3.33 \(\int \frac{(a+a \sin (e+f x))^2 (A+B \sin (e+f x))}{(c-c \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=112 \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac{2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac{a^2 B x}{c^3}-\frac{2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

[Out]

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

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Rubi [A]  time = 0.277027, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {2967, 2859, 2680, 8} \[ \frac{a^2 c^2 (A+B) \cos ^5(e+f x)}{5 f (c-c \sin (e+f x))^5}+\frac{2 a^2 B \cos (e+f x)}{f \left (c^3-c^3 \sin (e+f x)\right )}-\frac{a^2 B x}{c^3}-\frac{2 a^2 B \cos ^3(e+f x)}{3 f (c-c \sin (e+f x))^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^3,x]

[Out]

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

Rule 2967

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a^m*c^m, Int[Cos[e + f*x]^(2*m)*(c + d*Sin[e + f*x])^(n - m)*(A + B
*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && I
ntegerQ[m] &&  !(IntegerQ[n] && ((LtQ[m, 0] && GtQ[n, 0]) || LtQ[0, n, m] || LtQ[m, n, 0]))

Rule 2859

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

Mathematica [B]  time = 0.695633, size = 278, normalized size = 2.48 \[ \frac{a^2 (\sin (e+f x)+1)^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (24 (A+B) \sin \left (\frac{1}{2} (e+f x)\right )+2 (3 A+43 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^4-4 (3 A+8 B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^3-8 (3 A+8 B) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^2+12 (A+B) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )-15 B (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )^5\right )}{15 f (c-c \sin (e+f x))^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sin[e + f*x])^2*(A + B*Sin[e + f*x]))/(c - c*Sin[e + f*x])^3,x]

[Out]

(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(12*(A + B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - 4*(3*A + 8*B)*(
Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^3 - 15*B*(e + f*x)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^5 + 24*(A + B)*S
in[(e + f*x)/2] - 8*(3*A + 8*B)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])^2*Sin[(e + f*x)/2] + 2*(3*A + 43*B)*(Cos
[(e + f*x)/2] - Sin[(e + f*x)/2])^4*Sin[(e + f*x)/2])*(1 + Sin[e + f*x])^2)/(15*f*(Cos[(e + f*x)/2] + Sin[(e +
f*x)/2])^4*(c - c*Sin[e + f*x])^3)

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Maple [B]  time = 0.127, size = 249, normalized size = 2.2 \begin{align*} -2\,{\frac{A{a}^{2}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-2\,{\frac{B{a}^{2}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) }}-{\frac{32\,A{a}^{2}}{5\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-5}}-{\frac{32\,B{a}^{2}}{5\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-5}}-16\,{\frac{A{a}^{2}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{3}}}-{\frac{32\,B{a}^{2}}{3\,f{c}^{3}} \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) ^{-3}}-16\,{\frac{A{a}^{2}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-16\,{\frac{B{a}^{2}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{4}}}-8\,{\frac{A{a}^{2}}{f{c}^{3} \left ( \tan \left ( 1/2\,fx+e/2 \right ) -1 \right ) ^{2}}}-2\,{\frac{B{a}^{2}\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))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x)

[Out]

-2/f*a^2/c^3/(tan(1/2*f*x+1/2*e)-1)*A-2/f*a^2/c^3/(tan(1/2*f*x+1/2*e)-1)*B-32/5/f*a^2/c^3/(tan(1/2*f*x+1/2*e)-
1)^5*A-32/5/f*a^2/c^3/(tan(1/2*f*x+1/2*e)-1)^5*B-16/f*a^2/c^3/(tan(1/2*f*x+1/2*e)-1)^3*A-32/3/f*a^2/c^3/(tan(1
/2*f*x+1/2*e)-1)^3*B-16/f*a^2/c^3/(tan(1/2*f*x+1/2*e)-1)^4*A-16/f*a^2/c^3/(tan(1/2*f*x+1/2*e)-1)^4*B-8/f*a^2/c
^3*A/(tan(1/2*f*x+1/2*e)-1)^2-2/f*a^2/c^3*B*arctan(tan(1/2*f*x+1/2*e))

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Maxima [B]  time = 1.57944, size = 1538, normalized size = 13.73 \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))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

-2/15*(B*a^2*((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*a^2*(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) - 6*A*a^2*(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*s
in(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) - 3*B*a^2*(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) + 2*A*a^2*(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) + 4*B*a
^2*(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 = 1.42677, size = 655, normalized size = 5.85 \begin{align*} \frac{60 \, B a^{2} f x -{\left (15 \, B a^{2} f x -{\left (3 \, A + 43 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{3} - 12 \,{\left (A + B\right )} a^{2} -{\left (45 \, B a^{2} f x -{\left (9 \, A - 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (5 \, B a^{2} f x -{\left (A + 11 \, B\right )} a^{2}\right )} \cos \left (f x + e\right ) -{\left (60 \, B a^{2} f x + 12 \,{\left (A + B\right )} a^{2} -{\left (15 \, B a^{2} f x +{\left (3 \, A + 43 \, B\right )} a^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \,{\left (5 \, B a^{2} f x +{\left (A - 9 \, B\right )} a^{2}\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))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

1/15*(60*B*a^2*f*x - (15*B*a^2*f*x - (3*A + 43*B)*a^2)*cos(f*x + e)^3 - 12*(A + B)*a^2 - (45*B*a^2*f*x - (9*A
- 11*B)*a^2)*cos(f*x + e)^2 + 6*(5*B*a^2*f*x - (A + 11*B)*a^2)*cos(f*x + e) - (60*B*a^2*f*x + 12*(A + B)*a^2 -
(15*B*a^2*f*x + (3*A + 43*B)*a^2)*cos(f*x + e)^2 + 6*(5*B*a^2*f*x + (A - 9*B)*a^2)*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))**2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.21705, size = 215, normalized size = 1.92 \begin{align*} -\frac{\frac{15 \,{\left (f x + e\right )} B a^{2}}{c^{3}} + \frac{2 \,{\left (15 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 15 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 60 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 30 \, A a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 170 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 100 \, B a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 3 \, A a^{2} + 23 \, B a^{2}\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))^2*(A+B*sin(f*x+e))/(c-c*sin(f*x+e))^3,x, algorithm="giac")

[Out]

-1/15*(15*(f*x + e)*B*a^2/c^3 + 2*(15*A*a^2*tan(1/2*f*x + 1/2*e)^4 + 15*B*a^2*tan(1/2*f*x + 1/2*e)^4 - 60*B*a^
2*tan(1/2*f*x + 1/2*e)^3 + 30*A*a^2*tan(1/2*f*x + 1/2*e)^2 + 170*B*a^2*tan(1/2*f*x + 1/2*e)^2 - 100*B*a^2*tan(
1/2*f*x + 1/2*e) + 3*A*a^2 + 23*B*a^2)/(c^3*(tan(1/2*f*x + 1/2*e) - 1)^5))/f