3.234 \(\int \frac{a+a \sin (e+f x)}{(c-c \sin (e+f x))^4} \, dx\)
Optimal. Leaf size=92 \[ \frac{2 a \cos ^3(e+f x)}{105 c f (c-c \sin (e+f x))^3}+\frac{2 a \cos ^3(e+f x)}{35 f (c-c \sin (e+f x))^4}+\frac{a c \cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5} \]
[Out]
(a*c*Cos[e + f*x]^3)/(7*f*(c - c*Sin[e + f*x])^5) + (2*a*Cos[e + f*x]^3)/(35*f*(c - c*Sin[e + f*x])^4) + (2*a*
Cos[e + f*x]^3)/(105*c*f*(c - c*Sin[e + f*x])^3)
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Rubi [A] time = 0.17018, antiderivative size = 92, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used =
{2736, 2672, 2671} \[ \frac{2 a \cos ^3(e+f x)}{105 c f (c-c \sin (e+f x))^3}+\frac{2 a \cos ^3(e+f x)}{35 f (c-c \sin (e+f x))^4}+\frac{a c \cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^4,x]
[Out]
(a*c*Cos[e + f*x]^3)/(7*f*(c - c*Sin[e + f*x])^5) + (2*a*Cos[e + f*x]^3)/(35*f*(c - c*Sin[e + f*x])^4) + (2*a*
Cos[e + f*x]^3)/(105*c*f*(c - c*Sin[e + f*x])^3)
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)}{(c-c \sin (e+f x))^4} \, dx &=(a c) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^5} \, dx\\ &=\frac{a c \cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}+\frac{1}{7} (2 a) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^4} \, dx\\ &=\frac{a c \cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}+\frac{2 a \cos ^3(e+f x)}{35 f (c-c \sin (e+f x))^4}+\frac{(2 a) \int \frac{\cos ^2(e+f x)}{(c-c \sin (e+f x))^3} \, dx}{35 c}\\ &=\frac{a c \cos ^3(e+f x)}{7 f (c-c \sin (e+f x))^5}+\frac{2 a \cos ^3(e+f x)}{35 f (c-c \sin (e+f x))^4}+\frac{2 a \cos ^3(e+f x)}{105 c f (c-c \sin (e+f x))^3}\\ \end{align*}
Mathematica [A] time = 0.466238, size = 109, normalized size = 1.18 \[ \frac{a \left (7 \sin \left (2 e+\frac{5 f x}{2}\right )+70 \cos \left (e+\frac{f x}{2}\right )-21 \cos \left (e+\frac{3 f x}{2}\right )+\cos \left (3 e+\frac{7 f x}{2}\right )+35 \sin \left (\frac{f x}{2}\right )\right )}{210 c^4 f \left (\cos \left (\frac{e}{2}\right )-\sin \left (\frac{e}{2}\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^7} \]
Antiderivative was successfully verified.
[In]
Integrate[(a + a*Sin[e + f*x])/(c - c*Sin[e + f*x])^4,x]
[Out]
(a*(70*Cos[e + (f*x)/2] - 21*Cos[e + (3*f*x)/2] + Cos[3*e + (7*f*x)/2] + 35*Sin[(f*x)/2] + 7*Sin[2*e + (5*f*x)
/2]))/(210*c^4*f*(Cos[e/2] - Sin[e/2])*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^7)
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Maple [A] time = 0.086, size = 116, normalized size = 1.3 \begin{align*} 2\,{\frac{a}{f{c}^{4}} \left ( -4\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-2}-8\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-6}-14\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-4}-{\frac{28}{3\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-{\frac{68}{5\, \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{5}}}- \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{-1}-{\frac{16}{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))/(c-c*sin(f*x+e))^4,x)
[Out]
2/f*a/c^4*(-4/(tan(1/2*f*x+1/2*e)-1)^2-8/(tan(1/2*f*x+1/2*e)-1)^6-14/(tan(1/2*f*x+1/2*e)-1)^4-28/3/(tan(1/2*f*
x+1/2*e)-1)^3-68/5/(tan(1/2*f*x+1/2*e)-1)^5-1/(tan(1/2*f*x+1/2*e)-1)-16/7/(tan(1/2*f*x+1/2*e)-1)^7)
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Maxima [B] time = 1.25502, size = 757, normalized size = 8.23 \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))/(c-c*sin(f*x+e))^4,x, algorithm="maxima")
[Out]
2/105*(a*(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/(c
os(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 - 13)/(
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*(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 - 2
