)/(c^4*f*(tan(1/2*
f*x + 1/2*e) - 1)^7)