3.280 \(\int \frac{(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=124 \[ -\frac{7 c^4 \cos (e+f x)}{a^3 f}-\frac{2 a^3 c^4 \cos ^7(e+f x)}{5 f (a \sin (e+f x)+a)^6}-\frac{7 c^4 x}{a^3}+\frac{14 a c^4 \cos ^5(e+f x)}{15 f (a \sin (e+f x)+a)^4}-\frac{14 c^4 \cos ^3(e+f x)}{3 a f (a \sin (e+f x)+a)^2} \]

[Out]

(-7*c^4*x)/a^3 - (7*c^4*Cos[e + f*x])/(a^3*f) - (2*a^3*c^4*Cos[e + f*x]^7)/(5*f*(a + a*Sin[e + f*x])^6) + (14*
a*c^4*Cos[e + f*x]^5)/(15*f*(a + a*Sin[e + f*x])^4) - (14*c^4*Cos[e + f*x]^3)/(3*a*f*(a + a*Sin[e + f*x])^2)

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Rubi [A]  time = 0.221114, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2736, 2680, 2682, 8} \[ -\frac{7 c^4 \cos (e+f x)}{a^3 f}-\frac{2 a^3 c^4 \cos ^7(e+f x)}{5 f (a \sin (e+f x)+a)^6}-\frac{7 c^4 x}{a^3}+\frac{14 a c^4 \cos ^5(e+f x)}{15 f (a \sin (e+f x)+a)^4}-\frac{14 c^4 \cos ^3(e+f x)}{3 a f (a \sin (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sin[e + f*x])^4/(a + a*Sin[e + f*x])^3,x]

[Out]

(-7*c^4*x)/a^3 - (7*c^4*Cos[e + f*x])/(a^3*f) - (2*a^3*c^4*Cos[e + f*x]^7)/(5*f*(a + a*Sin[e + f*x])^6) + (14*
a*c^4*Cos[e + f*x]^5)/(15*f*(a + a*Sin[e + f*x])^4) - (14*c^4*Cos[e + f*x]^3)/(3*a*f*(a + a*Sin[e + f*x])^2)

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 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2682

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e
+ f*x])^(p - 1))/(b*f*(p - 1)), x] + Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2), x], x] /; FreeQ[{a, b, e, f, g
}, x] && EqQ[a^2 - b^2, 0] && GtQ[p, 1] && IntegerQ[2*p]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int \frac{(c-c \sin (e+f x))^4}{(a+a \sin (e+f x))^3} \, dx &=\left (a^4 c^4\right ) \int \frac{\cos ^8(e+f x)}{(a+a \sin (e+f x))^7} \, dx\\ &=-\frac{2 a^3 c^4 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^6}-\frac{1}{5} \left (7 a^2 c^4\right ) \int \frac{\cos ^6(e+f x)}{(a+a \sin (e+f x))^5} \, dx\\ &=-\frac{2 a^3 c^4 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^6}+\frac{14 a c^4 \cos ^5(e+f x)}{15 f (a+a \sin (e+f x))^4}+\frac{1}{3} \left (7 c^4\right ) \int \frac{\cos ^4(e+f x)}{(a+a \sin (e+f x))^3} \, dx\\ &=-\frac{2 a^3 c^4 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^6}+\frac{14 a c^4 \cos ^5(e+f x)}{15 f (a+a \sin (e+f x))^4}-\frac{14 c^4 \cos ^3(e+f x)}{3 a f (a+a \sin (e+f x))^2}-\frac{\left (7 c^4\right ) \int \frac{\cos ^2(e+f x)}{a+a \sin (e+f x)} \, dx}{a^2}\\ &=-\frac{7 c^4 \cos (e+f x)}{a^3 f}-\frac{2 a^3 c^4 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^6}+\frac{14 a c^4 \cos ^5(e+f x)}{15 f (a+a \sin (e+f x))^4}-\frac{14 c^4 \cos ^3(e+f x)}{3 a f (a+a \sin (e+f x))^2}-\frac{\left (7 c^4\right ) \int 1 \, dx}{a^3}\\ &=-\frac{7 c^4 x}{a^3}-\frac{7 c^4 \cos (e+f x)}{a^3 f}-\frac{2 a^3 c^4 \cos ^7(e+f x)}{5 f (a+a \sin (e+f x))^6}+\frac{14 a c^4 \cos ^5(e+f x)}{15 f (a+a \sin (e+f x))^4}-\frac{14 c^4 \cos ^3(e+f x)}{3 a f (a+a \sin (e+f x))^2}\\ \end{align*}

Mathematica [B]  time = 0.601386, size = 270, normalized size = 2.18 \[ \frac{(c-c \sin (e+f x))^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (96 \sin \left (\frac{1}{2} (e+f x)\right )-105 (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5-15 \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5+464 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+128 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-256 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-48 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{15 f (a \sin (e+f x)+a)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^8} \]

Antiderivative was successfully verified.

