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

Optimal. Leaf size=106 \[ \frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac{2 a^3 \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac{a^3 x}{c^3}-\frac{2 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

[Out]

-((a^3*x)/c^3) + (2*a^3*c^2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^5) - (2*a^3*Cos[e + f*x]^3)/(3*f*(c - c*
Sin[e + f*x])^3) + (2*a^3*Cos[e + f*x])/(f*(c^3 - c^3*Sin[e + f*x]))

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Rubi [A]  time = 0.188967, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {2736, 2680, 8} \[ \frac{2 a^3 c^2 \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac{2 a^3 \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac{a^3 x}{c^3}-\frac{2 a^3 \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^3*x)/c^3) + (2*a^3*c^2*Cos[e + f*x]^5)/(5*f*(c - c*Sin[e + f*x])^5) - (2*a^3*Cos[e + f*x]^3)/(3*f*(c - c*
Sin[e + f*x])^3) + (2*a^3*Cos[e + f*x])/(f*(c^3 - c^3*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 8

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

Rubi steps

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

Mathematica [B]  time = 0.455322, size = 249, normalized size = 2.35 \[ \frac{(a \sin (e+f x)+a)^3 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (48 \sin \left (\frac{1}{2} (e+f x)\right )-15 (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5+92 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-44 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-88 \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+24 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )}{15 f (c-c \sin (e+f x))^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(24*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 44*(Cos[(e + f*x)/2] - Sin[
(e + f*x)/2])^3 - 15*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + 48*Sin[(e + f*x)/2] - 88*(Cos[(e + f*
x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 92*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(a
+ a*Sin[e + f*x])^3)/(15*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(c - c*Sin[e + f*x])^3)

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Maple [A]  time = 0.092, size = 143, normalized size = 1.4 \begin{align*} -{\frac{64\,{a}^{3}}{5\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-5}}-32\,{\frac{{a}^{3}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-{\frac{80\,{a}^{3}}{3\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-8\,{\frac{{a}^{3}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-4\,{\frac{{a}^{3}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-2\,{\frac{{a}^{3}\arctan \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) }{f{c}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-64/5/f*a^3/c^3/(tan(1/2*f*x+1/2*e)-1)^5-32/f*a^3/c^3/(tan(1/2*f*x+1/2*e)-1)^4-80/3/f*a^3/c^3/(tan(1/2*f*x+1/2
*e)-1)^3-8/f*a^3/c^3/(tan(1/2*f*x+1/2*e)-1)^2-4/f*a^3/c^3/(tan(1/2*f*x+1/2*e)-1)-2/f*a^3/c^3*arctan(tan(1/2*f*
x+1/2*e))

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Maxima [B]  time = 2.66346, size = 1060, normalized size = 10. \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))^3/(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(a^3*((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)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e) +
1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 15*arctan(sin(f*x + e)/(cos(f*x + e) +
1))/c^3) + a^3*(20*sin(f*x + e)/(cos(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)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e)
+ 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) - 9*a^3*(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)/(c^3 - 5*c^3*sin(f*x
+ e)/(cos(f*x + e) + 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) +
1)^3 + 5*c^3*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5) + 6*a^3*(5*sin(f*x
+ e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 1)/(c^3 - 5*c^3*sin(f*x + e)/(cos(f*x + e)
+ 1) + 10*c^3*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 10*c^3*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 5*c^3*sin(f*
x + e)^4/(cos(f*x + e) + 1)^4 - c^3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5))/f

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Fricas [B]  time = 1.35291, size = 551, normalized size = 5.2 \begin{align*} \frac{60 \, a^{3} f x -{\left (15 \, a^{3} f x - 46 \, a^{3}\right )} \cos \left (f x + e\right )^{3} - 24 \, a^{3} -{\left (45 \, a^{3} f x + 2 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (5 \, a^{3} f x - 12 \, a^{3}\right )} \cos \left (f x + e\right ) -{\left (60 \, a^{3} f x + 24 \, a^{3} -{\left (15 \, a^{3} f x + 46 \, a^{3}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (5 \, a^{3} f x - 8 \, a^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \,{\left (c^{3} f \cos \left (f x + e\right )^{3} + 3 \, c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} f -{\left (c^{3} f \cos \left (f x + e\right )^{2} - 2 \, c^{3} f \cos \left (f x + e\right ) - 4 \, c^{3} 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))^3/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(60*a^3*f*x - (15*a^3*f*x - 46*a^3)*cos(f*x + e)^3 - 24*a^3 - (45*a^3*f*x + 2*a^3)*cos(f*x + e)^2 + 6*(5*
a^3*f*x - 12*a^3)*cos(f*x + e) - (60*a^3*f*x + 24*a^3 - (15*a^3*f*x + 46*a^3)*cos(f*x + e)^2 + 6*(5*a^3*f*x -
8*a^3)*cos(f*x + e))*sin(f*x + e))/(c^3*f*cos(f*x + e)^3 + 3*c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c
^3*f - (c^3*f*cos(f*x + e)^2 - 2*c^3*f*cos(f*x + e) - 4*c^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((a+a*sin(f*x+e))**3/(c-c*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 2.35508, size = 150, normalized size = 1.42 \begin{align*} -\frac{\frac{15 \,{\left (f x + e\right )} a^{3}}{c^{3}} + \frac{4 \,{\left (15 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 30 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 100 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 50 \, a^{3} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 13 \, a^{3}\right )}}{c^{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((a+a*sin(f*x+e))^3/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/15*(15*(f*x + e)*a^3/c^3 + 4*(15*a^3*tan(1/2*f*x + 1/2*e)^4 - 30*a^3*tan(1/2*f*x + 1/2*e)^3 + 100*a^3*tan(1
/2*f*x + 1/2*e)^2 - 50*a^3*tan(1/2*f*x + 1/2*e) + 13*a^3)/(c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f