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

Optimal. Leaf size=135 \[ \frac{35 c^4 \cos (e+f x)}{2 a^2 f}-\frac{2 a^3 c^4 \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}+\frac{14 a^4 c^4 \cos ^5(e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )^3}+\frac{35 c^4 \cos ^3(e+f x)}{6 f \left (a^2 \sin (e+f x)+a^2\right )}+\frac{35 c^4 x}{2 a^2} \]

[Out]

(35*c^4*x)/(2*a^2) + (35*c^4*Cos[e + f*x])/(2*a^2*f) - (2*a^3*c^4*Cos[e + f*x]^7)/(3*f*(a + a*Sin[e + f*x])^5)
+ (14*a^4*c^4*Cos[e + f*x]^5)/(3*f*(a^2 + a^2*Sin[e + f*x])^3) + (35*c^4*Cos[e + f*x]^3)/(6*f*(a^2 + a^2*Sin[
e + f*x]))

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Rubi [A]  time = 0.224106, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {2736, 2680, 2679, 2682, 8} \[ \frac{35 c^4 \cos (e+f x)}{2 a^2 f}-\frac{2 a^3 c^4 \cos ^7(e+f x)}{3 f (a \sin (e+f x)+a)^5}+\frac{14 a^4 c^4 \cos ^5(e+f x)}{3 f \left (a^2 \sin (e+f x)+a^2\right )^3}+\frac{35 c^4 \cos ^3(e+f x)}{6 f \left (a^2 \sin (e+f x)+a^2\right )}+\frac{35 c^4 x}{2 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*c^4*x)/(2*a^2) + (35*c^4*Cos[e + f*x])/(2*a^2*f) - (2*a^3*c^4*Cos[e + f*x]^7)/(3*f*(a + a*Sin[e + f*x])^5)
+ (14*a^4*c^4*Cos[e + f*x]^5)/(3*f*(a^2 + a^2*Sin[e + f*x])^3) + (35*c^4*Cos[e + f*x]^3)/(6*f*(a^2 + a^2*Sin[
e + f*x]))

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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
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))^2} \, dx &=\left (a^4 c^4\right ) \int \frac{\cos ^8(e+f x)}{(a+a \sin (e+f x))^6} \, dx\\ &=-\frac{2 a^3 c^4 \cos ^7(e+f x)}{3 f (a+a \sin (e+f x))^5}-\frac{1}{3} \left (7 a^2 c^4\right ) \int \frac{\cos ^6(e+f x)}{(a+a \sin (e+f x))^4} \, dx\\ &=-\frac{2 a^3 c^4 \cos ^7(e+f x)}{3 f (a+a \sin (e+f x))^5}+\frac{14 a c^4 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac{1}{3} \left (35 c^4\right ) \int \frac{\cos ^4(e+f x)}{(a+a \sin (e+f x))^2} \, dx\\ &=-\frac{2 a^3 c^4 \cos ^7(e+f x)}{3 f (a+a \sin (e+f x))^5}+\frac{14 a c^4 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac{35 c^4 \cos ^3(e+f x)}{6 f \left (a^2+a^2 \sin (e+f x)\right )}+\frac{\left (35 c^4\right ) \int \frac{\cos ^2(e+f x)}{a+a \sin (e+f x)} \, dx}{2 a}\\ &=\frac{35 c^4 \cos (e+f x)}{2 a^2 f}-\frac{2 a^3 c^4 \cos ^7(e+f x)}{3 f (a+a \sin (e+f x))^5}+\frac{14 a c^4 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac{35 c^4 \cos ^3(e+f x)}{6 f \left (a^2+a^2 \sin (e+f x)\right )}+\frac{\left (35 c^4\right ) \int 1 \, dx}{2 a^2}\\ &=\frac{35 c^4 x}{2 a^2}+\frac{35 c^4 \cos (e+f x)}{2 a^2 f}-\frac{2 a^3 c^4 \cos ^7(e+f x)}{3 f (a+a \sin (e+f x))^5}+\frac{14 a c^4 \cos ^5(e+f x)}{3 f (a+a \sin (e+f x))^3}+\frac{35 c^4 \cos ^3(e+f x)}{6 f \left (a^2+a^2 \sin (e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.503876, size = 243, normalized size = 1.8 \[ \frac{(c-c \sin (e+f x))^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (128 \sin \left (\frac{1}{2} (e+f x)\right )+210 (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+72 \cos (e+f x) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-3 \sin (2 (e+f x)) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3-640 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-64 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{12 f (a \sin (e+f x)+a)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*(c - c*Sin[e + f*x])^4*(128*Sin[(e + f*x)/2] - 64*(Cos[(e + f*x)/2] + S
in[(e + f*x)/2]) - 640*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 + 210*(e + f*x)*(Cos[(e + f*x)
/2] + Sin[(e + f*x)/2])^3 + 72*Cos[e + f*x]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 - 3*(Cos[(e + f*x)/2] + Si
n[(e + f*x)/2])^3*Sin[2*(e + f*x)]))/(12*f*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^8*(a + a*Sin[e + f*x])^2)

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Maple [A]  time = 0.093, size = 229, normalized size = 1.7 \begin{align*}{\frac{{c}^{4}}{{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{3} \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+12\,{\frac{{c}^{4} \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}}{{a}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{{c}^{4}}{{a}^{2}f}\tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \left ( 1+ \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2} \right ) ^{-2}}+12\,{\frac{{c}^{4}}{{a}^{2}f \left ( 1+ \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2} \right ) ^{2}}}+35\,{\frac{{c}^{4}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{{a}^{2}f}}-{\frac{64\,{c}^{4}}{3\,{a}^{2}f} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) ^{-3}}+32\,{\frac{{c}^{4}}{{a}^{2}f \left ( \tan \left ( 1/2\,fx+e/2 \right ) +1 \right ) ^{2}}}+32\,{\frac{{c}^{4}}{{a}^{2}f \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))^2,x)