10*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/(cos(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))
/f
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Fricas [B] time = 1.40681, size = 517, normalized size = 5.62 \begin{align*} \frac{2 \, a \cos \left (f x + e\right )^{4} + 8 \, a \cos \left (f x + e\right )^{3} - 9 \, a \cos \left (f x + e\right )^{2} + 15 \, a \cos \left (f x + e\right ) -{\left (2 \, a \cos \left (f x + e\right )^{3} - 6 \, a \cos \left (f x + e\right )^{2} - 15 \, a \cos \left (f x + e\right ) - 30 \, a\right )} \sin \left (f x + e\right ) + 30 \, a}{105 \,{\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))/(c-c*sin(f*x+e))^4,x, algorithm="fricas")
[Out]
1/105*(2*a*cos(f*x + e)^4 + 8*a*cos(f*x + e)^3 - 9*a*cos(f*x + e)^2 + 15*a*cos(f*x + e) - (2*a*cos(f*x + e)^3
- 6*a*cos(f*x + e)^2 - 15*a*cos(f*x + e) - 30*a)*sin(f*x + e) + 30*a)/(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 [A] time = 48.9284, size = 1061, normalized size = 11.53 \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))/(c-c*sin(f*x+e))**4,x)
[Out]
Piecewise((-30*a*tan(e/2 + f*x/2)**7/(105*c**4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c
**4*f*tan(e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2)**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*ta
n(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) - 210*a*tan(e/2 + f*x/2)**5/(105*c**4*f*tan(e/2
+ f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan(e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2)*
*4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**
4*f) + 140*a*tan(e/2 + f*x/2)**4/(105*c**4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*
f*tan(e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2)**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/
2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) - 350*a*tan(e/2 + f*x/2)**3/(105*c**4*f*tan(e/2 + f*
x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan(e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2)**4 +
3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f)
+ 84*a*tan(e/2 + f*x/2)**2/(105*c**4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan
(e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2)**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/2 + f
*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) - 98*a*tan(e/2 + f*x/2)/(105*c**4*f*tan(e/2 + f*x/2)**7 -
735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan(e/2 + f*x/2)**5 - 3675*c**4*f*tan(e/2 + f*x/2)**4 + 3675*c**
4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2 + f*x/2) - 105*c**4*f) - 16*a/(
105*c**4*f*tan(e/2 + f*x/2)**7 - 735*c**4*f*tan(e/2 + f*x/2)**6 + 2205*c**4*f*tan(e/2 + f*x/2)**5 - 3675*c**4*
f*tan(e/2 + f*x/2)**4 + 3675*c**4*f*tan(e/2 + f*x/2)**3 - 2205*c**4*f*tan(e/2 + f*x/2)**2 + 735*c**4*f*tan(e/2
+ f*x/2) - 105*c**4*f), Ne(f, 0)), (x*(a*sin(e) + a)/(-c*sin(e) + c)**4, True))
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Giac [A] time = 1.7361, size = 154, normalized size = 1.67 \begin{align*} -\frac{2 \,{\left (105 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{6} - 210 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 455 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 350 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 273 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 56 \, a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 23 \, a\right )}}{105 \, 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))/(c-c*sin(f*x+e))^4,x, algorithm="giac")
[Out]
-2/105*(105*a*tan(1/2*f*x + 1/2*e)^6 - 210*a*tan(1/2*f*x + 1/2*e)^5 + 455*a*tan(1/2*f*x + 1/2*e)^4 - 350*a*tan
(1/2*f*x + 1/2*e)^3 + 273*a*tan(1/2*f*x + 1/2*e)^2 - 56*a*tan(1/2*f*x + 1/2*e) + 23*a)/(c^4*f*(tan(1/2*f*x + 1
/2*e) - 1)^7)