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]
1/7*a^2*c^2*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^6+1/35*a^2*c*cos(f*x+e)^5/f/(c-c*sin(f*x+e))^5
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Rubi [A] time = 0.14, antiderivative size = 67, normalized size of antiderivative = 1.00,
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 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]
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 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]))
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*}
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Mathematica [A] time = 0.63, 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]
-1/140*(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]))/(c^4*f*(-1 + Sin[e + f*x
])^4)
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fricas [B] time = 0.47, size = 222, normalized size = 3.31 \[ -\frac {a^{2} \cos \left (f x + e\right )^{4} + 4 \, a^{2} \cos \left (f x + e\right )^{3} + 13 \, a^{2} \cos \left (f x + e\right )^{2} - 10 \, a^{2} \cos \left (f x + e\right ) - 20 \, a^{2} - {\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 10 \, a^{2} \cos \left (f x + e\right ) + 20 \, a^{2}\right )} \sin \left (f x + e\right )}{35 \, {\left (c^{4} f \cos \left (f x + e\right )^{4} - 3 \, c^{4} f \cos \left (f x + e\right )^{3} - 8 \, c^{4} f \cos \left (f x + e\right )^{2} + 4 \, c^{4} f \cos \left (f x + e\right ) + 8 \, c^{4} f + {\left (c^{4} f \cos \left (f x + e\right )^{3} + 4 \, c^{4} f \cos \left (f x + e\right )^{2} - 4 \, c^{4} f \cos \left (f x + e\right ) - 8 \, c^{4} f\right )} \sin \left (f x + e\right )\right )}} \]
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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giac [A] time = 0.22, size = 128, normalized size = 1.91 \[ -\frac {2 \, {\left (35 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{6} - 35 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 140 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 70 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 91 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 7 \, a^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, a^{2}\right )}}{35 \, c^{4} f {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}^{7}} \]
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)
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maple [A] time = 0.27, size = 118, normalized size = 1.76 \[ \frac {2 a^{2} \left (-\frac {1}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1}-\frac {5}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {14}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{3}}-\frac {32}{7 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{7}}-\frac {16}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{6}}-\frac {128}{5 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{5}}-\frac {24}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{4}}\right )}{f \,c^{4}} \]
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*(-1/(tan(1/2*f*x+1/2*e)-1)-5/(tan(1/2*f*x+1/2*e)-1)^2-14/(tan(1/2*f*x+1/2*e)-1)^3-32/7/(tan(1/2*f*
x+1/2*e)-1)^7-16/(tan(1/2*f*x+1/2*e)-1)^6-128/5/(tan(1/2*f*x+1/2*e)-1)^5-24/(tan(1/2*f*x+1/2*e)-1)^4)
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maxima [B] time = 0.57, size = 816, normalized size = 12.18 \[ \text {result too large to display} \]
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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mupad [B] time = 7.30, size = 99, normalized size = 1.48 \[ \frac {\sqrt {2}\,a^2\,\cos \left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {5\,\cos \left (3\,e+3\,f\,x\right )}{8}-\frac {105\,\sin \left (e+f\,x\right )}{8}-\frac {27\,\cos \left (2\,e+2\,f\,x\right )}{4}-\frac {121\,\cos \left (e+f\,x\right )}{8}+\frac {7\,\sin \left (2\,e+2\,f\,x\right )}{2}+\frac {7\,\sin \left (3\,e+3\,f\,x\right )}{8}+\frac {109}{4}\right )}{280\,c^4\,f\,{\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f\,x}{2}\right )}^7} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + a*sin(e + f*x))^2/(c - c*sin(e + f*x))^4,x)
[Out]
(2^(1/2)*a^2*cos(e/2 + (f*x)/2)*((5*cos(3*e + 3*f*x))/8 - (105*sin(e + f*x))/8 - (27*cos(2*e + 2*f*x))/4 - (12
1*cos(e + f*x))/8 + (7*sin(2*e + 2*f*x))/2 + (7*sin(3*e + 3*f*x))/8 + 109/4))/(280*c^4*f*cos(e/2 + pi/4 + (f*x
)/2)^7)
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sympy [A] time = 25.12, size = 1074, normalized size = 16.03 \[ \text {result too large to display} \]
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]
Piecewise((-70*a**2*tan(e/2 + f*x/2)**6/(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*
c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*ta
n(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) + 70*a**2*tan(e/2 + f*x/2)**5/(35*c**4*f*tan(e/2
+ f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**
4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f
) - 280*a**2*tan(e/2 + f*x/2)**4/(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*
tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 +
f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) + 140*a**2*tan(e/2 + f*x/2)**3/(35*c**4*f*tan(e/2 + f*x/
2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 12
25*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) - 18
2*a**2*tan(e/2 + f*x/2)**2/(35*c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/
2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2
)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) + 14*a**2*tan(e/2 + f*x/2)/(35*c**4*f*tan(e/2 + f*x/2)**7 - 24
5*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan(e/2 + f*x/2)**4 + 1225*c**4*f*
tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x/2) - 35*c**4*f) - 12*a**2/(35*
c**4*f*tan(e/2 + f*x/2)**7 - 245*c**4*f*tan(e/2 + f*x/2)**6 + 735*c**4*f*tan(e/2 + f*x/2)**5 - 1225*c**4*f*tan
(e/2 + f*x/2)**4 + 1225*c**4*f*tan(e/2 + f*x/2)**3 - 735*c**4*f*tan(e/2 + f*x/2)**2 + 245*c**4*f*tan(e/2 + f*x
/2) - 35*c**4*f), Ne(f, 0)), (x*(a*sin(e) + a)**2/(-c*sin(e) + c)**4, True))
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