[Out]

1/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)^3+12/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*
x+1/2*e)^2-1/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^2*tan(1/2*f*x+1/2*e)+12/f*c^4/a^2/(1+tan(1/2*f*x+1/2*e)^2)^2+3
5/f*c^4/a^2*arctan(tan(1/2*f*x+1/2*e))-64/3/f*c^4/a^2/(tan(1/2*f*x+1/2*e)+1)^3+32/f*c^4/a^2/(tan(1/2*f*x+1/2*e
)+1)^2+32/f*c^4/a^2/(tan(1/2*f*x+1/2*e)+1)

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Maxima [B]  time = 2.35734, size = 1219, normalized size = 9.03 \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))^2,x, algorithm="maxima")

[Out]

1/3*(c^4*((75*sin(f*x + e)/(cos(f*x + e) + 1) + 97*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 126*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + 98*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 63*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 21*sin(
f*x + e)^6/(cos(f*x + e) + 1)^6 + 32)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 5*a^2*sin(f*x + e)^2/(cos
(f*x + e) + 1)^2 + 7*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 7*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 5*a
^2*sin(f*x + e)^5/(cos(f*x + e) + 1)^5 + 3*a^2*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + a^2*sin(f*x + e)^7/(cos(f
*x + e) + 1)^7) + 21*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 16*c^4*((12*sin(f*x + e)/(cos(f*x + e) + 1
) + 11*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 9*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^4/(cos(f*x
+ e) + 1)^4 + 5)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 4*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4
*a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*a^2*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + a^2*sin(f*x + e)^5/(cos
(f*x + e) + 1)^5) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/a^2) + 12*c^4*((9*sin(f*x + e)/(cos(f*x + e) + 1
) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 4)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x +
e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3) + 3*arctan(sin(f*x + e)/(cos(f*x + e) + 1
))/a^2) - 2*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 2)/(a^2 + 3*a^2*s
in(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e)
+ 1)^3) + 8*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + 1)/(a^2 + 3*a^2*sin(f*x + e)/(cos(f*x + e) + 1) + 3*a^2*s
in(f*x + e)^2/(cos(f*x + e) + 1)^2 + a^2*sin(f*x + e)^3/(cos(f*x + e) + 1)^3))/f

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Fricas [A]  time = 1.43884, size = 513, normalized size = 3.8 \begin{align*} -\frac{3 \, c^{4} \cos \left (f x + e\right )^{4} - 30 \, c^{4} \cos \left (f x + e\right )^{3} + 210 \, c^{4} f x - 32 \, c^{4} -{\left (105 \, c^{4} f x - 193 \, c^{4}\right )} \cos \left (f x + e\right )^{2} +{\left (105 \, c^{4} f x + 194 \, c^{4}\right )} \cos \left (f x + e\right ) +{\left (3 \, c^{4} \cos \left (f x + e\right )^{3} + 210 \, c^{4} f x + 33 \, c^{4} \cos \left (f x + e\right )^{2} + 32 \, c^{4} +{\left (105 \, c^{4} f x + 226 \, c^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{6 \,{\left (a^{2} f \cos \left (f x + e\right )^{2} - a^{2} f \cos \left (f x + e\right ) - 2 \, a^{2} f -{\left (a^{2} f \cos \left (f x + e\right ) + 2 \, a^{2} 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))^2,x, algorithm="fricas")

[Out]

-1/6*(3*c^4*cos(f*x + e)^4 - 30*c^4*cos(f*x + e)^3 + 210*c^4*f*x - 32*c^4 - (105*c^4*f*x - 193*c^4)*cos(f*x +
e)^2 + (105*c^4*f*x + 194*c^4)*cos(f*x + e) + (3*c^4*cos(f*x + e)^3 + 210*c^4*f*x + 33*c^4*cos(f*x + e)^2 + 32
*c^4 + (105*c^4*f*x + 226*c^4)*cos(f*x + e))*sin(f*x + e))/(a^2*f*cos(f*x + e)^2 - a^2*f*cos(f*x + e) - 2*a^2*
f - (a^2*f*cos(f*x + e) + 2*a^2*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))**2,x)

[Out]

Timed out

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Giac [A]  time = 2.22172, size = 205, normalized size = 1.52 \begin{align*} \frac{\frac{105 \,{\left (f x + e\right )} c^{4}}{a^{2}} + \frac{6 \,{\left (c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 12 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 12 \, c^{4}\right )}}{{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 1\right )}^{2} a^{2}} + \frac{64 \,{\left (3 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 9 \, c^{4} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 4 \, c^{4}\right )}}{a^{2}{\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}^{3}}}{6 \, 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))^2,x, algorithm="giac")

[Out]

1/6*(105*(f*x + e)*c^4/a^2 + 6*(c^4*tan(1/2*f*x + 1/2*e)^3 + 12*c^4*tan(1/2*f*x + 1/2*e)^2 - c^4*tan(1/2*f*x +
1/2*e) + 12*c^4)/((tan(1/2*f*x + 1/2*e)^2 + 1)^2*a^2) + 64*(3*c^4*tan(1/2*f*x + 1/2*e)^2 + 9*c^4*tan(1/2*f*x
+ 1/2*e) + 4*c^4)/(a^2*(tan(1/2*f*x + 1/2*e) + 1)^3))/f