[In]

Integrate[(c - c*Sin[e + f*x])^4/(a + a*Sin[e + f*x])^3,x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(96*Sin[(e + f*x)/2] - 48*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) - 256*S
in[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 128*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + 464*Si
n[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4 - 105*(e + f*x)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5
- 15*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)*(c - c*Sin[e + f*x])^4)/(15*f*(Cos[(e + f*x)/2] -
Sin[(e + f*x)/2])^8*(a + a*Sin[e + f*x])^3)

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Maple [A]  time = 0.099, size = 145, normalized size = 1.2 \begin{align*} -2\,{\frac{{c}^{4}}{f{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) }}-14\,{\frac{{c}^{4}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{a}^{3}}}-{\frac{128\,{c}^{4}}{5\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-5}}+64\,{\frac{{c}^{4}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{4}}}-{\frac{128\,{c}^{4}}{3\,f{a}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}-16\,{\frac{{c}^{4}}{f{a}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x)

[Out]

-2/f*c^4/a^3/(1+tan(1/2*f*x+1/2*e)^2)-14/f*c^4/a^3*arctan(tan(1/2*f*x+1/2*e))-128/5/f*c^4/a^3/(tan(1/2*f*x+1/2
*e)+1)^5+64/f*c^4/a^3/(tan(1/2*f*x+1/2*e)+1)^4-128/3/f*c^4/a^3/(tan(1/2*f*x+1/2*e)+1)^3-16/f*c^4/a^3/(tan(1/2*
f*x+1/2*e)+1)

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Maxima [B]  time = 1.96228, size = 1480, normalized size = 11.94 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(3*c^4*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 189*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 200*sin(f*x + e
)^3/(cos(f*x + e) + 1)^3 + 160*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 75*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 +
15*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 24)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 11*a^3*sin(f*x + e
)^2/(cos(f*x + e) + 1)^2 + 15*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*a^3*sin(f*x + e)^4/(cos(f*x + e) +
1)^4 + 11*a^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 5*a^3*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^3*sin(f*x +
e)^7/(cos(f*x + e) + 1)^7) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + 4*c^4*((95*sin(f*x + e)/(cos(f*
x + e) + 1) + 145*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 75*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 + 22)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin
(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^3) + c^4*(20*sin(f*x + e)/(co
s(f*x + e) + 1) + 40*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 30*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 15*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 + 7)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*
x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*s
in(f*x + e)^5/(cos(f*x + e) + 1)^5) + 12*c^4*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^2/(cos(f*x +
e) + 1)^2 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 10
*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^3*sin(f*x + e)^5/(cos
(f*x + e) + 1)^5) - 12*c^4*(5*sin(f*x + e)/(cos(f*x + e) + 1) + 5*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 5*sin(
f*x + e)^3/(cos(f*x + e) + 1)^3 + 1)/(a^3 + 5*a^3*sin(f*x + e)/(cos(f*x + e) + 1) + 10*a^3*sin(f*x + e)^2/(cos
(f*x + e) + 1)^2 + 10*a^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*a^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^
3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [B]  time = 1.3706, size = 633, normalized size = 5.1 \begin{align*} -\frac{15 \, c^{4} \cos \left (f x + e\right )^{4} - 420 \, c^{4} f x - 48 \, c^{4} +{\left (105 \, c^{4} f x + 277 \, c^{4}\right )} \cos \left (f x + e\right )^{3} +{\left (315 \, c^{4} f x - 134 \, c^{4}\right )} \cos \left (f x + e\right )^{2} - 6 \,{\left (35 \, c^{4} f x + 74 \, c^{4}\right )} \cos \left (f x + e\right ) +{\left (15 \, c^{4} \cos \left (f x + e\right )^{3} - 420 \, c^{4} f x + 48 \, c^{4} +{\left (105 \, c^{4} f x - 262 \, c^{4}\right )} \cos \left (f x + e\right )^{2} - 6 \,{\left (35 \, c^{4} f x + 66 \, c^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \,{\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f +{\left (a^{3} f \cos \left (f x + e\right )^{2} - 2 \, a^{3} f \cos \left (f x + e\right ) - 4 \, a^{3} f\right )} \sin \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/15*(15*c^4*cos(f*x + e)^4 - 420*c^4*f*x - 48*c^4 + (105*c^4*f*x + 277*c^4)*cos(f*x + e)^3 + (315*c^4*f*x -
134*c^4)*cos(f*x + e)^2 - 6*(35*c^4*f*x + 74*c^4)*cos(f*x + e) + (15*c^4*cos(f*x + e)^3 - 420*c^4*f*x + 48*c^4
+ (105*c^4*f*x - 262*c^4)*cos(f*x + e)^2 - 6*(35*c^4*f*x + 66*c^4)*cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x
+ e)^3 + 3*a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x + e) - 4*a^3*f + (a^3*f*cos(f*x + e)^2 - 2*a^3*f*cos(f*x +
e) - 4*a^3*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((c-c*sin(f*x+e))**4/(a+a*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 2.05805, size = 182, normalized size = 1.47 \begin{align*} -\frac{\frac{105 \,{\left (f x + e\right )} c^{4}}{a^{3}} + \frac{30 \, c^{4}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )} a^{3}} + \frac{16 \,{\left (15 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 60 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 130 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 80 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 19 \, c^{4}\right )}}{a^{3}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sin(f*x+e))^4/(a+a*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/15*(105*(f*x + e)*c^4/a^3 + 30*c^4/((tan(1/2*f*x + 1/2*e)^2 + 1)*a^3) + 16*(15*c^4*tan(1/2*f*x + 1/2*e)^4 +
60*c^4*tan(1/2*f*x + 1/2*e)^3 + 130*c^4*tan(1/2*f*x + 1/2*e)^2 + 80*c^4*tan(1/2*f*x + 1/2*e) + 19*c^4)/(a^3*(
tan(1/2*f*x + 1/2*e) + 1)^5))/f